Principles-of-communication-systems-by-taub-and-schilling_text.pdf

PRINCIPLES

OF COMMUNICATION SYSTEMS Second Edition

Herbert Taub

Donald L.

Schilling

Professors of Electrical Engineering

The City College of New York

McGraw-Hill Publishing Company New York St. Louis San Francisco Auckland Bogota Caracas Hamburg Lisbon London Madrid Mexico Milan Montreal New Delhi Oklahoma City Paris San Juan Sao Paulo Singapore Sydney Tokyo Toronto

CHAPTER

ONE SPECTRAL ANALYSIS

INTRODUCTION commuSuppose that two people, separated by a considerable distance, wish to one from extending wires conducting of pair there is a If nicate with one another. microphone and earlocation to another, and if each place is equipped with a microphone, at one end of piece, the communication problem may be solved. The signal voltage on the line, electric the wire communications channel, impresses an which voltage

is

have associated with

which

more

is

at the

then received it

an

other end.

erratic,

The

received signal, however, will

random, unpredictable voltage waveform

The origin of this Here we need but to note

described by the term noise.

fully in

the universe

Chaps. is

7

and

14.

in a constant state of agitation,

and

noise will be discussed that at the atomic level

that this agitation

is

the

the wire link, the source of a very great deal of this noise. Because of the length of comparison with its received message signal voltage will be greatly attenuated in the message signal voltage level at the transmitting end of the link. As a result, and the message will voltage, noise very large in comparison with the

may

not be amplifier at the receiving be perceived with difficulty or possibly not at all. An amplify signal and noise will amplifier the since problem, the solve not end will source matter of fact, as we shall see, the amplifier itself may well be a alike.

As

a

of additional noise.

A

principal concern of

extensively in this

book

is

communication theory and a matter which we discuss precisely the study of methods to suppress, as far as

possible, the effect of noise.

We

shall see that, for this purpose,

not to transmit directly the original signal (the

it

may

be better

microphone output

in

our

2 PRINCIPLES OF

SPECTRAL ANALYSIS 3

COMMUNICATION SYSTEMS while the coefficients A„ and B„ are given by

signal waveis used to generate a different line. This processing the on impressed then form which new signal waveform transmitted signal is called encoding or of the original signal to generate the called decoding or demodumodulation. At the receiving end an inverse process

example). Instead, the original signal is

lation

A.

required to recover the original signal. wire commay well be that there is a considerable expense in providing the naturally led to inquire whether we may use

munication the link link of

mission

link.

We

is

called multiplexing

and

of this book. It area of concern of communication theory and then, at employed, are links communications when wire

nications

An

alternative

form

questions:

v(t)

is is

C

,

(1.1-3)

dt

'0

T ° 12

2nnt

sin

>it)

J

,

(1.1-4)

dt

l>

= C0 + £c„

is

is

/2nm cos

(1.1-5)

4>n

again a principal to be noted that

least

in

where

C0

,

C, and ,

are related to

A0

principle,

C.

,

A„ and B„ by the equations ,

A0

(l.l-6o)

=

J At + Bl

(l.l-6b)

=

tan

Co

essential.

itself to the following then, communication theory addresses communication channel, how do we arrange to transmit as

and

tf>„

B.

(l.l-6c)

Given a

many simultaneous

and how do we devise to suppress the

signals as possible,

possible? In this book, after a few matheeffect of noise to the maximum extent ourselves precisely to these questions, first address shall we matical preliminaries, the discussion of noise in commuto the matter of multiplexing, and thereafter to nications systems.

which is of inestimable value in the study of comwith the analysis. Spectral analysis concerns itself spectral munications svstems is with the correspondence description of waveforms in the jrequency domain and description. It is between the irequencv-domain description and the time-domain

A branch

of mathematics

assumed that the reader has some

familiarity with spectral analysis.

The

The Fourier

series of a periodic function is thus seen to consist of a

remainder of

.

-

frequency nf0 A typical amplitude spectrum of a periodic waveform is shown in Fig. 1.1-la. Here, at each harmonic frequency, a vertical amplitude associated line has been drawn having a length equal to the spectral

C„ cos

(27tn/0

1

„)

at

.

lackint with each harmonic frequency. Of course, such an amplitude spectrum, the phase information, does not specify the wavelorm u(r).

presen-

c„

this text.

FOURIER SERIES

1

0

f0

2/„

3/„

4/c (0)

having a fundamental period T0 can be represent waveforms. This summation, called a t owner sinusoidal of sum infinite ed as an such form is the following: series, may be written in several forms. One -

A

summa-

coefficients C„ are tion of harmonics of a fundamental frequency /„ = 1/T0 The spectral component called spectral amplitudes; that is, C„ is the amplitude of the

will serve further to allow a tation in this chapter is intended as a review, and to use throughout the occasion have shall we which results compilation of

1.1

=2

— — 2nnt

cos


for the Fourier series

the commube used for individual messages. When, however, antenna to from communication radio is free space, as in

antenna, multiplexing

summary

/2

'0 J- To/2

is

may medium

r°

are, therefore,

simultaneous multiple transmission

In

Bm =

and

over the more effectively by arranging for the simultaneous transmission transmultiple such more than just a single waveform. It turns out that Such ways. of number in a indeed possible and may be accomplished

separate links

—2

'oJ-To/2

is

It

=

periodic function of time

^ 2nm

A0 + The constant A 0

is

lv„l

v(t)

£A

r

cos

the average value of

v(t)

Iwnt

+

X B,

(1.1-1)

_L...J -»4

t 4/0 -3fB -*i

1

-2f0 -f0

0

i

1

1

/„

2/„

3/0

L...1< 4/c

given by

CToi: 1

v{t) d\

'o J- To/2

f

(6)

(1.1-2)

Figure 1.1-1

(a)

two-sided plot.

A

one-sided plot of spectral amplitude of a periodic waveform.

(f rel="nofollow">)

The corresponding

SPECTRAL ANALYSIS 5 4 PRINCIPLES OF COMMUNICATION SYSTEMS

1.2

EXPONENTIAL FORM OF THE FOURIER SERIES 7i(( +

The exponential form

of the Fourier series finds extensive application in

nication theory. This form

is

2T0

)

1SU + T0

|/6U>

)

/{«-!•„)

7«((-25T„)

To

2TB

commu-

given by

V„e J2

£

v(t)=

n~ -

'- 17 "

d-2-1)

-2T0

cc

-T.

I

la)

where

V„ is

given by f r ° /2

— 1

Vn =

v(t)e-

To J-

The

coefficients K„

one another, that (1.1-5)

V„

= Vl„.

(1.2-2)

These

V„

and V-„ are complex conjugates of

coefficients are related to the C„'s in Eq.

by

V0 = C0 V" =

—

d.2-3fl)

-2TC

Observe that while V0 = C 0 otherwise each spectral line in 1.1-la at frequency / is replaced by the 2 spectral lines in The 1.1-lb, each of half amplitude, one at frequency /and one at frequency -/ amplitude spectrum in 1.1-la is called a single-sided spectrum, while the spectrum in Fig. 1.1-la.

spectrum.

T/2

J~

2TD

D

Figure

1

3-1 Examples of periodic functions,

pulses of duration

(a)

A

periodic train of impulses.

We

shall find

We

then find, using Eqs. (1.1-2) and

more convenient to use do so from this point on.

m

A0 =

EXAMPLES OF FOURIER SERIES shall

which has the property that

periodic train of

tt

4

<5(f)

= 0 except when = r

and, using Eq.

(1.3-3)

J0

cos

2nm

.

—^r

dt

=

21 (1.3-4)

T0

7o

- To/2

have occasion to have some special interest is shown in Fig. 1.3-1 a. The waveform consists of a periodic sequence of impulses of strength /. As a matter of convenience we have selected the time scale so that an impulse occurs at t = 0. The impulse at t = 0 is written as / 5(t). Here d(t) is the delta function

-

=

T °' : 5{t)

which we

dt

-To/2 "

in

A

(1.1-3),

it

21

A waveform

(i>)

i.

,

the two-sided amplitude spectrum and shall consistently

1.3

1

(6)

V„'s

in 1.1-16 is called a two-sided

t/2

(1.2-3b)

e-'*"

2 ilnKS '".

spectrum of the C„'s shown

-r0

0

The are the spectral amplitudes of the spectral components Vr e amplitude to the corresponds \.\-\b Fig. V„'s shown in of the spectrum amplitude

The

v{t)

To/2

have the property that is,

i2 *"" To dt

(1.1-4).

B

-'o

3(t) sin

—

-

dt

=

0

(1.3-5)

T0

- To/2

0 and further Further we have, using Eq.

(1.1-6).

d-3-1)

b(t)dt=\

(1.3-6)

The strength strength of

The

of an impulse

5{t) is 1.

equal to the area under the impulse.

is

and the strength of /

periodic impulse train

is

Thus the

<5(t) is 1.

and from Eq.

(1.2-3)

written 1

v(t)

=

l

t Mf-kT0

(1-3-2) )

v0 = y.=

(1.3-7)

'0

SPECTRAL ANALYSIS 7

6 PRINCIPLES OF COMMUNICATION SYSTEMS

Hence

v(t)

may

in the

be written

forms

' v(t)

=

l

£ «t-*T0 »=-„

°

21 -+1

=

)

'0

2nnt

I cos—o „=1 J

'o

(1.3-8)

As a second example, pulses of amplitude

A and

let

us find the Fourier series for the periodic train of

duration

shown

x as

in Fig. 1.3-16.

[To/

40

= C 0 = V0 =

^

v{t) di

-Toll

'o

"

A. =

=

Tal2 y(l)

cos

B„

and

=

(1.3-9)

at

To

-To/2

(nnT/T0 )

sin

T0

Ax

Y

2nm

C, = 2Vn =

2Ax

We find

(1.3-10)

wtt/7,,

=

0

(1.3-11)

0

Thus v(t)

=

— + —Y" /4t

2/4t

To

T„

= dl T0 Ax

a

is

2nm -—

(mn/r0

(1.3-12a)

i0

)

(1.3-12b)

n"t/T0

waveform of Fig. 1.3-lb we reduce t while adjusting A so constant, say Ax = /. We would expect that in the limit, as x-> 0, the

Suppose that that

cos

nJtT/y,

n=)

sin

f

sin(mtT/T0 )

J

in the

series for the Fourier series for the pulse train in Eq. (1.3-12) should reduce to the since impulse train in Eq. (1.3-8). It is readily verified that such is indeed the case

as

t-> 0 sin (nflt/T0 )

(1.3-13)

1

nnx/Tc

1.4

A

the value Sa(0)

THE SAMPLING FUNCTION

function frequently encountered in spectral analysis

is

the sampling function

Sa(x) defined by

Sa(x)

=

sin

closest to

x (1.4-D

x

[A

closely related function

tion Sa{x}

I

is

is

x defined by sine x = (sin nx)/nx.] The funcIt is symmetrical about x = 0, and at x = 0 has

sine

plotted in Fig. 1.4-1.

=

1. It

oscillates with

an amplitude that decreases with increasing

The function passes through zero at equally spaced intervals at values of = x ±nn, where n is an integer other than zero. Aside from the peak at x = 0, the maxima and minima occur approximately midway between the zeros, i.e., at x = ±(n + \)n, where |sinx| =1. This approximation is poorest for the minima x.

x

=

0 but improves as x becomes

imate value of Sa(x)

at these

extremal points

larger.

Correspondingly, the approx-

is

V^ + to'&^k

(L4 - 2)

SPECTRAL ANALYSIS 9 8 PRINCIPLES OF COMMUNICATION SYSTEMS

We

encountered the sampling function

A

Figure 15-1

preceding section in Eq. (1.3-121

in the

sinusoidal

of angular frequency

pulses. In that which expresses the spectrum of a periodic sequence of rectangular (1.3-12fc) are of Eq. amplitudes equation we have x = nnx/T0 The spectral components spectral The = = t/T 4 and I 0 plotted in Fig. 1.4-1 ft for the case A frequency fundamental the of multiples J0 are which appear at frequencies

input

to

network

a

characteristic

.

i„
at

differs

»„(t, a)„)

a>„

waveform is

(filter)

co.

is

vj,t, u)„)

applied

whose

H(a>„).

the

al

transfer

The output

from the input only

in

ampli

tude and phase

-

l/r0

that

,

is,

at frequencies

ponents of Sa(nxf)

shown

also

is

/=

x

1.5

=

nf0

= n/T0 The .

Here we have replaced x by

in the figure.

=

wit/To

envelope of the spectral com-

nxnj 0

=

The corresponding (1-4-3)

tit/

RESPONSE OF A LINEAR SYSTEM

The Fourier trigonometric

may

function

be expanded in terms of other (unctions.

of such possible alternative expansions

number

is

2

a sinusoidal excitation

in the

That

system

is

and has waveform preserves

that

distinctive

the

the

same frequency

its

waveshape.

w„)

v£t,

may

the

filter

is.

the output

t

M

If

the form of Eq. (1.1-5)

0<>(t)

=

is

«(0)C 0

H(u>„)V„e*

used, the response

+

t HK) |

Given a periodic waveform, the

may

if

|

C. cos

~

L

'«

-

-

0(wj] J

(1.5-9)

coefficients in the Fourier series

be written out formally as

in,

is

may

known,

be the

say, Eq. (1.5-8) or Eq. (1.5-9). Actually

(1.5-1)

°=V„e">'<

For, except in rather special and infrequent cases,

output v 0 (t,

o)„) is

an

electri-

related to the input

by a

(15-2:

|H(a,„)k-^

fflj

= H(co„HU

«°J

component

=

I

UK)

in Eq. (1.5-1)

the physical input voltage v ip(t) which

of this spectral

=

[

the transfer function H{w„) of a system

we should be hard-pressed is expressed as the sum of

indeed to recognize the waveform of a response which

an

infinite (or

even a large) number of sinusoidal terms.

may

On

the other hand, the

be written in the form of a linear superposition

responses to individual spectral components,

as, say, in

Eq.

(1.5-8), is of

ol

inestima-

ble value.

I

'- mm " )v« ***

1.6

NORMALIZED POWER

(1.5-3)

In the analysis of communication systems, Actually, the spectral


is

these equations are generally of small value from a computational point of view

= |H(w„)|F„^"'- ,,to"»

v ip(t,

(1.5-8)

°

is

v 0 (t,

sum

2 *"« T

= -a

»=i

concept that a response

HK)=

(1-5-7)

is

v 0 (t)=

response

be, say, the voltage applied to the input of

Then

d-5-6)

\

in Eq. (1.2-1), the response

single

complex transfer function

that

|

then,

since the wave-

Let the input to a linear system be the spectral component

cal filter as in Fig. 1.5-1.

|H(-bj.)|

as the excitation.

And

complex number il *"« 1

must be complex conjuH*{-w„). Therefore, since H{w„) =

must be an even function and ti(w n ) an odd function of w r an excitation is expressed as a Fourier series in exponential form as

H(a>„)

is,

evaluated. Thereafter,

vj,t,w n )=V„e

(1.5-5)

0K)=-0(-co„)

If,

being that

and very convenient to incorporate these two numbers into a

The waveform

in Eq. (1.5-5)

=

and

is

the response shape is preserved, then, in order to characterize the relationship of amplitude is related to response the how specify to the excitation, we need but to is related to the excitation phase response the how and amplitude the excitation (amplitude ratio and phase. Therefore with sinusoidal excitation, two numbers response. It turns out to be phase difference) are all that are required to deduce a possible

to„) must be real, the two terms and hence we must have that H(co n ) H{u>J\e~ m,a "\ we must have that

and

applied to a linear system, the response everyplace

similarly sinusoidal

the sinusoidal

is^

is

Since v op(t, gates,

|HK)| =

by

Hi-aW*"

+

As a matter of fact, the However, what makes

this characteristic

unique characteristic of the sinusoidal waveform,

when

v op (t, co„) given

tfK,)K„e"*"'

)

which a periodic

in

is

limitless.

Fourier trigonometric expansion especially useful

the

wn =

v op (t,

|

by no means the only way

series is

physical output voltage

component and

K^"

1

its

is

not a physical voltage. Rather,

gives rise to this spectral

complex conjugate, that

component

is

the

is,

form

v(t).

we

shall

J °>"' = 2Re(V„e + K..*"*- = Ke*" + v:e-

(1.5-4)

sh all o ften find that, given a wave2 (t)

where the bar indicates the is done

lime-average value. In the case of periodic waveforms, the time averaging

over one cycle. Ja-')

we

be interested in the quantity v

If,

in

our mind's eye, we were to imagine that the waveform t^n

appear across a 1-ohm

resistor,

then the power dissipated in that resistor would

SPECTRAL ANALYSIS

10 PRINCIPLES OF COMMUNICATION SYSTEMS

2

ohm =

2

W

W

would be numerical]) where the number 2 (t), the mean-square value of v{t). For this 2 reason v (t) is generally referred to as the normalized power of v(t). It is to be kept in mind, however, that the dimension of normalized power is volts^ and not

be

D

volts /l

(t)

equal

watts.

t

watts,

o the numerical value of

When, however, no confusion

We will

numerator and the denominator

evaluate, say, S 2 /S,. this ratio directly

if

the dimension "watts"

is

of the ratio will

ut

An

impedance

amplifier of input

It

K

is

in the

K

to specify

If it

should happen that R;

= Rc

defined by

^ are stating the

advantages of the use of the decibel are twofold.

may

be expressed in decibels by a

number. Second,

nient

may

power

if

much

First,

more conveif

the ratios

expressed in decibels. Suppose that S, and S, are, respectively, the nor-

V\/2, S,

may

real

be written as

in

Eq. (1.6-2)

if

Rj

f R0

then Eq. (1.6-4) does not apply.

,

power

So

far as the 1.6-1

K,„,

= K norm

=

,

the other hand,

if

we

calculate

101og^ = 20 log^

normalized power gain

are

Fig.

On

we have

gain, then

K„o,n,

(1.6-4)

absolutely irrelevant.

is

(1.6-5)

concerned, the impedances

If

should happen that

it

R and R 0 in = /?„, then ;

/?,

but otherwise they would be different.

1.7

NORMALIZED POWER

IN A

FOURIER EXPANSION

ratios are to be multiplied, such multiplication

Let us consider two typical terms of the Fourier expansion of Eq. take, say, the fundamental

and

(1.1-5).

If

we

harmonic, we have

first

V2 and V Then

malized power associated with sinusoidal signals of amplitudes

=

X

a very large power ratio

smaller and therefore often

be accomplished by the simpler arithmetic operation of addition

first

But

(1.6-1)

number S 2 /S, or the number K, the term decibel (abbreviated dB) is attached to the number K. Thus, for example, suppose S 2 /Sj = 100, then log S 2 /S, =2 and K = 20 dB. The

we

then

,

K relI = 201ogp°

the normalized

dimensionless. However, in order that one

whether, in specifying a ratio

J?,

applied to

more convenient not

but instead to specify the quantity

Like the ratio S 2 /S,, the quantity

may know

with load

cancel out.

frequently turns out to be

10 log

IS,

ratios

have been done, for the dimensional error

Ka

S2

V cos

Suppose that in some system we encounter at one point or another normalpowers S] and S 2 If the ratio of these powers is of interest, we need but to

ized

are

=

Figure 1.6-1

have occasion also to calculate

shall often

t'.(f)

from so doing, we shall often follow word "normalized" and refer

results

of normalized powers. In such cases, even

normalized power, no harm

o-

Amplifier

u

the generally accepted practice of dropping the

instead simply to " power."

1

t

.

= V 2J2, and

v'(t)

K = 101og^ = 201og^

(1.6-2)

To

=

C, cos

calculate the normalized

(jr - 0,) + C 2

power

S' of

v'{t),

cos

(jr -

we must square

4> 2

(1.7-1)

)

v'(t)

and evaluate

To! 2

The use

of the decibel

was introduced

in the early

days of communications

systems in connection with the transmission of signals over telephone

"bel"

comes from the name Alexander Graham

in decibel

decibel

was used

for the

purpose

ized powers. Because of this early history, occasionally

the

meaning of

fusion and,

a ratio expressed in decibels.

we hope,

Bell.) In

of specifying ratios of real

thereby to avoid

it, let

To

lines.

Here a waveform

=

l*(t)¥ dt

(The

those days the

powers, not normal-

some confusion occurs

in

point out the source of this con-

us consider the situation represented

= V, cos wt is applied to the input of a linear amplifier of input impedance R,. An output signal v 0 (t) = V„ cos (wr + 6) then appears across the load resistor R 0 A real power P, = Vf/2Ri is supplied to the 2 input, and the real power delivered to the load is P 0 = V J2R 0 The real power

in Fig. 1.6-1.

S'

1

When we

square

we

v'(t),

get the square of the

term, the square of the second

(1.7-1) are orthogonal.

period, the result

is

term corresponding to

is, when their product is integrated over a compleu Hence in evaluating the normalized power, we find no

That

zero.

this cross product.

We

find actually that S'

is

given by

v,{t)

.

PJP,

first

term, and then the cross-product term. However, the two cosine functions in Eq.

S

.

gain

(1-7-2)

-ToC

of the amplifier expressed in decibels

K--10 'eg

is

^

By extension Fourier series

d.6-3)

it is

= ~T + 2

(1.7-3)

2

apparent that the normalized power associated with the entire

is

S

=

C£+I?

d.7-4)

12 PRINCIPLES OF

SPECTRAL ANALYSIS 13

COMMUNICATION SYSTEMS

Hence we observe

that because of the orthogonality of the sinusoids used in a

Fourier expansion, the total normalized power

power due

to each term in the series separately.

which are not orthogonal,

of terms

We may

apply.

this

the normalized power

w

= Al+

S

E B=l

we

write a

of the normalized

waveform

as a

sum

very simple and useful result will not

note here also that, in terms of the

resentation of Eq. (1.1-1),

sum

the

is

If

,4's

and Fs of the Fourier rep

A2

w

R2

1

»=\

l

y+Iy

(1-7-5)

-nf

power and normalized power are to be associated with a complex waveform. Thus suppose we have a not with a real waveform and term A n cos (2nnt/T0 ) in a Fourier series. Then the normalized power contributed to be observed that

It is

2 by this term is A J2 quite independently of all other terms. And this normalized power comes from averaging, over time, the product of the term A„ cos (Innt/T) by itself. On the other hand, in the complex Fourier representation of Eq. (1.2-1)

we have terms term

zero.

is

of the form

We

Vr eJ2 *"" To The .

average value of the square of such a

find as a matter of fact that the contributions to normalized

power come from product terms

y The

iv„r

is

total

A

two-sided power spectrum

component. The height of each nience and because

we

_

v„F_„

= vn V*

(1-7-6)

is

it

throughout

We

vertical line is

|

Vn

2 . \

Because of

lends a measure of systemization to very

shall use the two-sided

its

greater conve-

many

calculations,

amplitude and power spectral pattern exclusively

this text.

shall similarly use a two-sided representation to specify the transmission

Thus, suppose we have a low-pass

filter which transmits components up to a lrequency fM and transmits nothing at a higher frequency. Then the magnitude of the transfer function will be given as in Fig. 1.7-2. The transfer characteristic of a bandpass filter will be given

characteristics of

-n*"tlTo ]2mtlToy_ e ne

normalized power

Figure 1.7-1

filters.

without attenuation

all

spectral

as in Fig. 1.7-3.

S

= "£V„Fr n

Thus,

in the

=

nf0 (f0

component

is

in the

the fundamental lrequency)

at nf0

the combination of spectral

one

o.

complex representation, the power associated with a particular

frequency n/T0 the spectral

— -

(1.7-7)

is

nor with the component

components, one

in the

negative-frequency range. This power

!«'/'!

real

associated neither with

at

-nf0

,

but rather with

positive-frequency range and

is

Vn V* + V_„V* r = 2KVr,

(1.7-8) Figure

nonetheless a procedure of great convenience to associate one-half of the power in this combination of spectral components (that is, V„V*) with the freIt is

quency n/0 and the other half with the frequency -nf0 always be valid provided that

we

power associated with the

transfer

characteristic

of

a procedure will

-nf0 Thus we may say component at nf0 is V„V* and the component at -nf0 is similarly V„V* .

to arrive at the result that the total

correspondence with the one-sided and two-sided spectral amplitude

one-sided and two-sided spectral spectral diagram is shown in power (normalized) power diagrams. A two-sided associated with each spectral power the labeled S„, axis is vertical Fig. 1.7-1. The pattern

The

spectral

power associated with the spectral (= K_„ V* „). If we use these associations only power is 2V„V*. we shall make no error. In

Such

1.7-2

idealized low-pass hliei

are careful to use the procedure to calculate

only the total power associated with frequencies n/0 and that the

.

/

of

Fig.

1.1-1

we may

construct

-I:

Figure 1.7-3

'I,

The

transfer characteristic of

I

I.

an idealized bandpass

filter

with passband from /, to /,

an

SPECTRAL ANALYSIS 15

14 PRINCIPLES OF COMMUNICATION SYSTEMS

1.8

To

POWER SPECTRAL DENSITY

Suppose start

that, in Fig.

at/= - oo

1.7-1

where S„

and then, moving

is

given for each spectral component, we

in the positive-frequency direction,

normalized powers contributed by each power spectral

line

typically will S(f), a function of frequency. S(f)

up

we add

the

range

df.

power

we must

spectral density,

S(f).

Thus we would

differentiate S(f) in Fig. 1.8-1.

=

At

0.

a

harmonic

to the size of the

fre-

jump

in

find

to the frequency/

G(/)=

have the appearance

sum is spectral line to shown in Fig. 1.8-1. It does not change as / goes from one spectral line is each power of another, but jumps abruptly as the normalized frequency the power at normalized / in a added. Now let us inquire about the

This

find the

Between harmonic frequencies we would have G(f) quency, G(/) would yield an impulse of strength equal

This quantity of normalized power dS(f) would be written

t

\VJ

2

d(f-nf0

(1.8-6)

)

n= - a If,

in plotting G(/),

we were

to represent

an impulse by a

vertical

arrow of height

proportional to the impulse strength, then a plot of G(f) versus/ as given by Eq. (1.8-6) would have exactly the same appearance as the plot of S„ shown in Fig. 1.7-1.

dS{f)

The quantity dS(f)/df is

=

^df

(1.8-1)

power spectral density G(f); thus

called the (normalized)

1.9

EFFECT OF TRANSFER FUNCTION

ON POWER SPECTRAL DENSITY Let the input signal

The power in the range d/at /is G(f) range/, to/2 i> S(/,

The power

df.


in the negative-frequency

range

in the positive-frequency

The power

£

to -J,

to a

filter

G{f)=

(1.8-3)

G(f)dJ

-f2

v,{t)

t

n= -

where, from Eq.

is

"
G,{f). If

Vin

are

ym\ 2 V-n/o)

\

(1-9-1)

oo

(1.2-2).

f

1

S(-/2
have a power spectral density

the spectral amplitudes of this input signal, then, using Eq. (1.8-6)

To/2

vW "" 2

K„=~

d-8-4)

'o j -Ton

7*

di

(1.9-2)

J"

The

quantities in Eqs. (1.8-3)

However, the total power in the significance, and this power S(/,

and

do not have physical

(1.8-4)

real frequency

< /
1

)

is

significance.

range/ to/: does have

physical

Let the output signal of the

)

v 0 (t). If V„„ are the spectral

amplitudes of

this

is

given b\

G(f)dj

G(f)df+

n

(1.8-5)

2

I

Go(/)=

J?

S(Z < l/l
be

filter

output signal, then the corresponding power spectral density

I

K„\

b\f- nf0 )

(1.9-3)

-

pro/2 l

Vm

where

v 'o J

As discussed

in Sec.

output coefficient

1.5, if

j^-n«uiT a

the transfer function of the

V„„ is related to

.

dl

(1

. 4)

To/2 filter is

H{f), then the

the input coefficient b\

K„=V H(f=nf0 in

)

(1.9-5)

Hence.

\VJ 2 \W=nfo)\ 2 Substituting Eq. (1.9-6) into Eq. (1.9-3) Figure

1*1 The sum

power

/= -ao

in

to

all i

S( f) of the

spectral

normalized

components

and comparing the

(1-9-6)

result with Eq. (1.9-1)

yields the important resuli

from

G 0 (f) = GU)\H(f)V

(1.9-7)

16 PRINCIPLES OF

importance since

(1.9-7) is of the greatest

Equation

SPECTRAL ANALYSIS 17

COMMUNICATION SYSTEMS

it

relates the

power

spec

-

one point in a system to the power spectral the density at another point in the system. Equation (1.9-7) was derived for signals and nonperiodic to special case of periodic signals; however, it applies

tral density,

and hence the power,

signals represented

by random processes

The

S

[iu<)]=

(Sec. 2.23) as well.

Vf ifj

A waveform v,{t) of transform Vj(f) is transmitted through a network of The output waveform vQ(t) has a transform Vc (f) = V{f)H{f

Figure 1.10-1

As a special application of interest of the result given in Eq. (1.9-7), assume a that an input signal v,{t) with power spectral density G&f) is passed through differentiator.

Vj\t>

at

differentiator output v 0 (t)

is

related to the input

tion H{f).

by amplitudes V(f)

more

where

x is a

constant.

The operation

transfer func-

I.

t

df.

The quantity V(f)

n/")

indicated in Eq. (1.9-8) multiplies each spec-

2 component of y,(t) by jlnfx = j'wt. Hence H(f)=jwx, and \H(f)\ = coV. is output of the Thus from Eq. (1.9-7) the spectral density

tral

in

is

generally the Fourier transform of

correspondence with

Vn

which

,

called the amplitude spectral density or

v(t).

The Fourier transform

=

j"

is

given by

o(l)e-

is

given by

JU "di

(1.10-3)

pro/:

G 0(/) =

toVGjt./

V =

d-9-9)

i

It

1

liCie'^"'

rr

(1.10-4)

dl

'o J - T 0 r.

Again, in correspondence with Eq.

1.10

a network.

THE FOURIER TRANSFORM

waveform may be expressed, as we have seen, as a sum of spectral components. These components have finite amplitudes and are separated by waveform is finite lrequency intervals 0 = V To- The normalized power of the

A

If

the input signal

is vj(t),

H(f)V(f)e>

periodic

finite,

as

also the normalized energy of the signal in an interval

is

limit the period T„ of the

suppose we increase without Fig. 1.3-lb the pulse centered

move outward away from

t

around

=

0 as

= 0 remains T0 — oc. Then t

T0 Now .

waveform. Thus, say.

in place,

but

eventually

all

Comparing Eq.

(1.10-5) with Eq. (1.10-2).

yo(f) = ^[vMl

is

let!

^WO] or

continuous variable with a one-to-one correspondence with the integers, becomes instead a continuous variable. The normalized energy of the nonperiodic wave form remains finite, but, since the waveform is not repeated, its normalized power

1.11

The hourier

The

spectral amplitudes similarly

series for the periodic

become

infinitesimal

waveform

v(t)=

jr

(1

In this sedion

we

10-1)

Example

shall evaluate the

(see

Prob. 1.10-1)

1.11-1

If v(t)

=

= j°

V(f)e J2

" J,

cos

w0

v(t)

=

The exponential Fourier dJ

b> (1-10-6) (1.10-7)

transforms of a number of functions both for

1,

have occasion to

shall

refer to the results.

find V(f).

cos

w0

1

is

periodic,

and therefore has a

spectral amplitudes

VK

series representation of v{t) is

(1.10-2)

v(t)

finite

v,{t)

Fourier series representation as well as a Fourier transform. „( t )

The

see that the Fourier transform

= H(fmf)

and also because we

Solution The function becomes

given by

EXAMPLES OF FOURIER TRANSFORMS

the sake of review

vn epnf '"

v 0 (t)

(1.10-5)

df

m/WMt)]

=

K(f)

as indicated in Fig. 1.10-1.

infinitesimal.

the transfer function of

in

with a single-pulse nonperiodic waveform As T0 -> oo. the spacing between spectral components becomes infinitesimal. The frequency of the spectral components, which in the Fourier series was a dis-

becomes

we

2 *i>

related to the transform K
other pulses

we would be

H(f) be

(1.5-8). let

then the output signal will be

are analogous to the infinitesimal spectral

=

\e

+ -I-

\e - "°

0>

w0 =

(1.11-1) '0

SPECTRAL ANALYSIS 19

18 PRINCIPLES OF COMMUNICATION SYSTEMS

Thus

K,-K.,-J

(l.H-2)

=

0

(1.11-3)

The Fourier transform V(f)

is

and

V(f)

K„

=

2 (Me"' ""

cos

n

+ ±1

found using Eq. *-' 2 ""-'*

*=

5 J°

J'

(1.10-3):

*+5

f-*""***

A

J" (l.ll-4o)

= W/-/o) + W/+/o) [See Eq.

(1.1 1-22)

(1

-

11 " 4 *)

below.]

Eqs. (1.11-1) and (1.11-3) we draw the following conclusion: The of a sinusoidal signal (or other periodic signal) consists of transform Fourier impulses located at each harmonic frequency of the signal, i.e., at/„ = n/T0 = nf0 The strength of each impulse is equal to the amplitude of the Fourier coefficient

From

.

of the exponential series.

Example

A

1.11-2

signal m(t)

quency fc The product .

multiplied by a sinusoidal

is

signal

=

v{t)

If

waveform of

fre-

is

the Fourier transform of m(t)

ntt) cos 2s/r

M(/), that

is

A

(1.11-5)

1

oidal.

case of special interest arises

is,

m(f)

M(f) =

m(t)e~

j2 *"

(1.11-6)

di

where

when

waveform

the

m(t) is itself sinus-

Thus assume

m

a constant.

is

=m

We then find

cos 2nfm

that V(f)

(1.11-10)

t

is

given by

J° find the Fourier transform of

v(t).

V{f)

= ^d{f+f +/J t

Solution Since m(t) cos

2nfc

t

=

then the Fourier transform K(/)

is

*m(,V"-'

(1.11-8) with Eq. (1.11-6).

the right

r

is

M(f)

we have

in Fig. 1.11-2.

(l.ii-U)

Observe that the pattern has +/„ and / -/„.

four spectral lines corresponding to two real frequencies/,. (1.1 1-8)

M(f)

(i.n-9)

i

m

to the

left,

we

see the spectral

same iorm. One is shifted to each by amount/.. Further, the amplitudes of is

i

m

m

m

of m(t) to the transform V(f) ol

replaced by two patterns of the

and one

M(j

shown

t

the result that

illustrated in Fig. 1.11-la. In Fig. 1.11-16

each of these two spectral patterns pattern

is

Hf-f +/J + j */-/, -/J

IV(/)|

relationship of the transform

pattern of

+ J */+/. -/J + j

(1.11-7)

J'

c

m(t) cos 2jt/r

""

mfrV-^-^" A

I

v(n=mf+L)+Wif-f The

2

This spectral pattern

(t

J°

Comparing Eq.

+

given b\

m >-^' A +

=1

K(/)

2 $m(t)e> "''

one-half the amplitude of the spectral

't

i

-/ -/ m f

!

-/.

r -/;+/„

t'._ 0 fr

Figure 1.11-2 The two-sided amplitude spectrum of the product waveform

c(t)

=m

cos

2)1/,

t.

SPECTRAL ANALYSIS 21

PRINCIPLES OF COMMUNICATION SYSTEMS

The waveform

itself is

Example

given by

(a) (b)

1.11-4

Find the Fourier transform of <5(r), an impulse of unit strength. Given a network whose transfer function is H(f). An impulse applied at the input.

Show

the inverse transform of H(f), that

+-

ry2«tr«-/-><

+

p-j2«(/ ( -/-)rj

is,

show

is

is

x

that

(l.ll-12a)

Solution

4

(a)

=^ 4

<5(t)

= h(t) at the output h(t) = F~ [H(f)\

that the response v 0 (t)

+fjt +

[cos 2n(fc

cos 2n(fc

-/Jr]

(1.1

The impulse

S(t)

=

0 except at

=

t

0 and,

further,

has the property that

M2fc)

«(t)A«l

f

(1.11-18)

J- o

Example

1.11-3

A

pulse of amplitude

^

extends from

f

= -t/2

to

r

=

+t/2.

Hence

series for a V(f). Consider also the Fourier the Compare T intervals separated by 0 periodic sequence of such pulses -» oc. T limit as in the transform with the V„ 0 coefficients Fourier series

Find

houner transform

its

2 HQe-* '"

=

K(/)

dt

=

(1.11-19)

1

J

.

Thus

(b)

Solution

We

have directly that

output

•t/:

V(f)

-, , ^-jz*/'

=

dt

—y-

sin t/1

=

At

n

< <

i>

=

0 (r)

h(t) is V„(f)

given by

W)=lxW)

ni

(1.11-13)

71/1

The

the spectral components of <5(f) extend with uniform amplitude and phase over the entire frequency domain Using the result given in Eq. (1.10-7), we find that the transform of the

since the transform of

V0 (f), which

given by Eq. Fourier series coefficients of the periodic pulse train are

cifically, for

is

S(t),

the function

^[S(ty]

=

(1.11-20)

Hence

I.

the inverse transform of

also the inverse transform of H(f). Spe-

h(t), is

an impulse input, the output

i>

(1.3-12b)a>

= The fundamental frequency /„

= Af in

At.

si n

(n7tT/7jj (J

in the Fourier series

order to emphasize that

f0 s Af is

=

l/T0

.

We

m=

j

We may shall set

1/T0

=

Af,

we may

5{t) is

If

«5(r) itself.

the impulse

itself.

H(/y2 *"

5(0

H(f)

=

1,

then the response

Hence, setting H(f)

=

e*

2 ""

=

1

in

df=

J*

Sin<™4/r)

(i.n-21)

df

h(t)

to an inpulse

Eq. (1.11-21),

we

find

rewrite

(1.1 1-14) as

dj

(1.11-22)

J"

(11M5)

Af

7

*

j

use the result given in Eq. (1.11-21) to arrive at a useful rep-

resentation of

the trequency interval between

spectral lines in the Fourier series. Hence, since

Eq.

is/0

n^4

nn A/t In the limit, as n

A/by

a

T0 -»

oo,

Af— 0. We

then

may

replace

replace

Af by
continuous variable /. Equation (1.11-15) then becomes

lim Vm

=

At

^

1.12

CONVOLUTION

Suppose that

V2 (f). What (l.H-16)

df

i>,(r)

then

is

has the Fourier transform V^f) and v 2 {t) has the transform the

waveform

This question arises frequently

v{t)

whose transform and

in spectral analysis

is is

V (f)V2 {f)'} answered by the convol-

the product

1

ution theorem, which says thai

Comparing

this result

with Eq.

(1.1 1-13).

we do indeed note

that

"(0

V(f)= Thus we confirm our density.

lim

£

earlier interpretation of

=

J

(1.1M7)

v t (t)v 2 (t

-

t)

dt

(1.12-1)

v 2 (x)v,(t

-

t)

dr

(1.12-2)

or equivalent!}

K(/) as an amplitude spectral p(0

= J

SPECTRAL ANALYSIS 23

22 PRINCIPLES OF COMMUNICATION SYSTEMS

convolution integrals, and the integrals in Eq. (1.12-1) or (1.12-2) are called referred to as taking the conis integrals these process of evaluating v(t) through

The

and v 2 (t) prove the theorem, we begin by writing

volution of the functions p,(t)

To

= ^- tF (/)K2 (/)J

(1.12-3a)

1

^()

1

(1.12-36)

2n

1.13

THEOREM

PARSEVAL'S

We saw that for periodic waveforms we may express the normalized power as a summation of powers due to individual spectral components. We also found for periodic signals that it was appropriate to introduce the concept of power spectral density. Let us keep in mind that we may make the transition from the periodic to the nonperiodic waveform by allowing the period of the periodic waveform to approach

infinity.

A

nonperiodic waveform, so generated, has a

finite

normalized

power approaches zero. We may therefore expect that for a nonperiodic waveform the energy may be written as a continuous summation (integral) of energies due to individual spectral components in a continenergy, while the normalized

By

definition

we have

W)= Substituting

as given

by Eq.

[°

(1.12-4)

r,(tV-*"*

(1.12-4) into the integrand of

Eq. (1.12-3b). we

uous distribution. Similarly we should expect that with such nonperiodic waveforms it should be possible to introduce an energy spectral density The normalized energy of a periodic waveform v{t) in a period T0 is T

have

£= Vj(t)e-

JO"

dzV2 (fy

a

'

dw

"

Mt)l

(1.12-5)

From

Eq. (1.7-7) we

may

write

Interchanging the order of integration, we find

£

E=T0 S=T0

V2 (fV"*-« dw^di

Bi (t)

recognize that the expression in brackets in Eq. (1.12-6)

We

£\ Vi

"

Again, as in the illustrative Example 1.11-3, v 2 (t

-

fundamental frequency, that x),

so thai

is,

Af

is

(1.13-2)

n

= -

o

(1.12-6)

is

(1.13-1)

dt

as

n

*)-]""

2

J- To>:

we

let

A/ = l/T0

= f0

,

where f0

the spacing between harmonics.

is

the

Then we

have

finalh v,(t)v 7 U

-

KVt =

(1.12-7)

t) di

and Eq. Examples of the use of the convolution 1.12-3. These problems will also serve to

integral are given in Probs. 1.12-2 recall to the

(1.13-2)

may

theorem and one of very great utility is Suppose a waveform vfr) whose considerations. following the through arrived at transfer function tf(/). The translorm is V^f) is applied to a linear network with then is the waveform of What transform of the output waveiorm is Vlf)H(f).

and v 2 (t) with the inverse transform that the inverse transform of H{f) is of H(f). But we have seen Tin Eq. (1.11-21)] Hence Eq. (1.12-7) becomes impulse response of the network. h(t).

we

the

=

j

•j(r)h[t »i

-

In the limit, as

A/—»0, we

77 AT

Af by

¥V

d- 13 " 41

-77

df,

we

replace

=

V(f)V*(f) df=

\

V(f)\

2

in

response of the network.

is

expressed in terms of the input

df =

Wt)Y

This equation expresses Parseval's theorem. Parseval's theorem

and the inpulse

dt

(1.13-5)

is

the extension to

the nonperiodic case of Eq. (1.7-7) which applies for periodic wavelorms. Both

(nonperiodic case) v,{t)

the translorm

Equation (1.13-4)

J°

may

the fact that the

power

(periodic case) or the energy

be written as the superposition of power or energy due to

components separately. The validity of these results depends components are orthogonal. In the periodic case the components are orthogonal over the interval T0 In the nonperiodic case

individual spectral

which the output v 0 (t)

VJAfby

integral.

then becomes

(1.12-8)

di

replace

I .= -»

V(f) [see Eq. (1.11-17)], and the summation by an

results simply express t)

13 ' 31

v V*

>

£=

identify w,(t) with vfc)

i>„(t)

-

be written

reader the relevance of the

special case of the convolution

In Eq. (1.12-7)

(1

and

term convolution.

A

i/^ (A/)5

on the

fact that the spectral

spectral

.

SPECTRAL ANALYSIS 29

28 PRINCIPLES OF COMMUNICATION SYSTEM?

Figure 1.15-2

and

The of

response

The

essential

network input

is

which

is

/,t

(6)

(01

distortion

=

rise

time

t,

a

high-pass

a

0.35r.

RC

(i>)

circuit

RC The for

o.o:.

limit.

From

when

Eq. (1.15-7)

the

If

very

cutoff/, alone determines the percentage

much more complicated

the input to the

RC

than the simple

high-pass circuit

output has the waveshape shown

and

v 2 (t) are

waveforms of

finite

energy

type waveforms), then the cross correlation

Ri2W =

(1.15-8)

drop

in

voltage

Again the importance of Eq. (1.15-8) is that, at least as a reasonable approximation, it applies to high-pass networks quite generally, even when the is

If U](f)

c,(t)D 2 (i)

=

0.

is

(for

example, nonperiodic pulse-

defined as

is

RC circuit

in Fig. 1.15-2b.

The output

The need

may

t,

then the

exhibits a

tilt

and

an undershoot. As a rule of thumb, we may assume that the pulse is reasonablx faithfully reproduced if the tilt A is no more than 0.1K0 Correspondingly, this .

parameter

for introducing the

(1.15-9)

A

(1.16-3)

i in

the definition of cross correlation

the integral of the product Vi(t)v 2 (t)

one or the other function shifted to the

R ]2 (x) Thus

left

"searching" or shift to reveal,

synonym

for correlation.

v 2 (t

is

+

zero since at

t)

is

all

times

the function v 2 (t)

that, while R i2 (0) = 0, maximum when t = t 0 "scanning" parameter which may be adjusted to a to the maximum extent possible, the relatedness or

by amount

x. It is

will increase as t increases t is a

The function

zero.

is

clear

from the figure

from zero, becoming a

correlation between the functions.

/,t*0.02

t)

while different, are obviously related. They have the same period and nearly the

proper time

condition requires that /j be no higher than given by the condition

+

be seen by the example illustrated in Fig. 1.16-1. Here the two wavelorms,

same form. However,

a pulse of duration

»,(t)» 2 (t

J

RC

di

per unit time.

network

The

related waveforms.

such that the product

we have

-J- =-2nj\ v.

Thus the low-frequency

is

introduced by the Irequency discrimination of this

that the output does not sustain a constant voltage level

asymptotic

Two

Figure 1.16-1 timing

constant. Instead, the output begins immediately to decay toward zero,

is its

A"7* m

1

rectangular pulse (solid)

the response (dashed) of a low-pass

circuit.

T-»

A

(a)

The term coherence

Functions for which

R 12 (t) -

0

.

is

sometimes used as

a

for all x are described as

being uncorrected or noncoherent In scanning to see the extent of the correlation

1.16

ary to specify which function

CORRELATION BETWEEN WAVEFORMS

The correlation between wavelorms

is

nor confined to a

finite

»j(t)

time interval.

and

Then

isR 12 (i) defined

v 2 (t), not nec-

and

it is

necess-

not equal to

*2iW = l™ r-oo

»,(t + r)v i1 J-t/: f

2 (t)

dt

= R l2 (-i)

(1.16-4)

and with identical results for periodic wavelorms or wavelorms of finite energ\

T

u,(0

v,{t)

as

R 12 (x) s hm If

is

readily verified (Prob. 1.16-2) thai

the correlation

between them, or more precisely the average cross correlation between v 2 (t),

It is

between functions,

being shifted. In general, Ri 2 (x)

a measure of the similarity or relatedness

between the waveforms. Suppose that we have wavelorms essarily periodic

R 21 {t).

is

^

Vt(t)v 2 (t

+

t) di

(1.16-1)

J7:

v 2 (t) are periodic with the

average cross correlation

:

same fundamental period T„, then

the

is

Let To/2

*1 2

1.17

M=T '0

»,(i)o 3 (t .

To/:

+

POWER AND CROSS CORRELATION i>,(t)

and

v 2 (i)

be waveforms which are not periodic nor confined to a

time interval. Suppose that the normalized power of t) di

t>,(t) is

(1.16-2)

ized

power of

v 2 (t)

is

S 2 What, .

then,

is

the normalized

finite

and the normal-

power of u,(t)

+

v 2 (t)l Or,

SPECTRAL ANALYSIS 31

30 PRINCIPLES OF COMMUNICATION SYSTEMS

more

generally,

what

the normalized

is

power S 12 of

+

y,(r)

v 2 (t

+

t)?

We have

That

=

"

1

lim r-oc

'

f

[17,(0

+

2 (t

t>

+

t)]

2

A

r/;

1

v

f

2

U-jv:

'

r/2

+

dt

(t)

+

[v 2 (t

t)]

2

A+

2

Vi(t)v 2 (t

+

S,

-ti:

J-r/2

power of

v 2 (t

+

t) is

the

between

shift in v 2 will clearly

that the normalized

we have taken account of the fact same as the normalized power of

v 2 (t). For, since the

extends eventually over the entire time axis, a time

not affect the value of the integral

Eq. (1.17-lc) we have the important result that if two waveforms are uncorrelated, that is, R l2 (r) = 0 for all x, then no matter how these waveforms are

From

power due

time-shifted with respect to one another, the normalized

to the super-

sum of the powers due to the waveforms individually. Similarly if a waveform is the sum of any number of mutually uncorrelated waveforms, the normalized power is the sum of the individual powers. It is position of the waveforms

readily verified that the

is

Vi(t)

=

R 12 (r)

that

same

result applies for periodic

and

K,

+ V and = K] V2 In most v\(t)

i

.

V2

waveforms. For

v 2 (t)

is

=

v 2 (t),

t)

be a

we would

maximum when

t

surely expect that simi-

=

0.

The student

is

guided

1.18-1.

= R(-x)

(1.18-4)

Thus the autocorrelation function is an even function of t. To prove Eq. (1.18-4), assume that the axis t = 0 is moved in the negative t direction by amount t. Then the integrand in Eq. (1.18-1) would become v(t - x)v(t), and R(x) would become R(-x). Since, however, the integration eventually extends from -oo to oo, such a shift in time axis can have no effect on the value of the integral. Thus

R(x)

= R(-x) The three

characteristics given in Eqs. (1.18-2) to (1.18-4) are features not only of R(x) defined by Eq. (1.18-1) but also for R{x) as defined by Eqs. (1.16-2) and (1.16-3) for the periodic case and the non-periodic case of finite energy. In the

=

latter case, of course, R(0)

1.19

£, the energy rather than the power.

AUTOCORRELATION OF A PERIODIC WAVEFORM

When

the function

=

2 (r)

i>'

+ V2

will

correlated

be

with

is

periodic,

we may

write

t

v(t)=

correlation

applications where the correlations between waveforms

is

K„ e

i2nMti

"

(1.19-1)

and, using the correlation integral in the form of Eq. (1.16-2),

we have

T ° 12

R(X)

is

called the autocorrelation.

R(x) given, in the general case,

=

lim T-oo

Thus with

v{t)v{t

^

R(0)

integration shall

be

2 „,, + t)/ T0 ]

dt

(U9. 2)

/

and summation may be interchanged in Eq. (1.19-2). If with a double summation over m and n of terms I m „

left

+

t) di

C

'(>

J-

2 ""' To T^

Since m and n are integers, we see m = -n or m + n = 0. To evaluate 3a)

^ 1

MO] -T/2

2

dt

=

S

(1.18-2)

Vm

= Vm V„ e32 "tlTB

(1.18-1)

and

To/2

l

'

T/2

lim T-oo

0

L., =

by

of the properties of R(x) are listed in the following

=

\( £-„ Ky ^\J-To/2 (\m=X-« Vmei2^!T /W

given by

of a function with itself

R 12 (x) becomes

=

J0

•ti:

(a)

+

rarely

R(x)

A number

(1.18-3)

v\(t)

AUTOCORRELATION

The correlation

v{t

R(z)

The order of we do so, we 1.18

R(r)

finite

any interest in the dc component. It is customary, then, to continue to refer to waveforms as being uncorrelated if the only source of the correlation is the dc components. of concern, there

and

and v'2 (t) are uncorrelated. If dc are added to the waveforms, then the waveforms

waveforms

two

power S of the waveform.

the

energy waveiorms, the result applies to the normalized energ\

Suppose components

t^t)

(c)

(1.17-lc)

integration in Eq. (1.1 7- lft)

>

rather intuitively obvious since

is

J

+ S 2 + 2K 12 (t)

In writing Eq. (1.17-lc),

the average

is

through a more formal proof in Prob.

x)ai> x)dt\

(1.17-lb)

=

0

«(0)

This result

J-Ti:

ff - \\

lim t-oo

=

(1.17-1*) larity

=

the autocorrelation for t

W

T

S l3

is,

Vy

2 *<"

+ "»' T

°

d:

(1.19-3a)

To/2

—~ +

-

n(m

from Eq.

(1.19-36) h)

(1.19-3fc) that l m „

l m _„ in this latter case,

we

=

0 except when

return to Eq. (1.19-

find

= '-„.„=

KV.^ ™ "' 2

1

(1.19-4)

SPECTRAL ANALYSIS 33

32 PRINCIPLES OF COMMUNICATION SYSTEMS

We

Finally, then.

use the convolution theorem.

the case where the

£

R( X )=

V.V-.e'

2 '""'*'

=

2 J2m" T ° \VJ e

I

n= - a

n= -

I

2 \

+

£

2

Vm

|

2 |

=

Vi(t)

cos 2n«

note from Eq. (1.19-56) that R(x)

same fundamental period T0

We its

shall

power

R(x).

We

now

«

For

find, using R(x) as in

V(-f)

is

=

V*(f)

= F[v(-t)l

^-»[K(/)F*(/)] =

also periodic with

this

waveform

to

The

purpose we compute the Fourier transform of

F~

Eq. (1.19-5a), thai

integral in Eq. (1.20-2) l

may

Eq. (1.20-1)

F~X\

=

Z

(

I

Vn

-

t) dx

(1.20-1)

be written

K(/)|

2

]

KtMt - 0

=

e

"nlTo )e' J2 'fx

= -a <x 00 \« \n~

dx

dx

(1.20-2)

e

Comparing Eq. (1.19-8) with Eq. periodic waveform

(1.8-6),

2

\K\

and

x.

We

If

t,

and hence

we want

&~

1

[V(f)V*(f)~\ as a

we need but

to inter-

1

[K(/)F*(/)]

=

v(t)iit

-

x) dx

(1.20-3)

yields

-J2'U-»IToH ix

The

integral in Eq. (1.20-3)

is

precisely R(x),

F[R(x)}

(1.19-7)

=

and thus

V(f)V*(f)

=

|

V(f)\-

which verifies that R(x) and the energy spectral density form pairs

s(f-^)

we have

equation expresses

this

to express

then have -

U-19-8)

the interesting result that for a

= #T*M]

d- 19 " 9 )

In the preceding sections

We

|

V(f)\

(1.20-4) 2

are Fourier trans-

AUTOCORRELATION OF OTHER WAVEFORMS

1.21

tion function

G(/)

t.

(1.19-6)

write Eq. (1.19-7) as

JW)]= i

I

/

\K\

a function of

function of x without changing the form of the function,

2 J2 \

is

\V(f)V*(f)~] as a function of

change

we may

oft)*

J"'

Wt)]= I (1.11-22),

get

.

Interchanging the order of integration and summation

Using Eq.

and

and we note as well

as anticipated,

relate the correlation function R(x) of a periodic

spectral density.

v(t),

combine Eqs. (1.12-1) and (1.12-3) for same waveforms, that is,

v 2 {t) are the

(1.19-5b)

the correlation R(x)

waveform

that for this case of a periodic the

= R(-x)

=

We

and

p-*mf)v(n\ =

-

Since

We

v 2 (t)

u,(l)

o

CO

= V0

(1.19-5fl)

waveforms

we

discussed the relationship between the autocorrela-

and power or energy

spectral density of deterministic waveforms.

use the term "deterministic" to indicate that at least, in principle,

it

is

pos-

which specifies the value of the function at all times. For such deterministic waveforms, the availability of an autocorrelation function is of no particularly great value. The autocorrelation function does not include within sible to write a function

and, of course, conversely R(i)

= ^-'[G(/)]

(1.19-10)

itself

Expressed in words, we have the following: T/ie power spectral density and the correlation junction oj a periodic wavejorm are a houner transjorm pair.

1.20

AUTOCORRELATION OF NONPERIODIC FINITE ENERGY

Eq. (1.19-9) for the periodic waveform. This

and the energy spectral density

with such waveforms that the concepts of correlation and autocorrelation find

For pulse-type waveforms of finite energy there is a relationship between the correlation function of Eq. (1.18-1) and the energy spectral density which correrelationship

is

in

tral components of the waveform. The waveform cannot be reconstructed from a knowledge of the autocorrelation functions. Any characteristic of a deterministic waveform which may be calculated with the aid of the autocorrelation function may be calculated by direct means at least as conveniently

On the other hand, in the study of communication systems we encounter waveforms which are not deterministic but are instead random and unpredictable in nature. Such waveforms are discussed in Chap. 2. There we shall find that for such random wavelorms no explicit function of time can be written. The waveforms must be described in statistical and probabilistic terms. It is in connection

WAVEFORM OF

sponds to the relationship given

complete inlormation about the function. Thus we note that the autocorrelis related only to the amplitudes and not to the phases of the spec-

ation function

that the correlation function R{x)

are a Fourier transform pair. This result

is

established as follows.

their true usefulness. Specifically,

it

turns out that even for such

random wave-

SPECTRAL ANALYSIS 35 34 PRINCIPLES OF COMMUNICATION SYSTEMS

transform pair. The proof

3

that such

spectral density are a Fourier

power

forms, the autocorrelation function and

the case

is

formidable and

is

not be

will

Because of the orthogonality,

become zero with a

all

of the terms on the right-hand side of Eq. (1.22-4) and we are left with

single exception X

undertaken here.

EXPANSIONS IN ORTHOGONAL FUNCTIONS

1.22

g„(x)

Let us consider a set of functions g,(x), g 2 (x),

x < any two different ones of the

val Xj

set satisfy the

is.

when we multiply two

=

A

•

,

way

(1-22-1)

and then integrate over the

.

is

vectors V, and encountered in dealing with vectors. The scalar product of two quantity scalar a is product) inner the (also referred to as the dot product or as

i}

"

The mechanism by which we use the orthogonality of the functions to " drain away all the terms except the term that involves the coefficient we are evaluating is

"

often called the " orthogonality sieve

Next suppose that each g„(x) is hand member of Eq. (1.22-6) (which

selected so that the

denominator of the

right-

a numerical constant) has the value

is

2

g

^

V,|

1 1

V, cos (V |

f ,

Vj)

=

(x)dx=)

(1.22-7)

d.22-2)

Vji

=

the respective vectors and In Eq. (1.22-2) |V,| and |V;| are the magnitudes of vectors. If it should turn the between angle the of cosine cosfVj, V,.) is the in which V, = 0 or V ; = 0) cases trivial = the (ignoring 0 then that out K, that the vectors V and V, cos (V„ \j) must be zero and correspondingly it means vectors whose scalar Thus another. are perpendicular (i.e., orthogonal) to one in correspondence, and, another one to orthogonal product is zero are physically orthogonal integrated product, as in Eq. (1.22-1) is zero are also ;

When

^

fix)g„„(x) dx

(1.22-8)

the orthogonal functions g„(x) are selected as in (1.22-7) they are described

as being normalized.

The use

of normalized functions has the merit that the C„'s

can then be calculated from Eq. 2

\*\ g „{x)

dx

in

(1.22-8)

each case as called for

both orthogonal and normalized

is

and thereby avoids the need to evaluate A set of functions which is

in Eq. (1.22-6).

called

an orthonormal

set.

whose

one another.

Now

we have some

consider that

interested infix) only in

arbitrary function f(x) and that x, to x 2

the range from

the set of functions g{x) are orthogonal.

write fix) as a linear

sum

fix)

=

,

i.e.,

are

which

Suppose further that we undertake to

of the functions g„{x). That

C,ff,(x)

we

in the interval over

+ C 2 g 7 (x) +

is.

•+C

n

we

g„(x)

write

+

which the C's are numerical

1.23 COMPLETENESS OF AN ORTHOGONAL THE FOURIER SERIES

Suppose on the one hand we expand a function fix) (1

'

-

22 " 3 )

fix)

coefficients.

interval of orthogonality.

We

= C lSl (x) + C 2 s 2 (x) + C 3 s 3 (x) +

and on the other hand, we expand fix)

have

That /(x)0„(x)

dx

=

C,

is,

in the

=C

l

it

as

s1

(x)+C 3 s 3 (x)+-

•

(1.23-1)

•

(1.23-2)

second case, we have deliberately omitted one term.

A moment's

review of the procedure, described in the previous section, for evaluating

g (x)g„(x)d>

coeffi-

t

cients X2

+ C2

terms of orthogonal tunc

tions

.

and integrate over the

in

SET:

'

Assuming that such an expansion is makes it very easy to compute the qs the of orthogonality the indeed possible, sides of Eq. (1.22-3) by g r (x) coefficients C„ Thus to evaluate C„ we multiply both

in

(1.22-6)

In this case

V =

to

fix)g„(x) dx

V

defined as

functions

becomes

qlix) dx

from x, to x 7 the result is x 2 The term described as being orthogonal over the interval from Xj to which is "orthogonal" is employed here in correspondence to a similar situation

interval

(1.22-5)

that

functions which has this property

set of

2 C„ \'~g (x) dx

are evaluating

defined over the inter-

0

different functions

zero.

•

we

condition

['"flMffjM dx

That

•

in the very special

x 2 and which are related to one another

<

so that the coefficient that

=

dx

'fix)g n (x)

\ J*)

gAx)g„(x) dx

+

+ C„

[ J*'

'g 2 ,(x)

makes

expansions

dx

+

-

-

(1

it

apparent that

will turn

.22-4)

other

is

in error.

We

all

the coefficients C,,

out to be the same. Hence

if

C3

,

etc.,

that appear in both

one expansion

might be suspicious of the expansion of Eq.

is

correct the

(1.23-2)

on the

48 PRINCIPLES OF COMMUNICATION SYSTEMS

SPECTRAL ANALYSIS 49

t he p ower P s and duration T. Thus the magnitude of ^/P S T. Furthermore, each vector is rotated ji/4 radians from the previous vector as shown in Fig. 1.26-3. The reader should verify (Prob. 1.26-1) that Fig. 1.26-3 is correct by finding the two orthonormal com-

Then each

signal has

each signal vector

ponents in a

is

and u 2 {t) as

Ujf')

form similar to Eq.

and (1.26-4), expanding Eq. and then plotting the result.

in Eqs. (1.26-3)

(1.26-5)

(1.26-6)

REFERENCES 1.

Javid, M.,

New

and

E. Brenner: "Analysis of Electric Circuits,"

2d

McGraw-Hill Book Company.

ed.,

York, 1967.

Papoulis, A.:

"The Fourier

Integral

and Applications," McGraw-Hill Book Company,

New

York,

1963. 2.

Churchil, R. V.: "Fourier Series," McGraw-Hill

3.

Papoulis, A.: " Probability,

Book Company, New York,

Random

Variables,

C„ and

to /)„

194!

and Stochastic Processes," McGraw-Hill Book

Company, New York, 1965 Figure

1 26-2

Signal vectors representing the four signals of Eq. (1.26-2|

important to note that the magnitude of the signal vectors shown in Fig. 1.26-2 is equal to the square root of the signal energy. Thus, in Eq. (1.262) each signal s,(t) has a power P s and a duration T. Thus the signal energy is

PROBLEMS

It is

Ps T, which

is

equal to the magnitude squared of the signal vectors as shown

in Fig. 1.26-2. This result simplifies the

drawing of the signal vectors

in signal

space

1.1-1. Verify the relationship of

1.1-2. Calculate

A„, B„, C„, and

which makes peak excursions transition occurs at

I

=

to

+|

s,{t)

= y/2P

signal

1.2-2.

[i= s

cos

C0 o

i

+ (2i-

volt,

is

(l.l-6i

a symmetrical square

and has a period

T=

_

<>
f'~"

}

elsewhere

(0

< T is

repeated every

T=

1

sec.

Thus, with

w(I)

tft)

the unit step function

=

£

-

p(r

n)u(t

-

n)

n= - »

Find 1.3-1.

K„

and the exponential Fourier

The impulse function can be

series for

u(t).

defined as

Ht)

=

" p- 1,1,1

lim

Discuss. 1.3.2.

In Eq. (1.3-8)

we

see thai *

*)=' I *

Set /

=

1,

T0 «

1.

Show by

=

~ kT0 = )

21 —+— 1

'0

a.

£

3

=

1

+

2

£ n~

cos

2nm

1

that the Fourier series does approximate the train of impulses.

2jim

±,

'0«=1

plotting

v(t)

Figure 1.26-3 Signal vectors representing the four signals of Eq. (1.26-6)

1

sec.

complex number K„to C„and 0„ as given by Eq.

(1.26-6) f

—i

which

v(t)

The function

1,2,...,'

1)

t0;£

and

volt

and B„ as given by Eq.

waveform

A

cos

_'0

wave and

positive

0

1.2- 1. Verify the relationship of the

For example, consider the

„

for a

<j>„

(1.2-3)

goinj.'

50 PRINCIPLES OF COMMUNICATION SYSTEM'

=

1.4-1. Sa(x)

A

Determine the maxima and minima maxima and minima obtained by letting x =

train of rectangular pulses,

and are separated by intervals of 10 (a)

Assume

and

Assume

+

and compare your

=

\)n/2, n 1

volt,

1,

side

is

that the

and draw

0,

located at

I

=

0.

Write the exponential Fourier

A

network

is

periodic waveform vff) u„(l)

=

series

also the envelope of these spectral amplitudes.

edge of a pulse rather than the center

left

is

located at

applied to the input of a network.

is

where

x[dv^l)ldt],

(

=

t

What

a constant.

is

Correct the

0.

Why

The output

not

A

!>„(()

=

under these circumstances, the output

ti„(l) is

approximately

u„(t) =:

T[du(t)/di].

and period T.

=

t

I/A,

and period

G(/) for a voltage source represented by an impulse

Comment on

7

and period

train of strength I

nT" loi

this limiting resul;

power spectral density of a square-wave voltage of peak-to-peak amplitude I and The square wave is filtered by a low-pass RC filter with a 3-dB frequency 1. The output is

1.9- 1. G|(/) is the 1.

taken across the capacitor (a)

Calculate G(,f).

(ft)

Find G„(/J.

A

1.9- 2. (a)

sec

(ft) A periodic waveform t^t) is applied to the RC network shown, whose time constant = RC. Assume that the highest frequency spectral component of v£l) is of a frequency/ « 1/t. Show

train of strength I

pulse train of amplitude A, duration

1, 2, 10, infinity.

of the

the transfer function fl(io) of this

is

n

?

is

that,

An impulse

(ft)

period

network 1

t

(a)

1.8- 2. Plot

us

of/ =

Find G(/) for the following voltages:

1.8-1.

have a duration of 2 us

Fourier expansion accordingly. Does this change affect the plot of spectral amplitudes? 1.5- 1. (a)

result with

2

plot the spectral amplitude as a function of frequency. Include at least 10 spec-

components on each (ft)

(2n

making excursions lrom zero to

that the center of one pulse

for this pulse train tral

of So(x)

(sin x)/x.

the approximate 1.4- 2.

SPECTRAL ANALYSIS 51

- 3.5

range

the


What

(ft)

is

outpui

filter

mean

symmetrical square wave of zero

applied to an ideal low-pass

is

3.5

The

filter.

value, peak-to-peak voltage

has a transfer function \H(f)

filter

\

volt,

1

=

and period

1

$ in the frequency

= 0 elsewhere. Plot the power spectral density of the filter output power of the input square wave? What is the normalized power of

Hz, and H(/)

the normalized

':

and

1.10- 1. In Eqs. (1.2-1)

to oo as n ranges from 0 to oo. differential df, Eq. (1.2-1)

f0 =

(1.2-2) write

between harmonics. Replace n A/by /

Show

1/TC

/0 by

Replace

.

as A/"—»0,/

i.e.,

becomes

that in the limit as

A/,

i.e.,

A/

is

the frequency interval

a continuous variable ranging

A/-»

0,

so that A/

may

oby-' 1 "

dt

from 0

be replaced by thr

becomes

»„«>

off)

i*!)

Fiyure PI. 5-1

o

A

1.5.2.

RC filter

Find the Fourier

(ft)

If the third

(ft)

A

(a) (ft)

1.7- 1.

A

and period

T

is filtered

which V{f)

series of the

to be attenuated

is 1 volt,

mw,

is

1

is

produced by an atlenuatoi

What What

f

t

is

the gain in decibels?

is

the

power gain

tKt)

=

ttt±T) =

(ft)

T

—<

lot

"J "

=

di

f

1

J-a

-

sin eo 0

l.

spectral densities of cos u> 0

1

Compare and

sin io?

has the Fourier transform V{f).

v\.i)

with the transform of cos

Show

that the

t.

Plot and

waveform delayed by

The waveform

»(»)

has the Fourier transform K(/).

Show

turn

that the time derivative (d/dt)v(l)

Show

that the transform of the integral of

t^r) is

given b)

mi)

« (-ir +

Derive the convolution formula

i

that

if

V(f)

=

in the

lrequency domain. That

is,

let

K,(/)

=

^[d,(()] and

&[v,{t)v 2 (lj]. thei

'

i

"

2_

« ».


r

has the transform V{f)e~'"''.

VJJ) = ?[v 2 Uj]. Show 2

2

l< T

1.12-1.

=-

v(t',e-'

7

and has the Fourier expansion

o(r)

'

has the transform (j2nf)V(f )

defined b>

-

power

The waveform

I,, i.e., c(i

2/

and

1.10-3.

the

1.10-4. (a)

t^t) is

f

Find the Fourier transform of

1.10-2.

compare

(not in decibels)'

periodic triangular waveform

'

lim

T.

calculate the output powei

power

the input

by 1000, find

calculate the output voltagi.

voltage gain of 0.1

If

=

/0-4/-0 J-l/J;,

a voltage amplifier indicate a gain of 20 dB.

input voltage

V(f\e™'dJ

is

output voltage across the capacitor is

j"

by a low-pass V(f)

harmonic of the output

Measurements on

(a) If the

1.6- 2.

/

having a 3-dB frequency /,

(n)

1.6-1.

in

voltage represented by an impulse train ol strength

=

sm

2>tn

— V(f)=

'

'

Calculate the fraction of the normalized power of this waveform which

is

contained in

its first

h

^ v^f-^

\° v

a-

threi

harmonics 1.7-2.

The complex

spectral amplitudes of a periodic

V= — e -'"«">-r->

wavelorm are given by 1.12-2. (a)

b=±1,

±2.

Show + 2/M

...

In

Find the ratio of the normalized power in the second harmonic to the normalized power harmonic.

in the first

A waveform

that the .

{Hint:

v[t)

waveform

Use

c

has a Fourier transform which extends over the range from

2 (t)

to

+J U

.

-2/M U

the result of Prob. 1.12-1.)

A waveform v{t) has a Fourier transform 2 elsewhere. Make a plot of the transform of v (t). (ft)

—fu

has a Fourier transform which extends over the range from

V(f)

-

1

in the

range

—/„

to

+/M and

V(f)

-0

SPECTRAL ANALYSIS 53

52 PRINCIPLES OF COMMUNICATION SYSTEMS

1.12-3.

A

filter

has an impulse response

amplitude extending from

f

=0

to

t

=

2.

h(t)

By

as shown.

input to the network

The

is

a pulse of unit

graphical means, determine the output of the

An

1.15-1. (a)

late the output

filter.

A

(i>)

r

«

Air)

l/f2

,

impulse of strength /

pulse of amplitude

tion to the circuit of

\

0

1.13-1.

The energy

1

of a nonperiodic

waveform

Show

Show

1.13-3.

V(f)

This

(1.13-5).

= AT

A waveform

sin

is

3-dB frequency f2

.

Calcu-

t is

applied to the low-pass

RC

approximately the response that would /'

=

circuit.

result

Show

that,

if

from the applica-

Ax. Generalize this result by considering any voltage

;

pulse extending from 0 to

A

volts

that the area under the response

and having

a duration x

waveform

zero

1.16- 1.

Find the cross correlation of the functions

1.16- 2.

Prove that

1.17- 1.

Find the cross-correlation function

sin tot

is

and cos

is

RC

applied to a high-pass


R 21 (r) = R 12 (-x> K 12 M of the two

periodic waveforms

shown

v (t)ai

1

it

we ha\ t

-T

-2T

1.13-2. If

circuit of

2

|

that by interchanging the order of integration

which proves Eq.

duration

is

an impulse of strength

and duration

that this can be written as

E-J"

RC

1>,M

K(/V"-"

(f>)

A

Show

/'

v(l) is

£ =

(a)

1.15- 2. circuit.

Figure Pl.12-3

t

A and

the response at the output

waveform of area 1

applied to a low-pass

is

waveform

V(f)V*(f)df=

['

T

C

2T

t

\V{f)\U)

an alternate proof of Parseval's theorem

2nJT/2nfT, find the energy £ contained

nit) has a Fourier transform

in

w

i

M(f) whose magnitude

is

as

1

shown

(a) Find the normalized energy content of the wavelorn

Calculate the lrequency

(fc)

such that one-half of the normalized energy

/,

is

in the

frequency

range -/, toy,.

-T

—T

-•f-i

|JH(/)I

1

Figure Pl.17-1

1.18-1.

_1

C

/

1

R(x)

a

The

signal

til)

3-dB frequency/, = (a)

Find

GU)

(b)

Find

G 0(/).

(c)

FindS,,.

1.14-2.

The wavelorm

equal to

=

cos

u> 0

(

+

2 sin 3cu 0

r

+

0.5 sin 4
i

is

filtered

by an

RC

low-pass

filter

£

R{x).

Is /

e~"'u(t)

is

v{t)v(t

+

t) di

Hint: Consider

I

=

T,:

J-ti:

with

2j,

lit)

r

i

lim

T-« T

Figure PI .13-3

Prove that R{0) 1.14-1.

=

passed through a high-pass

RC

circuit

> 0? Expand and

=

lim

-

integrate term by term.

having a time constant

/

=

Mr) -

Show

2[*(0)

-

i*t

+

t)]

2

dl

that

R(i)]

>

(i

:

(a)

Find the energy spectral density

(b)

Show

at the

that the total output energy

is

output of the

circuii

one-half the input energy

1.18-2.

Determine an expression

0 and a period T.

for the correlation function of a

square wave having the values

1

or

54 PRINCIPLES OF COMMUNICATION SYSTEMS

Find the power spectral density of a square-wave voltage by Fourier transforming the cor-

1.19-1. (a)

relation function.

Use

1.9-la) itself

1.19- 2. If

v(t)

=

sin to c

(a)

Find R(t)

(f>)

If

G(f)

G{f) directly and compare.

A

consists of a single pulse of amplitude

extending from

(b) (c)

Calculate

1.24-1.

G^f)

directly

Interchange the labels

and

s 2 (()

Gram-Schmidt procedure

ing apply the

t

= -t/2

to

I

=

is

nate system whose axes are

figure

shown

G^f) =

on the waveforms of

&[R(i\\.

waveforms

to find expansions of the

and with

this

new

label-

terms of orthonormal

in

Use the Gram-Schmidt procedure

in Eq. (1.25-13)

A

of signals

set

y/W sin to 0 b

tinguishability, 1.26-3.

what

Given two

1,

is

and u 2 (t)of the same

u,(t)

rotated 60^ counterclockwise from the coordi-

suffice to

(1.26-6).

allow a representation of any of the signals.

Verify that two

Show

that in a coor-

is

j 62

=

s„,(i)

=

s 3 (t)

b

as

t

+

t,

s„ 2

n/2). If

= -y/2P

the

two

c

w0

sin

!.

A

second

sets of signals are to

set

of signals

have the same

dis-

PJPp.

the ratio

that independently of the

0
s 2 (r)

0
=-/(!)

form of f(t) the distinguishability of the signals

the normalized energy of the

1.26-4. (a)

sin co 0

JlF sin (a>„

s,(r)=/W

£ is

is

signal.'

to express the functions in Fig. Pl.24-2 in terms of ortho-

normal component'

the

is

u\u'2 coordinate system.

in Fig. 1.26-

=

s t2

Show

MO

and the

Apply the Gram-Schmidt procedure to the eight signals of Eq.

1.26-2. is

Fig. 1.24-la

parameter a

u 2 coordinate system

the functions s,(t) and s 2 (t) of Fig. 1.24-1 in terms of orthonormal functions u\(t) and

functions. 1.24-2.

u„

dinate system of these two orthonormal signals, the geometric representation of the eight signals

t/2.

by Parseval's theorem and compare

s,(r)

Expand

(i)

orthonormal components

Find the autocorrelation function R(t) of this wavelorm Calculate the energy spectral density of this pulse by evaluating

(a)

(1.25-13). Verify that the

u 2 (t) which are the axes of a coordinate system which

1.26- 1.

A waveform

1.20- 1.

i.

= &[R(x)l find

and Eq.

1.25-4. (a) Refer to Fig. 1.25-2

tangent of the angle between the axes of the

the results of Prob. 1.18-2.

the answer to (a) with the spectral density obtained from the Fourier series (Prob.

Compare

(i>)

SPECTRAL ANALYSIS 55

is

given

by^/ZE where

waveform fil).

Use the Gram-Schmidt procedure

to express the functions in Fig. Pl.26-4 in terms of

orthonormal components. In applying the procedure, involve the functions in the order s,(l), s (i), 2 s 3 (l), and s 4 (l). Plot the functions as points in a coordinate system in which the coordinate axes are measured in units of the orthonormal functions. (/>)

Repeat part

s 2 (l). Plot the

3

2

1

2

1

t

3

1

t

3

2

t

(c)

A

set of signals (k

=

1, 2, 3,

*»(')

4) is given

by

Show

that the procedures of parts (a)

S|(0

= cos(a> 0 r + *^j

except that the [unctions are to be involved in the order

and

(fc)

yield the

s,l'l

Osist — = a

sj(i)

Use the Gram-Schmidt procedure

3

3

an orthonormal

set of

lunclions in which the functions

s,(r)

1.25-2. (a)

and

s 2 (r)

Show

that the pythagorean

(fc)

that s,(r)

Let

and

s,(l)

and

s 2 (l) is s 2 (r)

theorem applies to signal functions. That s,(r)

is.

sum s,(r) + s 2 (r) Draw the signal j,(t) +

Show

that

|

in Fig. Pl.25-2. 2 2 s 2 (r) |s,(()| s,(()

+

|

=

+

:

|s 2 (i)!

1

1.25-3. Verify

t

Eq. (1.25-131

show

that

if s,(()

equal to the square of the length of the

MO

+1

3

~

t

-1

-1

-2

-2

-3

-3

21

V

3,

0 1

-1 -J

'

2 1

-2

(defined as in Eq. (1.25-106)) plus the

be the signals shown

s 2 (r) are orthogonal.

1

2

t

1

I

are orthogonal then the square of the length of

square of the length of

-i

0 3

Eq. (1.25-1 Of

2

1

to find

A. Figure Pl.25-2

s 2 (().

and

distances between function

otherwm

can be expanded 1.25-1. Verify

same

s,Uj

2

o

s,(I), s 4 (l), s 3 ((),

points.

Figure Pl.24-2 1.24-3.

(a)

functiom

Show Figure Pl.26-4

-3

I

3|

t

PROBABILITY

2.1

RANDOM VARIABLES AND

PROCESSES 57

when we contemplate

the possible

1

CHAPTER The concept

TWO

of probability occurs naturally

outcomes of an experiment whose outcome is not always the same. Suppose that one of the possible outcomes is called A and that when the experiment is repeated N times the outcome A occurs N A times. The relative frequency of occurrence of A is NJN, and this ratio NJN is not predictable unless N is very large. For

RANDOM VARIABLES AND PROCESSES

example,

let

the experiment consist of the tossing of a die and

correspond to the appearance the

number

3

may

not appear at

times in between. Thus with

On

of,

the other hand,

say, the

6,

3

on

let

the die.

the

Then

outcome A in 6 tosses

may appear 6 times, or any number of NJN may be 0 or 1/6, etc., up to NJN = 1.

all,

N=

number

or

it

we know from experience

that

when an experiment, whose

outcomes are determined by chance, is repeated very many times, the relative frequency of a particular outcome approaches a fixed limit. Thus, if we were to toss a die very many times we would expect that NJN would turn out to be very close to 1/6. This limiting value of the relative frequency of occurrence

the probability of

outcome A, written P(A), so P(A)

In

many

=

is

^

lim

called

thai

(2.1-1)

cases the experiments needed to determine the probability of an

event are done

more

in

thought than

in practice.

Suppose that we have 10

balls

A waveform

in a container, the balls being identical in every respect except that 8 are white

of time v(t)

and 2 are

all

times in

which can be expressed, at least in principle, as an explicit function is called a deterministic waveform. Such a waveform is determined for that, if we select an arbitrary time t = t l5 there need be no uncertainty

about the value of

v(t) at

that time.

The waveforms encountered

in

communica-

on the other hand, are in many instances unpredictable. Consider, the very waveform itself which is transmitted for the purpose of communica-

tion systems, say, tion.

This waveform, which

dictable. If such

is

called the signal, must, at least in part, be unpre-

were not the case,

i.e.,

if

the signal were predictable, then

its

transmission would be unnecessary, and the entire communications system would

no purpose. This point concerning the unpredictability of the signal is explored further in Chap. 13. Further, as noted briefly in Chap. 1, transmitted signals are invariably accompanied by noise which results from the ever-present agitation of the universe at the atomic level. These noise waveforms are also not serve

black. Let us ask about the probability that, in a single draw,

select a black ball. If

we draw

is

shall

on the

that

is,

any of the 10

balls

may

be drawn; of these 10

outcomes, 2 are favorable to our interest. The only reasonable outcome we can imagine is that, in very many drawings, 2 out of 10 will be black. Any other

outcome would immediately suggest

that either the black or the white balls

had

been favored. These considerations lead to an alternative definition of the probability of occurrence of an event A, that is

number

p^

number

total It is

is

not possible, then

2.2

of possible favorable outco me' of possible

apparent from either definition, Eq.

of occurrence of an event

random process with a certain probability of being correct. Accordingly, in this we shall present some elemental ideas of probability theory and apply them to the description of random processes. We shall rather generally limit our discussion to the development of only those aspects of the subject which we shall

we

influence

absolutely nothing which favors one ball over another. There are 10 possible

outcomes of the experiment,

waveforms such as a signal voltage s(t) or a noise voltage n(t) are examples of random processes. [Note that in writing symbols like s(t) and n(t) we do not imply that we can write explicit functions for these time While random processes are not predictable, neither are they completely unpredictable. It is generally possible to predict the future performance of a

no

outcome, we would surely judge that the probability of drawing the black ball is 2/10. We arrive at this conclusion on the basis that we have postulated that there

predictable. Unpredictable

functions.]

blindly, so that the color has

P=

P

0,

is

a positive

while

if

j

(2.1-1) or (2.1-2), that the probability

number and

an event

^

equally likely outcome?

is

certain,

that

0

P=

1


1. If

an event

MUTUALLY EXCLUSIVE EVENTS

chapter

have occasion to employ 56

in this text.

Two if

possible outcomes of an experiment are defined as being mutually exclusive

the occurrence of one outcome precludes the occurrence of the other. In this

case,

if

the events are 4, and

A2

with probabilities P(A^) and P(A 2 ), then the

RANDOM VARIABLES AND

94 PRINCIPLES OF COMMUNICATION SYSTEMS

the shaded area in Fig. 2.22-2 measures the probability that

m2

was transmitted

we

In Fig. 2.22-3

represent the situation in which one of four messages

m„

s„ s 2 s 3 and s 4 The probability density function Mn) of the noise (multiplied by |) is shown in Fig. 2.223a. Also shown are the functions P(m k )f^r - s k ) for k = 1, 2, 3, and 4. We have taken all the probabilities P(m k ) to be equal at P(m k ) = i The boundaries of the received response R at which decisions change are also shown and are marked

m 2 ,m 3

,

and

m4

are sent yielding receiver responses

,

,

.

P12.P23.and p 34 The sum of the two shaded areas shown, each of area A, equals the probabilP(E, m ) that m is transmitted and that the message is erroneously read. Thus ity .

2

m2 =

P(E,

)

2A.

The remaining area under \f^r

transmitted and

and read asm,.

is

m3 =

of course that P(E,

=

P(E,

m4 =

2 P(E,

)

P(C,

is

m2 = )

P(C)

)

).

minimum

Thus

=

P(C, m,)

=

(i

=

1

-

A)

+

+

when we make boundary

average likelihood of error, the probability of

P(C,

m2 +

-

+

(i

2A)

-

(i

m3 +

2A)

+

m4

P(C,

)

(i

—

)

A) (2.22-4)

the probability of an error

is

=

1

-

=

P(C)

6A

(2.22-5)

RANDOM PROCESSES

2.23

To determine

the probabilities of the various possible

necessary to repeat the experiment

is

P(C,

)

this result applies for

- 6A

P(E)

it

2A.

it

).

making an error is not the same for all messages. Of course, any number of equally likely messages greater than two. The probability that a message is read correctly is

2

-

j

)

esting result that given four equally likely messages,

decisions which assure

m 7 is We find

the probability that

s 2 ) is

m 2 But is most interesting to note that when m 4 there is only a single area A involved. Hence m 2 = 2 P(E, m 3 Thus, we arrive at the most intei-

P(E,

)

we evaluate P(E, m,) or P(E, P(£, m^)

—

correctly read. This probability

PROCESSES 95

many

times.

outcomes of an experiment, Suppose then that we are

interested in establishing the statistics associated with the tossing of a die.

We

On

one hand, we might use a single die and toss it repeatedly. Alternatively, we might toss simultaneously a very large number of dice. Intuitively, we would expect that both methods would give the

might proceed

same

results.

in either of

two ways.

we would

Thus,

outcome, on the average, of

1

expect that a single die would yield a particular

time out of

6.

Similarly, with

many

dice

we would

expect that 1/6 of the dice tossed would yield a particular outcome.

Analogously,

mentioned noise,

let

at the

us consider a

beginning of

we might make

single noise source, or

random process such

this chapter.

To

n(t)

statistics of the

repeated measurements of the noise voltage output of a

we

might, at least conceptually,

surements of the output of a very large collection of sources.

waveform

as a noise

determine the

Such a collection of sources

is

make simultaneous mea-

statistically identical noise

called an ensemble,

noise wavelorms are called sample functions.

A

statistical

and the individual

average

may

be deter-

mined from measurements made at some fixed time r = r, on all the sample func2 tions of the ensemble. Thus to determine, say, n (t), we would, at t — t,, measure the voltages n(t,) of each noise source, square and add the voltages, and divide by the (large)

number

of sources in the ensemble.

The average so determined

is

the

ensemble average of n 2 (tj).

Now

random variable and The ensemble averages

n(t,) is a

density function.

will

have associated with

it

will

be identical with the

statistical aver-

ages computed earlier in Sees. 2.11 and 2.12 and

may

a probability

be represented by the same

RANDOM VARIABLES AND

96 PRINCIPLES OF COMMUNICATION SYSTEMS

T hus he statistical or ensemble average of n 2 (f,) may be = « 2 ('i)- The averages determined by measurements on a single

symbols.

E["

t

2

(fi)]

function at successive times will yield a time average, which


we

written

In a similar

sample

represent

a.'

(b)

Since 6

known,

is

In general, ensemble averages and time averages are not the same. Suppose, for example, that the statistical characteristics of the sample functions in the

ensemble were changing with time. Such a variation could not be reflected in measurements made at a fixed time, and the ensemble averages would be different

When

the statistical characteristics of the

random process

is

sample functions do

described as being stationary.

which

used to form the average.

is

When

random process

the nature of a

such that ensemble and time averages are identical, the process ergodic.

An

ergodic process

is

is

of the

moments and

all is

Example with a

V(t)

v(t)

=

For example,

deterministic.

is

=

V(t)

cos

(o> 0

+

+

if

6

=

30

constant

(2.23-5)

a gaussian ergodic

random process

1

30°)

not stationan

is

2.23-2

mean

A

voltage V(t\ which

is

rms meter, and component true

a meter

which

first

2

measured by a dc meter, a squares V(i) and then reads its dc

of zero and a variance of 4 volt

,

is

Find the output of each meter.

is

referred to as

stationary, but, of course, a stationary process

£{ Thus, V(t)

However, even the property of being stationary does not ensure that ensemble and time averages are the same. For it may happen that while each sample function is stationary the individual sample functions may differ statistically from one another. In this case, the time average will depend on the particular sample function

all

a stationary process

(t)}.

not change with time, the

can be established that

it

other statistical characteristics of V(t) are independent of time. Hence V(t)

2

at different times.

manner

PROCESSES 97

Solution

(a)

The dc meter reads

is



=

E{V(t))

not necessarily ergodic

Throughout

we

shall

the

same

this text

we

shall

assume

that the

random processes with which

have occasion to deal are ergodic. Hence the ensemble average £{"(')} 2 as the time average , the ensemble average E{n (t)} is the same as

s

the time average
2

since V(t) (b)

The

2.23-1 Consider the

random

V{t)

where

©

is

a

random

Show

that the

the

random

(b) If

will the

rms meter

=

cos

process (to 0

1

first

0

=

(c)

(2.23-1)

0

ensemble mean of

in

Eq. (2.23-1)

V(t)

is

is

ergodic. Since V(t) has a zero

(a)

Choose

a fixed time

The square and average meter



(2.23-21

V(t) are

2.24

=

f,.

(a

full-wave rectifier meter) yields a deflec-

.

= £{K 2 (t)}=<7' =

process

n(t),

=

by Eq.

(1.18-1).

?

J"

Ei^di)} =

and

j

We

0 t,

lim

Yn

2

+

6)

d6

=

0

(2.23-3)

significance to the concept of a

(2.24-1)

we were

able to give a physical

+

power

spectral density G(f)

(w 0 i, +

6)

dd

=

I

(2.23-4)

and

to

show

that

As an extension of that result of a random process in the same way. pair.

we shall define the power spectral density Thus for a random process we take G(f) to be

note that these moments are independent oft, and hence independent of

G(f) time.

x)di

n(t)n{t

- 7/2

G(f) and R(i) constitute a Fourier transform cos

energy has an autocor-

Thus 'TP

=

Then

^ cos «»

finite

being neither periodic nor of

relation function defined

In connection with deterministic waveforms

E{V( tl )}

4

AUTOCORRELATION

A random

independent of time.

replaced by a fixed angle 6 0

be time independent

t

mean, the true rms meter reads

2 volts.

R{x)

Solution

dc meter reads zero.

tion proportional to

elsewhere

and second moments of

variable

0, the

J
+ &)

=

read;;

variable with a probability density

= (a)

ergodic. Since E{ V{t)}

(t)>- e' c

since V(i)

Example

is

true

=

#TK(t)]

=

-I

R(x)e- J °" d-

(2.24-2)

RANDOM VARIABLES AND

98 PRINCIPLES OF COMMUNICATION SYSTEMS for a random is of interest to inquire whether G(f) defined in Eq. (2.24-2) process has a physical significance which corresponds to the physical significance It

G(f) which

of G(f) for deterministic waveforms. For this purpose consider a deterministic

colors, that

- oo

sharply.

which extends trom waveform waveform which extends from - 7/2 to range, and otherwise i^f) = 0. The wave-

to ao. Let us select a section of this

T/2. This

waveform

Vj(t)

=

v(t)

in this

2 has a Fourier transform V^f). We recall that Vj{f)\ is the energy 2 spectral density; that is, Vj(f)\ df is the normalized energy in the spectral range 2 normalized power density is |Fr(/)| /T. As df. Hence, over the interval T the

form

vj{t)

|

|

T-> the

oo, v-jit)-* v(t),

power

and we then have the

result that the physical significance of

waveform,

spectral density G{f), at least for a deterministic

is

that

there

lim T-tr

Correspondingly,

we

state,

i |W)!

3

(2.24-3)

1

is

defined for a

is

that

|

2 I

is,

colors of

all

frequencies. Actually as

Now,

=

0.

that n(t)

Since, then, for white noise, R(x)

and

+

n(t

x)

write, instead of Eq. (2.24-1

R(i)

=

E{n(t)n(t

+

=

0 except

Eq. (2.24-1), a tim<

replace the time average by an

shall occasionally

of a sequence of

+

there

is

assume If

seval's

times of occurrence.

waveforms are time invariant. Correspondingly,

an invariant average time of separation that there

no overlap between

is

T

s

about the noise, which is

is

between pulses.

the Fourier transform of a single sample pulse P,(r)

theorem [Eq.

We

further

pulses.

(1.13-5)] states that the

is

P^f)

then Par-

normalized energy of the pulse

(2.24-5)

t)}

£i

=

T

Pit/VW) df=

[" |P,(/)| 2 df

nit>

x.

with communications systems,

no matter how

is indicated in Fig. 2.25-1. The pulses are random amplitudes and statistically independent The waveform (the random process) is stationary so

sample function, a knowledge of the value of n(r) at time t would be of no assistance in improving our ability to predict the value attained by that same sample fact

Eq. (2.24-5) says

pulses such as

Suppose then that we should find that lor some x, R(r) = 0. Then the random variables n(t) and n(t + x) are uncorrected, and (or the gaussian process of intersome est to us, n(t) and n(t + x) are independent. Hence, if we should select

t

0,

need to have information about the power spectral density

random

variables.

The physical

=

is

).

The averaging indicated in Eq. (2.24-5) has the following significance: At some the values fixed time t, n(t) is a random variable, the possible values for which are Simiensemble. the of n(t) assumed at time t by the individual sample functions appears then It variable. random also a t) is larly, at the fixed time t + x, n(t + that R(t) as expressed in Eq. (2.24-5) is the covanance between these two random

function at time

for x

2.25 POWER SPECTRAL DENSITY OF A SEQUENCE OF RANDOM PULSES

We

average of the product n(t) and n(t + x). Since we have assumed an ergodic process, we are at liberty to perform the averaging over any sample function of However, the ensemble, since every sample function will yield the same result.

ensemble average and

off for

(2-24-4)

}

as indicated in

we may

all

14.5,

i.

random

ergodic.

pointed out in Sec.

are uncorrected and hence independent,

that the statistical features of the

is

"white"

a combination of

and the power spectral density G(f) are Thus when G(f) extends over a wide frequency range, R(z) is restricted to a narrow range of x. In the limit, if G(f) = I (a constant) for all frequencies from — oo
random process n(t). The autocorrelation

again because the noise process

is

referred to as is

since the autocorrelation R(z)

of the same form but have

is,

is

an upper-frequency limit beyond which the spectral density falls However, this upper-frequency limit is so high that we may ignore it

where E{ } represents the ensemble average or expectation and Nj(f) represents the Fourier transform of a truncated section of a sample function of the function R(x)

Such noise

random

process, as in Eq. (2.24-2), as the transform of R(x). then G(f) has the significance

G(/)= llirnfj^ AM/)

frequencies.

a Fourier transform pair, they have the properties of such pairs.

small

without proof, that when G(f)

all

our purposes.

for x

G(/)=

uniform over

is

noise in analogy with the consideration that white light

PROCESSES 99

of principal

that such noise has a

concern

power

in

connection

spectral density

Figure 2.25-1 Pulses of random amplitude and time of occurrence.

(2.25-1)

RANDOM VARIABLES AND

108 PRINCIPLES OF COMMUNICATION SYSTEMS

PROBLEMS

2.8- 3. Refer to the Rayleigh density function given in

thrown simultaneously. What

2.1-1. Six dice are

the probability that at least

is

1

die

P(0.5

(b)

P(0.5

shows a 3?

The joint

2.9- 1.

A

2.1-2.

card

(a) ((>)

(c)

A

2.2- 1.

card

(b)

A

a deck of 52 cards.

the probability that a 2

is

the probability that a 2 of clubs

is

the probability that a spade

is

the probability of selecting at least one 6 of spades'/

is

the probability of selecting at least

is

card larger than an 8?

1

(i>)

If

is

is

the

first

and second cards are

hearts,

what

the probability that the third card

is

is

the king

Two

2.3-2.

produce identical clocks. The production of the

factories

The second

clocks of which 100 are defective. tive.

What

One box

2.3- 3.

contains two black

balls.

A

first

factory consists of 10,000

was produced

in the first factory?

second box contains one black and one white

ball.

Two

We

Find the probability of a 3 and a 4 appearinj Find the probability of a

A (f>)

A

2.6-1.

card

die

drawn from

is

What What

<

random

the probability of

first

as a function ol

Let

is n,.

toss defined by the specifications

N=

N

appears.

Make

each with probability

Find the joint density fx

(fc)

Show that/^r) = \* m f„bt,

The

K (x,

i

co,

=

0,

=

o\

1.

V

is

a

random

=

xe'*'*'

)

0
=

y)

random

variables

X

and Y

is

fix, y)

A

coin

which occur

X

(a)

Find fix) and

(b)

Are the random variables dependent or independent?

fiy), the probability density

of

Y and Y

independently of

is

tossed four times. Let

Make

a plot of the probability

2.6- 3.

A

coin

number of P{T < t) as 2.7- 1.

An

is

It is

P(H <

H

P(r 0 |m,)

=

communication channel

P(m,

P(m 0

|r 0 ),

and

0.4,

r,) 1

P(r,

|

m,)

and P(m,

a single

function of

we choose m 0

no mailer what

is

|m 0 )

as represented in Fig. 2.10-1 P(r 0

On

0.6.

r,) as a

|

of Eq. (2.10-1) prescribes that

m0

=

set

P(m 0

if r 0

is

).

of coordinate axes

Mark

received

the range of

and m,

m0

if r,

received and the range for which

=

0.9, P|r,

make is

for

h) as a

H=

tossed until a head appears. Let

tosses

t

where h

h,

function of

T

up

I

to

I

=

the

head.

Make

received, the range for

Calculate and plot the probability of error as a function of

P(m 0

i

Suppose the decision-making apparatus

at the receiver

ceiver nothing could be determined except that a message

strategy for determining

were inoperative so that

had been

received.

What would

is

appeal.

a plot of the probability

the Rayleigh density

'2

Figure P2.10-2 (a)

Prove that fix)

(fc)

Find the distribution function F(x\

satisfies

Eqs. (2.7-2)

and

(2.7-3).

2.11-1. If/Jx)

Find P(2

<

n

(b)

< FindP(2
(c)

Find P(2

<

n

(d)

Find

(a)

2.8-2.

P(x,

Refer

<x<

replace

<

(b)

the

I

+

x,,

for all x.

show

that

E(X 1 ")= 1-3-5 ••• («- l),n= E{X l "-') = 0,n= 1,1

1.2

11).

2.11-2.

Compare

the most probable [fix)

is

a

1

(a)

Rayleigh

x 2 ), where x,

x 2 by

(a)

11)

F(9).

to

= -4= f""' 2

J*

2.8-1. Refer to Fig. 2.6-1

-

x,

=

density 1,

and maximize

so that P(x,

P

given

function

<

x

< x2

with respect to x,.

)

in is

a

Prob.

2.7-1.

Find

the

probabilitx

maximum. Hint: Find P(x, < x < x 2 ),

fx

(x)

= -J= e -"- m)

'

:2

for

JTr, (xe-"> ....

<»WH,.

best deci-

at the re-

be the best

what message had been transmitted and what would be the corresponding

*

important probability density function

).

we choose m, no matter what

be the random variable which identifies the first

0.1

received.

a plot of the prob-

number of heads which

=

P(m 0 \r 0

which the algorithm

h.

required for the appearance of this

a function of

is

|m 0 )

plots of

be the random variable which identifies the number of heads

defined by

in the

independently of

i.,

in these four tosses.

1

0 otherwise.

error probability'' 2.6-2.

variable

A

«) dx.

joint probability density of the

<x<

(c)

n,

1,

independently of Y.

For the channel and message probabilities given in Fig. P2.10-2 determine the sions about the transmitted message for each possible received response (b) With decisions made as in part (a) calculate the probability of erroi

be the random variable identifying the

when

n,

-

X

3).

variable having a gaussian density. E(X)

or

(a)

(b)

number appearing

1

X < 2; 2 < Y <

2.10-2. (a)

and a second card drawn same card twice' drawing a 3 of hearts and then drawing a 4 of spades?

is

«,)

a

<

k

X.

is

a deck of 52 cards, then replaced,

the probability of drawing the

outcome of the ability P|/V

the

7 being rolled.

is

tossed and the

is

is

which we choose

(a)

(a)

is

are

dice are tossed

(b)

2.4-2.

Are the random variables dependent or independent ?

2.10- 1. (a) In a

was withdrawn from one of the boxes and that it turned out to be black. What probability that this withdrawal was made from the box that held the two black balls' told that a ball

2.4- 1.

(d)

range 0

factory produces 20,000 clocks of which 300 are defec-

the probability that a particular defective clock

is

Find the probability P(0

2.9- 3.

of clubs''

X and Y is fix, y) = ke''**'* in the range

(c)

X

first

variables

Find the probability density fix), the probability density of

having the values

(a)

card

drawn. The

random

0 otherwise.

Find the value of the constant

2.9-2.

drawn from a 52-card deck, and without replacing the first card a second card and second cards are not replaced and a third card is drawn " If the first card is a heart, what is the probability of the second card being a heart

2.3- 1.

2).

probability density of the

(a)

drawn?

picked from each of four 52-card decks of cards

is

Find

2.7-1.

(b)

drawn?

is

Prob.

2)

0<x
drawn ?

is

is

What What

(a)

drawn from

is

What What What

<x< <x <

(a)

PROCESSES 109

forx>0 elsewnere

all

x

maximum] and

the average value of

X when

110 PRINCIPLES OF COMMUNICATION SYSTEMS

random

2.12-1. Calculate the variance of the

(a)

The gaussian

variables having densities

—— e -«-"» l

=

density fXl {x)

RANDOM VARIABLES AND in

which

(c)

all x.

,

The Rayleigh density fXl (x) = xe-* m x > 0. The uniform density fXj {x) = 1/a, -a/2 < x S

f{x)

A —— +

=

If

Find

K

(fc)

Find

£(Af).

(c)

Find the variance of X.

2.12- 3.

so that f(x)

2.17-

X by y = aX + A

2.13- 1.

i),

(6)

If£(X

variable

X

Comment on

(a)

Calculate and plot the probability P[ x

(b)

Calculate and plot the quantity a /e

|

2

2

<

|

and

and

0

random

a

is

variable with

,

p

;

j

and

)

e] as a

function of

(2)
is

= £{(Z -

2

m,)

,

>

otherwise

2 .

The independent random

= xe-"

X

variables

12

0

<x<

and Y are added and

co

form Z.

to

=

/^y)

If

w

|

y

|

<

cr

related to

find/^z).

The independent random

2.18-2.

-L to +i is

?

= = i±l

X and y have the probability densities

variables

i

verify that Tchebycheff's rule

1

from the definition a 2

+ X K ,E[X,) = m

|1

X )=

fox)

mean m. The random variable V Find the mean and variance of 1

uniform over the range from

is

2

the significance of this result

are constants.

fc

probability density function f(x)

••

|0

2 has a variance o and a

where a and

1

Ra

the probability that

is

t

Find £(Z).

2.18- 1.

The random

what

(a)

find(l)£(Z

a density function,

is

1,

x'

1

(a)

=

2
1 Z = X +*, +

a/2.

-co<:x
}

variable with a Rayleigh probability density

2.17-1. Derive Eq. (2.17-6) directly

,

Consider the Cauchy density function

2.12-2.

random

a

is

uniform densitj

112

(fc)

(b)

R

PROCESSES 111

fix)

=

e ~*

0

f(y)

=

2e- 2 '

0
valid in this case.

<

x

<

o:

2.14- 1. Refer to the gaussian density given in Eq. (2.14-1).

Show Show

(a) (i)

2.14-2,

that E((X

that E((X

-

2 m) "~')

m)

2

")

=

Find and plot the probability density of the variable

= 0 1

•

3-5

As

(< rel="nofollow">)

Calculate and plot the quantity o

a function of a, plot the probability

A random

2 0 and variance a and b

is

,

mean b and variance c 2A4-4.

The joint

V =

variable

b

+

2

/e

2

X, where

Show

a constant.

P[ x |

A*

|

is

V

is

a

1

where. Find and plot the density of is

The

2.18-4.

valid in this case

random variable with mean gaussian distributed random variable with

a gaussian distributed

that

£

of

A', is

1

.

density function of two dependent variables

X

and V

N

and

Two

2.19- 1.

which there

= -±-e- 2t '-''+>"<

gaussian variable,

Find

(b)

<s\

X

when

that,

i.e.,/(x)

its

variance

gaussian

is 1,

find

random

and

a

2.15-2.

Verify that the

On

set of

axes used in Prob. 2.15-1 plot e'"' and erfc u versus

u.

Compare your

+ ka

< X < m+

—

ka)--

Change

variables by letting u

=

x

-

m/ x /2

voltage

K

is

a lunction of time V(t)

which

a< is

=

I

and

2 2 fx - 2pxy + y ~l

L

s

that the result

= 0 corresponds

The random

variables A",,

Show

—

.

that a\

+

a]

variance a

2 .

Show

is

A"

J- -

2ff ('

is

is

to the circumstance

by Eq.

X

may

sin

has a

mean

value

R -

+

X2 X} ,

cos

communication system used is

4 x 10~ 5

messages must be included 2.22- 1.

tui

and y are independent gaussian variables each

+ 0)

m

while

A communication

2

generates

r2

which extends from n (a)

be written (cut

,

are dependent but uncorrected.

...

Z=

that

A

+ X2 + A

Af,

?

oi -

.

A

to transmit a sequence of messages,

sample survey of N messages

ability not to exceed 0.05, the error rate in the

for

= R

(2.19-5).

X and Y are independem variable 0 by A" = sin © and

where

The random variables A", X 2 ,X } are independent and each has a uniform probability < x < 1. Find and plot the probability density of A", + X 7 and of AC, + X 2 + A

2.21-2. In a

+ V

J

)

indeed the correlation coefficient

as required

p

P

range 0

error probability

given b>

cos wl

that V(t) V{t)

a\

in the

densilx .

m

sample

is

is

made.

It is

it

5 not to be larger than 5 x 10~

channel transmits, in random order, two messages m, and

m

= +\V. The channel is = -1.5Kton = +1.51

Find the probability that m,

is

2

.

Message m, generates

What

is

that the

How mam

.

m

2

.

The message

a receiver response

r,

= — V 1

corrupted by noise n with a uniform probability density

mistaken for m, and the probability that

,

(fc)

known

is

required that, with a prob-

sample'

m, occurs three times more frequently than

a constant angular frequency and

mean and

exp-

2.21- 1. Verify Eq. (2.21-16)

,

with zero

•

in the

o.

Show that P ±l „ = erf (k/^/2 2. 16-1. Show that a random variable with a Rayleigh density as in Eq. (2.16-1) 2 2 2 •J(n/2) a, a mean square value J? = 2a and a variance a = (2 — n/2)y

A

•

2.20- 1.

d>

(fc)

2.16-2. (a)

that the case p

fT^

^

in the expression for fix, y)

E{XY)/a 2 and show

probability

\/2nc (a)

1

2

cos 0.

2.19- 3.

The

Show

symbol p

mean

h-

The random variables X and Y are related to the random The variable 0 has a uniform probability density in the range from 0 to 2n. Show and y are not independent but that, nonetheless, £(AT) = 0 so that they are uncorrelateu.

y =

erf u versus u

P tk , = P\m -

in

evaluate

is, (f>)

results.

2.15-3.

That

the

If

,

2.19-2.

and plot

for

same

is

are added to form Z.

have a joint probability density given

= 2na

and Y are each considered without reference to the other, each

X„

and y, each with mean zero and variance a 2 between

variables A"

-

o;.

Obtain values the

A„

f^zl

a correlation coefficient p,

is

:

and /(>') are gaussian density functions.

2.15- 1.

zero else-

)'

is

(o)

Show

{a)

Z=X+

independent gaussian random variables

fix,y) fix, y)

Y.

.

at].

verify that Tchebycheff's rule

and

Z=X+

The random variable X has a probability density uniform in the range 0 < x < and elsewhere. The independent variable y has a density uniform in the range 0 < x < 2 and zero 2.18-3.

l)
: Given a gaussian probability function f{x) of mean value zero and variance c

(a)

2.14- 3.

-

(n

•

the probability that the receiver will determine a message correctly

'!

m2

is

mistaken

112 PRINCIPLES OF COMMUNICATION SYSTEM?

A communication

2.22-2.

+

The channel

V.

channel transmits, in random order, two messages m, and

Message m, generates response

likelihood.

A

m2

generates

What

(a)

What

(ft)

m2

channel

ib)

is

to be

Message m, generates

.

that

m 2 was

transmitted

=

X

t

cos


r

—X

sin

2

=

M(t) cos

+

(cu 0 r

0). M(t)

is

random

a

message

a receiver response r,

= + V

.

w0

t

is

mean and

a

random

process. If

variance a

2 ,

X, and

X2

are

find

= 0 and £(M 2 (t)) =

M;,

''

(cu 0

r

+

M

=

0))

2

such that

/2. Is Z(t)

now

/„(())

=

-i<Jii,

l/2n,

show

that

stationary

R ; (t;

Find

has a power spectral density G(f) - if/2 for — oo <,f<, oo. The random passed through a low-pass filter which has a transfer function H(/) = 2 for -fu u process

n(t)


is

mf) =

0 otherwise. Find the power spectral density of the wavelorm at the output of the

White noise

n[t)

=

with G(f)

RC

passed through a low-pass

ij/2 is

filler

network with a 3-dB

fre-

.

(a)

Find the autocorrelation R(t) of the output noise of the networl

(i>)

Sketch p(t)

(c)

Find

cu r t

=

One

such that p(t)

£

0.

feth

pulse has a width t and a height

10 with equal probability.

...

At

Assuming

independence between amplitudes and assuming that the average separation between pulses

A

spectral density G„(/) of the pulse

.

/4, is a

statistical is 7",,

1

jts

or 2

fis

The mean time between

with equal probability.

2.26- 1.

shown

Consider the power spectral density of an in Fig. 2.26-3.

By consulting

reduced by only 10 percent

if

the

pulses

is

5 ps.

Find the power

NRZ

a table of integrals,

waveform

is

wavelorm as given by Eq. (2.26-4) and as show that the power of the NRZ waveform is

passed through an ideal low-pass

filter

with cutoff

power

shown

By consulting

waveform as given by Eq. (2.26-7) and as show that, if the wavelorm is passed through an ideal low-pass filter with cutoff at 2fb 95 percent of the power will be passed. Show also that, if the filter cutoff is set at fb then only 70 percent of the power is transmuted in Fig. 2.26-4.

spectral density of a biphase a table of integrals, ,

multiplexing.

the design and

A method

by which such

be achieved consists

a different position in the frequency

called frequency

in translating

spectrum. Such multiplexing

The individual message can eventually be

Frequency multiplexing involves the use of an auxiliary separated by waveform, usually sinusoidal, called a carrier. The operations periormed on the signal to achieve frequency multiplexing result in the generation of a waveform filtering.

which may be described as the or phase, individually or

ai

J-h2.26- 2. Consider the

each message to is

spectral density G„(/) of the pulse train

may

multiple transmission, called multiplexing,

find

trail,

pulse train consists of rectangular pulses having an amplitude of 2 volts and widths which

are either

many

is

individual messages to be transmitted

simultaneously over a single communication channel.

1

Consider a train of rectangular pulses. The random variable which can have the values 1, 2, 3,

power

of the basic problems of communication engineering

analysis of systems which allow

R(i)//t(0

2.25- 1.

2.25- 2.

AMPLITUDE-MODULATION SYSTEMS

).

= E(M 2 (r))£(cos 2

quency/,

the

THREE

'.'

process, with £(M(«))

0 = 0, find E(Z 2 Is Z(t) stationary If 0 is an independent random variable

A random

2.24- 3.

m 2 The

two messages m, and

,

2.24- 1. Refer to Prob. 2.23-1.

and

order,

E(Z 2 ), a 2 and

E(Z),

(fc)

process

CHAPTER

f^n) = 0 at f^n) = - 2 K and at f^n)= +21 which the decision is to be made that m, was transmitted and

for

r

made

of time Z(()

(a) If

2.24.2.

=

Find the probability

.

fM

2.23- 2. Z(t)

£(Z 2 (t))

2

with equal

corrupted by noise n whose probability density has the

is

independent gaussian random variables each with zero (a)

volt

1

m2

generates r 2

the probability that the message will be read correctly"

is

The function

2.23- 1.

=

m2

in Fig. 2.22-2a with

are the ranges of

which the decision

for

random

transmits, in

= — IK. The

r2

shown

triangular form

and message

message correctly.

communication channel

m, occurs three times more frequently than while

at the receiver

corrupted with gaussian noise with variance a 2

is

that the receiver will determine a 2.22- 3.

= - IV

r,

carrier

is

in

carrier modified in that

some

called a modulated carrier. In

carrier

we

discuss the generation

waveform?

amplitude, frequency,

cases the modulation

simply to the message; in other cases the relationship chapter,

its

combination, varies with time. Such a modified

is

is

related

quite complicated. In thif

and characteristics of amplitude-modulated

1

.

2.27- 1.

An

NRZ

waveform

pass

filter

of time constant

details of the

An

and I's. The bit interval is us and the waveThe waveform is transmitted through an RC highoutput waveform and calculate numerical values of all

consists of alternating 0's

form makes excursions between +1 V and 1

fis.

Draw

-

the

1

V.

FREQUENCY TRANSLATION

It is

often advantageous

waveform

NRZ

waveform

The bit interval is 1 ps and the waveform makes excursions between + V and - V. The waveform is transmitted through an RC lowpass filter of time constant 1 us. Draw the output waveform and calculate numerical values of all details of the waveform 2.27-2.

3.1

consists of alternating 0's 1

and

and convenient,

in processing a signal in a

communica-

l's.

tions system, to translate the signal from

another region. Suppose that a signal

is

one region

in the

frequency domain to

bandlimited, or nearly so, to the

fre-

quency range extending trom a frequency /, to a frequency f2 The process of frequency translation is one in which the original signal is replaced with a new .

113

PRINCIPLES OF COMMUNICATION SYSTEMS

Z

A communication channel transmits, in random order, two messages m and m 2 with equal Message m, generates response r, = - IV at the receiver and message m generates r = 2 2 The channel is corrupted with gaussian noise with variance a 2 = 1 volt 2 Find the probability

2-2.

l

•lihood -

1

t

.

CHAPTER

.

!hc receiver will determine a

THREE

message correctly.

A communication

channel transmits, in random order, two messages m, and m 2 The message occurs three times more frequently than m 2 Message m, generates a receiver response r = + V (

2-3.

.

.

m, generates r 2 = -IV. The channel is corrupted by noise n whose probability density has ngulir form shown in Fig. 2.22-2a with /„(n) = Oat fN (n) = -2Kand at/„(n) = +2V. ile

What

tt>)

are the ranges of

which the decision

iH What

is

which the decision

r for

made

to be

that

m 2 was

is

to be

made

AMPLITUDE-MODULATION SYSTEMS

the

m, was transmitted and

that

transmitted?

the probability that the message will be read correctly?

is

The function of time Z(t) = A cos a> 0 t - X 2 sin o> 0 t is a random process. e pendent gaussian random variables each with zero mean and variance a 2 find 2 ««» BZV. HZ ). <•], and <»*//*" -

J-t.

,

If

X

X, and

are

2

,

'

VI

7ii\

r^»«

=

Mfit cos (m 0

+

1

©). Af(r)

(
t

+

0))

=

Refer to Prob. 123-1. Find R^x).

«

E(M 2 (t)) =

0 and

AfJ.

Mj/2.

such that fjfi)

Is Z(t)

now

=

—n

l/2n,

<,

8

<,

it,

show

that

stationary?

has a power spectral density G(f) = i;/2 for - eo
n(t)

White noise Mi) with .(

=

passed through a low-pass

!>/) »

ncj

process, with £(M(t))

is

- £tV
A random

l-\

random

).

12.

!

a

2

If

M. cess

is

0 = a find E(Z Is Z(t) stationary? If 0 an independent random variable

im (M

C,(f)

=

t//2 is

passed through a low-pass

RC

network with a 3-dB

fre-

.

W Find

rtie

jotocorrelation

IM

Sk'«cfc i*t)


Find »,

t

-

/?(t)

fuch that p(t)

s

0.1.

Consider a train of rectangular pulses. The kth pulse has a width t and a height ,4 A k is a t *ms» tasHe wtuch can have the values 1, 2, 3, 10 with equal probability. Assuming statistical rpesAsswr hetweta amplitudes and assuming that the average separation between pulses is T find .

.

.

.

s

jw««r spectral density GJf) of the pulse

,

train.

A psfaie traiB consists of rectangular pulses having an amplitude of 2 cnha ! a* or 2 as with equal probability. The mean time between pulses Mra! dewmv GJf\ of the pulse train. i-1

volts is

and widths which power

5 us. Find the

NRZ waveform as given by Eq. (2.26-4) and as By consulting a table of integrals, show that the power of the NRZ waveform is iced by oarj iO percent if the waveform is passed through an ideal low-pass filter with cutoff at wis

Consider the power spectral density of an

m

of the basic problems of

analysis of systems

l-l.

(-1.

communication engineering is the design and which allow many individual messages to be transmitted simultaneously over a single communication channel. A method by which such multiple transmission, called multiplexing, may be achieved consists in translating each message to a different position in the frequency spectrum. Such multiplexing is called frequency multiplexing. The individual message can eventually be separated by filtering. Frequency multiplexing involves the use of an auxiliary waveform, usually sinusoidal, called a carrier. The operations performed on the signal to achieve frequency multiplexing result in the generation of a waveform

One

of the output noise of the network.

R(t)/R(0).

Fig. 2.26- J.

which may be described as the carrier

UConsider the power soectral density of a biphase waveform as given by Eq. (2.26-7) and as an in Fig 2.26-4. By consulting a table of integrals, show that, if the waveform is passed through deal low-pass filter with cutoff at 2/, , 95 percent of the power will be passed. Show also that, if the r cutoff is set at/ f then only 70 percent of the power is transmitted.

carrier modified in that

its

amplitude, frequency,

or phase, individually or in combination, varies with time. Such a modified is

some

called a modulated carrier. In

cases the modulation

simply to the message; in other cases the relationship chapter,

we

discuss the generation

carrier waveforms.

and

is

is

related

quite complicated. In this

characteristics of amplitude-modulated

1

.

n

An

NRZ

waveform consists of alternating 0's and l's. The bit interval is 1 /ts and the wavemakes excursions between j- V and - V The waveform is transmitted through an RC highfilter of time constant jis. Draw the output waveform and calculate numerical values of all

'-1.

3.1

FREQUENCY TRANSLATION

'-2.

It is

often advantageous

n

tions system, to translate the signal

i

I

tils

i

1

.

1

of the waveform.

An NRZ waveform consists of alternating 0's and l's. The bit interval is 1 jis and the wavemakes excursions between + V and - V. The waveform is transmitted through an RC lowfilter of time constant 1 ps. Draw tbe output waveform and calculate numerical values of all

tils

1

of the waveform.

1

and convenient, in processing a signal in a communicafrom one region in the frequency domain to

another region. Suppose that a signal

is

bandlimited, or nearly so, to the

fre-

quency range extending from a frequency/! to a frequency f2 The process of frequency translation is one in which the original signal is replaced with a new .

113

AMPLITUDE-MODULATION SYSTEMS 115

114 PRINCIPLES OF COMMUNICATION SYSTEMS

signal

whose

spectral range extends

from f \ tof2 and which new

signal bears, in

recoverable form, the same information as was borne by the original signal. discuss

now

a

number of

useful purposes

which

may

We

be served by frequency

translation.

Common Processing may happen

that we may have to process, in turn, a number of signals similar general character but occupying different spectral ranges. It will then be necessary, as we go from signal to signal, to adjust the frequency range of our It

in

processing apparatus to correspond to the frequency range of the signal to be processed. If the processing apparatus is rather elaborate, it may well be wiser to

Frequency Multiplexing

leave the processing apparatus to operate in

Suppose that we have several spectral range. Let single

it

different signals, all of

be required that

communications channel

in

signals be separately recoverable

channel

may

which encompass the same

all these signals be transmitted along a such a manner that, at the receiving end, the

and distinguishable from each

be a single pair of wires or the

free

other.

The

some

fixed frequency range

and

instead to translate the frequency range of each signal in turn to correspond to this fixed frequency.

single

space that separates one radio

antenna from another. Such multiple transmissions, i.e., multiplexing, may be achieved by translating each one of the original signals to a different frequency range. Suppose, say, that one signal is translated to the frequency range to 2 the second to the range f\ to f"2 and so on. If these new frequency ranges do not

f

,

3.2

A

METHOD OF FREQUENCY TRANSLATION

A

signal may be translated to a new spectral range by multiplying the signal with an auxiliary sinusoidal signal. To illustrate the process, let us consider initially

,

overlap, then the signal

bandpass

filters,

may

be separated at the receiving end by appropriate and the outputs of the filters processed to recover the original

that the signal

is

f

sinusoidal in

waveform and given by

wm = A m

JO =

A. m

cos

=

y

(e*-'

cos 2nfm

t

(3.2-la)

t

signals.

+

= 4= (^-' + e'^"')

e->»')

(3.2-16)

Practicability of Antennas in

When

free

space

is

the communications channel, antennas radiate

signal. It turns out that

and

receive the

antennas operate effectively only when their dimensions

are of the order of magnitude of the wavelength of the signal being transmitted.

A

signal of frequency 1 kHz (an audio tone) corresponds to a wavelength of 300,000 m, an entirely impractical length. The required length may be reduced to the point of practicability by translating the audio tone to a higher frequency.

which

Am

is

the constant amplitude

v c (t)

which

in

metric

of highest to lowest frequency

would be only

1.01.

Thus

is

The twoThe pattern

the frequency.

in Fig. 3.2-la.

signal

Returning to the matter of the antenna, just discussed, suppose that we wanted to

audio frequency to the lowest is 200. Therefore, an antenna suitable for use at one end of the range would be entirely too short or too long for the other end. Suppose, however, that the audio spectrum were translated so that it occupied the range, say, from (10 6 + 50) to (10 6 + 10 4 ) Hz. Then the ratio

shown

is

two lines, each of amplitude AJ2, located at/=/m and at/= Consider next the result of the multiplication of v m (t) with an auxiliary sinusoidal

transmit an audio signal directly from the antenna,

ratio of the highest

= cojln

consists of

Narrowbanding

and that the inordinate length of the antenna were no problem. We would still be left with a problem of another type. Let us assume that the audio range extends from, say, 50 to 10* Hz. The

and fm

sided spectral amplitude pattern of this signal

A

c

is

= Ac

cos

=

(**""

y

coc

t

+

= Ac

e~ J °"')

the constant amplitude

identity

cos a cos

cos 2nfc

P = \ cos

=

^ (y

and fc

(a

+

t

/?)

2 *'*'

+ e-J2 "'' ) 1

(3.2-2fc)

the frequency. Using the trigono-

is

+

(3.2-2a)

£ cos (a

-

ft),

we have

for

the

product uJ0»c(0

»

J0»

c

(0

A A = -y-*

[cos

K + wJt +

cos

(coc

-

co„)t]

(3.2-3a)

A A

the processes of fre-

may be used to change a "wideband" signal into a " narrowband " signal which may well be more conveniently processed. The terms

The new

wideband " and " narrowband " are being used here to refer not to an absolute range of frequencies but rather to the fractional change in frequency from one

tion by amount fe and also in the negative-frequency direction by the same amount. There are now four spectral components resulting in two sinusoidal waveforms, one of frequency fc + fm and the other of frequency/, Note that

quency translation "

band edge to the

other.

spectral amplitude pattern

original spectral lines

is

shown

in Fig. 3.2- It.

have been translated, both

Observe that the two

in the positive-frequency direc-

—fm

.

116 PRINCIPLES OF COMMUNICATION SYSTEMS

AMPLITUDE-MODULATION SYSTEMS 117

Amplitude of spectral

component

<<•)

o

-L

L Amplitude

I f.-L

-f.+L

-f.-L Figure 3.2-1

(a)

Spectral pattern of the

waveform A m A c cos

wm

t

cos


waveform A„ cos


t.

(fc)

Spectral pattern of the product

t.

while the product signal has four spectral components each of amplitude there are only is

two

frequencies,

and

A„ AJ4, component

the amplitude of each sinusoidal

A„AJ2.

A generalization of Fig. 3.2-1 is shown in Fig. 3.2-2. Here a signal is chosen which consists of a superposition of four sinusoidal signals, the highest in frequency having the frequency fM Before translation by multiplication, the two.

sided spectral pattern displays eight After multiplication,

we

components centered around zero frequency.

find this spectral pattern translated both in the positive-

and the negative-frequency

directions.

The

16 spectral

components

in this

two-

sided spectral pattern give rise to eight sinusoidal waveforms. While the original signal extends in range

plication fc

up

to a frequency fM

,

the signal which results from multi-

has sinusoidal components covering a range 2fM , from

fc - fM

to

+/M-

we consider in Fig. 3.2-3 the situation in which the signal to be transnot be represented as a superposition of a number of sinusoidal components at sharply defined frequencies. Such would be the case if the signal were Finally,

lated

may

of finite energy

and nonperiodic. In

this case the signal is represented in the freFourier transform, that is, in terms of its spectral density. Thus let the signal m(f) be bandlimited to the frequency range 0 to/M Its Fourier transform is M(jw) = #"[m(t)]. The magnitude M(jm)\ is shown in Fig.

quency domain

in

terms of

its

.

|

The transform M(jw) is symmetrical about = 0 since we assume that m(t) f The spectral density of the signal which results when m(t) is multiplied by cos m c t is shown in Fig. 3.2-36. This spectral pattern is deduced as an extension of the results shown in Figs. 3.2-1 and 3.2-2. Alternatively, we may easily verify (Prob. 3.2-2) that if M(jw) = ^[m(t)], then 3.2-3a.

is

Amplitude of spectral

components Spectrum

of signal

before translation

X

a real signal.

F\m(i) cos

The

=

KMC/'co

+ jcoc ) +

M(ja rel="nofollow">

- ja>c )}

(3.2-4)

spectral range occupied

by the original signal is called the baseband frequency range or simply the baseband. On this basis, the original signal itself is referred to as the baseband signal. The operation of multiplying a signal with an auxiliary sinusoidal signal is called mixing or heterodyning. In the translated signal, the part of the signal which consists of spectral components above the auxiliary signal, in the range fc to fc +fM is called the upper-sideband signal. The

a

u

coc t]

-L

,

Figve 3.2-1 An

original signal consisting of four sinusoids of differing frequencies

is

translated

through multiplication and becomes a signal containing eight frequencies symmetrically arranged about/,

part of the signal which consists of spectral components below the auxiliary signal, in the range fc M tofc is called the lower-sideband signal. The two sideband signals are also referred to as the sum and the difference frequencies, respec-

-f

,

118 PRINCIPLES OF COMMUNICATION SYSTEMS

AMPLITUDE-MODULATION SYSTEMS 119

The

lively.

aux.liary signal of frequency/,

is variously referred to as the local oscillator signal, the mixing signal, the heterodyning signal, or as the carrier signal

depending on the application. The student will note, as the discussion proceeds the various contexts is which the different terms are appropriate.

We may

note that the process of translation by multiplication actually gives somewhat different from what was intended. Given a signal occupying a baseband, say, from zero to/ M and an auxiliary signal / it would often be entirely adequate to achieve a simple translation, giving us a signal occupying the range / to / +fM that is, the upper sideband. We note, however that translation by multiplication results in a signal that occupies the range / -fM to/ + /M This feature of the process of translation by multiplication us something

,

,

,

.

on the application, be a nuisance, a matter of indifference, Hence,

this feature of the process

advantage.

which

will

of

may' depending or even an advantage

neither an advantage nor a dishowever, to be noted that there is no other operation so simple

It is,

is,

itself,

accomplish translation.

this case the signal strength will wax and wane, in addition, possibly, to disappearing entirely from time to time.

Alternatively, suppose that the recovery auxiliary signal is not precisely at is instead at/ Af In this case we may verify (Prob. 3.3-2) that the recovered baseband signal will be proportional to m(t) cos 2n Aft, resulting

frequency / but

in

a signal which will

m(f) has

tion with cos co c

t.

How

been translated out of is

its

baseband through multiplica-

the signal to be recovered?

than/,. In telephone or radio systems, an

larger

baseband. Alternatively,

[m(t) cos

w

c

t]

cos


t

we may simply note

=

m(t) cos

=

—+—

We

such a synchronous or coherent system a fixed initial phase discrepof no consequence since a simple phase shifter will correct the matter. Similarly it is not essential that the recovery auxiliary signal be sinusoidal (see Prob. 3.3-3). What is essential is that, in any time interval, the number of cycles executed by the two auxiliary-signal sources be the same. Of course, in a physical system, where some signal distortion is tolerable, some lack of synchronism may be allowed. is

m(t)

2

w

c t

= Mt\\ +

When

A

\ cos 2a>c

3.3-1.

t)

(3.3- la)

To

illustrate

baseband signal

cos

wm

w

cos

Thus, the baseband signal m(r) reappears.

- fx

is

We

c

t,

cos 2(oc

t

(3.3-16)

=

whose

A2 = -J

spectral range extends

from

.

+ The

n(r)

cos

ecovery ntirely.

0.

the

recovered

Therefore, unless

0,

then, as

will

possible to maintain 0

=

be

may

be verified proportional to

suffer.

If

it

is

s,(f)

=

wm t

+

i cos

[1

cos

+

cos

2

e»c t

2wm fXi

\ cos 2(coc

2wm t +

+

(3.3-2a)

-I-

cojt

cos 2
2mc t)

i cos

+

i cos 2(tuc

(3.3-26)

- wjt

(]

(3.3-2c)

2

the spectral component (A /A) cos 2
Received

DSB-SC

signal

Synchronizing Filter

i, (t)

-A

cos

the signal strength at n/2, the signal will be lost 0,

should happen that 0 = consider, for example, that 0 drifts back and forth with time. Then in

will

Or

it is

phase angle is baseband waveform

cos 2

filter selects

the auxiliary signal

3.3-1),

= A2

note, of course, that in addition to

he double-frequency signal is easily removed by a low-pass filter. This method of signal recovery, for all its simplicity, is beset by an important nconvenience when applied in a physical communication system. Suppose that he auxiliary signal used for recovery differs in phase from Prob.

signal

A

.

:ulty.

used in the initial translation. If this

One commonly employed scheme is indicated in we assume that

the operation of the synchronizer, a sinusoidal cos w m t. The received

a constant amplitude. This signal s,
to 2/ +fM As a matter of practice, this latter signal need cause no diffiFor most commonly / >/M> and consequently the spectral range of this louble-frequency signal and the baseband signal are widely separated. Therefore Vc

is not feasible, it is necessary to to provide a synchronous auxiliary signal at

circuit is

m(t)

a signal

is

with

auxiliary signal

means

the location of the receiver.

the

that

common

the use of a

sf(t)

he recovered baseband signal there

Af <

note, therefore, that signal recovery using a second multiplication requires that there be available at the recovery point a signal which is precisely synchronous with the corresponding auxiliary signal at the point of the first multi-

Fig.

w

at

offset

be comparable or 30 Hz is deemed

acceptable.

spectral plots as in Fig. 3.2-2 or 3.2-3 and noting that the difference-frequency signal obtained by multiplying m(t) cos co t by cos c t is a signal whose spectral c

drawing

back

Af is compa-

/

Af

resort to rather complicated

is

if

rable to, or larger than, the frequencies present in the baseband signal. This latter is a distinct possibility in many an instance, since usually > fM so that a small percentage change in/ will cause a which may

The recovery may be achieved by a reverse translation, which is accomplished simply by multiplying the translated signal with cos co t. That such is the case may be seen by c

range

entirely unacceptable

plication. In

RECOVERY OF THE BASEBAND SIGNAL

Suppose a signal

wax and wane or even be

contingency

ancy

3.3

+

u>

m

I

cos

Squaring ufc

(

signal

centered

circuit

at

Figure 3.3-1

A

simple squaring synchronizer.

2fc

120 PRINCIPLES OF COMMUNICATION SYSTEMS

AMPLITUDE-MODULATION SYSTEMS 121

A c COS Uc

quency division may be accomplished by using, for example, a bistable multivibrator. The output of the divider is used to demodulate (multiply) the incoming signal and thereby recover the baseband signal cos co t. m

We

turn our attention

now

to a modification of the

t

method of frequency

which has the great merit of allowing recovery of the baseband signal by an extremely simple means. This technique is called amplitude modulation. translation,

AMPLITUDE MODULATION

3.4

A

frequency-translated signal from which the baseband signal is easily recoveris generated by adding, to the product of baseband and carrier, the carrier

able

signal

Such a

itself.

signal with

signal

amplitude

A

lated signal (Fig. 3.4- lc)

c

is

is

shown

we

Figure 3.4-la shows the carrier

see the

baseband

signal.

The

trans-

given by v(t)

We

in Fig. 3.4-1.

in Fig. 3.4-lfc

,

=A

c

[\

+

m(t)] cos
t

(3.4-1)

observe, from Eq. (3.4-1) as well as from Fig. 3.4-lc, that the resultant waveis one in which the carrier A cos w t is modulated in amplitude. The process c c

form

waveform is called amplitude modulation, and a communicawhich employs such a method of frequency translation is called an

of generating such a tion system

amplitude-modulation system, or auxiliary signal

A

AM

for short.

The designation

" carrier " for the

seems especially appropriate in the present connection since this signal now " carries " the baseband signal as its envelope. The term "carrier" probably originated, however, in the early days of radio when this relacos

c

co c

t

high-frequency

signal was viewed as the messenger which actually baseband signal from one antenna to another. The very great merit of the amplitude-modulated carrier signal is the ease with which the baseband signal can be recovered. The recovery of the baseband signal, a process which is referred to as demodulation or detection, is accomplished tively

" carried " the

with the simple circuit of Fig. 3.4-2a, which consists of a diode D and the resistorRC combination. We now discuss the operation of this circuit briefly

capacitor

and

For

qualitatively.

carrier

which

is

simplicity,

we assume

applied at the input terminals

zero internal impedance.

We

that

the amplitude-modulated

supplied by a voltage source of assume further that the diode is ideal, i.e., of zero or is

depending on whether the diode current

infinite resistance,

is

positive or the

diode voltage negative.

assume that the input is of fixed amplitude and that the not present. In this case, the capacitor charges to the peak positive

shown in Fig. 3.4-2i>. The capacitor charges to the peak of each carrier cycle and decays slightly between cycles. The time constant RC is selected so that the

voltage of the carrier. The capacitor holds this peak voltage, and the diode would not again conduct. Suppose now that the input-carrier amplitude is increased. The diode again conducts, and the capacitor charges to the new higher carrier

change in v c between cycles is at least equal to the decrease in carrier amplitude between cycles. This constraint on the time constant RC is explored in Probs.

peak. In order to allow the capacitor voltage to follow the carrier peaks when the carrier amplitude is decreasing, it is necessary to include the resistor R, so that

seen that the voltage v c follows the carrier envelope except that v also c has superimposed on it a sawtooth waveform of the carrier frequency. In Fig.

Let us resistor

R

initially

is

the capacitor

may

discharge. In this case the capacitor voltage v has the c

form

3.4-1

and

3.4-2.

It is

3.4-26 the discrepancy between ve

and the envelope

is

greatly exaggerated. In

122 PRINCIPLES OF COMMUNICATION SYSTEMS

AMPLITUDE-MODULATION SYSTEMS 123

D -w-

In Fig. 3.5-lfo adjust

Amplitudemodulated

m>

1.

we have shown the situation which results when, in Eq. (3.5-1), we Observe now that the diode demodulator which yields as an output

the positive envelope (a negative envelope if the diode is reversed) will not reproduce the sinusoidal modulating waveform. In this latter case, where m > 1, we may recover the modulating waveform but not with the diode modulator. Recovery would require the use of a coherent demodulation scheme such as was

carrier

(a)

employed

in connection with the signal furnished by a multiplier. therefore necessary to restrict the excursion of the modulating signal in the direction of decreasing carrier amplitude to the point where the carrier ampliIt is

tude

is

tion

is

(3.5-1),

just reduced to zero.

No

such similar restriction applies when the modulaWith sinusoidal modulation, as in Eq.

increasing the carrier amplitude.

we

maximum

require that |m|

<

1.

More

negative excursion of m(t) be

generally in Eq. (3.4-1)

we

require that the

— 1.

The extent to which a carrier has been amplitude-modulated is expressed in terms of a percentage modulation. Let A c , A c (tmlx), and A c (min), respectively, be

>(<)

normal situation is one in which the time interval between carrier extremely small in comparison with the time required for the envelope to a sizeable change. Hence v c follows the envelope much more closely than is

practice, the

cycles

make

is

suggested in the figure. Further, again because the carrier frequency is ordinarily higher than the highest frequency of the modulating signal, the sawtooth

much

distortion of the envelope

3.5

waveform

is

very easily removed by a

filter.

MAXIMUM ALLOWABLE MODULATION

we are to avail ourselves of the convenience of demodulation by the use of the simple diode circuit of Fig. 3.4-2a, we must limit the extent of the modulation of If

is the case may be seen from Fig. 3.5-1. In Fig. 3.5-la is modulated by a sinusoidal signal. It is apparent that the envelope of the carrier has the waveshape of the modulating signal. The modulating signal is sinusoidal; hence m(t) = m cos co m t, where m is a constant. Equation (3.4-1) becomes

the carrier. That such

shown a

carrier

Figure 3.5-1 v(t)

= Ac (l + m

cos

co m t)

cos

a> c t

(3.5-1)

(a)

A

sinusoidally

modulated carrier (m

sinusoidal modulating waveform.

<

1).

(fc)

A

carrier

overmodulated (m

>

1)

by a

124 PRINCIPLES OF COMMUNICATION SYSTEMS

AMPLITUDE-MODULATION SYSTEMS 125

unmodulated carrier amplitude and the maximum and minimum carrier Then if the modulation is symmetrical, the percentage modulation is defined as P, given by the

levels.

—

/f c (max)

100%

•

—

— A (min) _ A (max) — A (min) _A — - = -

— Ac

C

—

c

c

In the case of sinusoidal modulation, given by Eq. (3.5-1) 3.5-la.

P = m x 100

A c [\ + that

a

nil) cos

similar

shows the

c

t.

c

and shown

THE SQUARE-LAW DEMODULATOR

be recovered from the waveform

t

easy

carrier signal.

through zero.

in Fig.

3.6

may

m(r) passes

(3.5-2)

by the simple circuit of Fig. 3.4-2a, it is of interest to note recovery of m(t) is not possible from the waveform That such is the case is to be seen from Fig. 3.5-2. Figure 3.5-2a

m(f)] cos co c

w

that the signal m(t)

absolute value of m(t). Observe the reversal of phase of the carrier in Fig. 3.5-2c

whenever

percent.

Having observed

3.5-2b, and the product m(t) cos a c t is shown in Fig. 3.5-2c. We note that the envelope in Fig. 3.5-2c has the waveform not of m(t) but rather of \m(t)\, the

The modulation or baseband

signal m(t)

is

shown

in Fig.

An

alternative method of recovering the baseband signal which has been superimposed as an amplitude modulation on a carrier is to pass the signal through a nonlinear device. Such demodulation is illustrated in Fig. 3.6-1. We assume here for simplicity that the device has a square-law relationship between input signal x (current or voltage) and output signal y (current or voltage). Thus 2 y = kx with k a constant. Because of the nonlinearity of the transfer character-

AM

,

istic

of the device, the output response

excursions of the carrier

away from

is

different for positive

and

the quiescent operating point

O

for negative

of the device.

Average output

Time

Figure 3.5-2

(a)

A

band signal m(t). and its envelope.

carrier cos oi c

(c)

The product

t.

(ft)

A

base-

m(r) cos
t

Figure 3.6-1 Illustrating the operation of a square-law demodulator.

averaged over

many

carrier cycles.

The output

is

the value of y

126 PRINCIPLES OF COMMUNICATION SYSTEMS

AMPLITUDE-MODULATION SYSTEMS 127

As a

result, and as is shown in Fig. 3.6- lc, the output, when averaged over a time which encompasses many carrier cycles but only a very small part of the modulation cycle, has the waveshape of the envelope.

The applied

signal

Amplitude of

component

spectral

Carrier

is /n 2 Ac

x

= A0 + A

Thus the output of the squaring

c

[\

circuit

+

m(t)] cos

w

c

Lower sideband

(3.6-1)

t

m, A c

Upper sideband

i

is

\

Ac

m,A c m.A,

y

=

+A

k{A 0

[l c

+

m(t)] cos o)c t} 2

,

filter

located after the squaring circuit, s„(t)

Observe that the modulation

is

1

0

/,

f2

1

f,

fc+k

is

fc-h

u

Spectral density

Thus the total recovered signal is a distorted version of the original modulaThe distortion is small, however, if jm 2 (t) -4 m(t) or if m(t) 4 2. There are two points of interest to be noted in connection with the type of

well.

|

|

f

(3.6-3)

indeed recovered but that m\t) appears as

tion.

\

fc+f,

-fl

(o)

= kA 2 im(t) + V(t)]

m(t)

m 3 Ae I

Squaring, and dropping dc terms as well as terms whose spectral components are located near co c and 2co c we find that the output signal s„(r), that is, the signal

output of a low-pass

2

(3.6-2)

[

Carrier

|

demodulation described here; the first is that the demodulation does not depend on the nonlinearity being square-law. Any type of nonlinearity which does not have odd-function symmetry with respect to the initial operating point will similarly accomplish demodulation. The second point is that even when demodulation

is

not intended, such demodulation

modulated

some

signal

is

may appear

incidentally

when

the

passed through a system, say, an amplifier, which exhibits

0

f*-f*

fc

fc+fM

nonlinearity. (i>)

Figure 3.7-1

3.7

(a) At left the one-sided spectrum of mft)^, where m(t) has three spectral components. At right the spectrum of A c [\ + m(t)] cos 2nf t. (6) Same as in (a) except m(r) is a nonperiodic signal t and the vertical axis is spectral density rather than spectral amplitude.

SPECTRUM OF AN AMPLITUDE-MODULATED SIGNAL

The spectrum signal

which

of an amplitude-modulated signal

results

similar to the spectrum of a

from multiplication except, of course, that

a carrier of frequency three sinusoidal

is

fc

is

present. If in Eq. (3.4-1) m(t)

components

m(t)

=

cos Wit

+ m2

cos

is

co 2

in the

former case

1

+ m3

cos

co 3

left

1,

then

in Fig.

The spectrum of the modulated carrier is shown at the right. The spectral the sum frequencies fc +fu fc + f2 and fc + f3 constitute the uppersideband frequencies. The spectral lines at the difference frequencies constitute the 3.7- la.

,

lower sideband.

The spectrum of

3.8

MODULATORS AND BALANCED MODULATORS

the superposition of

the (one-sided) spectrum of this baseband signal appears as at the

lines at

,

We have described a "multiplier" as a device that yields as an output a signal which is the product of two input signals. Actually no simple physical device now exists

which

On the contrary, all such devices yield, at a the product but the input signals themselves. Suppose, then,

yields the product alone.

minimum, not only

wc t and a modulating baseband The device output will then contain the product m(t) cos w c t and also the signals m(t) and cos w c t. Ordinarily, the baseband signal will be bandlimited to a frequency range very much smaller than fc = mjln. Suppose, for example, that that such a device has as inputs a carrier cos

signal m(t).

the baseband signal

Fig. 3.7- lb for the case of a

and modulated

carrier are

bandlimited nonperiodic signal of

this figure the ordinate is the spectral density,

finite

shown

in

energy. In

i.e., the magnitude of the Fourier transform rather than the spectral amplitude, and consequently the carrier is represented by an impulse.

the baseband signal extends from zero frequency to 1000 Hz, while/. = 1 MHz. In this case, the carrier and its sidebands extend from 999,000 to 1,001,000 Hz,

and the baseband

signal

is

easily

removed by a

filter.

128 PRINCIPLES OF COMMUNICATION SYSTEMS

COSUc

i4 c

AMPLITUDE-MODULATION SYSTEMS 129

f

i4 r fl

Amplitude modulator

"HI

+mi

fi]cosajc

i

Since

+ mu)

2m(()

A c cosu

a single amplitude-modulation system or a double-sideband suppressed-carrier system. The baseband signal may not be recovered from a single-sideband signal by the use of a diode modulator. That such is the case is easily seen by considering, for example, that the modulating signal is a sinusoid of frequency / In this case

Adder

Amplitude modulator

t

-A c [l -ml

i)]

cos

uic

t-m[t)

Figwc 3.8-1 Showing how the outputs of two amplitude modulators are combined to produce a double-sideband suppressed-carrier output.

The

overall result

is

that the devices available for multiplication yield

an

output carrier as well as the lower- and upper-sideband signals. The output is therefore an amplitude-modulated signal. If we require the product signal alone,

we must take steps to cancel or suppress the carrier. Such a suppression may be achieved by adding, to the amplitude-modulated signal, a signal of carrier frequency equal in amplitude but opposite in phase to the carrier of the amplitudemodulated signal. Under these circumstances only the sideband signals will remain. For this reason, a product signal is very commonly referred to as a dduble-sideband suppressed-carrier signal, abbreviated DSB-SC.

An

alternative arrangement for carrier suppression

is

shown

in Fig. 3.8-1.

Here two physical multipliers are used which are labeled in the diagram as amplitude modulators. The carrier inputs to the two modulators are of reverse polarity, as are the modulating signals. The modulator outputs are added with consequent

We observe a cancellation not only of the carrier but of the baseband signal m(t) as well. This last feature is not of great import, since, suppression of the carrier. as noted previously, the

baseband signal is easily eliminated by a filter. We note two modulators reinforce. The arrangement of Fig.

that the product terms of the 3.8-1

3.9

is

called a balanced modulator.

if only one sideband is available. For suppose a spectral component of the baseband signal is multiplied by a carrier cos co c t, giving rise to an upper sideband at co c + co and a lower sideband at co c - co. Now let us assume that we have filtered out one sideband and are left with, say, only the upper sideband at co c

now

this

sideband signal

+

and the

is

again multiplied by cos

co c

t,

we

shall generate

baseband spectral component at co. If we had used the lower sideband at coe — co, the second multiplication would have yielded a signal at 2co c - co and again restored the baseband spectral component. co

principle, that the synchronism be exact and, in practice, that synchronism be maintained to a high order of precision. The effect of a lack of synchronism is different in a double-sideband system and in a single-sideband system. Suppose that

the carrier received

with

DSB-SC,

is

cos

co c t

and that the

is cos (co t + 9). Then c component cos cot will, upon In SSB, on the other hand, the spectral

local carrier

as noted in Sec. 3.3, the spectral

demodulation, reappear as cos

cot

cos

6.

component, cos cot will reappear (Prob. 3.9-2) in the form cos (cot 6). Thus, in one case a phase offset in carriers affects the amplitude of the recovered signal and, for



=

jt/2,

may

result in a total loss of the signal. In the other case the

produces a phase change but not an amplitude change. Alternatively, let the local oscillator carrier have an angular frequency offset Aw and so be of the form cos (coc + Aco)t. Then as already noted, in DSB-SC, the recovered signal has the form cos cot cos Acot. In SSB, however, the recovered signal will have the form cos (co + Aco)t. Thus, in one case the recovered spectral offset

A

even

co. If

to

which the diode modulator can respond. Baseband recovery is achieved at the receiving end of the single-sideband communications channel by heterodyning the received signal with a local carrier signal which is synchronous (coherent) with the carrier used at the transmitting end to generate the sideband. As in the double-sideband case it is necessary, in

cot reappears with a " warble," that is, an amplitude fluctuation at the rate Aco. In the other case the amplitude remains fixed, but the frequency of the recovered signal is in error by amount Aco.

We have seen that the baseband signal may be recovered from a double-sideband suppressed-carrier signal by multiplying a second time, with the same carrier. It can also be shown that the baseband signal can be recovered in a similar manner

+

the single-sideband signal will consist also of a single sinusoid of frequency, say, fc + f, and there is no amplitude variation at all at the baseband frequency

component cos

SINGLE-SIDEBAND MODULATION

a signal at 2co c

possible to recover the baseband signal from a single sideband, there is in doing so, since spectral space is used more eco-

nomically. In principle, two single-sideband (abbreviated SSB) communications systems can now occupy the spectral range previously occupied by

t

-m{t)

A c COS w c

it is

an obvious advantage

original

phase offset between the received carrier and the local oscillator will cause baseband signal. In such a case each spectral component in the baseband signal will, upon recovery, have undergone distortion in the recovered

the

phase

shift.

Fortunately,

when SSB

same

used to transmit voice or music, such phase distortion does not appear to be of major consequence, because the human ear seems to be insensitive to the phase distortion. A frequency offset between carriers in amount of Af will cause each recovered is

component of the baseband signal to be in error by the same amount Af. had turned out that the frequency error were proportional to the frequency of the spectral component itself, then the recovered signal would sound like the original signal except that it would be at a higher or lower pitch. Such, however, is not the case, since the frequency error is fixed. Thus frequencies in the original signal which were harmonically related will no longer be so related after spectral

Now,

if it

130 PRINCIPLES OF COMMUNICATION SYSTEMS

AMPLITUDE-MODULATION SYSTEMS 131

8

iS

th3t 3 freqUenCy

between adv«sely ° Xn.^r-?!-" "? spoken communication and not tolerated U mUS,C A & matt6r ex erience ™ out that an error A/of Z than th* Tn h acceptable l to the ° P 30 Hz ffset

carriers

affects the intelligibility of

f

'

< less

is

is

>

il

well

100.3 kHz

tu

ear.

Oscillator

The need

/

to keep the frequency offset A/ between carriers small normally imposes severe restrictions on the frequency stabilities of the carrier signal generators at both ends of the communications system. For suppose that we require to keep AJ to 10 Hz or less, and that our system uses a carrier frequency of 10

2 VZ

MHz

qUen Cy

drift in thC tW ° Carfier generators ma y not exceed , eq " ahty 6 in C3rrier frequenc be maintained y thi use off quartz crystal through the oscillators using crystals cut for the same frequency at transmitter and receiver. The receiver must use as many crystals (or equivalent signals derived from crystals) as there are channels in the communicalions system.

1

I

It is

I

Audio baseband

Oscillator

- 100kHz

10 MHz

Balanced modulator

-

an SSB receiver manually and thereby reduce the the operator manually adjusts the frequency of the receiver earner generator until the received signal sounds "normal." Experienced operators are able to tune correctly to within 10 or 20 Hz. However, because of oscillator drift, such tuning must be readjusted periodically When the carrier frequency is very high, even quartz crystal oscillators may ,be hard pressed to maintain adequate stability. In such cases it is necessary to transmit the carrier itself along with the sideband signal. At the receiver the earner may be separated by filtering and used to synchronize a local carrier generator. When used for such synchronization, the carrier is referred to as a "pilot carrier and may be transmitted at a substantially reduced power level It is interesting to note that the squaring circuit used to recover the frequency and phase information of the DSB-SC system cannot be used here. In any event it is clear that a principal complication in the way of more widespread use of single sideband is the need for supplying an accurate carrier frequency at the

To do

offset.

this,

Figure 3.10-1 Block diagram of the

However, to system,

it is

alleviate

common

Balanced modulator or mixer

Filter

«,(/)

*T

also possible to tune

frequency

103 kHz

in

filter

Filter

method of generating a single-sideband

the sideband

filter

signal.

requirements in an

selectivity

SSI

SSB

to limit the lower spectral limit of speech to about 300 Hz.

found that such restriction does not materially affect the intelligibility of it is found that no serious distortion results if the upper limit of the speech spectrum is cut off at about 3000 Hz. Such restriction is advantageous for the purpose of conserving bandwidth. Altogether, then, a typical sideband It

is

speech. Similarly,

filter

3000

has a passband which, measured from

Hz and

response

falls

in

which range

its

off sharply, being

unwanted sideband also be

response

is

/

c

quite

,

extends from about 300 to

flat.

down about 40 dB

at least

40 dB. The

at

filter

Outside

4000

may

this

Hz and

passband the rejecting the

also serve, further, to

itself. Of course, in principle, no carrier should appear at the output of a balanced modulator. In practice, however, the modulator may not balance exactly, and the precision of its balance may be subject to some variation with time. Therefore, even if a pilot carrier is to be transmitted, it is well to

suppress the carrier

suppress

it

at the

output of the modulator and to add

it

to the signal at a later

point in a controllable manner.

Now 10

3.10

METHODS OF GENERATING AN

Filter

Method

A

straightforward

method of generating an SSB signal is illustrated in Fig 3 signal and a carrier are applied to a balanced modulator

10-1

The

output of the balanced modulator bears both the upper- and lower-sideband

One

or the other of these signals

is then selected by a filter The filter is a whose passband encompasses the frequency range of the sideband selected. The filter must have a cutoff sharp enough to separate the selected sideband from the other sideband. The frequency separation

bandpass

filter

the frequency of the lowest frequency spectral signal.

Human

speech contains spectral

of the sidebands

we

desire to generate

require a passband

an SSB

signal with a carrier of, say,

with a selectivity that provides 40 dB at a frequency of 10 MHz, a percentage frequency filter

of attenuation within 600 Hz change of 0.006 percent. Filters with such sharp selectivity are very elaborate and difficult to construct. For this reason, it is customary to perform the translation of the baseband signal to the final carrier frequency in several stages. Two such

SSB SIGNAL

Here the baseband signals.

consider that

MHz. Then we

is

twice

components of the baseband components as low as about 70 Hz

shown in Fig. 3.10-1. Here we have selected the first be of frequency 100 kHz. The upper sideband, say, of the output of the balanced modulator ranges from 100.3 to 103 kHz. The filter following the balanced modulator which selects this upper sideband need now exhibit a selectivity of only a hundredth of the selectivity (40 dB in 0.6 percent frequency change) stages of translation are

carrier to

10-MHz carrier. Now let the filter output be applied to a second balanced modulator, supplied this time with a 10-MHz carrier. Let us again select the upper sideband. Then the second filter must provide 40 dB of attenuation in a frequency range of 200.6 kHz, which is nominally 2 percent of required in the case of a

the carrier frequency.

132 PRINCIPLES OF COMMUNICATION SYSTEMS

AMPLITUDE-MODULATION SYSTEMS 133

We

have already noted that the simplest physical frequency-translating a multiplier or mixer, while a balanced modulator is a balanced arrangement of two mixers. A mixer, however, has the disadvantage that it presents at its output not only sum and difference frequencies but the input frequencies as well. device

Still,

is

when

it is

feasible to discriminate against these input signals, there

is

a merit

of simplicity in using a mixer rather than a balanced modulator. In the present if the second frequency-translating device in Fig. 3.10-1 were a mixer rather

case,

than a multiplier, then in addition to the upper and lower sidebands, the output would contain a component encompassing the range 100.3 to 103 kHz as well as the 10-MHz carrier. The range 100.3 to 103 kHz is well out of the range of the

second realistic

Similarly the baseband signal, before application to the modulators, is passed through a 90° phase-shifting network so that there is a 90° phase shift between any spectral component of the baseband signal applied to one modulator and the

component applied to the other modulator. most simply how the arrangement of Fig. 3.10-2 operates, let us assume that the baseband signal is sinusoidal and appears at the input to one modulator as cos co m t and hence as sin w m t at the other. Also, let the carrier be cos wc t at one modulator and sin co c t at the other. Then the outputs of the balanced modulators (multipliers) are like-frequency

To

see

cos

intended to pass the range 10,100,300 to 10,103,000 Hz. And it is to design a filter which will suppress the 10-MHz carrier, since the carrier

filter

frequency

separated from the lower edge of the upper sideband (10,100,300) by nominally a 1 -percent frequency change.

co m t

sin co m

t

cos sin

(o e t

=

w

= ^[cos

c

t

-

$[cos (»c (coc

+

cos

(co c

+

cojf]

(3.10-1)

- wjt -

cos

(co c

+

cojt]

(3.10-2)

co m )t

is

Altogether, then,

we note

summary

in

when a

that

single-sideband signal

is

which has a carrier in the megahertz or tens-of-megahertz range, is to be done in more than one stage frequently two but not uncommonly three. If the baseband signal has spectral components in the range of hundreds of hertz or lower (as in an audio signal), the first stage invariably employs a balanced modulator, while succeeding stages may use mixers.

waveforms are added, the lower sideband results; if subtracted, the upper sideband appears at the output. In general, if the modulation m(t) is given by If these

to be generated

m

—

the frequency translation

m(t)

we

then, using Fig. 3.10-2,

=

£ A, cos

(to, t

+

see that the output of the m(t) cos

w

c

t

±

(3.10-3)

0,)

SSB modulator

m(t) sin toc

is

in general

(3.10-4)

t

m

where

Phasing Method

m(t)

=

£ A, sin

(to, t

+

0 t)

(3.10-5)

(=i

An

scheme for generating a single-sideband signal is shown in Fig. Here two balanced modulators are employed. The carrier signals of angular frequency w c which are applied to the modulators differ in phase by 90°. alternative

3.10-2.

The single-sideband generating system of Fig. 3.10-2 generally enjoys less filter method. The reason for this lack of favor is that the present phasing method requires, for satisfactory operation, that a number of constraints be rather precisely met if the carrier and one sideband are adequately popularity than does the

to be suppressed.

It is

required that each modulator be rather carefully balanced

to suppress the carrier.

90" phase

Aid

Balanced modulator

It requires also that the baseband signal phase-shifting network provide to the modulators signals in which equal frequency spectral components are of exactly equal amplitude and differ in phase by precisely 90°. Such a network is difficult to construct for a baseband signal which extends over

sina!c r

difference sin

ur

I

many

Baseband signal

90° phase-

90° phase-

shift

shift

network

network

Carrier

Adder

Output

90° of phase

or

input

Subtractor

mi()COSuj c r 1 mii)sinuc<

90° phase difference

Balanced modulator

octaves. It

Figure 3.10-2

A method

also required that each

modulator display equal sensitivity to network must provide exactly

any of these constraints is not satisfied, the suppression of and of the carrier will suffer. The effect on carrier and sideband suppression due to a failure precisely to meet these constraints is explored in Probs. 3.10-3 and 3.10-4. Of course, in any physical system a certain level of carrier and rejected sideband is tolerable. Still, there seems to be a general inclishift. If

the rejected sideband

nation to achieve a single sideband by the use of passive

mUicosuc /

method which

of generating a single-sideband signal using balanced modulators

and phase

requires

many

filters

rather than by a

exactly maintained balances in passive

and

active

There is an alternative single-sideband generating scheme 2 which avoids the need for a wideband phase-shifting network but which uses four balanced

circuits.

shifters.

is

the baseband signal. Finally, the carrier phase-shift

m (A

modulators.

138 PRINCIPLES OF COMMUNICATION SYSTEMS

AMPLITUDE-MODULATION SYSTEMS 139 Carrier

,

fx

Carrier

,

/j

2.

_J_

\

m,

Modulator

Filter

{I,

A

:

Third Method of Generation and Detection of Single-sideband Signals, Proc.

IRE, December, 1956. 3.

Demodulator

Baseband

1

filter

Fitter

Demodulator

Baseband

h

2

fitter

1

Norgaard, D. E. Voelcker,

H: Demodulation

of Single-sideband Signals Via Envelope Detection.

IEEE

Trans, on

Communication Technology, pp. 22-30, February, 1966.

»>, («)

u \ Modulator

m 2 {t).

2

1

•m 2 (0

PROBLEMS

Sum 3.2-1.

plying

A

signal i>Jr)

bandlimited to the frequency range 0 to/„.

is

by the signal

it

i>

t

percent of the frequency fc

Common

3.2-2.

=

(t)

cos 2nfc

Find

t.

frequency-translated by multi-

It is

so that the bandwidth of the translated signal

fc

is

1

.

The Fourier transform

of m(()

^[mfl)]

is

=

M(f).

Show

that

communication channel Modulator

Fitter

A

^[m(f) cos 2nft

_^ Demodulator _^

Baseband

At)

filter

Figure 3.13-1 Multiplexing

many baseband

signals over a single

The

3.2- 3.

As a matter of fact, to facilitate this separation of the individual signals, the carrier frequencies are selected to leave a comfortable margin (guard band) between the limit of one frequency range and the beginning of the next. The combined output of all the modulators, i.e., the composite signal, is applied

channel. In the case of radio transmission, the

is free space, and coupling to the channel antenna. In other cases wires are used.

is

made by means

a bandpass

lator

1

flt f2 ,...,fx

.

The

to individual demodulators

which extract the baseband signals from the carrier. The carrier inputs to the demodulators are required only for synchronous demodulation and are not used otherwise.

The

final

baseband

filter.

The baseband

filter is

a low-pass

(f>)

modulator and

+

cos 2a>,r

2 cos 2co 2

1

>

v(t)

The baseband

low-pass

but

is

=

nit) cos

2nfc

recovered by

t is

which

filter

rejects the

double-

if

the recovered signal

bandlimited to 10 kHz, what

is

waveform

the

minimum

cos (2nf0

1

+

to be 90

is

value of fc for

0)?

signal mil) in the frequency-translated signal v[t) = m(t) cos 2nfc t is recovered by by the waveform cos 2n[fc + A/)t. The product waveform is transmitted through a which rejects the double-frequency signal. Find the output signal of the filter.

u(t)

filter

The baseband

3.3-3. (a)

is

value for the phase 8

possible to recover m(f) by filtering the product

multiplying

2to,,

the output signal of the filter?

is

the baseband signal m(t)

3.3-2.

recovered. There

is

signal m(t) in the frequency-translated signal

Show

waveform

available a

be recovered by appropriately that m{t)

may

filtering the

p(t)

which

is

product waveform

be recovered as well

if

lit)

=

m(t) cos 2irfc

periodic with period l/fc

.

Show

is

t

to be

that m(r)

may

p(t)v{t).

waveform has a period nffc where n kHz and let fc = 1 MHz. periodic waveforms acceptable?

the periodic

,

an integer. Assume m[t) bandlimited to the frequency range from 0 to 5

Find the largest n which

The

3.3- 4.

plying

no function in the system as described up to however, see in Chaps. 8 and 9 that such baseband are essential to suppress the noise which invariably accompanies the signal.

We

1

2a>,.

is the maximum allowable maximum possible value?

it is

(b)

=

a> 2

What

(c) If

in this sense serves

the present point.

+

a>,f

a> 2

signal m(t) in the frequency-translated signal

What

which

is

filter

with cutoff at the frequency fM to which the baseband signal is limited. This baseband filter will pass, without modification, the baseband signal output of the

filters

frequency signal.

operation indicated in Fig. 3.13-1 consists in passing the demodu-

lator output through a

cos

v(t)

filter

filter

They are then applied

2 cos

=

by the waveform cos (2nft r + 9). The product waveform is transmitted through a low-pass

multiplying (a)

Repeat for


The baseband

3.3- 1.

percent of the

which passes only the spectral range of the output of moduand similarly for the other bandpass filters. The signals have thus been

separated.

not a harmonic of

of an

At the receiving end the composite signal is applied to each of a group of bandpass filters whose passbands are in the neighborhood is

=

v 2 (t)

are multiplied. Plot the resultant amplitude-frequency characteristic, assuming that
to

channel

fi

u,(t)

communications channel.

multiplexing.

common communications

c)

signals

and

a

= $M{f+f + M(/-/c )]

t]

v[t)

(a)

shall,

will

allow m(t) to be recovered. Are

signal m[t) in the

DSB-SC

by a signal derived from

Show

v{t)

=

all

m(t) cos (co c

t

+

9) is to

be reconstructed by multi-

2 (t).

2

(l) has a component at the frequency 2fc Find bandlimited to /„ and has a probability density

that u

{b) If m(t) is

u

signal

.

— Am) = ^—e-" 11

its

amplitude.

-oosmsco

REFERENCES find the expected value of the amplitude of the

Telephone Laboratories: "Transmission Systems for Communications," Western Electric Company, Tech. Pub., Winston-Salem, N.C., 1964. Bell

3.4- 1.

The envelope detector shown

signal

v(t)

=

[1

having a period

+

m(()] cos

Tp

l/fc

.

m

c

t,

component of v 2 (t)

in Fig. 3.4-2a

where

m(t) is a

is

at 2fc

.

used to recover the signal m(f) from the

AM

square wave taking on the values 0 and —0.5 volt and

Sketch the recovered signal

if

RC =

T/20 and 4T.

AMPLITUDE-MODULATION SYSTEMS 141

140 PRINCIPLES OF COMMUNICATION SYSTEMS

The waveform

3.4- 2. (a)

v(t)

=

(1

diode demodulator of Fig. 3.4-2a.

+m

required that at any time r 0

lit), it is

cos

Show

co m t)

that,

cos

if

co c

m

with

t,

S

a constant (m

the demodulator output

is

1), is

applied to the

The input

:

3.10-3. In the

/

1

m sin o> m 0 + m cos co m

"\1

which

shift

t


RC ~

respects in t

0

input

Using the

{b)

m

times then

show

result of part (a),

must be

that

if

the demodulator

m0

than or equal to the value of

less

is

to follow the envelope at all

determined from the equation

,

is

3.10- 4.

phase

ized

is m(t),

given by Eq. (3.10-3). If/,

qualitatively, the

form of the demodulator output when the condition specified

in part

SSB

i)(r)

=

(1

(ft>

c

.

waveform

frequency range 0 <,/ <,fM

in the

3.6-2.

Repeat Prob.

3.6-3.

The

3.6-1

=

.

the square-law demodulator

if

[1

+

m(f)] cos co c

Fourier transform of v 0 (t)

in

the frequency range

istic

=

t)„

domain

v,

=

v

2u

2

2 .

M

If

signal

Show

3.9-1. (a)

=

centered at the origin so that v a

v

1 .

t is

-fM

t
=

+

+0.1 cos

(1

2 (r).

,


.

Hint: Convolution in the frequency

See Prob. 1.12-2.

0.1 cos 2
Plot the amplitude-frequency characteristic of

.

is

square-law detected by a detector having the characterthe Fourier transform of m(t) is a constant 0 extending from -/„ to +/M sketch the v(t)

needed to find the Fourier transform of m

is

The

3.6-4.

signal

r is

detected by a square-law detector,

t>„(r).

that the signal IV

°(0

=

Z i=

[ cos

a

c

1

c °s

(a>, t

+

0,)

-

sin
t

sin (
t

+

0,)]

i

is

an SSB-SC signal (f>)

(c)

(co c > u)„). Is it the upper or lower sideband? Write an expression for the missing sideband. Obtain an expression for the total DSB-SC signal.

The SSB

3.9-2.

signal in Prob. 3.9-1

multiplied by cos to t

is

t

and then low-pass

filtered to

recover the

modulation. (a) filter

Show

is/M

that the modulation


completely recovered

is

(f>)

2fr If the multiplying signal were cos

(c)

If

0

if

the cutoff frequency of the low-pass

.

the multiplying signal were cos

(
+ 6), find the recovered signal. + A
that Atu

<«

CO,.

3.9- 3.

Show

that the squaring circuit

oscillator signal capable of

3.10- 1.

on

A

baseband

signal,

bandpass

which

is

about

each point 3.10-2.

filters 1

in the

40

not permit the generation of a local

signal.

MHz as

a single-sideband modulation using the

are available which will provide 40

percent of the

center frequency.

dB Draw a

is

to be

filter

superimposed

method. Assume

of attenuation in a frequency interval

block diagram of a suitable system. At system draw plots indicating the spectral range occupied by the signal present there.

The system shown

ideal 90° phase-shifting shift is

in Fig. 3.3-1 will

bandlimited to the frequency range 300 to 3000 Hz,

a carrier of frequency of

that

shown

demodulating an SSB-SC

filter

in Fig. 3.10-2 is

network which

approximated by a

lattice

used to generate a single-sideband signal. However, an is unattainable. The 90° phase

independent of a frequency network having the transfer function is

f]{

_

e

-J«rcun(//30)

Hz and/M = 3000 Hz show

that

from 90° by a small angle a. Calculate the output waveform and point out the which the output no longer meets the requirements for an SSB waveform. Assume that the

a single spectral component cos co m t. Repeat Prob. 3.10-3 except, assume instead, that the baseband phase-shift network produces a

A

received

power of

from 90° by a small angle

SSB

0.5 volt

signal in

2 .

A

carrier

percent of the normalized power

a.

which the modulation is

is

Find the carrier amplitude required.

+ m cos
300

generating system of Fig. 3.10-2, the carrier phase-shift network produces a phase

satisfied.

The

3.5- 1.

=

differs

added to the

diode demodulator. The carrier amplitude

Draw,

not

(/>) is

network

shift differing

3.11- 1.

(c)

to this

3 e -«V J0", for/,
to follow the envelope of

in the

is

signal,

a single spectral

and the

component has a normal-

carrier plus signal are applied to a

is to be adjusted so that at the demodulator output 90 recovered modulating waveform. Neglect dc components.

we have noted that, at least in the special case of single sideband using modulation with a single sinusoid, there is no amplitude variation at all. And even more generally, when the amplitude of the modulated signal does with some reservation, for

CHAPTER

FOUR

vary, the carrier envelope need not have the

FREQUENCY-MODULATION SYSTEMS

We now

turn our attention to a

new

waveform of the baseband

type of modulation which

signal.

not charac-

is

by the features referred to above. The spectral components in the moduwaveform depend on the amplitude as well as the frequency of the spectral components in the baseband signal. Furthermore, the modulation system is not linear and superposition does not apply. Such a system results when, in connecterized

lated

tion with a carrier of constant amplitude, the phase angle

some way

to a

baseband

=A

v(t)

in

which A and

co c

cos

[_w c

t

+

Modulation of

this

type

quency modulation for reasons to be discussed

amplitude and in which the signal information carrier through variations of the carrier frequency. stant

is

superimposed on the

review

some elementary

recall that the function

A

ANGLE MODULATION

The

up

to the present point have

two

prin-

common.

In the first place, each spectral component of the baseone or two spectral components in the modulated signal. These components are separated from the carrier by a frequency difference equal to the frequency of the baseband component. Most importantly, the nature of the modulators is such that the spectral components which they produce depend

band signal gives

Dnly

on the

the spectral af the

rise to

carrier frequency

components of

the

and the baseband frequencies. The amplitudes of modulator output may depend on the amplitude

input signals; however, the frequencies of the spectral components do not. second place, all the operations performed on the signal (addition, subtracand multiplication) are linear operations so that superposition applies.

With respect

if a baseband signal m^t) introduces one spectrum of components into the nodulated signal and a second signal m 2 (t) introduces a second spectrum, the

Thus,

ipplication of the

sum

m,(t)

+ m 2 (()

will

introduce a spectrum which

is

the

sum

the spectra separately introduced. All these systems are referred to under the iesignation " amplitude or linear modulation." This terminology must be taken )f

142

co c

cos

t

(t)

is

the phase

another designation

is

fre-

in the next section.

let

us

can be written as

a> c

=

t

real part

(AeJ '°'')

(4.2-1)

in the

counterclockwise direction with an angular velocity

4> is

,

actually not time-dependent but

precisely in the

manner

is

a constant, then

just described.

But suppose

time and makes positive and negative excursions.

v(t),

A

cos

w

c

t.

We

may,

therefore, a representation of a signal

+

the phasor of angle 9

v(t) is

=


.

to be represented

does change with

would be represented behind the phasor rep-

v(t)

falls

therefore, consider that the angle o) c

undergoes a modulation around the angle 6 If

d>

Then

by a phasor of amplitude A which runs ahead of and resenting

co c

which also rotates in the counterclockwise the phasor will be stationary. If in Eq. (4.1-1)

to a coordinate system

direction with angular velocity co c

In the :ion,

Still

function Ae is represented in the complex plane by a phasor of length A and an angle 9 measured counterclockwise from the real axis. If 9 = w c t, then the

1

the modulation schemes considered

cipal features in

a function of

19

phasor rotates All


called angle modulation for

ideas in connection with sinusoidal waveforms,

cos

A 4.1

is

PHASE AND FREQUENCY MODULATION

4.2

To

to respond in

(4.1-1)

also referred to as phase modulation since

It is

angle of the argument of the cosine function.

In the amplitude-modulation systems described in Chap. 3, the modulator output consisted of a carrier which displayed variations in its amplitude. In the present chapter we discuss modulation systems in which the modulator output is of con-

made

4>(t)1

are constant but in which the phase angle

the baseband signal.

obvious reasons.

is

Such a signal has the form

signal.

d>(t)

which

is

=w

+

c

t

=w

c t.

modulated
t

The waveform in

+


of

of

v(t) is,

phase.

alternately runs

ahead of and

behind the phasor 9 = a> c t, then the first phasor must alternately be rotating more, or less, rapidly than the second phasor. Therefore we may consider that the falls

v(t) undergoes a modulation around the nominal angular velocity co c The signal u(t) is, therefore, an angular-velocitymodulated waveform. The angular velocity associated with the argument of a sinusoidal function is equal to the time rate of change of the argument (i.e., the

angular velocity of the phasor of .

Thus we have that the instantaneous and the corresponding frequency / = m/2n is

angle) of the function.

d=

d(6

+


/-£ 5 The waveform

u(t) is,

radial frequency

(4

-

2 - 2)

.

tion refers only to the

therefore, modulated in frequency.

In initial discussions of the sinusoidal

waveform

customary

it is

we have generalized these concepts somewhat. To acknowledge is

it

not

uncommon

instantaneous frequency and

to refer to the frequency

(t)

to consider

/

this

gener-

in Eq. (4.2-2) as the

w

c

is

small, that

is, if d(t)/dt
wc

,

the

will

modulation changes

Among

We

might arrange that the phase

4>(t)

in Eq. (4.1-1)

be

modulating signal, or we might arrange a direct proportionality between the modulating signal and the derivative, d(t)/dt. From Eq. directly proportional to the

(4.2-2), with, fe

is

it

of

is,

involved simply

from a visual examination of the waveform or from an analytical expression for the waveform. We would also have to be given the waveform of the modulating signal. This information is, however, provided in any practical communication system.

RELATIONSHIP BETWEEN PHASE AND FREQUENCY MODULATION

4.3

The

level.

the possibilities which suggest themselves for the design of a modula-

tor are the following.

the basis of these definitions

then

have an appearance which is readily recognizable as a " sine wave," albeit with a period which changes somewhat from cycle to cycle. Such a waveform is represented in Fig. 4.2-1. In this figure the modulating signal is a square wave. The frequency-modulated signal changes frequency whenever

waveform

On

as the instantaneous phase. If the frequency

variation about the nominal frequency the resultant

second type.

course, not possible to determine which type of modulation

such a waveform as having a fixed frequency and phase. In the present discussion alization,

where / is the instantaneous frequency. Hence in this latter case the proportionality is between modulating signal and the departure of the instantaneous frequency from the carrier frequency. Using standard terminology, we refer to the modulation of the first type as phase modulation, and the term frequency modula-

and frequency modulation may be visualized

relationship between phase

further by a consideration of the diagrams of Fig. 4.3-1. In Fig. 4.3-la the phase-

modulator block represents a device which furnishes an output carrier, phase-modulated by the input signal m,(t). Thus v{t)

=w /2n

= A

cos

[co c

+

t

v(t)

which

is

a

(4.3-1)

k'mffl]

c

k'

(4.2-3)

being a constant. Let the waveform

modulating signal

m,(t)

be derived as the integral of the

m{t) so that

m,(t)

m.(l)

=

m{t)dt

fc"

(4.3-2)

J

J- 00 in

which

k"

is

also a constant.

u(t)

Modulating s,gnal

... m (t)

Then with k

=A

Integrator

cos

n»i

signal

Figure 4.2-1

An angle-modulated waveform,

soidal carrier signal.

(a)

Modulating

signal, (b)

o m(<)

o—

t

+

k'k"

k

Phase modulator

we have m(t) dt]

v(°t)

-

* frequencymodulated signal

{g) v

-

A phase modulated signal

(b)

0 Differentiator

m (0 f

Frequency modulator

(4.3-3)

I

o

o

o

Modulating

(0

[wc

—

v{t)

0

Frequently-modulated sinuFigure 4.3-1 Illustrating the relationship between phase and frequency modulation.

The instantaneous angular frequency

W=

Wc lit

1

+

k

\_

The maximum frequency deviation

is

d

\

=

+

°

c c

*\

(

Equation

(4.4-1) can, therefore,

is

directly proportional to the

modulating signal, the combination of integrator and phase modulator of Fig. :he

combination

a signal whose phase departure from the

phase-modulated output,

proportional to the modulating signal.

i.e.,

=A

In summary, we have referred generally to the waveform given by Eq. (4.1-1) an angle-modulated waveform, an appropriate designation when we have no nterest in, or information about, the modulating signal. When
:he

instantaneous frequency deviation

sider next the spectral pattern of

4.5

find that such precision of language

is

±

Af,

it

should not be

in this range.

We

con-

not

is

we

In this section

shall look into the frequency

=

v(t)

which

is

cos (w c

t

+

spectrum of the signal

P sin m m t)

and without

common. Very

the signal of Eq. (4.4-1) with the amplitude arbitrarily set at unity as a

We have

matter of convenience. cos

(
t

+

p

sin co m

t)

=

cos

co c

t

cos

(/?

sin co m

frequently

reference to, or even interest

in,

the

now

Consider

the expression cos (P sin

(4.1-1) the

maximum

value attained by

4>(t),

phase deviation. Similarly the

maximum

that

(oc

t,

is

is,

departure of the instantaneous

quency from the carrier frequency is called the frequency deviation. When the angular (and consequently the frequency) variation with frequency fm we have, with w m = 2vfm

the

called

wm

Therefore

.

which

it

is

an

t

sin

(/?

sin oi m

t)

(4.5-2)

w m t)

which appears as a factor on the an angular expand this expression in a Fourier

even, periodic function having

mJ2n

is

possible to

the fundamental frequency.

result is

fre-

cos (P sin is

(4.5-2). It is

sin co c

We shall not undertake the evaluation of the coefficients in the Fourier expansion of cos (J? sin m m t) but shall instead simply write out the results. The coefficients are, of course, functions of /?, and, since the function is even, the coefficients of the odd harmonics are zero. The series in

deviation of the total angle from the carrier angle

t)

— right-hand side of Eq.

PHASE AND FREQUENCY DEVIATION waveform of Eq.

(4.5-1)

directly proportional

signal.

maximum phase

(4.4-4)

t^j

SPECTRUM OF AN FM SIGNAL:

frequency

(n the

sin co m

such an angle-modulated waveform.

terms angle modulation, phase modulation, and frequency modulation are

modulating

the

+ j-

t

While the instantaneous frequency /lies in the range fc concluded that all spectral components of such a signal lie

the signal m(t) which appears explicitly in the expression. In general usage,

ised rather interchangeably

4.4

cos (co c

SINUSOIDAL MODULATION

is

we

be written

producing a frequency-modulated output. Similarly, and frequency modulator gen-

carrier is

lowever,

(4.4-3)

in Fig. 4.3- If) of the differentiator

;rates a

:o

given by

H-3-5)

Since the deviation of the instantaneous frequency

4.3-la constitutes a device for

is

is

v(t)

y-f-f*

Af and

¥=PL

4 -3-4)

The deviation of the instantaneous frequency from the carrier frequency cojln

defined as

is

o) m

t)

=

J 0 (p)

+

2J 2 (P) cos

sinusoidal

2co M

+

t

+

IJJ^P) cos 4
+

2J 2 Jlfi) cos 2ruo m t

t

+

(4.5-3)

,

v(t)

where P

is

deviation,

quency

=A

the peak amplitude of is

(/? sin co m t), which is an odd function, only odd harmonics and is given by

while for sin

cos (t).

{coc t

+P

sin co m

In this case

/?

(4.4-1)

t)

which

usually referred to as the modulation index.

is

the

maximum

phase

The instantaneous

sin (P sin co m

fre-

t)

=

2J 1 (/?)

sin co m

t

+

+

is

/=g + ^cos
t

The functions

(4.4.2a)

are

= fc + PL

COS

(O m

t

(4.4-26)

known

J„(P)

2J 3 (P)

+

we

sin 3tu m

2J 2n . 1 (p)

find the expansion contains

t

sin (2n

-

\)w m

t

+

(4.5-4)

occur often in the solution of engineering problems. They first kind and of order n. The numerical

as Bessel functions of the

values of J„(p) are tabulated in texts of mathematical tables. 2

Putting the results given in Eqs. (4.5-3) and (4.5-4) back into Eq. (4.5-2) and using the identities

cos sin

we

find that

v(t)

A

B=

cos

A

B=

sin

j cos (A

-

B)

+

| cos (A

-

B)

-1

(-4

+

B)

(4.5-5)

cos (A

+

B)

(4.5-6)

{ cos

J„(n-

0,

1,2)

becomes

in Eq. (4.5-1)

1

J

=

v(t)

— J^lcos (o) — co m )t — cos (a> + + J 2 (/?)[cos (« - 2«Jl + cos (co + 2coJt] - J 3 (/8)[cos (co - 3coJt - cos (w + 3coJr]

J 0 (fi) cos

o) c

t

c

c

c

c

c

c

1

Pa rrtcr

am

litude

<»Jt]

,

First-or der

sideb and

com jonent

arnplitude

I

+

>econd or der

••

(4.5-7)

Observe that the spectrum

composed of a carrier with an amplitude J 0 (P) and on either side of the carrier at frequency

is

a set of sidebands spaced symmetrically

m 3
separations of

,

,

,

which

AM

a

sinusoidal modulating signal gives rise to only one sideband or one pair of side-

bands.

A

4.5-1), is

second difference, which is left for verification by the student (Prob. that the present modulation system is nonlinear, as anticipated from the

discussion of Sec.

4.6

4.1.

0

-0.2

SOME FEATURES OF THE BESSEL COEFFICIENTS

-0.4 0

2

6

4

Several of the Bessel functions which determine the amplitudes of the spectral

components in the Fourier expansion are plotted in Fig. 4.6-1. We note that, at P = 0, J o (0) = 1, while all other J„'s are zero. Thus, as expected when there is no modulation, only the carrier, of normalized amplitude unity,

is

is

JB (n -

,

10

12

14

16

modulation index

3,4, 5)

present, while all

departs slightly from zero, J^P) acquires significant in comparison with unity, while all higher-order

sidebands have zero amplitude. a magnitude which

When

ft

8

/?

comparison. That such is the case may be seen either from from the approximations 3 which apply when /? 4 1, that is,

J's are negligible in

Fig. 4.6-1 or

j 0 (/?)=n-(|y

J#) =

(4.6-D

n

^(f)"

+

0

(4.6-2) 0

Accordingly, for

fl

very small, the

FM

is

is,

a signal where

of significant magnitude,

/?

is

is

is

composed of a

carrier

and a

single

called a

.

2

4

6

0

8 ,

10

narrowband

FM

signal.

We

see further, in

Fig. 4.6-1, as /? becomes somewhat larger, that the amplitude J t of the first sideband pair increases and that also the amplitude J 2 of the second sideband pair

14

12

16

modulation index

FM

signal which is so constiw c ± co m An small enough so that only a single sideband pair

pair of sidebands with frequencies tuted, that

signal

Figure 4.6-1 The Bessel functions JJjl) plotted as a function of

/J

for n

=

0, 1,

2

5.

oo \o o v> os n oo - * — tNOrN©fN
becomes

p continues to increase, J 3

significant. Further, as

,

J4

,

etc.

begin to

r-

r--

<*">

CT\

rso v\

'

acquire significant magnitude, giving rise to sideband pairs at frequencies

±

2ct> m

,

w ± c

3&i m

co c

ooodooocidooo ociddd III I

,

etc.

which FM is unlike the linear-modulation schemes described earlier is that in an FM signal the amplitude of the spectral component at the carrier frequency is not constant independent of p. It is to be expected that

Another respect

in

such should be the case on the basis of the following considerations. The signal has a constant amplitude. Therefore the power of such envelope of an a signal is a constant independent of the modulation, since the power of a period-

u~>

m d

oo os

v~i

s m



r~-

ooooooooooo III

no

On 00 n n oo oo (N m r- o w — Q Q \C O O O O O O O O O O (N On

«*>

r-i

*

I

FM

waveform depends only on the square of its amplitude and not on its frequency. The power of a unit amplitude signal, as in Eq. (4.5-1), is P„ = \ and is independent of p. When the carrier is modulated to generate an FM signal, the power in the sidebands may appear only at the expense of the power originally in the carrier. Another way of arriving at the same conclusion is to make use of the 3 = 1. We calculate the power P„ by squaridentity J\ + 2J\ + 2J\ + 2J| + 2 ing v(t) in Eq. (4.5-7) and then averaging v (t). Keeping in mind that crossic

•

product terms average to zero, we P„

as expected.

We

these values of

/?

•

find,

independently of

= i(j* +

2

/?,

r^vOO— oo NO NO m m — a, o oo n m n
ooooooooo <

I

I

I

NO rN ^ n On NO r- r- ^ oo m oo »m Q ^ — O N O Q O O O o o o o o o o NT)

w*i

w~i

(*i

'

I

I

that

2 jCJ„ ) = ±

(4.6-3)

On OO n© ~h OO NO SrNO fN ^ r» (N On i/ir-^-^mNO^cN -nr-itN*— n ci fs h i/*n

OOOOOOOO i

=

observe in Fig. 4.6-1 that, at various values of /?, J 0 (P) all the power is in the sidebands and none in the carrier.

I

0.

I

At 00 —i
m m

FM signal

is

w ^ ^ o o S 8

dodo dodo

BANDWIDTH OF A SINUSOIDALLY MODULATED FM SIGNAL

4.7

when an

r-t

dodo

NO NO <0

In principle,

r*"i

modulated, the number of sidebands

is

infinite

— (S S O O m m oc
and the bandwidth required to encompass such a signal

is

similarly infinite in

As a matter of practice, it turns out that for any /?, so large a fraction of the total power is confined to the sidebands which lie within some finite band-

r-i

I

*

f*1

Tfr

I

extent.

width that no serious distortion of the signal results

bandwidth are

lost.

We

if

os

the sidebands outside this

see in Fig. 4.6-1 that, except for

J 0 (P), each

J„(P)

hugs the

no

m

oo 0.01.'

0.30!

0.04:

an:

zero axis initially and that as n increases, the corresponding J„ remains very close to the zero axis up to a larger value of /?. For any value of p only those J„ need

be considered which have succeeded in making a significant departure from the zero axis. How many such sideband components need to be considered may be seen from an examination of Table 4.7-1 where J n(P) is tabulated for various values of n

m

^ — dodo

and of p.

found experimentally that the distortion resulting from bandlimiting an FM signal is tolerable as long as 98 percent or more of the power is passed by the bandlimiting filter. This definition of the bandwidth of a filter is, admittedly, somewhat vague, especially since the term " tolerable " means different things in different applications. However, using this definition for bandwidth, one can proceed with an initial tentative design of a system. When the system is built, the It is

On r» OO On no
On tT m — Tt a> N Q h Q 9 r- 4 — o O c> d c> d> d> rs

>C

OHMm^irt\OM»OiO--(NWTtv)>o 151

thereafter be readjusted, if necessary. In each column of Table has been drawn after the entries which account for at least 98 percent of the power. To illustrate this point, consider B = 1. Then the power contained

may

bandwidth

4.7-1, a line

=

terms n

in the

0,1,

and 2

By way of example, when B = carrier

is

Emde 2

Eq. (4.7-2) holds

+ Jim + Jim

=

0.289

+

0.193

+

0.013

for

=

0.495

(4.7-1)

FM

FM

is

furthest

We

(4.7-2) in

a form which

is

more imme-

find

B = 2(A/+/J Expressed in words, the bandwidth ation and the

from the which

.

c

We

signal

components

to require consideration are those

it

which J m is tabulated there. Using Eq. (4.4-3), we may put Eq.

diately significant.

signal, which is {. is 99 percent of the power in the note that the horizontal lines in Table 4.7-1, which indicate the value of n for 98 percent power transmission, always occur just after n = B + 1. Thus, for sinusoidal modulation the bandwidth required to transmit or receive the

The sum 0.495

the sideband

f ± 6fm From the table of Bessel functions published in may be verified on a numerical basis that the rule given in without exception up to 0 = 29, which is the largest value of 6

occur at frequencies

Jahnke and

p=

5,

which have adequate amplitude

(4.7-3)

sum of the maximum frequency devimodulating frequency. This rule for bandwidth is called Carson's is

twice the

rule.

B=

2(B

+

(4.7-2)

l)/„

imation applies quite well even when 0

1.0

(a)

We deduced Eqs. (4.7-2) and (4.7-3) as a generalization from Table 4.7-1, which begins with B = 1. We may note, however, that the bandwidth approxB = 2fm which we know narrowband FM.

(4.7-2) gives

0-0.2

The spectra of

,

several

FM


For, in that case,

we

find that Eq.

to be correct from our earlier discussion of

signals with sinusoidal

modulation are shown in These spectra are constructed directly from the entries in Table 4.7-1 except that the signs of the terms have been ignored. The spectral lines have, in every case, been drawn upward even when the corresponding entry is negative. Hence, the lines represent the magnitudes only of the spectral components. Not all spectral components have been drawn. Those, far removed from the carrier, which are too small to be drawn conveniently to scale, have been omitted. Fig. 4.7-1 for various values of B.

1

<4 ->m>

1.0

4

<4+'«>

f

-

(*)

(3-1

,1

1, 4.8

f

P

0-5 1.0

EFFECT OF THE MODULATION INDEX

ON BANDWIDTH

-

plays a role in FM which is not unlike the role played by the parameter m in connection with AM. In the case, and for sinusoidal modulation, we established that to avoid distortion we must observe m = 1 as an upper limit. It was also apparent that when it is feasible to do so, it is advantageous to adjust m to be close to unity, that is, 100 percent modulation; by so

The modulation index B

AM

kill

lll.l

1

.

1

doing,

On

/J-10 1.0

-

sible.

AM

id)

,

Figure 4.7-1

1

1

The

1

1

1

1

.

1

,1.11

spectra of sinusoidally modulated

.

1

1

1

1

FM signals for various values of

1

/!.

.

.

For, again, the larger the constraint that

is fi,

m<

at a

maximum.

as large as pos-

the stronger will be the recovered signal. While in

imposed by the necessity to avoid distortion, on fi. There is, however, a constraint which needs to be imposed on fi for a different reason. From Eq. (4.7-2) for B > we have B » 2Bfm Therefore the maximum value we may allow for B is determined by the maximum allowable bandwidth and the modulation frequency. In comparing AM with FM, we may there

.

we keep the magnitude of the recovered baseband signal same basis we expect the advantage to lie with keeping fi

this

is

no

1

is

similar absolute constraint

1

.

then note, in review, that in

AM

the recovered modulating signal

may

1.0

made

be

manner which keeps is* no similar limit on the modula-

(o)

progressively larger subject to the onset of distortion in a

the occupied bandwidth constant. In FM there tion, but increasing the magnitude of the recovered signal is achieved at the expense of bandwidth. A more complete comparison is deferred to Chaps. 8 and 9, where we shall take account of the presence of noise and also of the relative

power required

0-5

!

..III.

!

i

i

.

i

I

i

fc

^

2*f

for transmission. 1.0

0-10

(6)

4.9

SPECTRUM OF "CONSTANT BANDWIDTH" FM ,

we

Let us consider that

are dealing with a modulating signal voltage v„ cos 2nfm

1

1

1

1

1

.

,

i

.

.

i

i

,

i

.

1

1

1

4

t

2Af

with v m the peak voltage. In a phase-modulating system the phase angle <j>(t) would be proportional to this modulating signal so that <j>(t) = k'v m cos 2nfm t, with k' a constant. The phase deviation is /? = k'v m and, for constant v m , the bandwidth occupied increases linearly with modulating frequency since B S ,

2f)fm

=

2k'v m

fm We may .

avoid

quency by arranging that


this variability of

=

bandwidth with modulating

{k/2nfm )v m sin 2nfm

t

(k

a constant). For,

0-20

(C)

fre-

in this

Illi

..1

iii.h.i.n.i. .l.ll.l.ll.lll.lllll

1...

fc

case

and the bandwidth

is

BS

(2k/2n)v m

,

independently

of/m

.

In this latter case,

is a> = co c + kv m cos 2nfm t. Since the proportional to the modulating signal, the initially angle-modulated signal has become a frequency-modulated signal. Thus a signal intended to occupy a nominally constant bandwidth is a frequently-modulated

however,

instantaneous

the

instantaneous frequency

2*f

Figure 4.9-1 Spectra o F sinusoidally modulated

FM

is

we have drawn the spectrum for three values of /J for the condition that fifm is kept constant. The nominal bandwidth B 3 2 A/ = 2fifm is consequently constant. The amplitude of the unmodulated carrier at/c is shown by a dashed line. Note that the extent to which the actual bandwidth extends beyond the nominal bandwidth is greatest for small 0 and large /„ and is least for large .

4.10

5.

there are

For

all

+

=

/?

1

With the aid of a phasor diagram we shall be able to arrive at a rather physically intuitive understanding of how so odd an assortment of sidebands as in Eq. (4.5-7) yields an FM signal of constant amplitude. The diagram will also make clear the difference between AM and narrowband FM (NBFM). In both of these cases there is only a single pair of sideband components. Let us consider first the case of narrowband FM. From Eqs. (4.4-1), (4.6-1), and (4.6-2) we have for ^ ^ 1 that

FM

other modulation frequencies

/?

is

larger than

6 significant sideband pairs so that at/m

2x6x15

=

15

When kHz the

5.

/?

=

,

v(t)

5,

=

cos (w c

S

cos

4.

10- la.

co e

t

t

+

fS

—-

sin
cos

(co c

(4.10-la)

t)

-

cujt

+-

cos (w c

+

cojt

(4.10-lfc)

band-

= 1 80 kHz, which is to be compared with width required is 5 = = there are 21 significant sideband pairs, and 20, 2 A/= 150 kHz. When /> B = 2 x 21 x 15/4 = 157.5 kHz. In the limiting case of very large 0 and correspondingly very small fm the actual bandwidth becomes equal to the nominal bandwidth 2 A/

2 tsf

PHASOR DIAGRAM FOR FM SIGNALS

broadcasting, the Federal Communications Commission allows a frequency deviation A/ = 75 kHz. If we assume that the highest audio frequency to be transmitted is 15 kHz, then at this frequency 0 = Af/fm =

=

The nominal bancIwidth B a 2ffm =

signals.

kept fixed.

is

In Fig. 4.9-1

75/15

»

frequency

rather than an angle-modulated signal.

and small fm In commercial

-

Refer to Fig.

Assuming a coordinate system which rotates counter-

clockwise at an angular velocity Eq. (4.10-1)

is

fixed

and oriented

co c

,

the phasor for the carrier-frequency term in

in the horizontal direction. In the

nate system, the phasor for the term

(fi/2)

cos

clockwise direction at an angular velocity

cu m

(co c ,

+

co m )t

same coordi-

rotates in a counter-

while the phasor for the term

and the individual terms are represented as phasors in Fig. 4.10-lb. Comparing Eqs. (4.10-1) and (4.10-3), we see that there is a 90° phase shift in the phases of the sidebands between the FM and AM cases. In Fig. 4. 10- lb the sum A of the sideband phasors is given by

A = m

sin

wm

(4.10-4)

t

The important difference between the FM and AM cases is that in the former the sum A, is always perpendicular to the carrier phasor, while in the latter the sum A is always parallel to the carrier phasor. Hence in the AM case, the resultant R does not rotate with respect to the carrier phasor but instead varies

between

1

m

-I-

and

1

—

amplitude

in

m.

Another way of looking NBFM where /?
at the difference

between

AM

and

NBFM

is

to note

that in

«

cos

u(t)

=

cos

the

first

v(t)

while in

Figure 4.10-1

(a)

Phasor diagram

for a

narrowband

Note

that in

— (/J/2) ity
cos .

—

cos

cojt rotates in a clockwise direction, also at the angular velocAt the time t = 0, both phasors, which represent the sideband com(co c

maximum

ponents, have parallel to,

that the

two

cancel.

The

carrier,

resultant R.

now

at the

reduced in amplitude, and A, combine to give rise to a of R from the carrier phasor is
4.

10- la that since

The small

to be expected.

appears in Fig.

4. 10- la is

<

1,

the

maximum

value of

tan


=

/?,

as

variation in the amplitude of the resultant which

only the result of the fact that

we have

neglected higher-

order sidebands.

Now let us consider the phasor diagram for AM. The AM (1

+m

sin

mm t)

cos

coc

t

=

cos

w t+— c

sin

(wc

+

cojt

signal

- ^ sin

is

(coc

-

(4.10-3)

if

sin coc

w

c

t

+m

sin co m

t

cos

term

is

cos

AM

/}

when A t =

—

and so

/?,

an additional sideband pair

(4.10-5)

t

coc

t,

both

coc

(4.10-6)

t

while the second term involves first

and second terms involve

on. is

On

to be

basis,

this

it

added to the

may

well be

to

make R

first

new pair must give rise to a resultant A 2 which varies 2w m Thus, we are not surprised to find that as the phase devi-

constant, this

frequency

.

comes into existence

increases, a sideband pair

As long as we depend on the 4.10-la that

at the frequencies co c


cannot exceed

first-order sideband pair only,

90°.

A

we

deviation of such magnitude

see

is

from

Fig.

hardly ade-

For consider, as above, that A/ = 75 kHz and that fm = 50 Hz. Then co m = = 1500 rad, and the resultant R must, in this case, spin completely about 1500/27: or about 240 times. Such wild whirling is made possible through the effect of the higher-order sidebands. As noted, the first-order sideband pair gives rise to a phasor A t = J^P) sin co m t, which phasor is perpendicular to the carrier phasor. It may also be established by inspection of Eq. (4.5-7) that the second-order sideband pair gives rise to a phasor A 2 = J 2 (P) cos 2w m t and that quate.

75,000/50

this

cojt

t

±2cv

(4.10-2)

The angular departure

seen from Fig. is

slightly

sin co„,t

NBFM

again

more nearly

'

0

sin a>„

an in-phase relationship.

maximum

ation

=

fi

a quadrature relationship. In

expected that

nitude

A,

t,

—

To return now to the FM case and to Fig. 4.10-la, the following point is worth noting. When the angle

projections in the horizontal direction. At this time one

and one is antiparallel to, the phasor representing the carrier, so The situation depicted in Fig. 4. 10- la corresponds to a time shortly after t = 0. At this time, the rotation of the sideband phasors which are in opposite directions, as indicated by the curved arrows, have given rise to a sum phasor Aj. In the coordinate system in which the carrier phasor is stationary, the phasor A! always stands perpendicularly to the carrier phasor and has the mag-

is

to c

t,

t

AM

FM signal, (b) Phasor diagram for an AM signal. sin coc

coc

all

phasor

is

parallel to the carrier phasor. Continuing,

odd-numbered sideband A«

we

easily establish that

pairs give rise to phasors

= Jn(fi)

sin

nw m t

n odd

(4.10-7)

which are perpendicular to the carrier phasor, while

even-numbered sideband

all

so that

pairs give rise to phasors

K = JJ.P) cos nco m which are

neven

t

parallel to the carrier phasor. Thus,

nately perpendicular

and

required, while maintaining

,

completely around as

added

at constant

magnitude.

many

,

to carry the

times as

Comparing Eq.

SPECTRUM OF NARROWBAND ANGLE MODULATION: ARBITRARY MODULATION 4.11

We

that, just as in

two sidebands at frequencies w c + w m and an arbitrary modulating waveform.

We may

a> c

which

just as

it

produced by sinus-

is

such modulation gives rise to w m We extend the result now to .

is, if /? x

sin (o^t

+

/J 2

sin a> 2

1

substituted in Eq. (4.10-la) in place of /? sin a> m t, the sidebands which result are sum of the sidebands that would be yielded by either modulation alone.

the

Hence even

if

a modulating signal of waveform

tion of spectral components,

is

used in either

m((),

AM

tion the forms of the sideband spectra will be the

More signal

formally,

we have

in

AM,

with a continuous distribu-

or narrowband angle modula-

same

two cases. when the modulating waveform is in the

m(t),

the

is

"amW =

PM (t),

+

"»(')]

cos

co c

t

=A

cos

co c

t

+

(4.1 1-2)

with Eq.

find that plots of

)

)~\

c

(4.1 1-4), 2

we observe

=

I^[>pm]|

that

2

(4.H-5)

power

spectral density

would be the same.

SPECTRUM OF WIDEBAND FM (WBFM):

In this section

FM

Am(t) cos

w

c t

(4.11-1)

signal.

we engage

We

possible in the

we

AM. That

does to

+ | e-^XMijo + jcoc - M(jw — j(ti

co c )]

ARBITRARY MODULATION 4

AM,

-

-

5(w

if we were to make plots of the energy spectral densities of v AM (t) and of we would find them identical. Similarly, if m{t) were a signal of finite power,

Thus,

4.12

readily verify (Prob. 4.11-1) that superposition applies in narrow-

band angle modulation is

NBFM,

the spectrum, in

found

+

I^["am]I

we would

we considered

co c )

as an exercise for

It is left

typical

trary angle (Prob. 4.10-2).

Previously

+

may be

i)

oidal modulation.

[<5(
etc., alter-

examples how the superposition of a carrier and swing a constant-amplitude resultant around through an arbi-

show by

may

sidebands

R

R

^W0] = \

(4.11-4)

A u A2 A3

phasors

parallel to the carrier phasor, are

end point of the resultant phasor the student to

(4.10-8)

NBFM

spectrum of a wideband deduce the spectrum with the precision that is

in a heuristic discussion of the

shall not be able to

case described in the previous section.

As a matter of

fact,

be able to do no more than to deduce a means of expressing approximately the power spectral density of a signal. But this result is imporshall

WBFM

tant

and

useful.

FM

signal as being narrowband or wideband, where Af is the frequency deviation and /„ the frequency of the sinusoidal modulating signal. The signal was then NBFM or depending on whether j? «^ 1 or 0 P- 1. Alternatively we distinguished one from the other on the basis of whether one or very many sidebands were produced by each spectral component of the modulating signal, and on the basis of whether or not superposition applies. We consider now still another alternative. Let the symbol v=f—fc represent the frequency difference between the instantaneous frequency / and the carrier frequency fc that is, v(t) = (k/2n)m(t)

Previously, to characterize an

we had used

the parameter

/?

=

Af/fm

>

WBFM

;

Let us assume, for simplicity, that m(t)

a finite energy waveform with a Fourier transform M(joo). We use the theorem that if the Fourier transform F[m(t)~\ = M(jw), then ^[m(t) cos w e t] is as given in Eq. (3.2-4). We then find is

[see Eq. (4.3-5)]. will v

its

^I>am(«)]

=\

[<5(»

+

co c )

+

5(a)

-

coc )]

+ | [A/O +jwc + )

M(jco

- jo)

and T. The frequency

in Fig. 4.10-1 rotates in the

In

that

WBFM

Since, the resultant c

)~\

(4.11-2)

The narrowband angle-modulation signal of Eq. (4.10-1), except of amplitude A and with phase modulation m(t), may be written, for m(t) < 1, t)

PM (t)

=A

cos

coc

t

—

Am(i) sin

|

w

c t

this resultant

(4.11-3)

v is the

to v

is

T=

1/v.

As /

varies, so also

frequency with which the resultant phasor

coordinate system in which the carrier phasor

phasor rotates through

many complete

is

R

fixed.

revolutions,

and

speed of rotation does not change radically from revolution to revolution.

R

is

constant, then

R

if

we were

to examine the plot as a func-

we would recwaveform because its frequency would be changing very slowly. No appreciable change in frequency would take place during the course of a cycle. Even a long succession of cycles would give the appearance of being of rather constant frequency. In NBFM, on the other hand, the phasor R simply oscillates about the position of the carrier phasor. Even though, in this case, we may still formally calculate a frequency v, there is no corresponding time interval tion of time of the projection of

ognize

|

The period corresponding

it

as a sinusoidal

in, say,

the horizontal direction,

» 200 kHz

fc Crystal

f

90"

oscillator

Phase

200 kHz

shifter

/c - 12.8

Uf -25 Hz (at

fm

A/*

-50

Hi

)/

MHz

-1.6 kHz

MHz

fc - 2.0

FM DEMODULATORS

4.20

4/ -1.6 kHz

»„,-32

from a

»„-32

With a view toward describing how we can recover the modulating

Multiplier

frequency-modulated carrier we consider the situation represented in Fig. 4.20-1. Here a waveform of frequency f0 and input amplitude A- is applied to a frequency selective network which then yields an output of amplitude A 0 The ratio of amplitudes A A, is the absolute value of the transfer function of the network,

-m-0-5

signal

t

.

Baseband

Multiplier

Adder

signal

Mixer

X 64

that

fc - 96

Balanced modulator

Integrator

Oscillator

10.8

J

X 48

\

.

This output waveform

MHz

FM

signal using multipliers to

increase the frequency deviation.

=

Thus, at fm = 50 Hz, we have n A/ = 25 Hz. Note that at higher modulating frequencies, 4> m is less than 0.5 rad. The carrier frequency before multiplication has been selected at 200 kHz, a fre-

quency

which very stable

at

0.5.

crystal oscillators

and balanced modulators are

readily constructed.

As already noted, if we require that A/ « 75 kHz, then a multiplication by a factor of 3000 is required. In Fig. 4.19-1 the multiplication is actually 3072(= 64 x 48). The values were selected so that the multiplication may be done by factors of 2 and 3, that is, 64 = 2 6 48 = 3 x 2 4 Direct multiplication would ,

.

This

signal

614.4

-

96.0

then

518.4

be

FM

sum and

614.4

heterodyned with

MHz. The

respondingly the diode demodulator output will follow the variation of A 0 In short, a fixed amplitude, frequency-modulated input will generate, at the output of the frequency selective network, a waveform which is not only fre.

quency modulated but also amplitude modulated. The diode demodulator will ignore the frequency modulation but will respond to the amplitude modulation. (In general, the frequency-selective network will not only give rise to an amplitude change but will also generate a frequency-dependent phase change, as noted in Fig. 4.20-1. But such a phase change is simply additional angle modulation

which the diode demodulator will ignore.) What we require of an FM demodulator

MHz. Note

a

signal

of frequency,

no

effect

on

its

that the instantaneous output

is

indicated in the plot of Fig. 4.20-lb.

cos (2a/. f

+

0)

say,

would be

particularly that a mixer, since

it

Frequency

A, cos 2n/"„r

difference frequencies, will translate the frequency spectrum of an

signal but will have

is

A„ be proportional to the instantaneous frequency deviation of the received signal from the carrier frequency. If the carrier frequency is /0 then we require a linear relationship between A 0 and (f-f0 ) where /is the instantaneous

signal

MHz

difference signal output of such a mixer

a signal of carrier frequency 96 yields

,

/I,

might

=

=

demodu-

.

frequency. Such a linear relationship

yield a signal of carrier frequency

200 kHz x 3072

AM

,

t

before multiplication, to the extent of

then applied to a diode

A 0 Suppose now that the input waveform, instead of being of fixed frequency f0 is actually a frequency modulated waveform. Then even for a fixed input amplitude A the output amplitude A 0 will not remain fixed but instead be modulated because of the frequency selectivity of the transmission network. Corequal to

1.546

Figure 4.19-1 Block diagram of an Armstrong system of generating an

is

|

lator (see Fig. 3.4-2). The diode demodulator generates an output which is equal to the peak value of the sinusoidal input so that the diode demodulator output is

4/ -77 kHz

MHz

H(jco)

is,

selective

network

frequency deviation. In the system of Fig. heterodyning at a frequency in the

4.19-1, in order to avoid the inconvenience of

range of hundreds of megahertz, the frequency translation has been accomplished where the frequency is only in the neighborhood of approximately 10 MHz. at a point in the chain of multipliers

A

feature, not indicated in Fig. 4.19-1 but

which

may

be incorporated,

is

Linear section, slope

to

MHz mixing signal not from a separate oscillator but rather through multipliers from the 0.2 MHz crystal oscillator. The multiplication

=

a

=

d(AJA *

)

t'

derive the 10.8

=

=

2 x 3 3 Such a derivation of the 10.8-MHz signal will suppress the effect of any drift in the frequency of this signal (see Prob. 4.19-1). required

is

10.8/0.2

54

(b)

f.

.

Figure 4.20-1

(a)

FM demodulation,

(fc)

Frequency

selective network, typically

an

LC

circuit.

We

require actually that the linearity extend only as far as

date the

maximum

As indicated

is

necessary to

frequency deviation to which the carrier Fig. 4.20-16 let

in

is

accommo-

subject.

us consider that the frequency selective

network has a linear transfer characteristic of slope a over an adequate range A J A, = R 0 the neighborhood of the carrier frequency f0 and that, at f = f0 ,

|

H(f =/0 )

Then we

I

,

\

\

shall have, as required

A.-R0 A, + «Atf-fd the term

R 0 Ai

modulated

We signal

is

(4.20-1)

Diode Demodulator

is a term of fixed value which displays no response and the second term aA,(f — f0 ) provides the required

in-Eq. (4.20-1)

to frequency deviation

response to

in

=

instantaneous frequency deviation

the

Modulated

slope

=

A',

= R 0 A, +

zAU-h) + PAHf-f0

2 )

a

input

Difference

of the input frequency-

—Demodulated output 2*Aif-f„)

signal.

Frequency

observe however from Eq. (4.20-1) that

not fixed then the demodulator output

if

the amplitude

will

A

t

selective

of the input

= —a

slope

tude variations as well as to frequency deviations. Ordinarily in a frequency-

modulated communication system the amplitude of the transmitted signal will not be modulated deliberately so that any such modulation which does appear will be due to noise. Hence it is of advantage, for the purpose of suppressing the noise, to reduce the dependence of A 0 on A, in Eq. (4.20-1). This is accomplished by passing the incoming signal through a hard limiter as shown in Fig. 4.20-2. The purpose of the hard limiter (or comparator) is to insure that variations of Aj(t) are removed. Figure 4.20-2 shows the original FM waveform with amplitude variations at the output of the hard limiter. Since the waveform has been reduced to a frequency modulated " square wave " a bandpass filter is inserted to extract to first harmonic frequency f0 The resulting FM waveform is now applied to the

Diode Demodulator

network

respond to the input ampli-

^=R

0

J-

A,

aAHf-f0 + PAtf-h) 2 )

\AJA, slope

«o

= —a

t

f

fo

Figure 4.20-3

A

balanced

FM demodulator.

.

FM demodulator. Unfortunately, one cannot construct a network having the precisely linear shown in Eq. (4.20-1). Indeed, in practical networks the

transfer characteristic

output amplitude appears as

(a)

A(,t]cos(m 0

t

+

tf(t))e

Hard

Bandpass

Limiter

Filter

,

To FM Demodulate*

A 0 = R 0 A, + aAtf-f0 + fSAif-fo? + )

The balanced

FM

constant term

R 0 Ai and

demodulator shown all

in Fig. 4.20-3

—

(4.20-2)

can be used to remove the

even harmonics, thereby reducing the distortion pro-

duced by the nonlinearity of the bandpass filters. Here two demodulators are employed, differing only in that in one case the frequency-selective network has a slope a and in the other the slope is —a. The output provided by this balanced modulator is the difference between the two individual demodulators. These individual outputs are (assuming a second order nonlinearity):

FM waveform

A'0

= R 0 A + ttAfJ-fa + (lAlf-fo) 1

(4-20-3)

A",

= R 0 A, - olAU-U) +

(4.20-4)

t

and Figure 4.20-2

(a)

Hard

and output of hard

limiter (or

limiter.

comparator) input to

FM

demodulator,

(b)

FM

waveform

at input

PAM-U) 1

The

difference output

is

then

arrange to amplitude-modulate a carrier in such manner that its envelope faithfully reproduces the modulating signal. Is it then possible to also angle-modulate

A 0 = 2aA{f-f0 ) and we

see that the linearity of the output of the

(4.20-5)

FM

demodulator has been

imation, the answer

improved.

LC

tuned circuit is used as the frequency selective network The relationship between the center frequency of each of the two tuned circuits and their bandwidths on the linearity of the demodulator is explored in Prob. 4.20-4.

FM

It has also been shown in the literature that a hard limiter is not required when using a balanced discriminator and that any overdriven amplifier (i.e., an

amplifier which

4.21

is

yes!

In the system described by

In practice an

is

driven betweeen cutoff and saturation) can be used.

good approx-

that carrier so that a single sideband results? It turns out that, to a

Kahn 5

the modulating signal modulates not only

the amplitude but the carrier phase as well.

modulating signal

is

nonlinear

The

relationship between phase

and has been determined,

and

at least in part experi-

An analysis of the system is very involved because of the two nonlinearities involved: the inherent nonlinearity of

mentally, on the basis of system performance.

FM

and the additional nonlinearity between phase and modulating signal. The is not strictly single sideband in the sense that a modulating tone gives rise not to a single side tone but to a spectrum of side tones. However, all the side tones are on the same side of the carrier. The predominant side tone is separated from the carrier by the tone frequency, and the others are separated by multiples system

of the modulating-tone frequency.

APPROXIMATELY COMPATIBLE SSB SYSTEMS

The system

is

able to operate, in effect, as a single-sideband system because of

the characteristics of speech or music for which

SSB-AM We noted

earlier in Sec. 3.12 the advantages that would accrue from the availcompatible single-sideband systems. While precisely compatible systems are presently impractical, approximately compatible systems are feasible, such systems are in use, and commercial equipment for such systems is available. In a strictly compatible system a waveform would be generated whose envelope exactly reproduces the baseband signal. Additionally if the highest baseband freability of

AM

quency component isfM the frequency range encompassed by the compatible signal would extend from/,, the carrier frequency, to, say,/ An approxc +fM imately compatible system is one in which there is some relaxation of these ,

.

spe-

cifications

concerning envelope shape or spectral range.

discussion of

saw

On

are now able briefly and qualitatively one type of approximately compatible

have the form of the modulating signal, provided the percentage modulation was kept small. Consider now the phasor diagram of the signal shown in

AM

bands

is

From

this

resultant

phasor diagram

suppressed, the resultant

amplitude and phase.

R

will

outside the allowed spectral range, are of very small energy.

that there

The amplitude

is

a

will

it is

apparent that when one of the side-

waveform which vary between

1

+

may

well be spectral

components

that

fall

The

overall result

outside the spectral range

allowable in a spectrum-conserving single-sideband system. However, it has been shown experimentally that such components are small enough to cause no interference with the signal in an adjacent single-sideband channel.

SSB-FM A

compatible

if

still

Fig. 4.10-lfc.

is

to

AM

we suppressed one sideband component of an amplitude-modulated waveform, the envelope of the resultant waveform would in Sec. 3.11 that

fall

the basis of our

FM-type waveforms we

discuss the principle of operation of system.

We

it is intended, and because the predominant one, fall off sharply in power content with increasing harmonic number. Most of the power in sound is in the lowerfrequency ranges. High-power low-frequency spectral components in sound may give rise to numerous harmonic side tones, but because of the low frequency of the fundamental, the harmonics will still fall in the audio spectrum. On the other hand, high-frequency tones, which may give rise to harmonic side tones that may

side tones, other than the

is modulated in both m/2 and 1 - m/2. The

no longer always be parallel to the carrier phasor. Instead it will by an angle

Thus we find that a carrier of fixed frequency (or phase), when amplitudemodulated, gives rise to two sidebands. But an angle-modulated carrier, when amplitude-modulated, may give rise to a single sideband. Suppose then that

we

components extending either above can be demodulated using a stanThe compatible FM signal is constructed by adding

signal has frequency

it

dard limiter-discriminator. amplitude modulation to the frequency-modulated signal. Since the limiter removes any and all amplitude modulation, the addition of the does not affect the recovery of the modulation by the discriminator. By properly adjusting the amplitude modulation, either the upper or lower sideband can be removed.

AM

rotate clockwise

clockwise by a similar angle.

SSB-FM

or below the carrier frequency. In addition

4.22

STEREOPHONIC FM BROADCASTING

In monophonic broadcasting of sound, a single audio baseband signal ted from broadcasting studio to a is

home

is

transmit-

At the receiver, the audio signal applied to a loudspeaker which then reproduces the original sound. The origreceiver.

Turning now to the composite signal M(t) in Eq. (4.22-1), we note that t oscillates rapidly between + 1 and - 1, and, ignoring temporarily the pilot carrier, we have that M(t) oscillates rapidly between M(t) = 2L(t) and

cos 2t$x

=

M

M

The maximum attained by M(t) is then m = 2L m or m = 2R n From Eq. (4.22-2) m = Vsm Hence, in summary, we find that the addition of the JVf(t)

2R(t).

M

difference signal

Vd

sum

signal

Vs

If

=

6k

K=

0 and

each B k

is

1, 2,

maximum

find the

frequency deviations.

an independent random variable, uniformly distributed between

—

it

and

%, find

rms frequency deviation.

the

Under

(c)

rms phase deviation.

the condition of (ft) calculate the

.

.

to the

(a) If (ft)

4.4-2. If v{t)

=

cos

j^a) c

+

t

k j"

m(k)

rf^J,

where

m(() has a probability density

does not increase the peak signal excur-

sion. f(m)

=

e-" 2

Effect of the Pilot Carrier

calculate the

Unlike the DSB-SC signal, the pilot carrier, when added to the other components of the composite modulating signal, does produce an increase in peak excursion.

A carrier which attains a peak voltage of 5 volts has a frequency of 100 MHz. This carrier is frequency-modulated by a sinusoidal waveform of frequency 2 kHz to such extent that the frequency deviation from the carrier frequency is 75 kHz. The modulated waveform passes through zero and is

Hence the addition of the pilot carrier calls for a reduction in the sound signal modulation level. A low-level pilot carrier allows greater sound signal modulation, while a high-level pilot carrier eases the burden of extracting the pilot carrier at the receiver. As an engineering compromise, the FCC standards call for a pilot carrier of such level that the peak sound modulation amplitude has to be reduced to about 90 percent of what would be allowed in the absence of a carrier. This 10 percent reduction corresponds to a loss in signal level of less than 1 dB.

rms frequency deviation.

4.4-3.

increasing at time

=

t

Write an expression for the modulated carrier waveform.

0.

4.4-4.

A

carrier of frequency 10"

4.5- 1.

A

carrier

Hz and amplitude 3 volts is frequency-modulated by a sinusoidal modulating waveform of frequency 500 Hz and of peak amplitude 1 volt. As a consequence, the frequency deviation is 1 kHz. The level of the modulating waveform is changed to 5 volts peak, and the modulating frequency is changed to 2 kHz. Write the expression for the new modulated waveform. is

angle-modulated by two sinusoidal modulating waveforms simultaneously so that lit)

=A

cos

(a>, t

+

jJ,

sin a>,t

+

fi 2

sin co 2

1)

Show

REFERENCES

a> 2

that this waveform has sidebands separated from the carrier not only at multiples of but also has sidebands as well at separations of multiples of cu, + a> and of to, - co 2 2

4.6- 1. Bessel functions are said to be almost periodic with a period of almost 1.

2. 3.

Telephone Laboratories: "Transmission Systems for Communications," Western Company, Tech. Pub., Winston-Salem, N.C., 1964. Jahnke, E., and F. Emde: "Tables of Functions," Dover Publications Inc., New York, 1945. Bell

Pipes, L. A.: "Applied

New 4.

Mathematics

for Engineers

and

Physicists,"

Electric

4.6-2.

N:

Calculation of the Spectrum of an

FM

Kahn,

L. R.:

Compatible Single Sideband. Proc. IRE,

The primary

fi,

for J 0 (fi)

and J ,(/(), required

McGraw-Hill Book Company

(a)

Signal Using

Woodwards Theorem, IEEE

vol. 49, pp.

+

<£(()]

where

fat) is

a square wave taking on the values ±n/3

sec.

Sketch cos

(ft)

Plot the phase as a function of time. Plot the frequency as a function of time.

[coc t

Show

that the

magnitude decreases as

FM

carrier

(i.e.,

no power

An

sidebands

A

bandwidth

and the

sine

wave

is

that the envelope of

in part (a) versus

fi

and draw a smooth curve

-—.

4.7-2.

A

carrier

sinusoidally modulated.

is

rule

For what values of

fi

does

all

the

power

in the

lie

in the carrier)?

sometimes used

power

fraction of the signal

0(t)].

is

for space communication systems included in that frequency band. Consider j? =

is 1

B= and

(2fi

+

l)fu

.

What

10.

frequency-modulated by a sinusoidal modulating signal of frequency 2 kHz, kHz. What is the bandwidth occupied by the modulated wave-

is

resulting in a frequency deviation of 5

waveform cos signal.

by

Tabulate the magnitude of all peak values of J 0 {fl), positive and negative peaks, as a function

(ft) Plot the magnitude of the peak values obtained through the points.

4.7- 1.

+

(a)

modulating

this

•J?

Consider the signal cos [m,t

4.2-2. If the

Demonstrate

1503-1527, October, 1961.

4.6- 3.

(c)

2it.

these functions equal to zero.

of/8.

(c)

every 2/fc

make

difference between the Bessel functions

PROBLEMS 4.2-1.

to

the Bessel function decreases.

Trans. Communication Technology, August, 1969. 5.

recording the values of

York, 1958.

Blachman,

and of

co,

.

(ojc t + k sin
form? The amplitude of the modulating sinusoid 1 kHz. What is the new bandwidth?

is

increased by a factor of 3 and

its

frequency

lowered to

modulated. 43-1.

What

and k

that appear in Eqs. (4 3-1) (4 3-2)

(4.3-3)? 4.4-1.

An

spectrum of cos (In x 4t + 5 sin 2nt). Note that the spectrum indicates the presence of a dc component. Plot the waveform as a function of time to indicate that the dc component is to have been expected. 4.7-3. Plot the

are the dimensions of the constants k, k",

FM signal is given by

and

4.10-1. tit) r{t)

=

cos jju c

t

+

£

p k cos (fao 0

1

+

=

cos coc

t

+

0.2 cos co m

t

sin o> c

(a)

Show

(ft)

Sketch the phasor diagram at

that

e(!) is

a combination t

t.

AM-FM signal, =

0.

Consider the angle-modulated waveform cos

4.10-2.

+

(o>c t

2 sin
0 =

i.e.,

2,

so that the wave-

may be approximated by a carrier and three pairs of sidebands. In a coordinate system in which the carrier phasor A 0 is at rest, determine the phasors A„ A and A 3 representing respectively the 2 first, second, and third sideband pairs. Draw diagrams combining the four phasors for the cases - °. k/2, 3)t/4, and n. For each case calculate the magnitude of the resultant phasor. form

,

Show

4.10-3. (a)

with a phasor diagram that

=

v{t)

represents a carrier which

is

cos

v(t)

x 10 6 ()

(2ti

modulated both

,

given by

Express

+

10 3 )t]

m and

Find

cos 2n x 10 3 t) cos (In x 10 6 t

+0

sin

2n x 10 3 t)

so that 4.12- 1.

cos

<j>(t)

|

+


|

<M«)]

Show

The frequency

in Sec. 4.11 that

where

that

each

for

=

(t)

lit) =:

cos

superposition applies in

/?,

sin oj,r

coc

t

-

+ 02

sin cu 2

1.

+ 02

(0, sin cu,t

NBFM. To do

this

con-

Let 0, and 0 2 be sufficiently small

sin a> 2

sin ai,

f)

+

t

cos

fl k

J]

(a> k

+

t

et

k.

(a)

Find Biffin 2[(A/),

(b)

Find

B

if is

B=

t.

B=

is varied back and forth extremely slowly and at a uniform rate between the frequencies of 99 and 101 kHz. The amplitude of the oscillator output is constant at 2 volts. Make a plot of the two-sided power spectral density of the oscillator output wave-

form.

(A/) 2

+

+

•

•

+

+ •

•

(A/)J.

+

•

(A/),, J.

given by Eq. (4.13-1), find the rms bandwidth B.

is

A/rms given

4.6

•

(A/) 2> _

capacitance

C0 =

is

adjusted for resonance at 5

is

is

1

by C„

MHz

The modulating voltage

=

(100/^/1

when a m(l)

is

=

4

+

2v)

pF. The capacitance

fixed reverse voltage v

=

4 volts

+

If

the oscil-

0.045 sin 2n x 10 3 t.

is

an expression for the angle-modulated output waveform which

volt, write

appears across the tank

a reversed-bia sed pn junction diode whose

is

related to the reverse-biasing voltage v

200 pF and L

Compare your

in Eq. (4.13-5).

4.15- 1. In Fig. 4.15-1 the voltage-variable capacitor

circuit.

4.17- 1. (a) In the multiplier circuit of Fig. 4.17-1

and

of a laboratory oscillator

+

2[(A/) I>m

gaussian and

result with the value

lator amplitude

comment made

[co,. t

1

applied to the capacitor C„.

phasor A, and the resultant phasor.

=

=

to,

4.14- 2. If G(f)

,

lit)

cos |ju t

)J

Consider the angle-modulated waveform cos (
=

i)(t)

and

4.10- 4.

sider

FM signal

amplitude and frequency.

in

Write an expression for the instantaneous frequency as a function of time.

/?.

1

Let p t

+m

lit) as (1

1

4.14- 1. Consider the

the form

lit) in

.

density of the output of the frequency-modulated source. 0.02 cos [2n x (10 6

+

(b) Show that, on the basis of the relative magnitudes of the two terms in i
The two independent modulating signals m,(t) and m 2 (t) are both gaussian and both of zero mean and variance 1 volt 2 The modulating signal m,(!) is connected to a source which can be frequency-modulated in such manner that, when m,(f) = 1 volt (constant), the source frequency, initially MHz, increases by 3 kHz. The modulating signal m 2 (t) is connected in such manner that, when m 2 (t) = volt (constant), the source frequency decreases by 4 kHz. The carrier amplitude is 2 volts. The two modulating signals are applied simultaneously. Write an expression for the power spectral 4.13- 2.

assume that the

transistor acts as a current source

so biased and so driven that the collector current consists of alternate half-cycles of a sinusoidal waveform with a peak value of 50 mA. The input frequency of the driving signal is 1 MHz, and is

the multiplication by a factor of 3

to be accomplished. If

is

C=

pF and

200

the inductance of the inductor and calculate the amplitude of the third

the inductor

Q=

30, find

harmonic voltage across the

tank. 4.12-2. If the probability density of the

amplitude of m(f) -""' 2

is

Rayleigh:

m£

multiplication by 10

If

(f>)

Assume

that the resonant

0

narrowband waveform

4.18- 1. (a) Consider the

elsewhere

Find the power spectral density G„(/) of the

lit)

=

|^» 4.12-3.

Repeat Prob. 4.12-2

if


t

t

+

k

|

mWittJ

the probability density of m(t) is/(m)

=

je

-1 "

1

.

The frequency of a laboratory oscillator is varied back and forth extremely slowly and in such a manner that the instantaneous frequency of the oscillator varies sinusoidally with time between the limits of 99 and 101 kHz. The amplitude of the oscillator output is constant at 2 volts. Find a function of frequency g(f) such that g(f) df is the fraction of the time that the instan-

taneous frequency (b)

Make

is

a gaussian

a plot of the

filter

between/ and/+ df. two-sided power spectral density of the

in the range

4.13- 1. Consider that the

oscillator output

that

v{t),

having the bandpass characteristic

filtered

by

i>

B

2

(a)

Sketch

(b)

Calculate the 3-dB bandwidth of the

(e)

Find

B

|

H(f)

|

as a function

of/

It is

4.20- 1. filter in

terms of B.

so that 98 percent of the signal power of the

WBFM signal

passed.

sin
+

ta c

t

/J/2

cos

{to,

-

,

cujl

drift in the carrier

of the resulting as

shown

0/2 cos

(a rel="nofollow">,

t),

with

[i

1

and with

written approximately as

+ wjt

quency

by

frequency

is

200

maximum angular deviation to 4> m = 0.2. down to 40 Hz. At the output of the modula-

desired, in order to avoid distortion, to limit the is

to

accommodate modulation frequencies is

to be 108

MHz and

the frequency deviation 80

.

kHz.

Select multiplier

and

this end.

The narrowband phase modulator of co m

oscillator

FM signal.

in Fig. 4.19-1, the crystal-oscillator

Fig. 4.16-1

preceding the balanced modulator with an integrator. is

fl

-

cos

mixer oscillator frequencies to accomplish

:

+

=

Armstrong modulator,

tor the carrier frequency

Assume fc

(
A/ = ffm may be

that the

Find the

4.19- 2. In an

The system

e -<j+f,imn

cos

a frequency deviation

10.8-MHz signal in Fig. 4.19-1 is derived from the 200-kHz multiplying by 54 and that the 200-kHz oscillator drifts by 0.1 Hz. (a) Find the drift, in hertz, in the 10.8-MHz signal.

kHz.

= e -(/-f.w +

=

which has

.

Assume

4.19- 1.

(b) is

v{t)

amplitude of the tank voltage.

in part (a).

and that this approximation is consistent with the general expansion for an angle-modulated waveform as given by Eq. (4.5-7). Use the approximations of Eqs. (4.6-1) and (4.6-2). 2 (fr) Let v{t) be applied as the input to a device whose output is D (t) (i.e., the device is nonlinear and is to be used for frequency multiplication by a factor of 2). Square the approximate expression for 2 lit) as given in part (a). Compare the spectrum of v (t) so calculated with the exact spectrum for an angle-modulated waveform with frequency deviation 2fifm

waveform.

WBFM signal having the power spectral density of Eq. (4.13-1) |H(/)| 2

Show

v{t)

4.12- 4.

(a)

.

FM signal

cos

to be accomplished, calculate the

is

impedance of the tank remains the same as

is

converted to a frequency modulator by

The input

signal

is

a sinusoid of angular

fre-

Show

(a)

demodulated,

unless the frequency deviation

power associated with

ized

is

not only the input signal but also

the modulation frequency

If

(fc)

that,

will yield

the third

kept small, the modulator output, when its

odd harmonics.

CHAPTER

50 Hz, find the allowable frequency deviation

is

harmonic

is

to be

no more than

the normal-

if

FIVE

percent of the fundamental

1

power.

RC

FM

Consider the

4.20-2. (a)

of the tivity

FM

waveform

f

how

,

With f2

Fig. 4.20-1. Let the frequency selective

c

selected for

is

f2 (= l/2nRQ.

If

ANALOG-TO-DIGITAL CONVERSION

network be an

the carrier frequency

should f2 be selected so that the demodulator has the greatest sensi-

greatest change in output per

(i.e., (fc)

is

demodulator of

The 3-dB frequency of the network

integrating network.

maximum

change

in input frequency)?

sensitivity

and with fc

=

MHz,

1

find the

change

in

demodu-

lator output for a 1-Hz change in input frequency.

A

4.20-3.

"zero-crossing"

FM

discriminator operates in the following manner.

The modulated wave-

form

v(t)

=A

cos

applied to an electronic circuit which generates a narrow pulse on each occasion when v(t) passes through zero. The pulses are of fixed polarity, amplitude, and duration. This pulse train is applied to

is

a low-pass

say an

filter,

baseband waveform

RC low-pass network

m(t)

is

/„ Discuss .

the output of the low-pass network

A

4.20-4. (a)

is

of 3-dB frequency

2

.

Assume

the operation of this discriminator.

that the

Show

bandwidth of the

that

fc

if

indeed proportional to the instantaneous frequency of

frequency selective network

shown

is

in Fig. P4.20-4. Calculate the ratio

|

t>

/j

> /„

t(t).

the ratio of the amplitude of the output to the amplitude of the input, as a function of frequency.

i.e.,

Verify that VJLTI

in

which /„

=

l/2n s/\/LC

the resonant network at

f0

is

+

^

the resonant frequency

i

1

['

and

Q0 =

R/2nT0 L

is

PULSE-MODULATION SYSTEMS

V0 (f)/V^f)\,

the energy storage factor of

.

In Chaps. 3 and 4 we described systems with which we can transmit many signals simultaneously over a single communications channel. We found that at the transmitting end the individual signals were translated in frequency so that each

occupied a separate and distinct frequency band. It was then possible at the receiving end to separate the individual signals by the use of filters. In the present chapter we shall discuss a second method of multiplexing. This

second method depends on the fact that a bandlimited signal, even if it is a continuously varying function of time, may be specified exactly by samples taken sufficiently frequently. Multiplexing of several signals is then achieved by interleaving the samples of the individual signals. This process is called time-division multiplexing. Since the sample is a pulse, the systems to be discussed are called pulse-amplitude modulation systems. Pulse-amplitude modulation

That Find the frequencies

(fc)

quency. at

Show

that, for large

Q0

,

which

is

higher by 3

these frequencies are / = 0 (1

dB than

±

its

value at the resonant fre-

1/Q 0 ). Calculate the slope of

|

VJV,\

fK = /0 (1 + 1/Q 0 A frequency modulated waveform with carrier frequency fH is What is the ratio of the change in amplitude AVc (f) to the change, Af, of the instanta-

the frequency

applied as V,{f).

at

).

neous frequency of the input? (c)

shown and

= fc (\ —

/,

Make

In the discriminator of Fig. 4.20-1

in Fig. P4.20-4.

1/Qo),

fc

The resonant

two frequency selective networks be the network two networks are to be /„ = fc (\ + 1/Q 0 ) frequency of an input frequency modulated waveform. let

the

=

with the

AM

noise present in the receiver. In this chapter we consider how to convert the analog signal into a digital signal prior to modulation. Such a conversion is

We shall see in Chap. 12 that when a digital signal is modulated and transmitted, the received signal is far less sensitive to receiver called source encoding.

noise than

is

analog modulation.

frequencies of the

being the carrier

a plot of the discriminator output as a function of the instantaneous frequency of the input.

Use Go

is,

(PAM)

systems are analog systems and share a and FM modulation systems studied earlier. each of these analog modulation systems are extremely sensitive to the

common problem

Noisy Communications Channels

50-

We consider a basic problem associated with the transmission of a signal over a noisy communication channel. For the sake of being specific, suppose we require that a telephone conversation be transmitted from New York to Los Angeles. If 183

Show

(a)

demodulated, (b)

that,

unless the frequency deviation

the modulation frequency

If

power associated with

ized

its

odd harmonics.

CHAPTER

50 Hz, find the allowable frequency deviation

is

harmonic

the third

kept small, the modulator output, when

is

not only the input signal but also

will yield

is

to be

no more than

if

the normal-

FIVE

percent of the fundamental

1

power.

RC

FM

Consider the

4.20-2. (a)

FM

of the tivity

waveform

With f2

A

c

selected for

lator output for a 4.20-3.

how

,

Fig. 4.20-1. Let the frequency selective

maximum

1-Hz change

"zero-crossing"

the network

is

2

(= l/2nRQ.

If

ANALOG-TO-DIGITAL CONVERSION

network be an

the carrier frequency

should f2 be selected so that the demodulator has the greatest sensi-

greatest change in output per

(i.e.,

(b)

f

is

demodulator of

The 3-dB frequency of

integrating network.

change

in input frequency)?

sensitivity

=

and with f€

MHz,

1

find the

change

in

demodu-

in input frequency.

FM

discriminator operates in the following manner.

The modulated wave-

form

v(t)

is

=A

cos ^2nfc

t

+

k

|

m(
dX

applied to an electronic circuit which generates a narrow pulse on each occasion

through zero. The pulses are of fixed polarity, amplitude, and duration. This pulse a low-pass

say an

filter,

baseband waveform

RC low-pass network

m(f)

is

/M

.

the output of the low-pass network

A

4.20-4. (a) i.e.,

of 3-dB frequency

2

.

Assume

Discuss the operation of this discriminator. is

that the

Show

when

train

is

passes

v(t)

applied to

bandwidth of the

that

if fc t>f2 indeed proportional to the instantaneous frequency of t(t).

frequency selective network

shown

is

in Fig. P4.20-4. Calculate the ratio

|

> fu

the ratio of the amplitude of the output to the amplitude of the input, as a function of frequency.

Verify that

+

wk 2

van in which /„ = l/2n s/\/LC the resonant network at f0

is

the resonant frequency

[(///o)2

and

"

Q0 =

R/2irf0

L

is

PULSE-MODULATION SYSTEMS

V0 (f)IV^f)\,

the energy storage factor of

.

In Chaps. 3 and 4 we described systems with which we can transmit many signals simultaneously over a single communications channel. We found that at the transmitting end the individual signals were translated in frequency so that each

occupied a separate and distinct frequency band. It was then possible at the receiving end to separate the individual signals by the use of filters. In the present chapter we shall discuss a second method of multiplexing. This

second method depends on the fact that a bandlimited signal, even if it is a continuously varying function of time, may be specified exactly by samples taken sufficiently frequently. Multiplexing of several signals is then achieved by interleaving the samples of the individual signals. This process is called time-division multiplexing. Since the sample is a pulse, the systems to be discussed are called pulse-amplitude modulation systems. Pulse-amplitude modulation

common problem That Find the frequencies

(i>)

quency.

Show

that, for large

at the frequency

applied as V,{f).

/„

Q0

= /0 (1 +

What

is

at ,

which

|

VJV

i

|

is

higher by 3

dB than

these frequencies are / = 0 (1

1/Q 0 ).

A

its

value at the resonant fre-

± 1/Q 0 ). Calculate the slope of VJVt frequency modulated waveform with carrier frequency /„ is

the ratio of the change in amplitude

AVJJ)

|

\

to the change, Af, of the instanta-

neous frequency of the input ? (c)

shown and

= fc (\ —

f,

Make

In the discriminator of Fig. 4.20-1

in Fig. P4.20-4.

1/Qo),

fc

The resonant

two frequency selective networks be the network two networks are to be fH = fc (\ + 1/Q 0 ) frequency of an input frequency modulated waveform. let

the

=

each of these

AM

noise present in the receiver. In this chapter we consider how to convert the analog signal into a digital signal prior to modulation. Such a conversion is

We shall see in Chap. 12 that when a digital signal is modulated and transmitted, the received signal is far less sensitive to receiver called source encoding.

noise than

is

analog modulation.

frequencies of the

being the carrier

a plot of the discriminator output as a function of the instantaneous frequency of the input.

Use G„

is,

(PAM)

systems are analog systems and share a and FM modulation systems studied earlier. analog modulation systems are extremely sensitive to the

with the

Noisy Communications Channels

50.

We consider a basic problem associated with the transmission of a signal over a noisy communication channel. For the sake of being specific, suppose we require that a telephone conversation be transmitted from New York to Los Angeles. If 183

the signal

is

transmitted by radio, then,

when

the signal arrives as

its

destination,

be greatly attenuated and also combined with noise due to thermal noise present in all receivers (Chap. 14), and to all manner of random electrical disturbances which are added to the radio signal during its propagation across it

will

country.

(We

neglect as irrelevant, for the present discussion, whether such direct

radio communication

reliable over such long channel distances.) As a result, not be distinguishable against its background of noise. not fundamentally different if the signal is transmitted over wires.

the received signal

The

situation

is

is

may

Any physical wire transmission path will both attenuate and distort a signal by an amount which increases with path length. Unless the wire path is completely and

perfectly shielded, as in the case of a perfect coaxial cable, electrical noise and crosstalk disturbances from neighboring wire paths will also be picked up in

amounts increasing with the path

length. In this connection

it

is

of interest to

note that even coaxial cable does not provide complete freedom from crosstalk. External low-frequency magnetic fields will penetrate the outer conductor of the coaxial cable and thereby induce signals

coaxial cables are

wrap the coax

to

combined with

on the

cable. In telephone cable,

parallel wire signal paths,

it is

common

where

practice

Permalloy for the sake of magnetic shielding. Even the use of which are relatively immune to such interference, does not sigthe problem since receiver noise is often the noise source of largest in

fiber optic cables

nificantly alter

communications link, and each repeater improves the signal-to-noise ratio over what would result if the corresponding gain were introduced at the receiver.

we might, conceptually at least, use an infinite number of repeacould even adjust the gain of each repeater to be infinitesimally greater than unity by just the amount to overcome the attenuation in the infinitesimal section between repeaters. In the end we would thereby have constructed a In the limit

ters.

We

channel which had no attenuation. The signal at the receiving terminal of the channel would then be the unattenuated transmitted signal. We would then, in addition, have at the receiving end all the noise introduced at all points of the channel. This noise is also received without attenuation, no matter how far away from the receiving end the noise was introduced. If now, with this finite array of repeaters, the signal-to-noise ratio

but to raise the signal level or to

The

problem

is

simply to raise the signal

level at the

is

nothing to be done

somewhat more dismal than has just been intimated, some noise on its own accord. Hence, as more repeaters are cascaded, each repeater must be designed to more exacting standards with respect to noise figure (see Sec. 14.10). The limitation of the system we have been describing for communicating over long channels is that once noise has been introduced any place along the channel, we are " stuck " with situation

is

actually

it.

If

to resolve this

not adequate, there the channel quieter.

since each repeater (transistor amplifier) introduces

power.

One attempt

is

make

we now were

to transmit a digital signal over the

same channel we would power would be needed in order to obtain the the receiver. The reason for this is that the significant par-

find that significantly less signal

transmitting end to so high a level that, in spite of the attenuation, the received substantially overrides the noise. (Signal distortion may be corrected

same performance at ameter is now not the signal-to-noise ratio but the probability of mistaking a

separately by equalization.) Such a solution

digital signal for a different digital signal. In practice

the signal

ratios of

signal

is hardly feasible on the grounds that power and consequent voltage levels at the transmitter would be simply astronomical and beyond the range of amplifiers to generate, and cables to handle. For example, at 1 kHz, a telephone cable may be expected to produce

an attenuation of the order of 1 dB per mile. For a 3000-mile run, even if we were satisfied with a received signal of 1 mV, the voltage at the transmitting end would have to be 10 147 volts.

40-60 dB are required

for

we

find that signal-to-noise

analog signals while 10-12

dB

are required

for digital signals. This reason

resulted in a large

and others, to be discussed subsequently, have commercial and military switch to digital communications.

5.1

THE SAMPLING THEOREM. LOW-PASS SIGNALS

We

consider at the outset the fundamental principle of digital communications;

An

amplifier at the receiver will not help the above situation, since at this point both signal and noise levels will be increased together. But suppose that a repeater (repeater is the term used for an amplifier in a communications channel) is

located at the midpoint of the long communications path. This repeater will in addition, it will raise the level of only the noise intro-

raise the signal level

duced

as

Let m(t) be a signal which

;

in the first half of the

communications path. Hence, such a midway repeaan amplifier at the receiver, has the advantage of improving the received signal-to-noise ratio. This midway repeater will relieve the burden imposed on transmitter and cable due to higher power requirements ter,

the sampling theorem:

contrasted with

when the repeater is not used. The next step is, of course, to use additional repeaters, say initially at the one-quarter and three-quarter points, and thereafter at points in between. Each added repeater serves to lower the maximum power level encountered on the

component

tral

separated seconds. signal,

Then

.

,

these samples

and the

signal

may

bandlimited such that its highest frequency specvalues of m(t) be determined at regular intervals that is, the signal is periodically sampled every T

is

fM Let the by times Ts < l/2/M is

m(nTs), where

s

an integer, uniquely determine the be reconstructed from these samples with no distorn

is

tion.

The time Ts

is

called the sampling time.

the sampling rate be rapid

enough so that

Note

that the theorem requires that

two samples are taken during the course of the period corresponding to the highest-frequency spectral comat least

m(<)

\M(jw)\

Iff

(S(Om(0)|

A 0

T,

iT 2L

3L

5/„

t

Figure 5.1-2

(a)

The magnitude

plot of the spectral density of a signal bandlimited to /„

amplitude of spectrum of sampled

.

(b)

Plot of

signal.

(a) A signal m(r) which is to be sampled, (b) The sampling function S(t) consists of a train narrow unit amplitude pulses, (c) The sampling operation is performed in a multiplier, (d) The

Figure 5.1-1 of very

samples of the signal

m{t).

Let the signal m(t) have a spectral density M(ja>)

shown in Fig. The spectrum

ponent.

We

now prove

shall

reconstructed from

The baseband

its

the theorem by showing

how

the signal

may be

samples.

The

signal S(t)

Eq. (1.3-8) with 1

is

= ,

S(t)

which

is

to be

periodic, with period

dt

=

and

Ts

and has the Fourier expansion

,

[see

T0 = TJ

—

dt 2 dt ( (^cos 2n - + cos 2 +

—+

\

t

t

x 2n

(5.1-1)

•

J For the case S(t)m(t)

We now

7;

=

=

l/2/M

,

the product S(t)m(t)

j m(t) + ~ [_2m(t) cos 2n(2fM

observe that the

the signal m(t)

itself.

first

term

is

)t

+

2m(t) cos 2n(4fM )t

in the series

is,

+

•

•

]

(5.1-2)

aside from a constant factor,

Again, aside from a multiplying factor, the second term

is

the

product of m(t) and a sinusoid of frequency 2fM This product then, as discussed in Sec. 3.2, gives rise to a double-sideband suppressed-carrier signal with carrier frequency 2fM Similarly, succeeding terms yield DSB-SC signals with carrier fre.

.

quencies 4/M 6/M ,

,etc.

Then

m(t)

is

—

^lm(t)2 which

is

as

bandlimited to the frequency range below fM

.

term in Eq. (5.1-2) extends from 0 to/M The spectrum of the second term is symmetrical about the frequency 2fM and extends from 2fM —fm =/m to 2/M +fM = 3/M Altogether the spectrum of the sampled signal has the appearance shown in Fig. 5.1 -2b. Suppose then that the sampled signal is passed through an ideal low-pass filter with cutoff frequency at fM If the filter transmission were constant in the passband and if the cutoff were infinitely sharp at fM the filter would pass the signal m(t) and nothing else. The spectral pattern corresponding to Fig. 5.1 -2b is shown in Fig. 5.1 -3a for the case in which the sampling rate /, = 1/T, is larger than 2fM In this case there of the

first

.

.

sampled is shown in Fig. 5.1- la. A periodic train of pulses S(t) of unit amplitude and of period Ts is shown in Fig. 5.1 -lb. The pulses are arbitrarily narrow, having a width dt. The two signals m(t) and S(t) are applied to a multiplier as shown in Fig. 5.1-lc, which then yields as an output the product S(t)m(t). This product is seen in Fig. 5.1-la" to be the signal m(t) sampled at the occurrence of each pulse. That is, when a pulse occurs, the multiplier output has the same value as does m(t), and at all other times the multiplier output is zero. signal m(t)

5.1 -2a.

.

,

.

is

a gap between the upper limit

the lower limit of the

fs >

2fM

For

.

this

fM of the spectrum of the baseband signal and DSB-SC spectrum centered around the carrier frequency

reason the low-pass

filter

used to select the signal m(r) need not

attenuation may begin at fM but need not attain a high value until the frequency fs —fM This range from fM to fs — fin is called a guard band and is always required in practice, since a filter

have an

infinitely

sharp cutoff. Instead, the

filter

A

.

with infinitely sharp cutoff is

used

in

limited to fM 8.0

-

is,

of course, not realizable. Typically,

connection with voice messages on telephone

2 x 3.3

= =

3.3 1.4

kHz, while fs kHz.

is

selected at 8.0

lines, the

when sampling voice signal

kHz. The guard band

is

—\

-H

is

-2-5/i -if,

then

The situation depicted in Fig. 5.1-36 corresponds to the case where fs < 2fM Here we find an overlap between the spectrum of m(t) itself and the spectrum of the DSB-SC signal centered around fs Accordingly, no filtering operation will allow an exact recovery of m(t). We have just proved the sampling theorem since we have shown that, in principle, the sampled signal can be recovered exactly when Ts < l/2f M It has also been shown why the minimum allowable sampling rate is 2fM This minimum sampling rate is known as the Nyquist rate. An increase in sampling rate above the Nyquist rate increases the width of the guard band, thereby easing the problem of filtering. On the other hand, we shall see that an increase in rate extends the bandwidth required for transmitting the sampled signal. Accordingly an engineering compromise is called for.

2f,

2.5/„

3/,

|ff(S(«m(«H

.

.

.

-

2i

-I.

0

2f.

f.

.

An

interesting special case

is

the sampling of a sinusoidal signal having the

all the signal power is concentrated precisely at the cutoff frequency of the low-pass filter, and there is consequently some ambiguity about whether the signal frequency is inside or outside the filter passband. To remove

frequency fM Here, .

this

ambiguity,

condition the

is

moment

we

require that /,

necessary,

>

2fM rather than that / > 2fM To see that this = 2fM but that an initial sample is taken at .

assume that /,

the sinusoid passes through zero.

also be zero. This situation

is

Then

avoided by requiring/,

>

all

successive samples will

2fM

.

(6)

Figure 5.1-4

(a)

The spectrum

of a bandpass signal,

(b)

The spectrum of the sampled bandpass

the sampling frequency, while the sampling frequency

part b

is

shown

is

exactly fs

the spectral pattern of the sampled signal S(t)m(t).

—

2(fu

signal.

— fL

).

The product

In

of

and the dc term of S(t) [Eq. (5.1-1)] duplicates in part b the form of the spectral pattern in part a and leaves it in the same frequency range from/L to fM The product of m(t) and the spectral component in S(t) of frequency /( = l/Ts ) gives rise in part b to a spectral pattern derived from part a by shifting the pattern in part a to the right and also to the left by amount fs Similarly, the higher harmonics of /, in S(t) give rise to corresponding shifts, right and left, of the spectral pattern in part a. We now note that if the sampled signal S(t)m(t) is passed through a bandpass filter with arbitrarily sharp cutoffs and with passband from m(t)

.

.

fL

to

fM

,

the signal m(t) will be recovered exactly.

In Fig. 5.1-4 the spectrum of m(t) extends over the

Bandpass Signals

first

half of the frequency

between harmonics of the sampling frequency, that is, from 2.0/", to 2.5/,. As a result, there is no spectrum overlap, and signal recovery is possible. It may also be seen from the figure that if the spectral range of m(t) extended over the second half of the interval from 2.5/, to 3.0/, there would similarly be no overlap. Suppose, however, that the spectrum of m(t) were confined neither to the first half nor to the second half of the interval between sampling-frequency harmonics. In such a case, there would be overlap between the spectrum patterns, and signal recovery would not be possible. Hence the minimum sampling frequency allowable is /, = 2(/M — fL ) provided that either /M or fL is a harmonic of /,. If neither fM nor fL is a harmonic of£ a more general analysis is required. In Fig. 5.1 -5a we have reproduced the spectral pattern of Fig. 5.1-4. The positiveinterval

whose highest-frequency spectral component is fM the sampling be no less than fs = 2/M only if the lowest-frequency spectral component of m(r) is fL = 0. In the more general case, where fL # 0, it may be that the sampling frequency need be no larger than /, = 2(fM — fL ). For example, if the spectral range of a signal extends from 10.0 to 10.1 MHz, the signal may be recovered from samples taken at a frequency fs = 2(10.1 — 10.0) = 0.2 MHz. To establish the sampling theorem for such bandpass signals, let us select a sampling frequency /, = 2(fM —fL ) and let us initially assume that it happens that the frequency fL turns out to be an integral multiple of/,, that is,/L = nfs with n an integer. Such a situation is represented in Fig. 5.1-4. In part a is shown the two-sided spectral pattern of a signal m(i) with Fourier transform M(jco). Here it has been arranged that n = 2; that is,/L coincides with the second harmonic of For a

signal m(t)

frequency

fs must

,

,

,

frequency part and the negative-frequency part of the spectrum are called

NS

respectively. Let us, for simplicity, consider separately

PS and PS and NS and the

selection of a

sampling frequency

As/

overlap region.

N = 3,

where there

is is

overlap region, while

=

=

1.

where

When/ > IB we

/ > 2fM

;

that

is,

m(()

s

is

further increase in

(from/,

3.5B

an overlap region while at

enter

N

2 kHz brings us to a point in an a small range of/, corresponding to no overlap. Further increase in / again takes us to an

still

N=2

responding to

f = 2B =

increased there

/

/

provides a nonoverlap range cor-

to/s = 5B). Increasing/, further we again = IB we enter the nonoverlap region for

do not again enter an overlap

region. (This

we assume we have a lowpass

is

i

!-4«-
the region

rather than a bandpass

1

signal.)

-7;/2

i

1

TJ2

|

The Discrete Fourier Transform

'

TJ2

1

|

There are occasions when the only information we have available about a signal a set of sample values, ./V in number, taken at regularly spaced intervals Ts over a period of time T0 From this sampled data we should often like is

to be able to

.

some reasonable approximation of the spectral content of the signal. If sample period and the number of samples is adequate to give us some con-

arrive at

the

fidence that generally,

We

what has been observed of the

we may indeed estimate

pretend that the signal

that the sampling rate

is

its

signal

is

periodic with period

is

adequate to

representative of the signal

satisfy the

T0

Nyquist

and we pretend, as

criterion.

A

If

the

waveform

may

possible set of sample values of a

waveform

m((),

taken every

T,

over a time interval

As a purely mathematical exercise we could use Eq. (5.1-10) to calculate spectral components v„ for any value of n. But consistently with our assumptions, the largest value of n allowable is determined by the Nyquist criterion. The highest frequency component should have a period which is 2 7^ so that:

sampled is m(i), then after sampling, the waveform we where S(t) is the sampling function, i.e., S(t) = 1 during

to be

is m(t)S(t),

the sampling duration dt, as shown in Fig. 5.1-7, and S(t) = 0 elsewhere. We note from Figs. 5.1-2 and 5.1 -3a that the spectrum of m(t) and the part of the spectrum

of m(t)S(t)

A

well,

typical set of

sample values is shown in Fig. 5.1-7. Here, for simplicity we have assumed an even number of samples so that we can place them symmetrically in the period interval T0 and symmetrically about the origin of the coordinate system The sample values are located at ± TJ2, ±3TJ2, etc. If there are N samples then the samples most distant from the origin are at ±[(N — l)/2]7\ have available

Figure 5.1-7

spectral content.

up to/M

are identical in form. Hence, to find the spectrum of m(t) evaluate instead the spectrum of m(t)S(t). The spectral

we

/„(max)

The fundamental period is T0 so 1/T0 Since T0 = A/Ts we have: .

= -!-

(5.1-11)

that the fundamental frequency

is

therefore 0

=

,

amplitudes of m(t)S(t)

are:

/„(max)

M„ =

— i

r To '

2

m(t)e-

12 *"" T °

dt

there are

N samples in

all,

for

t

in the integrand,

M„ =

— fa

(5.1-12)

0

Hence, the highest value of n for which Eq. (5.1-10) should be used

is

n

=

N/2.

then the sample times are-

5.2

Using these values

|^ = |/

(5 1-9)

'0 J -To/2 If

=

k

we

find that Eq. (5.1-9)

= (N-U/2

£

i O*=-(W-l)/2

m(fer>-^"* r'/ r»

PULSE-AMPLITUDE MODULATION

becomes: (5.1.10)

A technique by which we may take advantage of the sampling principle for the purpose of time-division multiplexing is illustrated in the idealized representation of Fig. 5.2-1. At the transmitting end on the left, a number of bandlimited signals

Decommutator

Commutator

mechanical switches, electronic switching systems

m 2 (0

«i(0 » 3

m,(«)

(t)»-

be employed. In either

the switch at the

left

in

which samples the signals, is called the commutator. The switching mechanism which performs the function of the switch at the right in Fig. 5.2-1 is called the decommutator. The commutator samples and combines samples, while Fig. 5.2-1,

m2 (0»FB

may

the switching mechanism, corresponding to

event,

"»,(/)•-

the

decommutator separates samples belonging

may

signals

to individual signals so that these

be reconstructed.

is shown in Fig. 5.2-2. we have considered the case of the multiplexing of just two signals m,(f) and m 2 (t). The signal m (t) is sampled regularly at intervals of 7^ and at the times indicated in the figure. The sampling of m 2 (t) is similarly regular, but the samples are taken at a time different from the sampling time of m^t). The input waveform to the filter numbered 1 in Fig. 5.2-1 is the train of samples of m,(f), and the input to the filter numbered 2 is the train of samples of m 2 (t). The timing in Fig. 5.2-2 has been deliberately drawn to suggest that there is room to multiplex more than two signals. We shall see shortly, in principle, how many

The

interlacing of the samples that allows multiplexing

Here, for simplicity,

x

Figure 5.2-1 Illustrating

how

the sampling principle

may be

used to transmit a number of bandlimit-

ed signals over a single communications channel.

are connected to the contact point of a rotary switch.

are similarly bandlimited.

For example, they may

We assume

that the signals

be voice signals, limited to 3.3 kHz. As the rotary arm of the switch swings around, it samples each signal sequentially. The rotary switch at the receiving end is in synchronism with the all

The two switches make contact simultaneously at simiWith each revolution of the switch, one sample is taken

switch at the sending end. larly

numbered

contacts.

of each input signal and presented to the correspondingly numbered contact of the receiving-end switch. The train of samples at, say, terminal 1 in the receiver, pass through low-pass

filter 1,

and, at the

filter

output, the original signal m,(t)

appears reconstructed. Of course, if fM is the highest-frequency spectral component present in any of the input signals, the switches must make at least 2fM revolutions per second.

When pling rate

may

the signals to be multiplexed vary slowly with time, so that the

sam-

correspondingly slow, mechanical switches, indicated in Fig. 5.2-1, be employed. When the switching speed required is outside the range of is

be multiplexed. observe that the train of pulses corresponding to the samples of each signal are modulated in amplitude in accordance with the signal itself. Accordingly,

We

scheme of sampling

the

Multiplexing of several

PAM

signals

called frequency-division multiplex If the

no further

I

!

h-r.-H

possible because the various signals

be

(FDM) systems.

signal processing

necessary

TDM-PAM

to

need be undertaken. Suppose, however, we from one antenna to another. It would

signal

amplitude-modulate

frequency carrier with the |

is

multiplexed signals are to be transmitted directly, say, over a pair of

TDM-PAM

1

Channel 2

and abbreviated

and are separately recoverable by virtue of the fact that they are sampled at different times. Hence this system is an example of a time-division multiplex (TDM) system. Such systems are the counterparts in the time domain of the systems of Chap. 3. There, the signals were kept separable by virtue of their translation to different portions of the frequency domain, and those systems are

then

time

called pulse-amplitude modulation

are kept distinct

require to transmit the

Sampling

is

PAM.

wires,

Channel

may

signals

would be referred same terminology

to, respectively, is

as

or

frequency-modulate

a

high-

signal; in such a case the overall system

PAM-AM

or

used whether a single signal or

PAM-FM. Note many

signals

that the

(TDM)

are

transmitted. i-f

i3k m 2 (()

A sample

of

A sample

of

5.3

.Jfc-J-I-:;:

CHANNEL BANDWIDTH FOR A PAM SIGNAL

Suppose that we have N independent baseband signals m^f), m 2 (t), etc., each of which is bandlimited to fM What must be the bandwidth of the communications channel which will allow all N signals to be transmitted simultaneously using .

114(1)

PAM Figure 5.2-2

The

interlacing of two

baseband

signals.

time-division multiplexing?

We

shall

now show that, Nfu

the channel need not have a bandwidth larger than

.

in principle at least,

The baseband

signal, say m^t), must be sampled at intervals not longer than Between successive samples of m^t) will appear samples of the other 1 signals. Therefore the interval of separation between successive samples of different baseband signals is \/2fM N. The composite signal, then, which is presented to the transmitting end of the communications channel, consists of a

= N— Ts

1/2/jir

•

= m^O) dt, which is in turn proportional to the value of the sample m^O). This response persists indefinitely. Observe, however, that the response passes through zero at intervals which are multiples of n/coc = 1/2/.. Suppose, then, that a It

sample of m 2 (r)

sequence of samples, that is, a sequence of impulses. If the bandwidth of the channel were arbitrarily great, the waveform at the receiving end would be the same as at the sending end and demultiplexing could be achieved in a straightfor-

This response

ward manner.

ing

however, the bandwidth of the channel is restricted, the channel response to an instantaneous sample will be a waveform which may well persist with significant amplitude long after the time of selection of the sample. In such a case, If,

the signal at the receiving

end

any particular sampling time may well have sigfrom previous samples of other signals. Consequently the signal which appears at any of the output terminals in Fig. 5.2-1 will not be a single baseband signal but will be instead a combination of many or even all the baseband signals. Such combining of baseband signals at a communication system output is called crosstalk and is to be avoided as far as possible. Let us assume that our channel has the characteristics of an ideal low-pass at

nificant contributions resulting

filter

with angular cutoff frequency

co c

=

2nfc

,

unity gain, and no delay. Let a

sample be taken, say, of m^t), at t = 0. Then at t = 0 there is presented at the transmitting end of the channel an impulse of strength / = m^O) dt. The x response at the receiving end is s Rl (t) given by (see Prob. 5.3-1) .

I,wc

.

sin co,

t

=

shown

is

taken and transmitted at

n

is

done

for mt(f) at

is

shown by

t

=

co c (t

\/2fc

-

.

If I 2

= m 2 (t =

t

=

0 and

for

l/2fc ) dt,

l/2/c )

the dashed plot. Suppose, finally, that the demultiplex-

also by instantaneous sampling at the receiving

channel response, there

m 2 (t)

will

at

t

=

1/2/.

.

Then,

end of the channel,

in spite of the persistence of the

be no crosstalk, and the signals m^f) and

m 2 (f) may

be completely separated and individually recovered. Similarly, additional signals may be sampled and multiplexed, provided that each new sample is taken synchronously, every 1/2/ s. The sequence must, of course, be continually repeated every l/2/M s, so that each signal is properly sampled.

We have then the result that with a channel of bandwidth we need to separate samples by intervals 1/2/. The sampling theorem requires that the samples of an individual baseband signal be separated by intervals not longer

/

than l/2/M fe/fitt

or X

.

Hence

= NfM

the total

number

of signals which

may

be multiplexed

is

JV

=

as indicated earlier.

In principle then, multiplexing a requires

number

of signals by

no more bandwidth than would be required

PAM

time division

to multiplex these signals

by

frequency-division multiplexing using single-sideband transmission.

t

5.4

The normalized response ns R(t)/a>c

is

in Fig. 5.3-1

by the solid

plot.

NATURAL SAMPLING

At

0 the reponse attains a peak value proportional to the strength of the impulse

It

was convenient,

for the

purpose of introducing some basic ideas, to begin our

discussion of time multiplexing by assuming instantaneous

commutation and decommutation. Such instantaneous sampling, however, is hardly feasible. Even if it were possible to construct switches which could operate in an arbitrarily short

T7

we would be disinclined to use them. The reason is that instantaneous samples at the transmitting end of the channel have infinitesimal energy, and time,

when transmitted through a bandlimited channel

A / / *

give rise to signals having a peak value which is infinitesimally small. We recall that in Fig. 5.3-1 I t = m^O) dt. Such infinitesimal signals will inevitably be lost in background noise.

f\ /

A much more

\

shown

reasonable manner of sampling, referred to as natural sampling,

Here the sampling waveform S(t) consists of a train of pulses having duration x and separated by the sampling time T The baseband s is

\

V

in Fig. 5.4-1.

.

i

-4-

signal

^^^^ i

i

,

2

i

4

is

m(t),

and the sampled

The response

to a sample at

t

=

filter

l/2£ (dashed plot).

to

j

tops are not

flat

!

an instantaneous sample

is

shown

in Fig. 5.4- lc.

Observe that whose

the sampled signal consists of a sequence of pulses of varying amplitude

but follow the waveform of the signal

With natural sampling, Figure 5.3-1 The response of an ideal low-pass

signal S(t)m(t)

at

(

=

0

(solid plot).

the Nyquist rate

may

ideal low-pass filter

m(t).

as with instantaneous sampling, a signal

sampled at

be reconstructed exactly by passing the samples through an with cutoff at the frequency fM , where fM is the highest-

m(<)

amplitudes of the various harmonics are not the same. The sampled baseband signal S(t)m(t)

S(tMt)

is,

=

for

Ts =

l/2/M

,

2t

- m(t) + — [m(f)C! T

cos 2^(2/M )t

+

m(t)C 2 cos 27r(4/M )f

+

•

•

•]

(5.4-3)

Therefore, as in instantaneous sampling, a low-pass deliver

an output signal

s 0 (t)

with cutoff at

filter

fM

will

given by

S(t) s„(t)

which

is

"1

same as

the

replaced by

given by the

first

(5.4-4)

term of Eq.

(5.1-2) except

with dt

t.

.

.

is

UN

advantageous, for the purpose of increasing the

make

level of the output signal, to t as large as possible. For, as is seen in Eq. (5.4-4), s„(t) increases with t.

However, to help suppress

crosstalk,

it is ordinarily required that the samples be than TJN. The result is a large guard time between the end of one sample and the beginning of the next.

S(«)Xm(()

limited to a duration

r.l

1

H

ft.

1

5.5

Pulses of the type

( fl )

A

much

less

FLAT-TOP SAMPLING shown

in Fig. 5.4- lc, with tops contoured to follow the waveare actually not frequently employed. Instead flat-topped pulses are customarily used, as shown in Fig. 5.5-la. flat-topped pulse has a

form of the

Figure 5.4-1

is

^ m(t)

With samples of finite duration, it is not possible to completely eliminate the crosstalk generated in a channel, sharply bandlimited to a bandwidth fc signals are to be multiplexed, then the maximum sample duration is t = TJN. It (»)

The

=

__

1

signal,

A

baseband signal

m(t).

(fe)

A

sampling signal

with pulses of

S(t)

finite

naturally sampled signal S(t)m(t).

duration

(c)

ell)

frequency spectral component of the signal. pling

and

waveform

S(t)

shown

in Fig. 5.4-1

is

To prove

given by [see

r0 = TJ t S(t)

2t /

f + -f

(

c

we note that the samEq (1 3-12) with A = 1

t

i

c °s 2n

— + C 2 cos 2

-m(0

this,

—+

\

t

x 2*

•

(5.4-1)

with the constant C„ given by

M (o)

sin (mtct/TJ

1

T Figure 5.5-1 (fc)

This sampling waveform differs from the sampling waveform of Eq. (5.1-1) for instantaneous sampling only in that dt is replaced by t

and by the

fact that the

A

(a)

Flat-topped sampling,

network with transform H(ja>)

which converts a pulse of width

dt

into a rectangular pulse of like ampli-

(6)

tude but of duration

t.

constant amplitude established by the sample value of the signal at some point within the pulse interval. In Fig. 5.5- la we have arbitrarily sampled the signal at

M0

the beginning of the pulse. In sampling of this type the baseband signal m(f)

cannot be recovered exactly by simply passing the samples through an ideal lowpass filter. However, the distortion need not be large. Flat-top sampling has the merit that

simplifies the design of the electronic circuitry used to

it

(o)

perform the

sampling operation.

To show

the extent of the distortion, consider the signal m(t) having a

Fourier transform M(jw).

We

have seen

the transform of the sampled signal,

and

(see Figs. 5.1-2

when

the sampling

transform of the sampled signal for flat-top sampling

how

5.1-3)

is

to

deduce

instantaneous.

The

determined by consider-

is

ing that the flat-top pulse can be generated by passing the instantaneously

sampled signal through a network which broadens a pulse of duration dt (an impulse) into a pulse of duration

and width

dt

The transform of a

x.

^[impulse of strength

The transform of a

dt at

t

=

0]

pulse of unit amplitude and width x

pulse, amplitude

&\

pulse of unit amplitude

is

=

1,

extending from

= -|

t

=

is

to

dt

[see Eq. (1.10-20)]

t

=- =

2

L

(5.5-1)

sin

t

(^Z2)

cox/2

2J

(5.5-2)

Hence, the transfer function of the network shown „.

.

,

=

H(ja>)

in Fig. 5.5-16,

is

required to be

—

x sin (ax/2)

-r dt

(5.5-3) v

cox/2

'

Figure 5.5-2

Let the signal

with transform M(jw), be bandlimited to fM and be sampled at the Nyquist rate or faster. Then in the range 0 to fM the transform of the flat-topped sampled signal is given by the product H{jco)M{joo) or, from Eqs. (5.5-1), (5.5-2),

and

m(t),

(a)

An

idealized spectrum of a

instantaneous sampling,

(c)

The form

introduced by flat-topped sampling,

effect)

baseband

[(sin x)/x, (rf)

signal.

with x

=

The spectrum

(i>)

The spectrum of the

signal with

wt/2] of the distortion factor (aperture of the signal with flat-topped sampling.

(5.5-3) is not the case and that, as a result, distortion will result. This from the fact that the original signal was " observed " through a rather than an infinitesimal time "aperture" and is hence referred to as

however, that such

^[flat-topped sampled m(t)]

=^ /,

^^B. M

distortion results

0

(jco)

cox/1


(5.5-4)

finite

aperture effect distortion.

To

illustrate the effect of flat-top

sampling,

signal m(t) has a flat spectral density equal to

fM

,

as

is

shown

in Fig. 5.5-2a.

The form

we consider

M

0

over

its

entire range

from 0 to

of the transform of the instantaneously

sampled signal is shown in Fig. 5.5-2b. The sampling frequency /, = 1/Tj is assumed large enough to allow for a guard band between the spectrum of the baseband signal and the DSB-SC signal with carrier/,. The spectrum of the flattopped sampled signal is shown in Fig. 5.5-2o". We are, of course, interested only in the part of the spectrum in the range 0 to/M If, in this range, the spectra of the sampled signal and the original signal are identical, then the original signal may be recovered by a low-pass filter as has already been discussed. We observe, .

The

for simplicity that the

distortion results from the fact that the spectrum

sampling function Sa(x) pling function (see Sec.

=

(sin x)/x

1.4) falls

(with x

=

is

multiplied by the

The magnitude of

the samx in the neighborhood we approach x = n, at which point cox/2).

off slowly with increasing

= 0 and does not fall off sharply until = 0. To minimize the distortion due to the aperture effect, it is advantageous to arrange that x = n correspond to a frequency very large in comparison with fM Since x = nfx the frequency f0 corresponding to x = n is /„ = 1/t. If 1//~ M the aperture distortion will be small. fo ^/m> or correspondingly, if x of x

Sa(x)

,

.

>

The as x

distortion

—

,

becomes progressively smaller with decreasing

x.

And, of course,

0 (instantaneous sampling), the distortion similarly approaches zero.

Equalization 1

-

2 Input,

As

in the

pling,

it is

case of natural sampling, so also in the present case of flat-top samadvantageous to make t as large as practicable for the sake of increas-

ing the amplitude of the output signal. If, in a particular case, it should happen that the consequent distortion is not acceptable, it may be corrected by including an equalizer in cascade with the output low-pass filter. An equalizer, in the present instance, is a passive network whose transfer function has a frequency dependence of the form x/sin x, that is, a form inverse to the form of H(jw) given in Eq. (5.5-3). The equalizer in combination with the aperture effect will then yield a flat overall transfer characteristic between the original baseband signal and the output at the receiving end of the system. The equalizer x/sin x cannot be exactly synthesized, but can be approximated. If

x/sin

N

~

x

< l/2fM N, and hence for large t
signals are multiplexed, t 1.

N

results.

samples of one baseband

C

Amplifier

—

I

°

1

o

Output

signal

Figure 5.6-2 Illustrating a method of performing the operation of holding.

A shown

method,

in principle,

in Fig. 5.6-2.

by which

may be performed is synchronism with the occurrence of

holding operation

this

The switch S operates

in

somewhat after the occurrence and opens somewhat before the occurrence whose gain, if any, is incidental to the present

input samples. This switch, ordinarily open, closes of the leading edge of a sample pulse of the trailing edge.

The

amplifier,

discussion, has a low-output impedance. Hence, at the closing of the switch, the

C

capacitor

charges abruptly to a voltage proportional to the sample value, and

the capacitor holds this voltage until the operation

sample. In Fig. 5.6-1

we have

output waveform maintaining a perfectly constant 5.6

SIGNAL RECOVERY THROUGH HOLDING

pulse interval and

its

following holding interval.

is

repeated for the next

somewhat by showing

idealized the situation

the

throughout the sample have also indicated abrupt

level

We

transitions in voltage level from

We

have already noted that the

the sampling interval,

is

t/T,

=

maximum

l/N,

N

x/T„ of the sample duration to

ratio

being the

number

of signals to be multiplexed. As increases, x/T, becomes progressively smaller, and, as is to be seen from Eq. (5.4-4), so correspondingly does the output signal. We discuss now an alternative method of recovery of the baseband signal which raises the level of the output signal (without the use of amplifiers which may introduce

N

noise).

method has

The

the additional advantage that rather rudimentary filtering is often quite adequate, but has the disadvantage that some distortion must be accepted.

The method

transitions will

one sample to the next. In practice, these voltage be somewhat rounded as the capacitor charges and discharges

exponentially. Further,

if

the received sample pulses are natural samples rather

than flat-topped samples, there level

during the sample interval

will

be some departure from a constant voltage

itself.

As a matter of practice however, the sample

very small in comparison with the interval between samples, and the voltage variation of the baseband signal during the sampling interval is small interval

is

enough to be neglected. If the baseband signal

is

m(t) with spectral density M(jco)

=

^"[m(t)],

we may

where the baseband signal m(t) and its flat-topped samples are shown. At the receiving end, and after demultiplexing, the sample pulses are extended; that is, the sample value of each individual baseband

deduce the spectral density of the sampled and held waveform in the manner of Sec. 5.5 and in connection with flat-topped sampling. We need but to consider that the flat tops have been stretched to encompass the entire interval between

signal

instantaneous samples. Hence the spectral density is given as in Eq. (5.5-4) except with t replaced by the time interval between samples. We have, then,

signal.

is

is

illustrated in Fig. 5.6-1,

held until the occurrence of the next sample of that

This operation

shown in Fig. 5.6-1 as sample pulses. The output waveform consists then, staircase waveform with no blank intervals. is

same baseband

the dashed extension of the as shown, of an up and down

3F\_m{t),

sampled and held]

=

an

(

"^

2)

M{Jm)

coTJ2 In Fig. 5.6-3 Samples

of

m

(

/

we have again assumed

0


(5.6-1)

for simplicity that the band-limited

M

magnitude 0 In Fig. 5.6-3a is shown the spectrum of the instantaneously sampled signal. In Fig. 5.6-3b has been drawn the magnitude of the aperture factor (sin x)/x (with x = a>TJ2), while in Fig. 5.6-3c is shown the magnitude of the spectrum of the sampled-and-held signal. These plots differ from the plots of Fig. 5.5-2 only in the location of the signal m(t) has a flat spectral density of

.

nulls of the factor (sin x)/x. In Fig. 5.6-3 the first null occurs at the

quency

Figure 5.6-1 Illustrating the operation t

of holding.

f

We

observe that, as a consequence, the aperture

sampling

fre-

which is responsible for the (sin x)/x term, has accomplished most of the filtering which is required to suppress the part of the spectrum of the output signal above the s

.

effect,

Magnitude, instantaneously stL

sampled

signal

we

M

new

create a

signal

m q(t)

quantized signal

m q(t)

which

an approximation to

is

has the great merit that

it

is,

in large

However, the

m(f).

measure, separable

from the additive noise.

The operation

of quantization

have divided

example

is

M=

which

in

tion levels

range into

this total

the step size,

called

is

whose excursion

plate a signal m(f)

m 0 ,m u

.

.

.

,

S

=

(V H

represented in Fig. 5.7-1. Here

is

confined to the range from

M equal intervals each of - V L)/M.

In

.

to

V„ We .

Accordingly

S,

we show the specific steps we locate quantiza-

Fig. 5.7-1

In the center of each of these

8.

m 7 The

size S.

we contem-

VL

quantized signal m,(t)

is

generated in the following

way: Whenever m{t) is in the range A 0 the signal m (t) maintains the constant q level m 0 whenever m(t) is in the range A,, m (t) maintains the constant level m, q and so on. Thus the signal m q (t) will at all times be found at one of the levels m 0 ,

;

,

m„ ly

Magnitude, held samples

•

"

.

. .

,

m7

when

.

The

from

m,(t)

= m0

state the matter in

=

m, is made abruptmidway between m„ and m,

to m,(t)

L 01 which is an alternative fashion, we say

m{t) passes the transition level

and so on. To

T,

m q(t)

transition in

that, at every has the value of the quantization level to which m(t) is closest. Thus the signal m (t) does not change at all with time or it makes a q

m q {t)

instant of time,

"quantum" jump the range from

to

the separation of the is

Note the disposition of the quantization levels in levels are each separated by an amount S, but extremes V L and V H each from its nearest quantization level

of step size

VL

S.

V„ These .

only S/2. Also, at every instant of time, the quantization error m(t) is equal to or less than S/2.

—

m,(t)

has a

magnitude which Figure 5.6-3

shown

Spectrum of instantaneously sampled signal m(t) with The magnitude of the aperture effect factor,

(a)

in Fig. 5.5-2a. (b)

nit) (c)

having idealized spectrum Spectrum of sampled-and-

held signal.

We

see, therefore, that the

inal signal.

The quality of

size of the steps,

thereby increasing the

with small enough steps, the

bandlimit/M Of course, the

filtering

.

may

be required.

some

be

ponents

may be

We

not perfect, and some additional

filtering

also note that, as in the case of flat-top sampling, there will

by the unequal transmission of

distortion introduced

in the

is

range 0 to/M

.

If

the distortion

is

spectral

com-

not acceptable, then, as before,

it

in

human

is

may

number

an approximation to the origbe improved by reducing the

of allowable

levels.

Eventually,

ear or the eye will not be able to distinguish

the original from the quantized signal.

To

give the reader an idea of the

of quantization levels required in a practical system,

we note

that

256

number

levels

can

be used to obtain the quality of commercial color TV, while 64 levels gives only fairly good color TV performance. These results are also found to be valid when quantizing voice.

corrected by an x/sin x equalizer.

Most importantly we note

quantized signal

the approximation

comparing Eq.

(5.6-1)

with Eq. (5.5-4) that,

aside from the relatively small effect of the (sin x)/x terms in the

two

cases, the

sampled-and-held signal has a magnitude larger by the factor TJx than the signal of sample duration t. This increase in amplitude is, of course, intuitively to have been anticipated.

Now let us consider that our quantized signal has arrived at a repeater somewhat attenuated and corrupted by noise. This time our repeater consists of a quantizer and an amplifier. There is noise superimposed on the quantized levels of m q (t). But suppose that we have placed the repeater at a point on the communications channel where the instantaneous noise voltage is almost always less than half the separation between quantized levels. Then the output of the quantizer will consist of a succession of levels duplicating the original quantized signal

and with 5.7

QUANTIZATION OF SIGNALS

The

limitation of the system

the noise removed. In rare instances the noise results in

tization level.

A

noisy quantized signal

is

shown

quantizer output levels are indicated by the dashed lines

long channels

we

is

we have been

describing for communicating over

that once noise has been introduced

are " stuck " with

it.

We now

describe

how

any place along the channel,

the situation

jecting a signal to the operation of quantization.

When

is

modified by sub-

quantizing a signal

m(f),

The output of

an error

in

quan-

The allowable separated by amount S.

in Fig. 5.7-2a.

the quantizer is shown in Fig. 5.1-2b. The quantizer output is the which the input is closest. Therefore, as long as the noise has an instantaneous amplitude less than S/2, the noise will not appear at the output. One instance in which the noise does exceed S/2 is indicated in the figure, and, correlevel to

spondingly, an error in level does occur. that even

if

the average noise magnitude

The is

statistical

much

less

nature of noise

than S/2, there

is

such

is

always a

finite

probability that from time to time, the noise magnitude will exceed S/2.

Note

that

it is

never possible to suppress completely level errors such as the one

indicated in Fig. 5.7-2.

We

have shown that through the method of signal quantization, the

additive noise can be significantly reduced. repeaters,

By decreasing

effect of

the spacing of the

we decrease the attenuation suffered by m,(t). This effectively decreases power and hence decreases the probability P, of an error in

the relative noise

P, can also be reduced by increasing the step

size S. However, increasing S an increased discrepancy between the true signal m(t) and the quantized signal m (t). This difference m(t) - m {t) can be regarded as noise and is called q q quantization noise. Hence, the received signal is not a perfect replica of the transmitted signal m(t). The difference between them is due to errors caused by additive noise and quantization noise. These noises are discussed further in Chap. 12. level.

results in

5.8 It

QUANTIZATION ERROR

has been pointed out that the quantized signal and the original signal from it was derived differ from one another in a random manner. This difference

which

or error

may

be viewed as a noise due to the quantization process and

We now

is

called

mean-square quantization error e 1 where e is the difference between the original and quantized signal voltages. Let us divide total peak-to-peak range of the message signal m(t) into equal voltage intervals, each of magnitude S volts. At the center of each voltage quantization error.

calculate the

,

M

interval

we

The dashed a time

t.

locate a quantization level level represents the

Since, in this figure, m(t)

quantizer output will be e 206

—

m(t)

— mk

.

mk

,

m„ m 2

,

mM

as

shown

in Fig. 5.8- la.

instantaneous value of the message signal m(t) at

happens to be

closest to the level

the voltage corresponding to that level.

The

mt

,

the

error

is

Now f (1) S

is

the probability that the signal voltage m(t) will be in the

zation range, etc.

i2) f S

Hence the sum of terms

unity. Therefore, the

m

the probability that

is

in the

first

quanti-

second quantization range,

in the

is

parentheses in Eq. (5.8-2b) has a total value of

mean-square quantization error

is

(5.8-3)

Error,

Peak- to- peak

5.9

range of signal

-

PULSE-CODE MODULATION

MS

A

signal

which

is

The quantization

to be quantized prior to transmission

usually sampled as well.

is

used to reduce the effects of noise, and the sampling allows us

is

to time-division multiplex a

number of messages

if

we choose

to

do

bined operations of sampling and quantizing generate a quantized

(»)

form, that

is,

a train of pulses

whose amplitudes are

so.

The com-

PAM

restricted to a

wave-

number of

discrete magnitudes.

We

if we choose, transmit these quantized sample values directly. Alterwe may represent each quantized level by a code number and transmit the code number rather than the sample value itself. The merit of so doing will be developed in the subsequent discussion. Most frequently the code number is con-

may,

natively

before transmission, into

verted,

base-2 arithmetic. Figure 5.8-1

(a)

tization ranges

error voltage

A

range of voltage over which a signal mfr) makes excursions is divided into size S. The quantization levels are located at the center of the range as a function of the instantaneous value of the signal m(t)

M quan-

each of

e(t)

(b)

The

The

digits of the

m+

dm/2.

dm

Then

be the probability that m(t)

m-

the voltage range the mean-square quantization error is

mi +S/2

lies in

transmitted as pulses. Hence the system of transmission

We

(*m 2

/(m)(m

-m

2 t)

/(mXm - m 2 ) 2


+

(5.8-1)

Now, ord.nanly

the probability density function /(m) of the message signal m(() not be constant. However, suppose that the number of quantization is large, so that the step size S is small in comparison with the peak-to-peak range of the message signal. In this case, it is certainly reasonable to make the approximation that/(m) is constant within each quantization range. Then in the first term of Eq. (5.8-1) we set f(m) =/<", a constant. In the second term 2 J(m) =f etc. We may now remove (1 etc., from

M

f \f

we make

the substitution

x

= m - mk

,

Eq. (5.8-1) becomes

rs/2

? = (/n> + f(2)

dx J -S/2

=

inside the integral sign If

\

=

(/«"

...

digits,

0 and

,

•••

+

k2 2

+

•

•

•)

12

2

An

1.

k 2 k^k 0 in which the

+

fc's

arbitrary

number

N

represent-

is

are determined from the equation

M

+

1

k 0 2°

(5.9-1)

with the added constraint that each k has the value 0 or

1. The binary representnumbers 0 to 15 are given in Table 5.9-1. Observe that to represent the four (decimal) numbers 0 to 3, we need only two binary digits k l and k 0 For the eight (decimal) numbers from 0 to 7 we require only three binary — 1 are to be represented, places, and so on. In general, if numbers 0, 1, then an N binary digit sequence k N _ , = 2N fe 0 is required, where The essential features of binary PCM are shown in Fig. 5.9-1. We assume

+/«» +

•

-)

(5.8-2a)

.

M

(5.8-26)

•

.

.

,

M

M

•

.

that the analog

message signal m(t) is limited in its excursions to the range from We have set the step size between quantization levels at 1 volt. Eight quantization levels are employed, and these are located at —3.5, —2.5, + 3.5 volts. We assign the code number 0 to the level at — 3.5 volts, the code number 1 to the level at —2.5 volts, etc., until the level at +3.5 volts, which is

—4

to

+4

volts.

assigned the code

12

.

•

number

7.

Each code number has

arithmetic ranging from 000 for code

(/«»S +/< 2 >S

called (binary) pulse-

ations of the decimal

will certainly

,

ed by the sequence

N=

<mi-SI2

is

review briefly some elementary points about binary arithmetic. The

+S/2

dm +

mi -S/2

dm/2 to

i.e.,

binary representation of the code number are

code modulation (PCM). binary system uses only two

Let/(m)

representation in binary arithmetic,

its

number 0

to

1 1 1

its

for

and the

code number

7.

we specify the sample code number and its binary rep-

In Fig. 5.9-1, in correspondence with each sample, value, the nearest quantization level,

representation in binary

Table 5.9-1 Equivalent

numbers

decimal

in

and

binary representation Binary

resentation. If

we were

sample values

1.3, 3.6, 2.3, etc. If

we were

1.5, 3.5, 2.5, etc.

transmit the binary representations 101, 111, 110,

Decimal

we would

transmit the

transmitting the quantized signal,

would transmit the quantized sample values

In binary

PCM

we we

etc.

K

*3

k2

0

0

0

0

0

0

0

1

1

0

0

1

0

2

0

0

0

1

1

3

0

1

0

0

4

0

1

0

1

5

0

1

1

0

6

0

1

1

1

7

0 0

0

0

8

0

1

9

0

1

0

5.10

ELECTRICAL REPRESENTATIONS OF BINARY DIGITS

As intimated trical

in the previous section, we may represent the binary digits by elecpulses in order to transmit the code representations of each quantized level

over a communication channel. Such a representation is shown in Fig. 5.10-1. Pulse time slots are indicated at the top of the figure, and, as shown in Fig. 5.10la, the binary digit 1 is represented by a pulse, while the binary digit 0 is rep-

The row of three-digit binary numbers given binary representation of the sequence of quantized samples in

resented by the absence of a pulse.

10

1

1

11

in Fig. 5.10-1 is the

1

0

0

12

Fig. 5.9-1.

I

0

1

13

that

1

1

0

14

1

1

1

15

0

transmitting the analog signal,

Hence the pulse pattern in Fig. 5.10-la is the (binary) PCM waveform would be transmitted to convey to the receiver the sequence of quantized samples of the message signal m(t) in Fig. 5.9-1. Each three-digit binary number that specifies a quantized

words allow

sample value

is

called a word.

The spaces between

for the multiplexing of other messages.

At the receiver, in order to reconstruct the quantized signal, all that is is that a determination be made, within each pulse time slot, about whether a pulse is present or absent. The exact amplitude of the pulse is not required

ZoOa imber

Quantization level

important. There

mil), volts

is

an advantage

since the pulse energy

is

in

making

the pulse width as wide as possible

thereby increased and

it

becomes

easier to recognize a

pulse against the background noise. Suppose then that

we eliminate the guard time t 9 between pulses. We would then have the waveform shown in Fig. 5. 10- lb. We would be rather hard put to describe this waveform as either a sequence of positive pulses or of negative pulses. The waveform consists now of a sequence of transitions between two levels. When the waveform occupies the lower level in a

U—

—

Word time

slot

Pulse time slot

ii

i 1

i

Sample value

1.3

3.6

2.3

0.7

-0.7

-2.4

-3.4

Nearest quantization leve

1.5

3.5

2.5

0.5

-0.5

-2.5

-3.5

Code number

5

7

6

4

3

1

0

101

111

110

100

on

001

000

Binary representation

Figure 5.9-1 A message signal is regularly sampled. Quantization levels are indicated. For each sample the quantized value is given and its binary representation is indicated.

*

•

101

111

110

n n

nnn

nn

i

100

-

n

011

001

nn

n

ooo

-flfh-rv-pr-^fl------^-----fi Figure 5.10-1 (fc)

(a)

Pulse representation of the binary numbers used to code the samples in Fig. 5.9-1.

Representation by voltage levels rather than pulses.

particular time slot, a binary 0 resents a binary 1.

is

represented, while the upper voltage level rep-

Suppose that the voltage difference of 2V volts between the levels of the waveform of Fig. 5.10-1& is adequate to allow reliable determination at the receiver of which digit is being transmitted. We might then arrange, say, that the waveform make excursions between 0 and 2V volts or between - V volts and + V volts. The former waveform will have a dc component, the latter waveform will not. Since the dc component wastes power and contributes nothing to the reliability of transmission, the latter alternative is preferred

and

is

indicated in

Fig. 5.10-lfc.

When

has been added during the transmission along the channel. As noted previously, separation of the signal from the noise the signal. first

is

Such an operation

is possible because of the quantization of again an operation of requantization ; hence the

is

block in the receiver in Fig. 5.11-1

eases the burden

on

quantizer

this

is

termed a quantizer.

A

feature which

that for each pulse interval

has only to the relatively simple decision of whether a pulse has or has not been received or which of two voltage levels has occurred. Suppose the quantized sample pulses had been transmitted instead, rather than the binary-encoded

In the

PCM

m(i)

first

the separation of the signal from the noise which

is

is

it

codes for such samples. Then this quantizer would have had to have yielded, in each pulse interval, not a simple yes or no decision, but rather a more complicated determination about which of the many possible levels had been received.

The Encoder

A

the digitally encoded signal arrives at the receiver (or repeater), the

operation to be performed

make

THE PCM SYSTEM

5.11

The Decoder

communication system is represented in sampled, and these samples are subjected to

The quantized samples

are applied to an encoder.

Fig. 5.11-1.

The analog

signal

the operation of quantization.

The encoder responds

to each

such sample by the generation of a unique and identifiable binary pulse (or binary level) pattern. In the example of Figs. 5.9-1 and 5.10-1 the pulse pattern happens to have a numerical significance which is the same as the order assigned to the quantized levels. However, this feature is not essential. We could have

example of

Fig. 5.10-1,

a quantized

if

PAM

signal

would have to decide which of the

the receiver quantizer

PCM

mitted, while with a binary

between two possible

levels.

The

had been transmitted, 0 to 7 was trans-

levels

signal the quantizer need only distinguish

PCM

over the multivalued decision required for quantized important advantage for PCM.

The

no decision

relative reliability of the yes or

receiver quantizer then, in each pulse slot,

PAM

in

constitutes an

makes an educated and

sophisticated estimate and then decides whether a positive pulse or a negative pulse was received and transmits its decisions, in the form of a reconstituted or

assigned any pulse pattern to any level. At the receiver, however, we must be able to identify the level from the pulse pattern. Hence it is clear that not only does the encoder number the level, it also assigns to it an identification code. The combination of the quantizer and encoder in the dashed box of

(If repeater operation is intended, the simply raised in level and sent along the next section of the transmission channel.) The decoder, also called a digital-to-analog (D/A) con-

Fig. 5.11-1

verter,

is called an analog-to-digital converter, usually abbreviated A/D concommercially available A/D converters there is normally no sharp distinction between that portion of the electronic circuitry used to do the quantizing

verter. In

and that portion used to accomplish the encoding. In summary, then, the A/D converter accepts an analog signal and replaces it with a succession of code symbols, each symbol consisting of a train of pulses in which each pulse may be interpreted as the representation of a digit in an arithmetic system. Thus the signal transmitted over the communications channel in a system is referred to as a digitally encoded signal.

PCM

regenerated pulse train, to the decoder.

regenerated pulse train

is

performs the inverse operation of the encoder. The decoder output is the sequence of quantized multilevel sample pulses. The quantized signal is now reconstituted. It is then filtered to reject any frequency components lying

PAM

The

outside of the baseband.

final

m(t) except for quantization noise

making

5.12

at the receiver

due

output signal

i

Analog

i

lished at voltages to

channel

—

let

Sampler m{t)

-j-»-jQuantizerl»-

Quantizer » Decoder

1

»

Filter

accommodate a

VL

to a high voltage

V H We .

excursions beyond the bounds tage. Quantized

Figure 5.11-1

A

PAM

PCM communication system.

For, within

exceeds

± S/2

us consider that

M levels with step signal m(t)

i

signal

with the input

yes-no decision

COMPANDING

quantization process employing Communication

in

to the presence of channel noise.

Referring again to Figs. 5.7-1 and 5.8-1 lAnalog-to-dlgital converter!

m'(t) is identical

and the occasional error

established a

being estab-

which ranges from a low voltage

can readily see that

V„ and V L

we have

size S, the levels

if

the signal m(r) should

make

the system will operate at a disadvan-

these bounds, the instantaneous

while outside these bounds the error

is

quantization error never

larger.

Further, whenever m(t) does not swing through the full available range the is equally at a disadvantage. For, in order that m,(f) be a good approx-

system

imation to m(t) it is necessary that the step size S be small in comparison to the range over which m(t) swings. As a very pointed example of this consideration consider a case in which m(f) has a peak-to-peak voltage which is less than S and never crosses one of the transition levels in Fig. 5.7-1. In such a case m (t) will be a fixed (dc) voltage and will bear no relationship to m(f). To explore this latter point somewhat more quantitatively let us consider that m(t) is a signal, such as the sound signal output of a microphone, in which vh = L =V, i.e., a signal without dc components, and with (at least approximately) equal positive and negative peaks. Further, for simplicity,

~V

assume that

in the range

ability density.

The

±V

let us characterized by a uniform probthen equal to 1/2 V and the normalized

the signal m(f)

probability density

is

is

average signal power of the applied input signal

——

+"

>

S

i

=

=

m(t)

f y^m

(t)

we have

feL =ioiogio fe] Equation

=ioiogio22w=6N

(5i2 - 7)

(5.12-7) has the interpretation that, in a

system where the signal is N levels is 2 ), and where the signal amplitude is capable of swinging through all available quantization regions without extending beyond the outermost ranges, the output signalquantized using an N-bit code

(i.e.,

the

number of quantization

to-quantization noise ratio is 6N dB. In voice communication we use N = 8 corresponding to 256 quantization regions and SJN = (6)(8) = 48 dB. If the Q signal is reduced in amplitude so that not all quantization ranges are used then

becomes smaller, since N Q depending as it does only on step S is not by the amplitude reduction, while S 0 is reduced. For example, if the amplitude were reduced by a factor of 2, the power is then reduced by a factor of 4 reducing the signal-to-noise ratio by 6 dB It is interesting to observe that the effective number of quantization levels is also reduced by a factor of 2. Correspondingly, the number N of code bits is reduced by 1. In summary, the dependence of SJN Q on the input signal power S, is such that as the number of code bits needed decreases, SJN decreases by 6 dB/bit. Q

SJN Q

is

,

affected

yl

— dm = — 1

2

In decibels

( 5 .12-1)

.

The quantization

noise, as given

by Eq.

(5.8-3) is

S2

It is

If

the

number

of quantization levels

is

M,

v =

then

MS = 2V so that

generally required, for acceptable voice transmission, that the received

have a ratio SJN Q not less than 30 dB, and that this minimum 30 dB figure hold even though the signal power itself may vary by 40 dB. (The signal, in this case, is described as having a 40 dB dynamic range.) In an 8-bit system, at maximum signal level we have SJN Q s S-/N Q = 48 dB Now N Q is fixed, and depends only on step size. If we are to allow SJNq to drop to no lower than 30 dB then the dynamic range would be restricted to 48 - 30 = 18 dB The dynamic range can be materially improved by a process called companding (a word formed by combining the words compressing and expanding). As signal

MS — 2

(5.12-3)

.

Combining Eqs.

(5.12-1), (5.12-2),

quantization noise power ratio

and

(5.12-3)

we have

that the input signal-to-

is

~tQ

= M2

(

512 "4)

Eventually the received quantized signal will be smoothed out to generate an output signal with power S If we have a useful 0 communication system then pre.

sumably the

effect of the

quantization noise is not such as to cause an easily perceived difference between input and output signal. In such a case the output power S„ may be taken to be the same as the input power i e S ~ S so that finally we may replace Eq. (5. 12-4) by

JTQ

= M2

(512-5)

M

If there are quantization levels then the code which singles out the closest quantization to the signal m(t) will have to have = 2" Hence bits with Eq. (5.12-5) becomes

N

-^ = Q

(2")

2

= 2™

M

(5.12-6)

we have

seen, to keep the signal-to-quantization noise ratio high we must use a which swings through a range which is large in comparison with the step This requirement is not satisfied when the signal is small. Accordingly before

signal size.

applying the signal to the quantizer input-output characteristic as

shown

we

pass

it

through a network which has an Note that at low amplitudes

in Fig. 5.12-1.

the slope is larger than at large amplitudes. A signal transmitted through such a network will have the extremities of its waveform compressed. The peak signal which the system is intended to accommodate will, as before, range through all available quantization regions. But now, a small amplitude signal will range through more quantization regions than would be the case in the absence of compression. Of course, the compression produces signal distortion. To undo the distortion,

at

the

receiver

we

pass

the recovered signal through an expander has an input-output characteristic which is the inverse of the characteristic of the compressor. The inverse distortions of compressor and expander generate a final output signal without distortion. The determination of the form of the compression plot of Fig. 5.12-1 is a

network.

An expander network

somewhat

subjective matter. In the United States,

Canada, and Japan a

p.-law

Output

,

v„

this

region a change in analog signal of step size

A

change the code word correspondence between addresses and transmitted code applies as shown for the middle 64 addresses. For the next 64 addresses (32 on one side and 32 on the other side of presented for transmission by the

the original 64)

amount

will

A. This one-to-one

we arrange

that at the memory locations specified by pairs of two adjacent addresses the same code word is written into the memory. Hence the transmitted code word will change at every second addresses change and correspondingly the step size is 2A. Next we arrange that for the next

addresses,

i.e.,

for

12-bit input

codewords

8-bit

output

codewords

Figure 5.12-1

An input-output

characteristic which provides compression.

16

OUTPUT

CODEWORDS compandor is used (see Prob. 5.12-1) and it differs somewhat from the A-law compander (see Prob. 5. 1 2-2) used by the rest of the world. The " and the " A " refer to parameters which appear in the equations for the compression and

>

expansion characteristic. In implementing the compression characteristic, the analog signal m(f) is left unmodified and, instead, the step size is tapered so that the quantization levels are close together at low signal amplitudes and progressively further apart in regions attained as the signal increases in amplitude. To see one way in which the step sizes are altered consider again an 8-bit system employing 2 8 = 256 quantization levels. In such a system, the A/D converter shown in Fig. 5.11-1

16

would ordinarily be an

8-bit converter. Each input sample would generate an output code identifying the closest quantization level. If the total range of the input was ± V then the step size would be S = 2K/2 8 Let us start however with a 12-bit A/D converter so that the step size is 2V/2 11 This 12-bit PCM signal is not applied to the communication channel directly but is instead applied to the address pins of a read-only memory (ROM) whose content is to be described. The has 12 address pins and 8 output data pins. The signal 8-bit

.

.

ROM

transmitted

is

the 8-bit data output of the

The content cessive

of the

ROM

ROM.

As shown in Fig. 5.12-2, in each suclocation in the region where the step size is to be smallest (i.e., 2V/2 12 = A) there are written successive 8-bit code words. Hence in is

as follows:

memory

step size

=

OUTPUT

CODEWORDS

PCM

Figure 5.12-2

ROM characteristic for compression.

128 addresses (64 above tions of four successive

+

64 below) the same code word

memory

is

written into the loca-

locations so that the step size

is

4A.

We

proceed

manner, each time doubling the number of successive memory locations which contain the same code word. As can be verified, when we have used all 12 2 = 4096 addresses we shall have generated 2 8 = 256 code words. At the receiver we have a mechanism for generating the quantization corresponding to each code word. in this

It

is

interesting to observe that for the smallest 64 levels, the input

output signals are the same,

small signals are not companded.

i.e.,

One

and

reason for

PCM

this is that signals other than voice are often digitized using the same system employed for voice. If a non-voice signal is subjected to the compression algorithm the result is usually a degradation of performance. Therefore, to avoid

this possibility,

such non-voice signals are kept 40

dB below

the peak level of a

voice signal. In conclusion,

we

refer to Fig. 5.12-3

which compares the variation of output

power when companding is Note that the companded system has a far greater dynamic range than the uncompanded system and that theoretically the companded system has an output signal-to-noise ratio which exceeds 30 dB over a dynamic range of input signal power of 48 dB, while the uncompanded system has a dynamic range of 18 dB for the same conditions. It should be noted however, that the penalty paid using companding is approximately 10 dB. Thus an 8-bit uncompanded system, when operating at maximum amplitude, produces a signal-to-noise ratio of 48 dB. The same 8-bit system, using a compandor, yields a 38 dB SNR. signal-to-noise ratio as a function of input signal

used, to the case of an

S 0 /N 0 ,dB

5.13

uncompanded

system.

MULTIPLEXING PCM SIGNALS 3

We

have already noted the advantage of converting an analog signal into a PCM waveform when the signal must be transmitted over a noisy channel. When a large number of such PCM signals must be transmitted over a common channel, multiplexing of these

PCM signals is required. In this section we discuss PCM waveforms used in the United States by

multiplexing methods for

common 5.13-1

carriers

The 71

(AT&T, GTE,

the the

etc.)

Digital System

Figure 5.13-1 shows the basic time division multiplexing scheme, called the Tl

which is used to convey multiple signals over telephone lines using wideband coaxial cable. It accommodates 24 analog signals which we shall refer to as Sj through s 24 Each signal is bandlimited to approximately 3.3 kHz and is sampled at the rate 8 kHz which exceeds, by a comfortable margin, the Nyquist rate of 2 x 3.3 = 6.6 kHz. Each of the time division multiplexed signals (still analog) is next A/D converted and companded as described in Sec. 5.12. The resulting digital waveform is transmitted over a coaxial cable, the cable serving to minimize signal distortion and serving also to suppress signal corruption due to noises from external sources. Periodically, at approximately 6000 ft intervals, the signal is regenerated by amplifiers called repeaters and then sent on toward its eventual destination. The repeater eliminates from each bit the effect of the distortion introduced by the channel. Also, the repeater removes from each bit any superimposed noise and thus, even having received a distorted and noisy signal, it digital system,

.

-44 -40 -36 -32 -28 -24 -20 -16 -12 -8 Figure 5.12-3 Comparison of

companded and uncompanded

-4

systems.

0

S„dB

3—

Amplitude o-

PROBLEMS

Balanced modulator

„

5.1-1. It is required to transmit

signal level

telephone messages across the United States, a 3000 mile run.

not to be allowed to drop below

is

be allowed to be larger than 15 volts in

Voiced-

how many

are to be located with equal spacings,

unvoiced

5.1-2.

A

5.1-3.

A bandpass

signal

is

The

and the signal is not to order to avoid amplifier overload. Assuming that repeaters 1

millivolt before amplification

repeaters will be required?

bandpass signal has a spectral range that extends from 20 to 82 kHz. Find the acceptable range of the sampling frequency fs .

Pulse generator

Pitch

Noise Generator

Regenerated speech

Adjustable

kHz

sampled

The

=

at a rate/,

find the ranges of/„ for

5.1-4.

and extends from /„ - 5 kHz to/0 + 5 kHz. The 25 kHz. As the center frequency /„ varies from /„ = 5 kHz to/0 = 50

signal has a center frequency /„

signal

=

v(t)

which the sampling rate

cos

5ji(

filter

samples

Signals to adjust filter

+

0.5 cos lOnt

(a)

Find the

(b)

If

maximum

allowable value for

the sampling signal

is S(t)

12

and show that

,

(c)

simplified linear predictive decoder.

=

5

T

The

between

interval

.

s

£f_ _„

<5(<

- O.lfe),

of a train of impulses, each with a different strength v£t)

sists

A

adequate.

is T,.

parameters

Figure 5.19-2

is

instantaneously sampled.

is

minimum

To

the sampled signal v£t)

=

=

Ik

- O.lfe).

h

reconstruct the signal

filter

=

i
Find

v,(l) is passed through a rectangular low-pass bandwidth to reconstruct the signal without distortion.

/„,

filter.

con-

/„ and

Find the

We have the signal n(r) = cos 2nf0 t + cos 2 x 2nf0 t + cos 3 x 2nf0 t. Our interest extends, however, only to spectral components up to and including 2f0 We therefore sample at the rate 5/ 0 which is adequate for the 2fn component of the signal. 5.1-5.

pitch signals. These are used in connection with the modulator, and pulse and noise generators, as at the encoder, to provide an input to the adjustable filter. The filter-parameter adjusting signals are also received and are used, as at the encoder, to adjust the filter characteristics for optimum voice regeneration

Not

shown

explicitly

in Figs. 5.19-1

stood

is

cally,

18 filter-adjusting signals a

and

5.19-2, but nonetheless to

be under-

that, as in the simpler

vocoder of Fig. 5.18-1, the transmitted signal must be the time-division multiplex of the individual signals to be transmitted Typirate

are employed.

t

If,

as before

40 samples/s and encode each sample value into three

R=

+

(18

=

3) signals

x 40

M

^

/*

bits,

we sample

at the

R

the bit rate

.

(a)

(b)

,9.^

S(0

M

>

R

'

Sa

'

E

Z

u^,"oBook Company, McGraw-Hill '

.

-

'

J '

Weld ° n

New

Jr '

- :

" Princi P'<* of

pp. 61-87,

Telephone Laboratories: "Telecommunications Transmission Engineering," Vol Western Electric Company, Tech. Pub., Winston-Salem, N.C., 1977.

S

N 7*1994

"

? NO " '

:

A Meth °d

Di8i,al

?CM TranSmis

(r)

The sampled

result

signal

in (a),

u,(t)

The sampled

signal ojf)

.

The bandpass I

5.2- 1.

The

m 2(t) =

If

t^t)

=

having a bandwidth

On

TDM

1

Io> 0

1

+

cos 12co 0

this basis

we

is

r is

sampled by an impulse

train

m 3 (r) =

each signal

,

to ensure reproduction without error. rj,(r)

=

S(t)u(t).

by a rectangular low-pass

by a rectangular bandpass

filtered

filter

with a bandwidth

filter

extending from 2f0 to

output.

cos 10o) 0

t

+

Tne sampled

B = 2f0 = cos

v(t)

.

cos

llrj) 0 (

signal

2 cos 2to 0

1,

v,(t)

+ =

t

and

at the

sampled by an impulse

cos 12co 0

t

S(t)v(t) is

then filtered by a rectangular

Obtain an expression
is

for the filter output.

as a bandpass signal occupying an arbitrarily narrow

find that the required

sampled

is

s

output.

system shown in Fig. 5.2-1 1,

T

obtain an expression for

filter

filter

— Wo)-

view the waveform

0.5 cos
(a)

signal

Y*--*>

filter

cos

filtered

is

.

=

+

t

time between samples,

obtained

4/0 Obtain an expression for the

*'

on Usin « a

1 " unit

II

AT&T

is

sampling rate is/,

=

0.

Discuss.

used to multiplex the four signals m,(t)

m 4 (() =

cos 4co 0

same sampling

=

cos

(o 0

f,

(.

rate, calculate the

minimum sampling

" Prentice-Hall Inc.,

What is the commutator speed in revolutions per second. Design a commutator which will allow each of the four signals to be sampled at a rate than is required to satisfy the Nyquist criterion for the individual signal. (fc)

(c)

Code PkUiP' ««•

faster

-

Three signals m„ m 2 and m } are to be multiplexed, m, and m 2 have a 5-kHz bandwidth, and has a 10-kHz bandwidth. Design a commutator switching system so that each signal is sampled at Nyquist rate.

5.2-2.

COding ° f Waveforms

passed through a rec-

is

rate/,.

Bell

tel^pp^^"^^''

cos 10Vo 0

2f0 Obtain an expression for the

Data Communication," pp. 128-165,

York, 1968.

=

c{t)

maximum

Using the

frequency band. - :

-

which the impulses

.

signal

Find the

5.1- 8. Let us

'

train in

-kT,).

{b)

low-pass

R a 2, a " d E 1 Weldon Jr "Principles of Data Communication," J!rJ? McGraw-Hill Book Company, New York, 1968.

by an impulse

filter

(a)

train, S(t)

C

v(t)

recover the part of the signal of interest, the sampled signal

= /Xr=-„<5(<

5.1-7.

REFERENCES

To

The bandpass

5.1-6.

(d)

(5

asccomplished by multiplying

is

with passband extending from 0 to slightly beyond 2f0 Write an expression for the filter output. Is the part of the signal of interest recovered exactly? If we want to reproduce the first two terms of v(t) without distortion, what operation must be performed at the very outset?

B=

2.52 kb/s

sampling

tangular low-pass

is

x 3 bits/sample

If

are of unit strength, write an expression for the sampled signal.

Englewood

Cliffs,

m3 its

,

5.3-1.

Show

function

/

that the response of a rectangular low-pass

-

<5((

^

M=

S

k/2fc )

Assume 5.3-2. o> 0

that in

Four

its

f

c

to the impulse

,

m

c

{t

numbers 0

that the

to 7 can be written using 3 binary digits (bits).

How many

bits are

5.10- 1. Consider that the signal cos 2nt

~ WW,)

-

passband the

=

signals, m,(()

is quantized into 16 levels. The sampling rate is 4 Hz. Assume sampling signal consists of pulses each having a unit height and duration dt. The pulses occur

that the

k/2fc )

every

filter

cos

1

has H(f)


1,

m 2 (t)

= =

=

t

1.

1

sin

m0

m,{t)

t,

= -

1

cos

ai 0

and

(,

m t(t) = -

1

- oo <

k/4 sec,

<

k

oo.

Sketch the binary signal representing each sample voltage,

(a)

How many bits are required A D/A converter is shown in

(fc)

sin

are sampled every \/2f0 sec by the sampling function

f

Show

5.9- 1.

required to write the numbers 0 to 5?

sin ">,('

k

with a bandwidth

filter,

is

5.11- 1.

per sample?

Using the

Fig. P5.11-1.

set-reset flip-flops

shown explain

the

operation of the device.

An A/D

5.11- 2.

converter

is

shown

in Fig. P5.11-2.

Using

trigger flip-flops as indicated explain the

operation of the device.

The

signals are then time-division multiplexed.

filter

=

having a bandwidth fe

The

TDM

(a)

Sketch the four outputs of the decommutator.

(fc)

Each of the four output

width/,,.

Show

signal

is

filtered

by a rectangular low-pass

signals

filter

that the four signals are reconstructed without error.

The four signals of Prob. 5.3-2 are sampled, as indicated in that problem, and time-division The TDM signal is filtered by a rectangular low-pass filter having a bandwidth = 2/ ft 0 and then decommutated. Sketch the output at the decommutator switch segment where the samples of m,(t) should appear and show that m,(I) cannot be recovered.

and

(a)

signal o(f) = cos
(fc)

Sketch

S(t)

=

Show

S[t)v(i) if S(t) is

£ <, nT + = 1/320/0

t/2

(fc) if

z

The

pulse duration

is

t

=

l/32/0

A commonly

(a)

x

= +

used value

is

u

=

is,

-

1

is

y- ± t—i

given by

' ± ,

0S1SI

Here the

+ - vJV where r, and v. to + V. The parameter A

shown

characteristic

=

»„

-

v,

versus v

Fourier series for the error

=

S

sin co 0

that

if

the signal

If v,

(,

v„

-

v,

is

periodic,

it

.

t

Assume

that

S=

1,

that

is,

m, +

,

-

m,

=

can be expanded in a Fourier

series.

Write the

to

x

= +

component of the

find the

Show

5.8-3.

Consider a signal having a probability density

between

+V= +

ir

levels

error e at the angular frequency cu 0

uniformly distributed, Eq. (5.8-3) results even

if

M

is

=

.

{*""""

- 4 <"<

(0

elsewhere

1

+

Do

(a)

5.8-4.

(c)

Calculate the variance of the quantization error is

+

1

(6)

S

is

Calculate

= K(l-|»|)

(a)

= -

1

to

output to input by the relations

l

+

if

there are four quantization levels.

constant over each

level.

when

Compare your

and If a

plot of the

there are four quantization levels. result

with Eq.

-1S»<1

to (c) of Prob. 5.8-3.

1

A

5.12-4.

An

(5.8-3).

A

lo g^|x| + log^

is

A =

87.6.

signal

^l* 1

for

For

this value

1

make

a plot of

y versus x from x

10 volts and 256 quantization levels are employed, what

is no compression? For between levels?

there

= -

A =

87.6 what

1

the

is

the voltage interval

minimum and

the

maximum

-lsxs+l

quantized by a quantizer that has eight quantization

-

is

signal m{t) has a probability density function

levels.

which the quantization levels should be located so that there shall between any two adjacent levels or between the extreme levels and

at

is

extreme voltages.

quantizer

compander

is

used with equal spacings between quantization levels, make an input-output must precede the quantizer to satisfy the requirement of part (a).

that

analog signal m(i) has a probability density function

/(x)=l-|x| The

|x|<;-7

determines the degree of compression.

used value

Determine the voltages

Consider a signal having a probability density /(t>)

a plot of y versus x from x

x is positive and the - sign when x is negative. Also x = v./V and are the input and output voltages and the range of allowable voltage is - V

be equal probabilities that m(f)

Find K. Determine the step

not assume that f(v)

make

relates

for

.

log

/(x)=l-|x|

4

(fc)

size

when

An analog

5.12-3.

the (a)

value

when

not large. This signal

/(„)

this

1.

effective separation

e(t>,).

5.8-2.

is

A commonly

(a)

in Fig. 5.8-lfc.

(fc)

= e =

sign applies

y

otherwise

Since the error e

(c)

For

maximum effective separation between levels? 5.12-2. An ,4-law compander uses a compressor which

volt. (fc)

255.

levels

/ 6\t) can be stretched to have a width t by passing the impulse e' lm )/jm. Show that this operation is identical with integrating the

Plot the error characteristic e

(a)

/<)

A\x\

that the output v 0 (t)

S(t) dt

For the quantizer

+

.

=0 5.8-1.

(1

V = 40 volts and 256 quantization levels are employed what is the voltage interval when there is no compression? For n = 255 what is the minimum and what is the

If

l

= ^1

by the relation

"l*l>

1.

(fc)

.

+

o,.

y v 0 (t)

'° gtl

log

between

a train of pulses having unit height, occurring at the rate/,;

t/2.

t

filter (1

r sec, that

uses a compressor which relates output to input

Here the + sign applies when x is positive and the - sign applies when x is negative. Also x = vJV and y = vJV where and v 0 are the input and output voltages and the range of allowable voltage is - V to + V. The parameter /i determines the degree of compression.

an impulse function

that

function through a

impulse for

=

t>,(()

nT -

Repeat

(c)

5.5- 1.

for

1

compander

having a band-

multiplexed.

The

/i-law

y=+

by a rectangular low-pass

is filtered

5.3- 3.

5.4- 1.

A

5.12- 1.

4/0 and then decommutated.

for

-Isis-l

applied to a quantizer with two quantization levels at ± 0.5 volts. Calculate the mean square quantization error and compare with the result that would be given if Eq (5 12-2) were applied.

is

5.12- 5. Verify that the procedure described in the text

content of the

ROM

yields 2

the smallest quantization

An

s

=

A and

and represented in Fig. 5.12-2 for setting the 256 code words for 2 12 = 4096 addresses. What is the ratio between

the largest quantization level?

NRZ

waveform as in Fig. 5.13-3 consists of alternating l's and O's. The waveform swings between + V and — V and the bit duration is T„. The waveform is passed through a RC network as in Fig. 5.13-3h whose time constant is RC = T Calculate all the voltage levels of the k output waveform for the circumstances that the NRZ waveform has persisted for a long time. (ft) For the circumstances of part (a) calculate the area under the waveform during the 5.13- 1. (a)

.

bit time consider that a sequence of 10 successive l's appears. Calculate the area under the waveform during the 10th 1 bit and compare the area calculated for the case when l's and O's alternate.

7;

.

Now

5.13-2.

the

first

Draw an AMI waveform corresponding to a pulse in the waveform may be set arbitrarily.

binary

bit

stream 1001 10010101. The polarity of

M

Two of the Tl inputs to the 12 multiplexer of Fig. 5.13-5 are generated in systems whose clocks have frequencies which are different by 50 parts/million. In how long a time will the faster clock have generated one more time slot than the slower clock. In this time interval, how many 5.13- 3.

frames

have been generated ?

will

5.14- 1. Consider the delta (a)

Explain

its

PCM system shown in

(ft)

Sketch the receiver.

(c)

If m(t)

=

Fig. P5.14-1.

operation,

0.05 sin

In,

find m(t)

and

Aft) graphically.

P

Reset pulse

Figure P5.11-1

Figure P5.14-1 5.15-1. (a)

"W = size

is

5

mV.

(c)

Write and run a computer program to solve for Repeat (b) if the samples occur every 0.1 sec.

{d)

Compare

(ft)

Figure P5.1 1-2

Write the nonlinear difference equation of the delta modulator shown in Fig. 5.15-1. Let 3 sin 2nt and let the samples occur every 0.05 sec starting at r = 0.01 sec. The step

18 x 10"

the results of (ft) and

(c).

A(t).

The input

5.15-2.

DM

to a

has a step size of 2

is

mV. Sketch

The input to a occurs when k exceeds 5.15-3.

DM

m(t)

=

The

0.01 1.

DM operates

sampling frequency of 20

at a

Hz and

the delta modulator output, A(f) and m(i).

=

CHAPTER

Prove, by graphically determining m(t\ that slope overload a specified value. What is this value in terms of the step size S and the is

m(t)

kt.

SIX

sampling frequency/,? 5.15-4. If the step size

to m(t)

2M <

if

S,

This

sampling frequency is/< A) and m(()

is

S, the

is

called step-size limiting.

,

=

M sin

cot,

DIGITAL MODULATION TECHNIQUES

explain what happens

M

The signal m(() = sin co 0 1 is to be encoded by using a delta modulator. If the step size S and sampling frequency /f> are selected so as to ensure that neither slope overloads [Eq. (5.15-1)] nor 5.15- 5.

step-size limiting (Prob.

5.

1

show

5-4) occurs,

that

/™ >

3/0

.

The adaptive delta-modulation system described in Sec. 5.16 has an input which is m(t) = 0 until time t = 0 and thereafter m(t) = 250 sin Int. An inactive edge of the clock occurs at t = 0 and the clock period is 0.05 sec. On a single set of coordinate axes draw the clock waveform, the input m(t) 5.16- 1.

1

and the approximation 5.16-2.

An

m(t).

using the following logic.

N

Extend the plot through a

adaptive delta modulator

negative pulses occurs,

If p„(t)

K

is

shown

alternates between

increases by

N;

if

full

cycle of m(t).

+

after a

1

and

-

K is variable and adjusted K= a sequence of N positive or of N pulses the polarity changes, K

The gain

in Fig. P5.16-2.

1,

sequence

1

;

is

if

|

decreases by

2.

Thus, consider the sequence

Find m(t) 0.05 sec,

and

at

if (

=

m(t)

=

sin

0.01 sec a

1, 1, 1.

Then

K=

1, 2, 3,

|

1, 2.

In. Perform the analysis graphically. Consider a sampling time of sample occurs. At this time the step size S = volt when K = 1. 1

INTRODUCTION

6.T

p m(t)

+

1

-1*

—

pn «>

When

it becomes necessary, for the purpose of transmission, to superimpose a binary waveform on a carrier, amplitude modulation (AM), phase modulation

(PM), or frequency modulation (FM) may be used. Combination AM-PM systems such as quadrature amplitude modulation (QAM) are also commonly employed.

The selection of the particular modulation method used is determined by the application intended as well as by the channel characteristics such as available DM)

bandwidth and the cation

is

intended

susceptibility of the channel to fading.

we must

When

radio

communiwe have

take account of antenna characteristics. As

already noted (Sec. 3.1) an antenna

Amplifier with gain

and Figure P5.16-2

logic circuit

K

is a narrow-band device whose operating frequency is related to its physical dimensions. When data transmission using a telephone channel is intended, we need to take account of the fact that often the channel may not transmit dc and low frequencies because of transformers which

may

be included in the transmission path. A fading channel is one in which the received signal amplitude varies with time because of variabilities in the transmission medium. When we must contend with such channels it is useful to use

FM

which

is

relatively insensitive to

amplitude fluctuations.

we need a modulator at the transmitter and, at the demodulator to recover the baseband signal. Such a modulatordemodulator combination is called a MODEM. In this chapter we present a description of many of the available modulation/demodulation techniques and compare them on the basis of their spectral occupancy. In Chap. 11 we shall complete our comparison and determine in each case the probability of error for In each of these situations

receiver, a

each system as a function of signal-to-noise ratio and bandwidth available for transmission. 249

6.2

BINARY PHASE-SHIFT KEYING (BPSK)

In binary phase-shift keying

amplitude.

It

the transmitted signal

has one fixed phase when the data

other level the phase

is

at the

it

has a power

Ps - \A 2

is

different

A =

so that

by

is

at

one

is

a sinusoid of fixed

level

and when the data

180°. If the sinusoid

is

of amplitude

*JlP,. Thus the transmitted signal

is

A

either (6.2-1)

or

(6.2-2fl)

(6.2-2b)

In

BPSK

the data

we say it is at logic Hence u BPSK (0 can be

a stream of binary digits with voltage levels which,

b(t) is

take to be at + 1 V and - 1 V. When b(t) = 1 V and when b(t) = - V we say it is at logic level 0. written, with no loss of generality, as

we

as a matter of convenience, level

1

(6.2-3)

In practice, a carrier, to a

BPSK

signal

modulating waveform. In

Reception of

The

is

generated by applying the waveform cos

balanced modulator and applying the baseband signal this sense

BPSK

can be thought of as an

co 0

b(t)

1,

as a

as the

AM signal.

BPSK

received signal has the form

"bpskW

= W)>fiP,

cos

(co 0 1

+

0)

=

b(t\/2P, cos

w 0 (t +

9/co 0 )

(6.2-4)

Here 8 is a nominally fixed phase shift corresponding to the time delay 6/w 0 which depends on the length of the path from transmitter to receiver and the phase shift produced by the amplifiers in the " front-end " of the receiver preceding the demodulator. The original data b(t) is recovered in the demodulator. The demodulation technique usually employed is called synchronous demodulation and requires that there be available at the demodulator the waveform cos (
The

received signal

is

cos

2

squared to generate the signal (co 0 t

+

0)

=

|

+

i cos 2(co 0

1

+

6)

(6.2-5)

The dc component is removed by the bandpass filter whose passband is centered around 2/„ and we then have the signal whose waveform is that of cos 2(co 0 1 + 6). A frequency divider (composed of a flip-flop and narrow-band filter tuned to /0 ) is used to regenerate the waveform cos (a> 0 1 + 6). Only the waveforms of the signals at the outputs of the squarer, filter and divider are relevant to our discussion and not their amplitudes. Accordingly in Fig. 6.2-1 we have arbitrarily taken each amplitude to be unity. In practice, the amplitudes

will

DIGITAL MODULATION TECHNIQUES 153

be determined by features of these devices which are of no present concern. In any event, the carrier having been recovered, it is multiplied with the received signal to generate

b{t)y/2P, cos

which

2

(
f

+

0)

=

b(t\JTpt

[i

+

BPSK waveform is the NRZ waveform multiplied by v/2 cos a> 0 1. Thus following the analysis of Sec. 3.2 we find that the power spectral density of the The

BPSK i cos 2(w 0

1

+

0)]

signal

is

(6.2-6)

then applied to an integrator as shown in Fig. 6.2-1. We have included in the system a bit synchronizer. This device, whose operation is described in Sec. 10.15, is able to recognize precisely the moment which corresponds to the end of the time interval allocated to one bit and the beginning is

moment, it closes switch S c very briefly to discharge (dump) the integrator capacitor and leaves the switch S open during the entire course of c the ensuing bit interval, closing switch S c again very briefly at the end of the next bit time, etc. (This circuit is called an " integrate-and-dump " circuit.) The output

k(/)

Equations

Note

is the integrator output at the end of a bit interval but immediately before the closing of switch S c This output signal is made available by switch S s which samples the output voltage just prior to dumping the capacitor. Let us assume for simplicity that the bit interval T is equal to the duration b

(6.2-8)

and

2

n(f-f0 )T„

n(f-fo)Tb

.

of the next. At that

signal of interest to us

sin

=

sin

+ .

.

n(f + f0 )Th

n(f + f0 )T„

(6.2-9) _

(6.2-9) are plotted in Fig. 6.2-2.

that, in principle at least, the

quencies and correspondingly so does multiplex signals using

BPSK,

spectrum of

G BPSK (/).

G b(f)

extends over

Suppose then that we

all fre-

tried to

using different carrier frequencies for different

baseband signals. There would inevitably be overlap in the spectra of the various signals and correspondingly a receiver tuned to one carrier would also receive, albeit at a lower level, a signal in a different channel. This overlapping of spectra

.

number n of cycles of the carrier of frequency f0 that is, n 2n = In this case the output voltage v 0 (kTb ) at the end of a bit interval extending

of an integral co 0

Tb

.

from time v<,(kTh )

=

(k

,

-

kTb is,

])Tb to

using Eq. (6.2-6)

rkn

f*n b{kTb )J2P s

±dt

+ HkT^y/lP,

\ cos 2(w 0

J(*-i>rt

1

+

6)dt

{6.2-1 a)

J<»-i>r»

(6.2-76)

since the integral of a sinusoid over a

Thus we

whole number of cycles has the value zero. demodulator output the transmit-

see that our system reproduces at the

b(t). The operation of the bit synchronizer allows us to sense each independently of every other bit. The brief closing of both switches, after each bit has been determined, wipes clean all influence of a preceding bit and allows the receiver to deal exclusively with the present bit.

ted bit stream bit

Our discussion has been rather naive since it has ignored the effects of thermal noise, frequency jitter in the carrier and random fluctuations in propagation delay. When these perturbing influences need to be taken into account a phase-locked synchronization system is called for as discussed in Sec. 10.16.

causes interchannel interference. Since efficient spectrum utilization is extremely important in order to maximize the number of simultaneous users in a multi-user communication system,

FCC and CCITT require that the side-lobes produced in BPSK be reduced below certain specified levels. To accomplish this we employ a filter to restrict the bandwidth allowed to the NRZ baseband signal. For example, before modulation we might pass the bit stream b(t) through a low-pass filter which suppresses (but does not completely eliminate) all the spectrum except the principal lobe. Since 90 percent of the power of the waveform is associated with this lobe the suggestion is not unreasonable. There is, however, the difficulty that such spectrum suppression distorts the signal and as a result, as we shall see, there is a partial overlap of a bit (symbol) and its adjacent bits in a single channel. This overlap is called intersymbol interference (ISI). Intersymbol interference can be somewhat alleviated by the use of equalizers at the receiver. Equalizers are filter-type structhe

undo the adverse effects of filters introduced, intentionally or unavoidably, at other places in a communications channel. tures used to

Geometrical Representation of Referring to Sec. 1.9

we

one orthonormal signal

BPSK

Signals

BP SK

signal can be represented, in terms of

see that a u,(t)

=

y/{2/Th ) cos

co 0

1

as [see Eq. (6.2-1)]

Spectrum of BPSK COS

The waveform

b(t) is

spectral density

between

is

a

NRZ

waveform whose power a waveform which makes excursions

ff> n

t

=

.]«,(()

(6. 2-10)

(non-return-to-zero) binary

given in Eq. (2.25-4) for

+ y/Ps and -J~PS We have Gbif)

The binary PSK

signal can then be

distance d between signals

.

= p„' T

/sin

7i/ 71

{-Hfn>

d

(6 - 2" 8)

where E b

= Ps Tb

is

drawn as shown

in Fig. 6.2-3.

Note

that the

is

= 2^/PJb = 2jEb

the energy contained in a bit duration.

(6.2-11)

We

show

in Sec. 11.13

-

u,

t)

6.2-3 Geometrical

Figure

resentation of

that the distance d

we

determine which of the

try to

rep-

signals.

we make an

inversely proportional to the probability that

is

error when, in the presence of noise, is

BPSK

levels of b(t)

being received.

6.3

DIFFERENTIAL PHASE-SHIFT KEYING

We

observed

in Fig. 6.2-1 that, in

squaring b[t)JlP s cos

— b(t) s/lP

s

cos

a> 0

mitted signal

or

b(t)

BPSK,

Accordingly,

f.

the recovered carrier

r,

shall not be able to

ro 0

we start by were instead

to regenerate the carrier

the

if

received

would remain

signal

its

negative

we

as before. Therefore

determine whether the received baseband signal

the trans-

is

— b(i).

encoded PSK which have the merit that they eliminate the ambiguity about whether the demodulated data is or is not inverted. In addition DPSK avoids the need to provide the synchronous carrier required at the demodulator for detecting a BPSK signal.

(DEPSK) which

A means

is

for generating a

To

Th

DPSK

differential

is

signal

is

shown

in

BPSK

The data

Fig. 6.3-1.

applied to one input of an exclusive-OR logic

d(t), is

the other gate input

delayed by the time

(DPSK) and

discussed in Sec. 6.4 are modifications of

stream to be transmitted, gate.

keying

phase-shift

Differential

applied the output of the exclusive or gate

allocated to one

bit.

This second input

is

then

we have drawn logic waveforms to illustrate the response input d(t). The upper level of the waveforms corresponds to logic 1, level to logic 0. The truth table for the exclusive-OR gate is given in Fig. 6.3-2

b{t)

*)o-

Delay

"dpsk(')

\

©

Balanced modulator

(U 0

jic level

-1 -1

Figure 63-1

Means

voltage

-1 1

1

-1

1

1

of generating a

DPSK

signal.

logic level

voltage

In

to an

Fig. 6.3-1

cos

Ht)

voltage

b(t) ).

the lower

cos

1

m

254

b(t)

7;

^JlP, cos

logic level

= HtW2P,

* = ± yflP,

— Th

b(t

m„ I
I

Interval no.

= 1 sometimes dence between the

and sometimes b(t) = 0. Thus there is no corresponand b(t), and the only invariant feature of the system is that a change (sometimes up and sometimes down) in b{t) occurs whenever d(t) — 1, and that no change in b(t) will occur whenever d(t) = 0. Finally, we note that the waveforms of Fig. 6.3-2 are drawn on the assumption that, in interval 1, b(0) = 0. As is easily verified, if not intuitively apparent, if we had assumed b(0) = 1, the invariant feature by which we have characterized the system would continue to apply. Since 6(0) must be either f>(0) = 0 or b(0) — 1, there being no other possibilities, our result is valid quite generally. If, however, we had started with b(0) — 1 the levels b(l) and 6(0) would have been d(t)

0

2

1

3

4

5

10

11

12

13

14


-

Th 0

b(t)

=

1

levels of d(t)

inverted.

hit)

As

Figure 6.3-2 Logic waveforms to illustrate the response

is

seen in Fig. 6.3-1

b(t)

to an input

b(t)

this table

we can

easily verify that the

are consistent with one another.

indeed

delayed by one

We

waveforms

for

- Tb and b(t - Th is

d(t), b(t

observe that, as required,

)

bit

©

waveforms (interval in Fig. 6.3-2). We cannot determine b(t) in this first interval of our waveform unless we know b(k = 0). But we cannot determine b(0) unless we know both d(0) and f>(-l), etc. Thus, to justify any set of logic levels in an

Thus altogether when beginning of the nitude

in Fig. 6.3-3.

time

Tb

b(t)b(t

we have circumvented the problem by = 0. It is shown below that in

arbitrarily assuming that in the first interval b(0)

the demodulator, the data will be correctly determined regardless of our assump-

We now

observe that the response of b(t) to d(t) is that b(t) changes level at the beginning of each interval in which d(t) = 1 and b(t) does not change level when d(t) = 0. Thus during interval 3, d(3) = 1, and correspondingly b(3) changes at the beginning at that interval. During intervals 6 and 7, d(6) = d(l) = 1 and there are changes in d(t)

=

intervals. This

1

b(t) at

the beginnings of both intervals.

and there are changes

behavior

in b(t) at the

to be anticipated

During

bits 10, 11, 12,

beginnings of each of these

from the truth table of the exclusive= b(t - Tb) so that, whatever the 0, initial valu e of fc(t - T ), it reproduces itself. On the other hand h when d(i) = 1 then b(t) = b(t - Tb ). Thus, in each successive bit interval b(t) changes from its value in the previous interval. Note that in some intervals where d(t) = 0 we have b(t) = 0 and in other intervals when d(t) = 0 we have £>(t) = 1. Similarly, when

OR

gate.

(6.3-1)

is

BPSK

=

0 the phase of the carrier does not change at the when d(t) = 1 there is a phase change of mag-

while

signal

are applied to a multiplier.

-

T„)(2P S ) cos

(<w 0 1

+

9)

The

(D 0

-

Tb +

T„)

+

signal

is

For we note that when

d(t)

=

is

shown

by the

bit

is

ff]

+

COS ^2co 0 (t

applied to a bit synchronizer and integrator as

demodulator. The

DPSK

signal delayed

multiplier output

cos [eo 0 (t COS

b(t)b(t

and the received

shown

26

(6.3-2)

in Fig. 6.2-1 for the

term on the right-hand side of Eq. (6.3-2) is, aside from a multiplicative constant, the waveform b(t)b(t — Tb ) which, as we shall see is first

we require. As noted previously in connection with BPSK, and so here, the output integrator will suppress the double frequency term. We should select w 0 Tb so that w 0 Tb = 2nn with n an integer. For, in this case we shall have cos co 0 Tb = +1 and the signal output will be as large as possible. precisely the signal

tion concerning b(0).

and 13

is

the trans-

is

= W) V2P, cos a> 0 1 — ± yJlP, cos a> 0 1

of recovering the data bit stream from the

= and

d(t)

bit interval,

Here the received

initial bit interval we need to know the logic levels in the preceding interval. But such a determination requires information about the interval two bit times earlier

Fig. 6.3-2

The modulator output, which

71.

A method

1

and so on. In the waveforms of

(-

),

time and that in any bit interval the bit b(t) is given by h(t) = d(t) b(t - Tb ). In the ensuing discussion we shall use the symbolism d(k) and b(k) to represent the logic levels of d(t) and b(t) during the fcth interval. Because of the feedback involved in the system of Fig. 6.3-2 there is a difficulty in determining the logic levels in the interval in which we start to draw the b(t)

%

d(t).

»dpsk(0

and with

applied to a balanced modulator to which

b(t) is

also applied the carrier s/lP s cos mitted signal is

b(t\/2P, cos

(to 0

(

+

i>(t)

To

Synchronous demodulator

0)

o—

(multiplier) co 0

bit

integrator

and

synchronizer

Th = 2m Delay

Ht Figure 63-3

Method of recovering data from

the

- Tb )j2P,

cos [o) 0 (I

DPSK signal.

- Tb + )

ff]

Further, with this selection, the bit duration encompasses an integral number of clock cycles and the integral of the double-frequency term is exactly zero. The transmitted data bit d(t) can readily be determined from the product b(t)b(t - Tb ). If d(t) = 0 then there was no phase change and b(t) = b(t - Tb ) both being + IV or both being - IK. In this case b(t)b(t - T =

=

d(t)

- V 1

1.

b)

then there was a phase change and either or vice versa. In either case b(t)b(t - T ) = - 1. b 1

The

differentially coherent system,

b(t)

= V

with

1

If

b(t

0

Hk -

however,

- Th =

DPSK

some

bit interval the

received signal

is

= Wt) ®Mk-

0

(a)

j

/

10 110 10

1)

one

error

*

1)

111110 0 \ 0 111110

1):

10

0

b'(k)

consider so contaminated by noise that

h'(k

d(k)

-

= b'(k)®b\k-

0 0 0 >»vJ

two

1

0

(b)

0 Figure 6.4-2 Errors

encoded

errors

PSK

occur

in

differentially-

in pairs.

may

cause errors to two bit determinations. The error rate in DPSK is therePSK, and, as a matter of fact, there is a tendency for bit errors to occur in pairs. It is not inevitable however that errors occur in pairs.

DPSK

Single errors are

manner shown

fore greater than in

kth and (k (k

l

been describing

in a PSK system an error would be made in the determination of whether the transmitted bit was a 1 or a 0. In DPSK a bit determination is made on the basis of the signal received in two successive bit intervals. Hence noise in one bit inter-

val

1)

\

I

has a clear advantage over the coherent BPSK system in that the former avoids the need for complicated circuitry used to generate a local carrier at the receiver. To see the relative disadvantage of in comparison with PSK, that during

\

110 110 0 0 110 110

)

rf(A)

DPSK, which we have

time

/ '

-

+

and

still

possible.

l)st bit intervals

For consider a case in which the received signals in are both somewhat noisy but that the signals in the

+

2)nd intervals are noise free. Assume further that the kth interval signal is not so noisy that an error results from the comparison with the (k - l)st interval signal and assume a similar situation prevails in connection with the (k + l)st and the (k + 2)nd interval signals. Then it may be that only a single error will be generated, that error being the result of the comparison of the kth and (k + l)st interval signals both of which are noisy. l)st

(k

The

transmitter of the

system shown

applied directly is

applied

b(t

-

in

DEPSK

system

Fig. 6.3-1.

The

is

identical to the transmitter of the

signal b(t)

BPSK

is

recovered in exactly the

The recovered signal is then to one input of an exclusive-OR logic gate and to the other input Th (see Fig. 6.4-1). The gate output will be at on e or the other of in Fig. 6.2-1

for a

system.

)

depending on whether b(t) = f>(f - Tb ) or b(i) = b(t - T„). In the first case b(t) did not change level and therefore the transmitted bit is d(t) = 0. In the second case d{t) = 1 We have seen that in DPSK there is a tendency for bit errors to occur in pairs but that single bit errors are possible. In DEPSK errors always occur in its

levels

is that in DPSK we do not make a hard deciabout the phase of the received signal. We simply allow the received signal in one interval to compare itself with the signal in an adjoining interval and, as we have seen, a single error is not precluded. In DEPSK, a firm definite hard decision is made in each interval about the value of b(f). If we

pairs.

The reason

for the difference

sion, in each bit interval

DIFFERENTIALLY-ENCODED PSK (DEPSK)

6.4

As

is

noted

DPSK demodulator requires a device which operates and provides a delay of Tb Differentially-encoded PSK

in Fig. 6.3-3 the

at the carrier frequency

make

.

eliminates the need for such a piece of hardware. In this system, synchronous demodulation recovers the signal b(i), and the decoding of b(t) to generate d(t) is

done

at

baseband.

the error-free signals b(k), b(k

have assumed that error.

Hi)

6.5

L

J=€> -

=

6(1)

© b(t

-

1)

and

d(k)

= b(k)®b(k - 1). (k - 1) must

has a single error. Then b

In Fig. 6.4-2b

QUADRATURE PHASE-SHIFT KEYING

(QPSK)

- Tb )

We

have seen that when a data stream whose bit duration is Tb is to be transmitBPSK the channel bandwidth must be nominally 2fh where fb = \/Tb Quadrature phase-shift keying, as we shall explain, allows bits to be transmitted .

to use the type-D flip-flop as a d(t)

from

we

also have a single

We note that the reconstructed waveform d'(k) now has two errors.

using half the bandwidth. In describing the

T„)

Figure 6.4-1 Baseband decoder to obtain

d(t)

b'(k)

ted by

1 Ht

-O

a mistake, then errors must result from a comparison with the preceding bit. This result is illustrated in Fig. 6.4-2. In Fig. 6.4-2a is shown

and succeeding

one

bit

QPSK

system we shall have occasion

storage device.

We

therefore digress, very

b(t).

briefly, to

remind the reader of the essential characteristics of this

flip-flop.

Type-D

flip-flop ing:

The

type-/) flip-flop represented in Fig. 6.5-la has a single data input terminal (D) to which a data stream d(t) is applied. The operation of the flip-flop is such that at the "active" edge of the clock waveform the logic level at D is transferred to the output Q. Representative waveforms are shown in Fig. 6.5-\(b) assume arbitrarily that the negative-going edge of the clock waveform is the active edge At the active edge numbered 1 we find that d(t) 0. Hence, after a short delay

bit,

Once it

will

the flip-flop, in response to an active clock edge, has registered a data

hold that

bit until

updated by the occurrence of the next succeeding

active edge.

We

=

(the delay

that

g=

not shown) from the time of occurrence of this edge

is

The delay

from the fact that some time input data to propagate through the flip-flop to the output the

0.

waveform

results

is

Q.

shall find

required for the

(On

shown, we have no basis on which to determine At active edge 2 it appears that the clock edge occurs

time.)

we

d(t)

Q

the basis of at

an

earlier

when ,/(*) is changing. If such were indeed the case, the response of would be ambiguous. As a matter of practice, the change in d(t) will

Such

is

occur slightly

is itself

same

type-D

flip-flop.

etc.)

to the very

(The delays referred

normally not indicated on a waveform diagram as in Fig. 6.5-1 because these delays are ordinarily very small in comparison with the bit time T ) In any b event, the fact is that at active edge 2 = we to are

have d(t) 0 and Q remains at Q = 0 The remainder of the waveforms are easily verified on the same basis. Observe that the Q waveform is the d(t) waveform delayed by one bit interval T The releb

vant point about the flip-flop in the matter of our present concern

The mechanism by which a bit stream b(t) generates a QPSK signal for transmission is shown in Fig. 6.5-2 and relevant waveforms are shown in Fig. 6.5-3. In these waveforms we have arbitrarily assumed that in every case the active edge of the clock waveforms is the downward edge. The toggle flip-flop is driven by a clock waveform whose period is the bit time Tb The toggle flip-flop generates an odd clock waveform and an even waveform. These clocks have periods 2Tb The active edge of one of the clocks and the active edge of the other are separated by the bit time Tb The bit stream h(t) is applied as the data input to both type-D flip-flops, one driven by the odd and one driven by the even clock waveform. The flip-flops register alternate bits in the stream b(i) and hold each such registered bit for two bit intervals, that is for a time 2Th In Fig. 6.5-3 we have numbered the bits in b(t). Note that the bit stream b 0 (i) (which is the output of the flip-flop driven by the odd clock) registers bit 1 and holds that bit for time 2Th then registers bit 3 for time 2Tb then bit 5 for 2Tb etc. The even bit stream b e (t) holds, for times 2Tb each, the alternate bits numbered 2, 4, 6, etc. The bit stream b e (t) (which, as usual, we take to be b e (t) = + volt) is superimposed on a carrier V/P ~ cos m 0 t and the bit stream b 0 {t) (also +1 volt) is .

the flip-flop

the case because, rather inevitably, the change in d(t)

the response of some digital component (gate, flip-flop, clock waveform which is driving our

Transmitter

.

precisely at the time

after the active edge.

QPSK

is

the follow-

.

.

,

,

,

1

(«)

Clock

nnnrmsmnniumn 3

4

X


7^1

(h)

1

J

i_n

r

i

r

i

i

r Ps

Figure 6.5-1

(a)

Type-D

flip-flop

symbol,

(fc)

Waveforms showing

flip-flop characteristics.

Figure 6.5-2

An

offset

QPSK

transmitter.

sin
t

iJiJiJTJiJijajirL^^

««*

quadrature. Also signals v m

tude

(t)

=

/lP and

s

s

drawn

s„(t)

+

se

are the phasors representing the four possible output

These four possible output signals have equal ampli-

(t).

are in phase quadrature; they have been identified by their corre-

b„ and b e At the end of each bit interval (i.e., after each time can change, but both cannot change at the same time. Consequently, the QPSK system shown in Fig. 6.5-2 is called offset or staggered QPSK

sponding values of T,,)

.

either b 0 or b e

and abbreviated OQPSK. After each time Th the transmitted signal, changes phase by 90° rather than by 180° as in BPSK. ,

Non-offset

Figure 6.5-3 Waveforms for the

QPSK

transmitter of Fig. 6.5-2.

if it

changes,

QPSK

Suppose that in Fig. 6.5-2 we introduce an additional flip-flop before either the odd or even flip-flop. Let this added flip-flop be driven by the clock which runs at the rate fb Then one or the other bit streams, odd or even, will be delayed by one bit interval. As a result, we shall find that two bits which occur in time sequence (i.e., serially) in the input bit stream b(t) will appear at the same time (i.e., in parallel) at the outputs of the odd and even flip-flops. In this case b e (i) and b„(t) can change at the same time, after each time 2Tb and there can be a phase change of 180° in the output signal. There is no difference, in principle, between a staggered and non-staggered system. In practice, there is often a significant difference between QPSK and OQPSK. At each transition time, Tb for OQPSK and 2Tb for QPSK, one bit for OQPSK and perhaps two bits for QPSK change from IV to -IV or -IV to IV. Now the bits b e (t) and b„(t) can, of course, not change instantaneously and, in changing, must pass through zero and dwell in that neighborhood at least briefly. Hence there will be brief variations in the amplitude of the transmitted waveform. These variations will be more pronounced in QPSK than in OQPSK since in the .

superimposed on a carrier ^/P, sin co 0 1 by the use of two multipliers (i.e., balanced modulators) as shown, to generate two signals s e (t) and «„(/) These signals are then

added

to generate the transmitted output signal v m vm

W

= \fPs K (t)

sin co 0

1

+

/¥ b

s

s

e (t)

(t)

cos

which

oj 0

is

,

(6.5-1)

t

As may be verified, the total normalized power of v m (t) is P, As we have noted in BPSK, a bit stream with bit time Tb multiplies a carrier, the generated signal has a nominal bandwidth 2 x \/Tb In the waveforms b (t) 0 and b e (t) the bit times are each l/2Tb hence both s e (t) and s„(f) have nominal bandwidths which are half the bandwidth in BPSK. Both s e (() and s„ (f) occupy the same spectral range but they are nonetheless individually identifiable because .

,

of the phase quadrature of their carriers.

When b„ = the signal s„(t) = sin w 0 and s 0 (t) = -y/P, sin w 0 when = - 1. Correspondingly, for b e (t) = ± 1, s e (t) = + y/F cos w 0 These four 1

b0

signals

t,

1

3

have been represented as phasors

in Fig. 6.5-4.

They

1.

are in mutual phase

first

tude

case both b e (t) and

may

b„(t)

may

be zero simultaneously so that the signal ampli-

actually be reduced to zero temporarily. There

is a second mechanism through which amplitude variations are caused at the transmitter. In QPSK as in BPSK a filter is used to suppress sidebands. It turns out that when waveforms which exhibit abrupt phase changes, are filtered, the effect of the filter, at the time

of the abrupt phase changes, tude of the waveform.

changes of is

1

Here

to cause substantial

is

too,

we

80° are possible than in

changes again

expect larger changes in

OQPSK

where the

in the ampli-

QPSK

maximum

where phase phase change

90°.

The amplitude variations can cause difficulty in QPSK communication systems which employ repeaters, i.e., stations which receive and rebroadcast signals,

such as earth

satellites.

For such stations generally employ output power

stages which operate nonlinearly, the nonlinearity being deliberately introduced

because such nonlinear stages can operate with improved efficiency. However, precisely because of their nonlinearity, when presented with amplitude variations, they generate spectral components outside the range of the main lobe, thereby Figure 6.5-4 Phasor diagram for sinusoids in Fig. 6.5-2.

undoing the

effect of the

band

limiting filter

and causing interchannel

inter-

ference.

Further

filtering to

on the phase of the

effect

suppress the effect of amplitude variation has an

signal

and

it

of course, precisely the phase which

is,

conveys the signal message.

frequency by

at twice the carrier

raised to the fourth

power

the carrier frequency and

Symbol Versus In

BPSK we

two

Bit Transmission

deal individually with each bit of duration

form what

Th

.

In

QPSK we

lump

termed a symbol. The symbol can have any one of four possible values corresponding to the two-bit sequences 00, 01, 10, and 11. We therefore arrange to make available for transmission four distinct signals. At the receiver each signal represents one symbol and, correspondingly, two bits. When bits are transmitted, as in BPSK, the signal changes occur at the bit rate. When symbols are transmitted the changes occur at the symbol rate which is one-half the bit rate. Thus the symbol time is T = 2T s h is

.

The

A

QPSK

Receiver

receiver for the

QPSK

required and hence sin to 0

is

shown

it

The scheme

problem

(see

Synchronous detection

is

is

for

which

finally

we

and recovered the

carrier

by frequency

required that the incoming signal be

is

it

filtering recovers a

waveform

at four times

frequency division by four regenerates the require both sin co 0

t

and cos

w0

f.

It is left

as

Prob. 6.5-3) to verify that the scheme indicated in Fig. 6.5-5 does

and cos w 0 t. two synchronous demodulators consisting, as usual, of a multiplier (balanced modulator) followed by an integrator. The integrator integrates over a two-bit interval of duration Ts = 2Th and then dumps its accumulation. As noted previously, ideally the interval 2Tb = Ts should encompass an integral number of carrier cycles. One demodulator uses the carrier cos w 0 1 and the other the carrier sin co 0 t. We recall that when sinusoids in phase quadrature are multiplied, and the product is integrated over an integral number of cycles, the result is zero. Hence the demodulators will selectively respond to the parts of the incoming signal involving respectively b e (t) or b a (t). indeed yield the required waveforms sin

The incoming

Of course, in Fig. 6.5-5.

necessary to locally regenerate the carriers cos
1.

BPSK.

signal

after

carrier. In the present case, also,

a

bits together to

filtering,

dividing by two. In the present case,

signal

is

co 0

as usual, a bit synchronizer

and ends of the

bit intervals

can be established. The

bit

t

also applied to

is

required to establish the beginnings

of each bit stream so that the times of integration

synchronizer

is

needed as well to operate the sampling

switch. At the end of each integration time for each individual integrator,

and just dumped, the integrator output is sampled. Samples are taken alternately from one and the other integrator output at the end of each bit time Th and these samples are held in the latch for the bit time T Each indih vidual integrator output is sampled at intervals 2Tb The latch output is the before the accumulation

is

.

«(')

= s/T b,(t) s

sin ri>„f

.

recovered >.

s(r)siiwo 0

.

/

bit

stream

b(t).

The voltages marked on Fig. 6.5-5 are intended to represent the waveforms of the signals only and not their amplitudes. Thus the actual value of the sample voltages at the integrator outputs depends on the amplitude of the local carrier, the gain, if any, in the modulators and the gain in the integrators. We have however indicated that the sample values depend on the normalized power Ps of the received signal and on the duration T of the symbol. The mechanism used in Fig. 6.5-5 to regenerate the local carriers is a source of phase ambiguity of the type described in Sec. 6.4. That is, the carrier may be 180" out of phase with the carriers at the transmitter and as a result the demodulated signals may be complementary to the transmitted signal. This situation can be corrected, as before, by using differential encoding and decoding as in Figs. 6.4-1 and 6.4-2. s

Signal Space Representation In Sec. 1.26 sin
here, s(r)

Figure 6.5-5

A QPSK

we

investigated four quadrature signals. Equation (1.26-2), repeated

r

receiver.

cos
is

(

v„(t)

= ^/2PS

cos

\m 0 + t

(2m

+

1)

^

m=

0, 1, 2,

3

(6.5-2)

These u,

(t)

were

sig nals

=

then represented

w0

V(2/T) cos

repeated here,

t

and u 2

(t)

=

i

n term s of the two orthonormal signals

^/(2/T) sin

cu 0

r.

The

result in Eq. (1.9-22),

is

and

our ability to determine a

will verify in Sec. 11.14,

bit

without error

is

mea-

sured by the distance in signal space between points corresponding to the different values of the

We

bit.

note in Fig. 6.5-6 that points which differ in a single bit

are separated by the distance


T

cos (2m

+

COS

1)


- \yP, T The

QPSK

t

where E b sin

(2m

+

1)

J

/-sinw 0 r

(6.5-3)

for

= 2jE h

h

(6.5-6)

Tb This distance Hence, altogether, we have

the energy contained in a bit transmitted for a time

the

is

same

BPSK

as for

(see Eq. (6.2-11)).

.

two in the bandcomparison with BPSK, the noise immunities of the

the important result that, in spite of the reduction by a factor of signal

t>„(f)

Eq. (6.5-1) can be put in the form of Eq. (6.5-3) by

in

setting

width required by

QPSK

in

two systems are the same. be

=

yfe cos (2m

+1)7

(6.5-4a)

6.6

and

b„

=

y/l

sin

(2m

+

-

1) t

4

M ARY BPSK we

transmit each bit individually. Depending on whether b(t) is logic 0 we transmit one or another of a sinusoid for the bit time Tb the sinusoids differing in phase by 2n/2 = 180°. In QPSK we lump together two bits. Depending on which of the four two-bit words develops, we transmit one or In

or logic

(6.5-5)

T = 2Tb

=T

Now

Fig. 1.9-9

can be redrawn as shown

s in Fig. 6.5-6, to the geometrical representation of QPSK. The points in signal space corresponding to each of the four possible transmitted signals is indicated by dots. From each such signal we can recover two bits rather than one. The distance of a signal point from the origin is which is the square root of the signal energy S s associated with the symbol, that is E = P T, = P (2T„). As we have noted earlier s S S .

show

/F

PSK

(6.5-46)

Thus

where

is

QPSK

2j¥j

=

d

1,

,

another of four sinusoids of duration

amount

27t/4

=

90°.

2Tb

the sinusoids differing in phase by

,

The scheme can be extended. Let

that in this N-bit symbol, extending over the time

symbols.

which

NT

b

us ,

lump together

there are 2

N

=

so

NT

Now

differ

N bits

M possible

= Ts let us represent the symbols by sinusoids of duration b from one another by the phase 2n/M. Hardware to accomplish such

M-ary communication is available. Thus in M-ary PSK the waveforms used vm

(t)

= JlPs

cos

(co 0 t

to identify the symbols are

+ 4> J

(m

=

0, 1,

...,M-

1)

(6.6-1)

with the symbol phase angle given by

4>

=

m

(2m

+

1)

^

(6.6-2)

The waveforms of Eq. signal Ui{t)

=

(6.6-1) are represented by the dots in Fig. 6.6-1 in a which the coordinat e axe s are the orthonormal waveforms ^/{2/Ts ) cos w 0 t and u 2 (t) = y/(2/Ts ) sin w 0 t. The distance of each dot

sp ace in

from the origin

From

is

^/~E S

Eq. (6.6-1) v m (t)

=

*J

s

T

s

.

we have

= (JlP,

cos

4> m )

cos

m0 1

(y/2P, sin
m0

1

(6.6-3)

Defining p e and p 0 by p„

= y/2Ps

Po

=

cos

s/ lp s si"

4> m

m

(6.6-4a) (6.6-46)

are multiplied by a carrier, the resultant spectrum

centered at the carrier

is

fre-

quency and extends nominally over a bandwidth

5=

^=

2/,

=

2^

(6.6-10)

thus note that as we increase the number of bits N per symbol the bandwidth becomes progressively smaller. On the other hand as we can see from Fig. 6.6-1 the distance between symbol signal points becomes smaller. We readily calculate

We

(using the law of cosines) that this distance

d

where E,

is

= J*E

S

sin

the symbol energy

2

is

= J^NE*

(n/M)

sin

2

(n/2

N

(6.6-11)

)

P s x (NTb) = PS TS = NE b and E b = Ps Th is Thus as we increase N, i.e., as we increase

the

the energy associated with one bit. as and. decreases distance d the decreases, bandwidth the symbol, duration of the we shall see, the probability of error becomes higher. Such is the case for all increases in

N

except for the increase from

N=

1

(BPSK)

to

N=

2

(QPSK).

M-ary Transmitter and Receiver ;

The

sin oj 0 /

physical implementation of an

ately elaborate. Figure 6.6-1 Geometrical representation of M-ary

PSK

we signals.

Such hardware

shall describe the

As shown

is

M-ary

PSK

transmission system

M-ary transmitter-receiver somewhat

in Fig. 6.6-2, at the transmitter, the bit

b(t) is

applied to a

Pe cos co 0

-

1

p0

sin co 0

1

(6.6-5)

Both p e and p 0 can change every T = NT„ and can assume any of s possible values. The quantities p e p 0 and

M

,

G e(/) and

^/) =

N

on

N

output lines of the converter, that

verter output remains

is

they are presented in parallel. The con-

unchanging for the duration NTb of a symbol during which assembling a new group of N bits. Each symbol time the

time the converter

is

converter output

updated.

,

—-— =

^

is

The converter output

is

applied to a

2PS Ts

= 2PI TJ

cos

J

2

0,,

^

(6.6-6)

1

sin^:(^J

D/A

converter. This

D/A

converter gen-

M

different values in a onewhich assumes one of 2" = applied to its input. That is, symbols possible the to-one correspondence to

erates an output voltage

M

(6.6-7) Sinusoidal

However, since 0 m

is

uniformly distributed

signal Serial

Digital

to

cos

so that

bits of a

symbol. The N bits have been presented serially, that is, in time sequence, one after another. These N bits, having been assembled, are then presented all at once

Eq. (6.6-3) becomes

=

moder-

superficially.

stream

serial-to-par a\\e\ converter. This converter has facility for storing the

v m(t)

is

only of incidental concern to us in this text so

2 4> m

=

sin

2

G e (f) = G 0 (/) = Ps Ts

As we have already noted, when

=

\

(6.6-8)

signals with spectral density given

parallel

analog

phase

converter

converter

controlled

N-

(6.6-9)

(^jfj by Eq.

(6.6-9)

source

to

Figure 6.6-2 M-ary

PSK

by v[SJ I

transmitter.

Output

—

D/A

the

M—

1).

output

is

a voltage v(S m ) which depends on the symbol S m (m

Finally r>(SJ

=

0, i,

.

.

.

applied as a control input to a special type of constant-

is

amplitude sinusoidal signal source whose phase

change once per symbol time.

The The

receiver,

shown

in Fig. 6.6-3 is similar to the nonoffset

QPSK

receiver.

carrier recovery system requires, in the present case a device to raise the

Mth

received signal to the

M. As was

divide by

power,

filter

to extract the

Mf0

component and then

the case in Fig. 6.5-5 the signals indicated in Fig. 6.5-5 are

intended to represent the waveforms only and not the amplitude. Since there is no staggering of parts of the symbol, the integrators extend their integration over

same time

the

interval.

Of

grator outputs are voltages

is needed. The intewhose amplitudes are proportional to Ts p e and Ts p„ the symbol rate. These voltages measure the com-

course, again, a bit synchronizer

respectively and change at ponents of the received signal

and cos

w 0 t.

in the directions of the

Finally the signals

T p e and T p0 s

s

quadrature phasors sin

co 0

t

are applied to a device which

reconstructs the digital N-bit signal which constitutes the transmitted signal.

may

There fact,

or

may

not be a need to regenerate the

bit

stream. As a matter of

the idea of transmitting information one bit at a time by a bit stream

when we have a

b(t)

BPSK

which can handle only one bit at a time. If, on the other hand, our system handles M-bit symbols, then the data may originate as M-bit words. In such a case the serial to parallel converter shown in arises

Fig. 6.6-2

is

system, like

not needed.

Current operating systems are

common

in

which

M=

16.

In this case the

2fh/4 = fJ2 in comparison to B = fb for QPSK. PSK systems transmit information through signal phase and not through signal amplitude.

bandwidth

is

B=

Hence such systems have

great merit in situations where, on account of the vagaof the transmission medium, the received signal varies in amplitude (i.e., fading channels). ries

6.7

QUADRATURE AMPLITUDE SHIFT KEYING

(QASK)

JW-ary PSK we transmit, in any symbol interval, one which are distinguished from one another in phase but are all of the same amplitude. In each of these individual systems the end points of the signal vectors in signal space falls on the circumference of a circle. Now we have noted that our ability to distinguish one signal vector from another in the presence of noise will depend on the distance between the vector end points. It is hence rather apparent that we shall be able to improve the noise immunity of a system by allowing the signal vectors to differ, not only in their phase but also in amplitude. We now describe such an amplitude and phase shift keying system. Like QPSK it involves direct (balanced) modulation of carriers in quadrature (i.e., In

BPSK, QPSK, and

signal or another

270

symbol time symbol time synchronizers (not shown) are required. input bits are recovered by using A/D converters.

and, of course,

integrators have an integration time equal to the

7^

as usual,

Finally, the orig-

inal

Spectrum of

BFSK

In terms of the variables

»bfsk(0 6.8

BINARY FREQUENCY-SHIFT KEYING (BFSK)

In binary frequency-shift keying

the binary data

waveform

d(t)

generates

a binary signal

= +1

—1

Here

d(t)

form.

The transmitted

or

or

= JlP,

cos [co 0

+

t

corresponding to the logic levels signal

is

of amplitude

^/lP s and

signal

is

and 0 of the data wave-

1

is

s„(t)

=

^/2P S cos

(co 0

+

fl)t

(6.8-2)

v BFSK (t)

=

s L {t)

=

Jlf cos

(co 0

-

0)t

(6.8-3)

s

a> 0

—

shall co 0

—

fl

with

Q).

Q

a constant offset

the higher frequency

call

We may

generated in the manner indicated in Fig. 6.8-1.

t

+

+ JlP

e„)

is

p L cos (w L

s

t

+

6 L)

(6.8-4)

BPSK

case, b(t)

is

bipolar,

alternates between

i.e., it

+ 1 and —

1

is

(6.8-5) p'H

In Eq.

complementary.

and

"bfskM

= /y

modulators are used, one with carrier a>„ and one with carrier w L The voltage values of p„(t) and of p L (t) are related to the voltage values of d(t) in the following

(6.8-5a)

PlM - i +

(6.8-5?))

i

PlW

L are bipolar, alternating

+

is

r

balanced

PM = h + i P'M p'

When

conceive that

Two

(oj„

signal

while in the present case pH and p L are unipolar, alternating between + 1 and 0. We may, however, rewrite p H and p L as the sums of a constant and a bipolar variable, that

either

=

.

BFSK

1

"bfskM

or and thus has an angular frequency a> 0 + from the nominal carrier frequency co 0 We to H ( = w 0 + Q) and the lower frequency
(6.8-


Ph cos

s/2~P s

BFSK

where we have assumed that each of the two signals are of independent and random, uniformly distributed phase. Each of the terms in Eq. (6.8-4) looks like the signal s/lPs b(t) cos co 0 1 which we encountered in BPSK [see Eq. (6.2-3)] and for which we have already deduced the spectrum, but there is an important difference. In the

v bfskW

=

p„ and p L the

cos (w„t

1,

+

p'

L

6„)

= — 1 and

+

between

vice versa.

—

and

1

We have

1

and are

then

If

+

I

-j cos (w L t

+

0 L)

.

p + /y

manner

+ 1V

PhW + 1V

-IV

ov

d(i)

ov + 1V

Thus when d(t) changes from + 1 to - 1 p H changes from 1. At any time either p„ or p L is 1 but not both so that either at angular frequency co H or at w L

1

to 0

and p L from 0 to

+

0 H )+

/y

pi,

cos

(co^

t

+

f?

L)

(6.8-6)

The first two terms in Eq. (6.8-6) produce a power spectral density which consists of two impulses, one at fH and one at fL The last two terms produce the spectrum of two binary PSK signals (see Fig. 6.2-2a) one centered about fH and one about/,,. The individual power spectral density patterns of the last two terms in Eq. (6.8-6) are shown in Fig. 6.8-2 for the case fH —fL = 2fh For this separation between /„ and fL we observe that the overlapping between the two parts of the spectra is not large and we may expect to be able, without excessive difficulty, to distinguish the levels of the binary waveform d{t). In any event, with this separation the bandwidth of BFSK is

is

BW(BFSK) =

p„(f) cos
which

twice the bandwidth of

is

(6.8-7)

4f„

BPSK.

w„ t Adder

"bfsk(0

Receiver for COS

t

.

the generated signal

I cos

cos (co„

.

.

IP,

Ip P'H

C0 L

T

[

A BFSK

^PipiW cos
t

t

at

JP,Tb pL(t) Figure 6.8-1

A

representation of a

signal

is

BFSK

signal

is

Signal

typically

demodulated by a receiver system as in filters one with center frequency

applied to two bandpass

fL Here we have .

assumed, as above, that f„

—fL =

frequency ranges selected do not overlap and each manner

in

which a

BFSK

signal can be generated.

enough

to

encompass a main lobe

in the

spectrum of Fig.

The

at /„ the other

= 2fh The filter has a passband wide

2(Q/2n)

filter

Fig. 6.8-3.

6.8-2.

.

Hence one

filter

Power

spectral density

pass nearly

will

G FSK (/)

all

the energy in the transmission at/„ the other will perform

similarly for the transmission at/,..

The

filter

outputs are applied to envelope

detectors and finally the envelope detector outputs are

compared by a comparacomparator is a circuit that accepts two input signals. It generates a binary output which is at one level or the other depending on which input is larger. Thus at the comparator output the data d(t) will be reproduced. tor.

A

When noise is present, the output of the comparator may vary due to the systems response to the signal and noise. Thus, practical systems use a bit synchronizer and an integrator and sample the comparator output only once at the

end of each time interval

7^.

Geometrical Representation of Orthogonal

We

BFSK

M-ary phase-shift keying and in quadrature-amplitude shift keying, any signal could be represented as C,u,(f) + C u (t). There Ul (t) and u 2 (t) 2 2 are the orthonormal vectors in signal space, that is, u (t) = /2/f cos w 0 t and t s s »2(0 = \/2/Ts sin w 0 t. The functions u, and u 2 are orthonormal over the symbol interval Ts and, if the symbol is a single bit, T = T„. The coefficients C, and C s noted, in

that

Figure 6.8-2

The power

spectral densities of the individual terms in Eq. (6.8-6).

2

The normalized energies associated with C^^t) and with C u (t) 2 2 are respectively C\ and C\ and the total signal energy is C\ + C\. In M-ary PSK and QASK the orthogonality of the vectors u, and u results from their phase 2 are constants.

B=

quadrature. In the present case of BFSK it is appropriate that the orthogonality should result from a special selection of the frequencies of the unit vectors. Accordingly, with m and n integers, let us establish unit vectors 2/,

- cos 2nmf„t

"tW=

(6.8-8)

Envelope detector

and

" 2 (0=

cos 2nnf„t

(6.8-9)

Filter

IIP, cos

(<»„

r

+

rf(r)nr)

in

o———

which, as usual,/,,

=

1/7;.

The vectors u t and u 2 are the mth and nth harmofh As we are aware, from the principles of

nics of the (fundamental) frequency

.

Fourier analysis, different harmonics (m the fundamental period T b -\lfb

Filter

±

n)

are orthogonal over the interval of

a

BFSK

.

JU 1

1

1

1

1

1

now

Envelope

If

detector

(assuming

the frequencies /„

m>

and

fL

in

system are selected to be

n)

fa

=

mfb

(6.8-

A = nf„

Wa)

(6.8-1 Oft)

then the corresponding signal vectors are

Figure 6.8-3

A

receiver for a

BFSK

signal.

MO = y/% Mf)

(6.8-1 la)

= s/%

(6.8-1 lb)

«t(0

u 2 (t)

MO

Using Eq. (6.8- 13ft) we first determine s 12 by multiplying both tion by u,(t) and integrating from 0 < t < Tb The result is:

sides of the equa-

.

«i2

= \f^P _

s,(f)

\

\

E„ «11

\

\

\ \ = j2E d

\

\

\

To

Ui(t)

L

(

-

(w„

°H

f

-

cos

co L

t

dt

+ co L )Tb + m L )Tb

sin (co„

ft o>i)T b

(m„

(D L)T„

we have used

=

(6.8- 16a) \

Eq. (6.8- 13a) where

(-JlPJs,,) cos co„t

(6.8-166)

Finally, s 22 is found from Eq. (6.8-136) by squaring both sides of the equation and then integrating from 0 to Tb Since u, and u 2 are orthogonal, the result is:

\

.

MO

Figure 6.8-4 Signal

MO

»

j"

sin

h

A

VE

arrive at this result

["

s

ation of orthogonal

space represent2 s L{t) dt

BFSK.

= 2P

cos ,

S

2 ct)

L

t

dt

=

sl 2

+

(6.8- 17a)

s 22

Jo

o

Hence

The

space representation of these signals

signal

is

in

The

Fig. 6.8-4.

signal

end

therefore

= ^2E b

d

Note

shown

The distance between

signals, like the unit vectors are orthogonal.

points

is

that this distance

points of

BPSK

signals,

is

(6.8-12)

considerably smaller than the distance separating end

The distance d between s„ and s L given in Eq. (6.8-14) can now be determined by substituting Eqs. (6.8-15), (6.8-16a), and (6.8-176) into Eq. (6.8-14). The result is:

which are antipodal. 2

Geometrical Representation of Non-Orthogonal

When

the

two

procedure can

FSK

and

signals s„(t)

s L (t)

are not orthogonal, the

be used to represent the signals of Eqs. Let us represent the higher frequency signal s H (t) as s„(t)

= s/lP

and the lower frequency s L (f)

= ^/2P

s

w„t =

cos

<*bfsk

[" sin (w 2w„ T„ l _ H - w L )Tb Zt + 2w H Tb J "l (a> H -co L )Tb

sin

+

sin

Gram-Schmidt

and

(w H

+

(a>„

+

2o>,

Tk

a> L )Tb

co L )Tb

~\

J

(6.8-3).

(6.8-13a)

b

S

cos

co L t

=

we

s,

s i2 )

2

+

s 22

0

u 2 (t)

+

2

< < Tb t

shown in distance separating s H and s L is:

signals in signal space

see that the

= (sn -

ju^f)

s 22

= s 2„ -

2s„s, 2

+

is

2

s, 2

+

s\ 2

(6.8-136) signal s L

Fig. 6.8-5.

(6.8-14)

In order to determine d^ ¥SK when the two signals are not orthogonal we s 12 and s 22 using Eqs. (6.8-13). From Eq. (6.8-13a) we have:

evaluate s„,

[]

signal s L (t) as:

The representation of these two Referring to this figure

(6.8-2)

0
s, ,u,(t)

tb

l

BFSK

still

-

must

,

si,

= 2PS

P

cos

2

a> H

tdt

=

E„ [l

+

""^y* ]

(6-8-15)

u,

Figure 6.8-5 Signal

BFSK when

s H (()

and

space

representation

s L (I) are

of

not orthogonal.

Equation (6.8-18) can be simplified by recognizing that: sin

2wH Tb

<

2co H T„ sin 2co L

sin

and

(m H

(u)„

The

final result is

+

+ m L )Tb

Tb

<

sin

I

(o L )Tb

1

1

(w„

(m„

-

-

(o L )Tb

\

co L )T„

then

.^2EJi-

si

"^-y

i

(6 8 . 19) .

Note that when s„(i) and s L (f) are orthogonal (m„ - w )T„ = 2n(m L n)fb T„ 2n(m - n) and Eq. (6.8-19) gives d = /2E as expected. Note also s that b (r% - m L )Th = 3n/2, the distance d is increased and becomes

= if

11/2

(+£)] an increase

6.9

in

2 rf

(6.8-20)

by 20 percent.

COMPARISON OF BFSK AND BPSK

Let us start with the for the cosine of the sin 6

= -sin ( -

6)

"bfskW

BFSK

signal of Eq. (6.8-1).

Using the trigonometric

identity

-

6) while

sum of two angles and recalling that cos 9 = cos we are led to the alternate equivalent expression

= y/2P

s

cos

flr

cos

w0 1 -

^/IF,

d(t) sin fir sin co 0

(

1

(6.9-1)

Note

that the second term in Eq. (6.9-1) looks like the signal encountered in BPSK i.e., a carrier sin 0 1 multiplied by a data bit d(t) which changes the carrier phase. In the present case however, the carrier is not of fixed amplitude but rather the amplitude is shaped by the factor sin fir. note further the presence of a quadrature reference term cos Sit cos co 1 which contains no information.

w

We

0

Since this quadrature term carries energy, the energy in the information bearing term is thereby diminished. Hence we may expect that BFSK will not be as effective as BPSK in the presence of noise. For orthogonal BFSK, each term has the same energy, hence the information bearing term contains only one-half of the total transmitted energy.

6.10

M-ARY FSK 3

An M-ary FSK communications system extension of a binary

ed each

Ts

to

an

FSK

iV-bit

is

shown

in Fig. 6.10-1. It is

an obvious

system. At the transmitter an N-bit symbol is presentconverter. The converter output is applied to a fre-

D/A

283

quency modulator, i.e., a piece of hardware which generates a carrier waveform whose frequency is determined by the modulating waveform. The transmitted signal, for the duration of the symbol interval, is of frequency /„ or/, ... orfM _ = 2 N At the receiver, the incoming signal is applied to with paralleled bandpass filters each followed by an envelope detector. The bandpass filters have center frequencies f0 ,fi, ,/m - 1 The envelope detectors apply their outputs to a device which determines which of the detector indications is the largest and

M

•

•

•

transmits that envelope output to an N-bit

As we

,

M

.

A/D

converter.

shall see (Sec. 11.15) the probability of error

,fm-i

frequencies f0 ,fi,

so that the

commonly employed arrangement simply successive even harmonics of the

quency, say /„, is/0

=

One

provides that the carrier frequency be

=

1/TS Thus the lowest

(k

+

symbol frequency/,

= (k + 2)/,/2 =

kf„ while/,

minimized by selecting

is

M signals are mutually orthogonal. .

4)/s

,

etc.

fre-

In this case,

the spectral density patterns of the individual possible transmitted signals overlap

manner shown

in the

in Fig. 6.10-2,

which

is

an extension to M-ary

FSK.

pattern of Fig. 6.8-2 which applies to binary

FSK

M-ary

Since fs

Note

the required spectral range

= fb/N

that

and

M-ary

decreases as

increases with

FSK

of the

observe that to pass

is

B = 2Mfs

(6.10-1)

B = 2 N+l f„/N

(6.10-2)

M = 2* we have

FSK

requires a considerably increased

son with M-ary PSK. However, as we shall

FSK

We

M

increases, while for

bandwidth

in

compari-

see, the probability of error for

M-ary PSK,

M-ary

the probability of error

M.

Geometrical Representation of M-ary

FSK

we provided a signal space representation for the case of orthogonal FSK. The case of M-ary orthogonal FSK signals is clearly an extension of

In Fig. 6.8-4,

binary

this figure.

We

simply conceive of a coordinate system with

M mutually orthog-

onal coordinate axes. The signal vectors are then parallel to these axes. The best

we can do and as

is

pictorially

is

the three-dimensional case

normalized signal energy. Note that, as in Fig. points

in Fig. 6.10-3.

As

usual,

6.8-4, the distance

is

the

between signal

is

d

Note

shown

indicated in the figure, the square of the length of the signal vector

that this value of d

is

16-QASK.

(6.10-3)

greater than the values of d calculated for

with the exception of the cases case of

= s/2Es ^^/2NEh

M = 2 and M =

4. It is also

M-ary

PSK

greater than d in the

The

JP

spectral density of

S

w0

b e {t) cos

S

is

1

generated by translating the two-

+/„ and also to -/„, each such translated pattern being reduced in magnitude by 4 because the multiplication of ^/p] />„{() by cos o) 0 1 translates one-half of the voltage spectrum to /„ and the other half to —f0 (See Sec. 3.2.) Thus the voltage of each term is reduced by 2 and the power by 4. The power spectral density of the second term in Eq. (6.11-1), that is the term ^/F, b 0 (t) sin co 0 t is identical to the density of the first. Finally we note that the two terms are not correlated since b e(t) and b„(t) are quite independent of one sided pattern of Eq.

(6.1 1-3)

to

.

Hence

another.

Goqpsk(/)

we

MINIMUM

2n(f-f0 )/f„J 2n(f-f0 )/f„ J

FSK (M =

M-ary

when

3)

the frequencies

If,

fo as

say,

we took

we

minimum shift keying, we shall want to make a number of compariMSK and QPSK. One of these comparisons will concern the spectra of the two systems. For this reason we review briefly some matters consons between

^oqpskW

To

find the

power

= J~P

S

In offset

b e (t) cos

QPSK

density of the baseband

random sequence of

the transmitted signal

w 0 + y/F t

s

spectral density of this signal

we

b„(t) sin

Ts - 2T„, and having an amplitude + ^/F,. given by the general formula (Eq. (2.24-14))

G P(/) =

1

w0

is

Tsee

(6.1 1-1)

t

start with the

power

spectral

=

JT b

(()

waveform b e {t). The waveform

rectangular waveforms,

p(t)

is

a

topped for symbol duration power spectral density G(f) is

flat

Its

^

+

[sin 2n(f + f0 )fl

L 2n(f + f0 )/fh

(6.11-4)

the pattern for the

the attitude that

establish additional channels at carrier frequencies f'a

difficulty,

baseband

signal.

range

i.e.,

= f0 ±fb

,

then

channel, having a peak value at frequency

the wide spectrum of

components

rise to spectral

QPSK,

is

due to the character of the

at high frequencies. In short, the

baseband spectral

very large and multiplication by a carrier translates the spectral pattern

is

its form. We might try to alleviate the difficulty by passing the baseband signal through a low-pass filter to suppress the many side lobes. Such filtering will cause intersymbol interference. The problem of interchannel interference in QPSK is so serious that regulatory and standardization agencies such as the FCC and CCIR will not permit these systems to be used except with bandpass filtering at the carrier frequency (i.e., at the transmitter output) to

without changing

suppress the side lobes.

The

filtering

which we have just described does not, in certain situations, necproblem of interchannel interference. We shall discuss the

essarily resolve the

We

recall that

QPSK (staggered

or not)

is

a system in which

of constant amplitude, the information content being borne by phase changes. In both QPSK and there are abrupt phase changes in the the signal

where P(f) is the Fourier transform p(t). We readily calculate that the two-sided power spectral density of p(t) is, with/ = \/T s s =fb/2 and P, = EJTh

first

This signal consists of abrupt changes, and abrupt changes give

matter qualitatively: (6.11-2)

the

3fJ4, will be a source of serious interchannel interference. These side lobes, to be noted in Fig. 6.2-2, are smaller than the main lobe by only 14 dB.

is

This

In discussing

QPSK.

,

power spectral density (see bandwidth of QPSK (QPSK and OQPSK yield the same result) is B — fb because such a bandwidth is adequate to encompass the main lobe. The main lobe contains 90 percent of the signal energy. Still, the not inconsiderable power outside the main lobe is a source of trouble when QPSK is to be used for multichannel communication on adjacent carriers.

SHIFT KEYING (MSK)

cerning the spectrum of Eq. (6.5-1)]

twice the density generated by either

sin

- Eh

the side lobe associated with the

6.11

is

find

upon examining

Earlier,

Fig. 6.2-2)

Figure 6.10-3 Geometrical representation of orthogonal are selected to generate orthogonal signals.

power density

the total

term. Altogether then

is

OQPSK

QPSK

symbol rate 1/TS = l/2Th and can phase changes of 90° can occur at the bit rate. turns out that when such waveforms with abrupt phase changes, are fil-

signal. In

these changes can occur at the

be as large as 180°. In

Now

it

OQPSK

tered to suppress sidebands, the effect of the

phase changes,

is

filter,

at the times of the

abrupt

to cause substantial changes in the amplitude of the waveform.

Such amplitude variations can cause problems

in

QPSK

communication systems

The

6.5-1.

v "

Duobinary decoder

"(

)

I

Eq. 6.5-

1

stream

bit

Repeat Prob.

6.5-2.

be transmitted using

b(t) is to

OQPSK.

If h(t) is

00101001 1010 sketch

v

It)

(see

Assume/ = fJ2 = /„

).

odd

6.5-1 if the

6.5-3. Verify that the circuit

shown

stream

bit

is

delayed by 7; so that

in Fig. 6.5-5 yields sin a> 0

1

QPSK

and cos

ro„

is

generated.

each with the same phase

r,

as the received signal. "oil)

O-

Show

6.5- 4. Verify Eq. (6.5-5).

6.6- 1. 8-ary

PSK

that the

minimum

used to modulate the

is

bit

distance between signals

given by Eq.

is

(6.5-6).

stream 00101001 1010. Sketch the transmitted waveform

'W(f). Assume/, =fji =/„. Duobinary decoder

v Tr(t)

Repeat Prob.

6.6-2.

6.6-1

if

16-ary

PSK

is

used.

Assume/, =/,/4 =/„.

6.6-3. Verify

Eq. (6.6-9) taking note of the fact that <*„ has the discrete values given by Eq. (6.6-2) and does not take on all values from - n to n.

6.6-4. Verify Eq. (6.6-11). 6.6- 5. If

M becomes large, in

6.7- 1. Verify Eqs. (6.7-2)

The

6.7-2.

waveform.

Eq. (6.6-1

stream 001010011010

bit

1),

show

that d approaches zero.

(6.7-3). is

be transmitted using 16-QASK. Sketch the transmitted

to

Assume/ = /,/4 =/„.

6.7-3. Verify Eq. (6.7-5).

Narrowband

:17b-

and

filter

6.7-4. Verify Eq. (6.7-11). 6.7-5. Verify Eq. (6.7-12).

6.7- 6. Verify Eq. (6.7-15). Calculate the

quency COS

0) 0

The bit stream 001010011010 Assume/^ = /, and /, = 2f„.

6.8- 1. t

form.

Frequency sin

divide

tion R(t)

Figure 6.15-2

QPR

One way

6.8-2.

<•>„ t

by 4

average power contained

in the signal c QASK (f) at the fre-

4/„.

function

R(z).

= E[,HMt +

The

Find R(t) for the

(b)

If

=

(I,

to be transmitted using

waveform

spectral density of a

spectral

density

BFSK. Sketch

the

is

is

to

first

Fourier

the transmitted wave-

obtain the autocorrela-

transform

of

R(r)

Since

r)].

(o)

0„

power power

to obtain the

is

BFSK

find R(r)

signal of Eq. (6.8-6)

and the power

and

verify Fig. (6.8-2).

spectral density.

6.8-3. Verify Eq. (6.8-12).

decoder.

6.8-4. Veriry Eqs. (6.8-15).

takes place; each duobinary decoder being similar to the decoder shown in Fig. 6.14-3/7 except that they operate at /„/2 rather than at/,,. The reconstructed

data

d„(t)

and d e (t)

is

then combined to yield the data

6.8-5. Verify Eqs. (6.8- 16o)

and

6.8-6. Verify Eq. (6.8-19).

What

is

the relationship between

f„,fL and

f„

in order that

Eq

(6 8-19)

reduce to Eq. (6.8-12). 6.10- 1. Consider using 4-ary

d(t).

(6.8-176).

FSK.

(a)

Show

(fe)

Calculate the bandwidth

that

if

the frequencies are separated

B under

by / they are each orthogonal, ,

the condition of

(a).

Show

that the

bandwidth

is

five-eighths

of the value required by Eq. (6.10-1). Explain the difference.

PROBLEMS

(c)

Determine the

separated by /, 6.2-1

The data

consists of the bit stream 001010011010.

and sketch

t>

and

Assume

that the bit rate/t

plot the

in the

power

spectral density

The

Show

that b(t)b(t

6.4- 1.

The

Show

that after decoding, using the circuit of Fig. 6.4-1, the data d(t)

fourth bit in

- Th

stream

f>(r) is

d(t) is to

equal to the

be transmitted

G b {f)

bandwidths of 2-ary

FSK

and 4-ary

FSK when

the frequencies are

2/,

Eq. (6.2-8) as

yields the original data.

)

d(f) is to

in error,

6.11-4.

is

00101001 1010 sketch

Repeat Prob. 6.11-3 ifm

=

v MSK (t).

fifth

DEPSK.

data

If d{t) is

bits, d(f), will

001010011010, determine is

recovered.

Show

also be in error.

that

if

b(<).

the

Assume

in

Eq.

(6.1 1-12)

that

m=

5.

3.

6.11-5. Verify Eq. (6.1 1-18).

6.11- 6. Verify that in Fig. 6.11-5« the

be transmitted using

then the fourth and

6.11-2. Verify Eq. (6.1 1-8) using Eq. (6.11-12).

6.11-3. If b{t) in

when/= 0, P f = Oand when /is infinite Pt = P,. using DPSK. If d{t) is 001010011010, determine b(t).

6.3- 1.

stream

is

BPSK (r).

power P f contained a function of the frequency range 0 to/ Note that 6.2- 2. Calculate

bit

ratio of

Repeat for

6.11- 1. Verify Eq. (6.1 l-6a). b(t)

carrier frequency /„

bit

.

s

W)-

6.12- 1. Verify Table 6.12-1. 6.12-2. Verify Eqs. (6.12-7)

and

(6.12-8).

waveforms

x(f)

and

y(t)

are as

shown and

that the output

6.12-3. Verify Eqs. (6.12-10) 6.12-4. Plot

G^f) and

and

(6.12-11).

Gjif) as a function of frequency using Eqs. (6.12-10) and

(6.12-11).

CHAPTER

6.14- 1. Verify Eq. (6.14-5). 6.15- 1.

Show

that

if

two duobinary signals v Tt = b,(k) b e (k which are in time quadrature so that

1)

and

o r-

=

b,(k)b.(k

-

1)

are

SEVEN

modu-

lated using carrier signals

"oft)

= -fP, VT.(t)

then the carrier signals cos(o 0

(

and

cos

sin co 0

f

a> 0

1

+ JF, VTm(t)

sin

can be recovered

in

w0

1

MATHEMATICAL REPRESENTATION OF NOISE

the receiver by filtering the

signal Kj(f). 6.15-2.

Repeat 6.15-1

if

VTr (t) and FT

(()

are each filtered by a "brick wall" low-pass-filter prior to

being amplitude modulated.

communication systems discussed in the preceding chapters accomplish same end. They allow us to reproduce the signal, impressed on the communi-

All the

the

cation channel at the transmitter, at the demodulator output. Our only basis for comparison between systems, up to this point, has been bandwidth occupancy, convenience of multiplexing, and ease of implementation of the physical hardware. We have neglected, however, in our preceding discussions, the very important and fundamental fact that, in any real physical system, when the signal

voltage arrives at the demodulator, it will be accompanied by a voltage waveform which varies with time in an entirely unpredictable manner. This unpredictable voltage waveform

such a waveform

now

find that

is is

a

random process

called noise.

A

signal

accompanied by

described as being contaminated or corrupted by noise.

we have

a

new

basis for system comparison, that

is,

We

the extent to

which a communication system is able to distinguish the signal from the noise and thereby yield a \ow-distortion and low-error reproduction of the original signal.

In the present chapter noise which are discussed

we shall make only a brief more extensively in Chap.

reference to the sources of

14. Here we shall be concerned principally with a discussion of the mathematical representation and statistical characterizations of noise.

7.1

SOME SOURCES OF NOISE

One source of noise is the constant agitation which prevails throughout the universe at the molecular level. Thus, a piece of solid metal may appear to our gross view to be completely at rest. know, however, that the individual molecules

We

315

NOISE IN AMPLITUDE-MODULATION SYSTEMS 347

Local

CHAPTER

oscillator

EIGHT

'osc

NOISE IN AMPLITUDE-MODULATION

modulated RF

SYSTEMS

carrier

+

noise

Radio frequency

Intermediate

frequency

Mixer

(RF)

(IF)

amplifier

amplifier

Output signal,

IF

Demodulator

carrier

power

Baseband

-

s»

filter

Noise output

filter

white noise, spectral density

r _x_ n

<

'Synchronous' I

I

signal

source

I

Chap.

In

3

we

described a

number

communicompare the performance of these

of different amplitude-modulation

cation systems. In the present chapter

we

shall

systems under the circumstances that the received signal

is

Figure 8.1-1

A

|

I

receiving system for an amplitude-modulated signal.

corrupted by noise. explicitly in

Fig. 8.1-1

and may be considered to be part of the mixer. The

difference-frequency carrier 8.1

1

'

system for processing an amplitude-modulated carrier and recovering the baseband modulating system is shown in Fig. 8.1-1. We assume that the signal has suffered great attenuation during the course of its transmission over the communication channel and hence is in need of amplification. The input to the system

A

might be a signal furnished by a receiving antenna which receives

signal

its

from

called a radio-

a transmitting antenna. The frequency (RF) carrier, and its frequency is the radio frequency fr! The input signal is amplified in an RF amplifier and then passed on to a mixer. In the mixer the modulated RF carrier is mixed (i.e., multiplied) with a sinusoidal waveform The process of generated by a local oscillator which operates at a frequency carrier of the received signal

is

called the intermediate frequency (IF) carrier, that

—fosc —f,i- The modulated IF carrier is applied to an IF amplifier. The process just described, in which a modulated RF carrier is replaced by a modu-

AMPLITUDE-MODULATION RECEIVER

is

s '/if

lated

IF

oscillator

carrier, is

is

called conversion.

The combination

of the mixer and local

called a converter.

The IF amplifier output is passed, through an IF carrier filter, to the demodulator in which the baseband signal is recovered, and finally through a baseband filter. The baseband filter may include an amplifier, not explicitly indicated in Fig. 8.1-1. If synchronous demodulation is used, a synchronous signal source will be required.

.

The only absolutely essential operation performed by the receiver is the process of frequency translation back to baseband. This process is, of course, the inverse of the operation of modulation in which the baseband signal is frequency-

.

mixing

is

also called heterodyning,

and

be explained, the heteroselected to be above the radio frequency since, as is to

dyning local-oscillator frequency /os<; is /rf the system is often referred to as a superheterodyne system. The process of mixing generates sum and difference frequencies. Thus the ,

and a carrier OTC -/rf osc +/rf modulated by the baseband signal to the same extent as was the carrier. The sum frequency is rejected by a filter. This filter is not shown

mixer output consists of a carrier of frequency

Each

carrier

input

RF

34*

is

.

translated to a carrier frequency.

The process

of frequency translation

is

per-

formed in the system of Fig. 8.1-1 in part by the converter and in part by the demodulator. For this reason the converter is sometimes referred to as the first detector, while the demodulator is then called the second detector. The only other components of the system are the linear amplifiers and filters, none of which would be essential if the signal were strong enough and there were no need for multiplexing.

348 PRINCIPLES OF COMMUNICATION SYSTEMS

NOISE IN AMPLITUDE-MODULATION SYSTEMS 349

It is apparent that there is no essential need for an initial conversion before demodulation. The modulated RF carrier may be applied directly

to the

demodu-

lator. However, the superheterodyne principle, which is rather universally incorporated into receivers, has great merit, as is discussed in the next section.

ADVANTAGE OF THE SUPERHETERODYNE PRINCIPLE:

8.2

SINGLE CHANNEL A

signal furnished

Thus

by an antenna to a receiver

may have a power as low as some signal may be of the order of tens of

magnitude of the required gain is very large. In addition, to minimize the noise power presented to the demodulator, filters are used which are no wider than

the

is

necessary to

accommodate

the baseband signal. Such

should be rather fiat-topped and have sharp skirts. It is more convenient to provide gain and sharp flat-topped filters at low frequencies than at high. By way of example, in

commercial

100

MHz,

FM

broadcasting, the

RF

carrier frequency

FM receiver the IF frequency

while at the

Thus, in Fig. 8.1-1 the largest part, by the IF amplifier, and the critical filtering

far,

is

filters

in the

is

range of

of the required gain

done by the IF

filter.

is provided by While Fig. 8.1-1

critical. It

serves principally to limit the total noise power input to the mixer avoids overloading the mixer with a noise waveform of excessive

This

is

devices;

is

is

because of the i.e.,

an

signal

very small.

is

RF amplifiers, such as masers, are low-noise be designed to provide relatively high gain while

fact that

little

noise.

When RF

applied directly to the mixer.

amplification

we need but change

the frequency of the local

carrier frequency to another.

Whenever

set so

osc is

tnat /o S c -/rf =fif, the mixer will convert the input modulated RF carrier to a modulated carrier at the IF frequency, and the signal will proceed through the

demodulator

to the output.

Of

course,

it is

necessary to gang the tuning of the

amplifier to the frequency control of the local oscillator. But again this

ganging is not employed.

critical,

since only

we may note

Finally,

one or two

the reason for selecting

higher selection the fractional change in

range of

RF

frequencies

is

smaller than

/osc

RF

amplifiers

/osc

higher than

required to

and

filters

are

ft( With this accommodate a given

would be the case

.

for the alternative

selection.

8.3

SINGLE-SIDEBAND SUPPRESSED CARRIER

(SSB-SC)

The

receiver of Fig. 8.1-1 is suitable for the reception and demodulation of all types of amplitude-modulated signals, single sideband or double sideband, with

and without

carrier. The only essential changes required to accommodate one type of signal or another are in the demodulator and in the bandwidth of the IF

carrier

Hence, from

our interest will focus on the section of receiand through to the output. The signal input to the IF filter is an amplitude-modulated IF carrier. The normalized power (power dissipated in a 1-ohm resistor) of this signal is S The filter.

ver beginning with the

IF

this point, filter

f

employed whenever the incoming

RF amplifier can

generating relatively signal

and thereby

amplitude.

amplification

RF

go from one

MHz.

10.7

suggests a separate amplifier and filter, actually in physical receivers these two usually form an integral unit. For example, the IF amplifier may consist of a number of amplifier stages, each one contributing to the filtering. Some filtering will also be incorporated in the RF amplifier. But this filtering is not

RF

In a superhet receiver, however, oscillator to

RF

tens of picowatts, while the required output watts.

gang-tune the individual stages over a wide band, maintaining at the same time a reasonably sharp flat-topped filter characteristic of constant bandwidth.

is

The mixer provides

not employed, the relatively

little

gain

and generates a relatively large noise power. Calculations showing typical values of gain and noise power generation in RF, mixer, and IF amplifiers are presented in Sec. 14.14.

Multiplexing

amplified in the

RF

also sources of noise,

gain of the

RF

i.e.,

even greater merit of the superheterodyne principle becomes apparent when we consider that we shall want to tune the receiver to one or another of a number of different signals, each using a different RF carrier. If we were not to take advantage of the superheterodyne principle, we would require a receiver in which many stages of RF amplification were employed, each stage requiring

Such tuned-radio-frequency (TRF) receivers were, as a matter of fact, commonly employed during the early days of radio communication. It is difficult enough to operate at the higher radio frequencies; it is even tuning.

more

difficult to

thermal noise, shot noise,

etc.,

but this noise, lacking the

amplifier, represents a second-order effect. (See Sec. 14.11.)

We

assume that the noise is gaussian, white, and of two-sided power spectral density n/2. The IF filter is assumed rectangular and of bandwidth no wider than is necessary to accommodate the signal. The output baseband signal has a power S 0 and is accompanied by noise of total power JV„ shall

Calculation of Signal

An

.

Added, is the noise generated in the RF amplifier and amplifier and IF amplifiers. The IF amplifiers and mixer are

signal arrives with noise.

Power

With a single-sideband suppressed-carrier signal, the demodulator is a multiplier as shown in Fig. 8.3-la. The carrier is A cos 2nf t. For synchronous demoduc lation the demodulator must be furnished with a synchronous locally generated carrier cos 2nfc t. We assume that the upper sideband is being used; hence the carrier filter has a bandpass, as shown in Fig. 8.3- If), that extends from/ to/ f +/M where fM is the baseband bandwidth. The bandwidth of the baseband filter extends from zero to fM as shown in Fig. 8.3- lc. Let us assume that the baseband signal is a sinusoid of angular frequency ,

350 PRINCIPLES OF COMMUNICATION SYSTEMS

NOISE IN AMPLITUDE-MODULATION SYSTEMS 351

cos 2ir/

I

= Acos[2r{f *f )t] c m

«,(«)

s„(f)

S -**

S

Noise spectral

=

£ cos um

t

the baseband signal were quite arbitrary in waveshape.

-4'

density

- G„„

single or

-4-4,

0

-4

4+

4

o

-4,

4,

Figure 8.3-1

(a)

A

sideband

is

synchronous demodulator operating on a single-sideband single-tone of the carrier filter, (c) The passband of the lowpass baseband filter.

signal. (6)

=A

The output

of the multiplier

cos [2n(fc

+/Jt]

We now

o

=

Sl (t)

cos

(8.3-1)

is

=^

coc t

Only the difference-frequency term fore the output signal

cos [2tt(2/c

will

+ /Jt] + y

components are involved.

cos 2jr/m t

pass through the baseband

Power

calculate the output noise

when

.

a noise spectral

total noise s 2 (t)

spectral

N„ For this purpose, we recall from Sec. 7.8 component at a frequency f is multiplied by cos 2nfc t, the original noise component is replaced by two components, one at frequency L +/and one at frequency/, -/, each new component having one-fourth the power of the original. The input noise is white and of spectral density rj/2. The noise input to the multiplier has a spectral density G nI as shown in Fig. 8.3-2a. The density of the noise after multiplication by cos 2nf t is G„ as is shown in Fig. 8.3-2b. Finally c 2 the noise transmitted by the baseband filter is of density G„ as in Fig. 8.3-2c. The that

The carrier frequency is fc , and, since we have assumed that the upper being used, the received signal is sfr)

the single-sideband

this

Calculation of Noise

The bandpass range

fmifm ^/m)-

many

by

/

(c)

(6)

Then

baseband signal may be resolved into a series of harmonically related spectral components. The input power is the sum of the powers in these individual components. Next, we note, as was discussed in Chap. 3, that superposition applies to the multiplication process being used for demodulation. Therefore, the output signal power generated by the simultaneous application at the input of many spectral components is simply the sum of the output powers that would result from each spectral component individually. Hence S, and S in 0 Eqs. (8.3-4) and (8.3-5) are properly the total powers, independently of whether a signal generated

Noise spectral power

= G„ -

power density

We may readily see that, even though Eq. (8.3-6) was deduced on the assumption of a sinusoidal baseband signal, the result is entirely general. For suppose that

output

is

the area under the plot in Fig. 8.3-2c.

(8.3-2)

N = 2fM 0

filter.

We

have, then, that

\^

(g.3-7)

There-

is

s 0 (t)

= -cos

2nfm

(8.3-3)

t

(<•)

which

is

the modulating signal amplified by

The input

signal

power

^.

-U-tu

is

-4

-o

4

4 + '„

"

/

G„ 2

s while the output signal power

t

-

1

(8.3-4)

2

i—r

is

-24-4,

-24

8

-t—

1

15

-4,

*m

-r— 2

4

24 +

4,

f

<

°

(8.3-5)

2V2/

8

4

8

(e) 1

Thus -4, (8.3-6) S,-

4

o

4,

/

Figure 8.3-2 Spectral densities of noises in SSB demodulator, (a) Density G„, of noise input to multiplier, (b) Density G„ of noise output of multiplier, (c) Density G„„ of noise output of baseband filter. 2

352 PRINCIPLES OF COMMUNICATION SYSTEMS

NOISE IN AMPLITUDE-MODULATION SYSTEMS 353

Use of Quadrature Noise Components It is

of interest to calculate the output noise

power

N

0

an alternative manner

in

=

We now

apply Eq.

spectral density [see Eqs.

(7.

1

(7. 1

t

The

as in Fig. 8.3-2a.

and

We

-

n s (t) sin 2nfc

output of the IF

(8.3-8) to the noise

G nl

2-7)

n c (t) cos 2nfc

(8.3-8)

t

filter

so that

spectral densities of n c (()

has the

n(t)

We

observe that for 0
=

r,/2,

8.4

+f)

G.,(/c

t

=

n c (t) cos

= K(t) + The spectra of

L

2

-fni to 2fc

the second

+fM

while

G rl (fc -f) =

so that

0,

and

2

2nfc

t

-

n s(t) sin 2nfc t

-

and are outside the baseband

cos 2itfc

a baseband signal of frequency range

\nlt) sin 4nfc

t

(8.3-10)

t

filter.

The output

noise

= H(t)

is,

(8.3-11)

The spectral density of n„(t) is then G no = iG„, = shown in Fig. 8.3-2c, the spectral density G m is and the total noise is again N„ = rjfu/4.

we may

When

noise, rather than only in t

third terms in Eq. (8.3-10) extend over the range

Calculation of Signal-to-Noise Ratio

DOUBLE-SIDEBAND SUPPRESSED CARRIER

$(r//2)

=

i//8.

Hence as

q/8 over the range

before, as

-fM

tofM

Calculation of Noise This situation

fM

is

transmitted over a

DSB-SC

must be 2fM rather than fM

- fu to f + fM will contribute the range fc to fe + fu as in the SSB case. c

.

Thus,

to the output

Power

which shows the spectral density G nl (f) This noise is multiplied by cos co c t. The multiplication results in a frequency shift by ±fe and a reduction of power in the power spectral density of the noise by a factor of 4. Thus, the noise in region d of Fig. 8.4- la shifts to regions d shown in Fig. 8.4-16. Similarly regions a, b, and c of Fig. 8.4- la are translated by ±fc and are also attenuated by 4 as shown in is

illustrated in Fig. 8.4- la,

of the white input noise after the IF

filter.

,

(SNR)

calculate, using Eqs. (8.3-6)

and

1 (8.3-7),

the signal-to-noise ratio at

2

.

SJ4

The importance

of

SJN 0

is

that

it

(a)

a\ b

c |
Si

tt = -TT. = -j-

(8.3-12)

0

serves as a figure of merit of the performance of

communication system. Certainly, as SJN 0 increases, it becomes easier to tinguish and to reproduce the modulating signal without error or confusion. a

f

4

disIf

a

G.2 (6)

V

=

(DSB-SC)

SJN 0 We have S„

G„

filter

input noise in the frequency range fc

±u c (t) cos 4nfc

«„(f)

Finally

the receiver quieter.

system, the bandwidth of the carrier

therefore,

the output,

make

(8-3-9)

s

cos 2nfc

frequency range, or

and n s (t) are

2-8)]

GJf) = Gml (f) = G nl {fc -/) +

n(t)

compare demodulators. observe from Eq. (8.3-12) that to increase the output signal-to-noise power ratio, we can increase the transmitted signal power, restrict the baseband of merit with which to

using the transformation of Eq. (7.11-2): n(t)

system of communication allows the use of more than a single type of demodulator (say, synchronous or nonsynchronous), that ratio SJN, will serve as a figure

8

G„.

A

R*l |-

z<-

t

n

i_ n

4

8 o,

b,

c

d )

^~T7T71 24

t

f

| |

~fM

ffi

f

Figure 8J-3 Power spectral densities of

and

G„_

Figure 8.4-1 Spectral densities of noise in DSB demodulation, Density G„ 2 of noise output of baseband filter.

filter, (b)

(a)

Density G„, of noise at output of IF

354 PRINCIPLES OF COMMUNICATION SYSTEMS

Fig. 8.4- 1 fo.

Note

— fM and + fM

NOISE IN AMPLITUDE-MODULATION SYSTEMS 355

that the noise-power spectral density in the region between is

»j/4,

SSB

while the noise density in the

case, as

shown

in

= A 2 /8 = SJ4

rather than S 0

when

SSB case. Thus we see that two sideband components each

as in Eq. (8.3-5) for the

power

a received signal of fixed

is split

into

Fig. 8.3-2c is only r//8. Hence the output noise power in twice as large as the output noise power for SSB given in Eq. (8.3-5). The output noise for DSB after

of half power, as in

baseband

contributions from each sideband yield output signals which are in phase.

filtering is therefore

signal

doubling

(8.4-1)

DSB,

rather than being

power increases by a

factor of

left in a single sideband, the output This increase results from the fact that the

2.

A

amplitude causes a fourfold increase in power. This fourfold increase, due to the inphase addition of s'„ and s'^, is in part undone by the need to split the in

power

two half-power sidebands. Thus the overall improvement in is by a factor of 2. On the other hand, the noise outputs due to noise spectral components symmetrically located with respect to the carrier are uncorrelated with one another. The two resultant noise spectral components in the output, although of the same frequency, are uncorrelated. Hence the combination of the two yields a power which is the sum of the two powers individually, not larger than the sum, as is the input

into

output signal power

Calculation of Signal

We

might imagine that

Power SJN„ for DSB would now see that such is not

for equal received powers, the ratio

We

be only half the corresponding ratio for SSB.

shall

SJN„ is the same in the two cases. Again, without loss assume a sinusoidal baseband signal of frequency f„
and

that the ratio

in generality, let us

s£t)

=

.

yjl

= The

received

power

A

cos 2nfm

cos

is

cos 2nfc

t

|>(/c +/Jt] +

case with the signal.

t

Calculation of Signal-to-Noise Ratio

4? cos

\2n(fc

-fM

(8.4-2)

Returning now to the calculation of signal-to-noise ratio for DSB-SC, we find from Eqs. (8.4-1) and (8.4-7) that

then

SSB-SC.

exactly as for as in Eq. (8.3-4).

In the demodulator (multiplier),

sideband term in Eq. filter

(8.4-2) yields

is

s,(t)

multiplied by cos

w

c

t

.

The upper-

Arbitrary Modulating Signal

a signal within the passband of the baseband In the discussion in the present section concerning

given by

the baseband signal S;(t)

= -^cos2;r/m t

(8.4-4)

2y/2

The lower-sideband term

of Eq. (8.4-2) yields

= -^T= cos 2V 2

2nfm t

is

sinusoidal.

DSB we

have assumed that

As pointed out

in

Sec. 8.3, this

assumption causes no loss of generality because of the linearity of the demodulation. Nonetheless it is often convenient to have an expression for the power of a DSB signal in terms of the arbitrary waveform m(t) of the baseband modulating signal.

s:(t)

waveform

Hence

let

the received signal be

(8.4-5)

sit)

=

m(t) cos

2nfc

(8.4-9)

t

The power of s,{t) is Observe, most particularly

in Eqs. (8.4-4)

phase and that hence the output signal

s„(r)

=

s'(t)

+

and

(8.4-5),

that

s^(t)

and

s'i(t)

are in Si

is

s"(t)

=

Now

—

z=

cos 2nfm

t

(8.4-6)

V2

[Of that

which has a power

m(i)

A2

if

range


= m 2 (t)

sf(t)

m(t)

2

2nfc

t

= V(t) + V(t)

cos

(4ji/c

(8.4-10)

t)

no

special relevance in the present discussion,

is

the fact

bandlimited to fM , m 2 (t) is bandlimited to 2fM See Prob. 8.4-1.] cos Anfc t consists of a sum of sinusoidal waveforms in the frequency

is

.

m 2 (t) 2f ± 2fu The c

cos

can always be represented as a sum of sinusoidal spectral components.

interest, albeit of

Hence

S,™f

=

.

average value of such a

(8.4-7) S, EE sf(t)

=

sum

VW

is

zero,

and we therefore have (8.4-11)

356 PRINCIPLES OF COMMUNICATION SYSTEMS

NOISE IN AMPLITUDE-MODULATION SYSTEMS 357

When

the signal s,(t) in Eq. (8.4-9) is demodulated by multiplication by and the product passed through the baseband filter, the output is m(t)/2. The output signal power is

cos 2nfc s„(t)

=

8.5

DOUBLE SIDEBAND WITH CARRIER

t,

(8.4-12)

Let us

now

signal.

Demodulation

used as a transmitted reference to obtain the reference signal cos m t c 8.5-1). We note that the carrier increases the total input-signal power but makes no contribution to the output-signal power. Equation (8.4-8) applies

carrier (see

so that, from Eqs. (8.4-1

and

1)

(8.4-12),

consider the case where a carrier accompanies the double-sideband is achieved synchronously as in SSB-SC and DSB-SC. The

is

Prob.

we

directly to this case, provided only that

S,

power

=|

in the

is.

same

the

modulating

result as given in Eq. (8.4-7) for

an assumed

TT =

single sinusoidal

to Calculate

s,(t)

N0

=

n(r)

n c (t) cos 2nfc

spectral density of n c (t)

t

-

where « s (t) sin 2nfc

and n s(t) are

(8.4-14)

t

[see Eqs. (7.12-7)

and

(7.12-8)]

=

signal

A{\

= A

again interesting to calculate N„ using the transformation of Eq. (7.11-2):

s!

GMc

= GJJ) =

+f) + G Hl (fc ~f)


In the frequency range \f\

c

,

c

n/2.

(8.4-15)

m(f)

(8.5-1)

m(t)] cos

cos 2nfc

baseband

the

is

-(-

is

t

+

2nfc

t

Am(t) cos 2ufc

was

which amplitude-modulates the carrier

signal

t.

Example

result of multiplying n(t) n(t)

Baseband

cos 2nfc

t

=

by cos

±nc (t)

filtering eliminates the

is

S.

The output

noise

power

N

^

0 is,

27t/f

t

is,

=

y+

SB >

=

y

[1

+

==

Si

^W]

(8.5-3)

,

cos 4nfc

t

- K(t)

sin 4;t/c

= K(0

+ m 2 (t)

1

t

(8.4-17) or,

with Eq.

(8.5-4)

(8.5-1),

(8.4-18)

S„

m\t)

No

2

+m

1

=

5

-fM
power Pc

= A 2 /2, we

S

t

(8.5-5)

ifu

(t)

therefore,

\

Vu =

^

the modulation

(8.4-20)

of course, identical with Eq. (8.4-1), which was obtained by conon a noise spectral component of a multiplication by

from Eqs.

get,

(8.5-3)

and (8.5-51 that

S0

(8.4-19)

is

sinusoidal, with m(t) s,(0

In this case

=

A(\

+m

-m

cos 2nfm

cos 2%fm

t)

t

(m a constant), then

cos 2nfc

t

(8.5-7)

m 2 (t) = m 2 /2 and

sidering directly the effect

S0

cos 2nfc

N0

t.

S!

yields

If

This result

then have that

then

G„ c (f)

=

We

given by

A 2 we have

In terms of the carrier

GJf) =

S, is

(8.4-16)

second and third terms, leaving

spectral density of n 0 (t)

power

7.12-1.)

+ H(r)

«„(')

The power

the total input

Eliminating also derived in

(8.5-2)

t

The carrier power is A 2 12. The sidebands are co ntained in t he term Am{t) cos 27i/c f. The power associated with this term is (A 2 /2)m 2 (t). where m 3tt) i*

A cos 2nfc

Thus

c

The

the

SB >

"V

the time average of the square of the modulating waveform.

G„Xf)

(This result

is

signal.

Use of Quadrature Noise Components

The power

SB)

(8.4-13)

Suppose that the received

It is

S!

sidebands alone. Then

S0

that

Sf*\ where

replace S, by

2

m2 + m2

S,

(85- 8)

M

r,f

NOISE IN AMPLITUDE-MODULATION SYSTEMS 359

358 PRINCIPLES OF COMMUNICATION SYSTEMS

When

the carrier

form cos 2nfc

m>

is

transmitted only to synchronize the local demodulator wave-

relatively

t,

m 2 /(2 + m 2 s

1,

)

1,

little

and the signal-to-noise

the presence of the carrier.

used (Sec.

A

is

On

the other hand,

ratio

transmitted. In this case is

not greatly reduced by

when envelope demodulation

this case

m 2 /(2

in the sidebands

power

nfM To give the product nfM some physical significance, we consider it to be the noise power N M at the input, measured in a frequency band equal to the baseband frequency. Thus the output signal

S,

NM = \ The

ratio

S /rifM t

is

overload suppression

and the

8.6

SQUARE-LAW DEMODULATOR

We

saw

is

differ-

carrier filter does indeed suppress noise.

in Sec. 3.6 that a double-sideband signal with carrier may be demoduby passing the signal through a network whose input-output characteristic is not linear. Such nonlinear demodulation has the advantage, over the linear synchronous demodulation methods, that a synchronous local carrier need not

lated

observe that in each demodulation system considered so far, the ratio S,/r//M in the expression for output SNR [see Eqs. (8.3-12), (8.4-8), and (8.5-5)]. is

are normally included, but the purpose

ent,

)

which contribute to signal power output.

appeared

This ratio

filters

rather than noise suppression. In single sideband, of course, the situation

is

m < 1. When m = 1, the carrier is 100 percent + m 2 = ^, so that of the power transmitted, only

Figure of Merit

We

such carrier

required that

3.4), it is

modulated. In one-third

power need be

carrier

divided by the product

2f„

=

.

nfu

be obtained. This eliminates the rather costly synchronizing circuits. In this

we discuss the performance and determine the output SNR of a nonlinear demodulator which uses a network whose output signal y (voltage or current) is related to the input signal x (voltage or current) by section

y

(8.5-9) in

which X

is

therefore, often referred to as the input signal-to-noise ratio

is,

Si/N M It needs to be kept in mind that N M is the noise power transmitted through the IF filter only when the IF filter bandwidth is fM Thus N„ is the true input noise power only in the case of single sideband.

shown

a constant. As

constitutes the demodulator,

is

=

kx 1

(8.6-1)

network, which

in Fig. 8.6-1, this nonlinear

preceded by a bandpass IF

filter

of bandwidth

.

.

For the purpose of comparing systems, we introduce

2/M and

We

is

followed by a baseband low-pass

discuss this square-law

the figure of merit

y,

it

above may now be summarized as follows:

H

1

SSB-SC

(8.5-11)

1

DSB-SC

(8.5-12)

DSB

(8.5-13)

1

m 2 (t) + m 2 (t)

2

m2 r 4- m

own

intrinsic interest,

is

We

observed [Eqs. (8.5-11) to (8.5-14)] that

SyN M

not a function of the input signal-to-noise ratio

.

SJN M

decreases, say,

IF

—

Input signal

Square-

X

filter

law

Baseband

/

««(/>

device

and

»,(')+ii,(0

filter

noise

DSB

with sinusoidal modulation

(8.5-14)

\H„if)\

I».f(/)| 1

A

point of interest in connection with double-sideband synchronous

demodu-

purpose of suppressing output-noise power, the carrier filter of Fig. 8.3-1 is not necessary. A noise spectral component at the input which lies outside the range fc ±/M will, after multiplication in the demodulator, lie outside the passband of the baseband filter. On the other hand, if the carrier filter is eliminated, the magnitude of the noise signal which reaches the modulator may be large enough to overload the active devices used in the demodulator. Hence, lation

is

this figure of

Therefore,

if

the

by a factor a, the output SJN 0 will also decrease by a. The nonlinear demodulator also has a range where the figure of merit y is indeinput

results given

in part for its

.

exhibits an important characteristic

the presence of noise.

merit

The

demodulator

of bandwidth fM

which is not displayed by the linear synchronous demodulators. We have previously adopted the quantity y = (SJN 0 )/(Sj/N M ) as a figure of merit for the performance of demodulators in but also because

defined by

filter

that, for the

ii i

4

i

ii i

Figure 8.6-1 The square-law

4

i

'

AM demodulator.

\1

Filter

stops dc

364 PRINCIPLES OF COMMUNICATION SYSTEMS

Equation

(8.6-22)

is

plotted (solid

NOISE IN AMPLITUDE-MODULATION SYSTEMS 365

plot)

in

expressed in terms of their decibel (dB) equivalents, in units of 10 log (PJN ). Above threshold, when M 22)

Fig. 8.6-4 i.e.,

with the variables

the abscissa

PJN M

is

is

marked

off

very large, Eq. (8.6-

chosen arbitrarily to be the point at which the performance curve falls away by as shown. On this basis it turns out that the threshold occurs when 1 dB

P,/Wm =

4.6

dB

or

becomes

PJN M


Eq. (8.6-22) becomes

We

again consider an

DSB (8.6-24)

For comparison, Eqs.

(8.6-23)

and

(8.6-24)

have also been plotted

in Fig. 8.6-4.

We

observe the occurrence of a threshold in that, as PJN M decreases, the demodulator performance curve falls progressively further away from the straight-line

plot corresponding to

PJN U

2.9JV

M

.

THE ENVELOPE DEMODULATOR

8.7

Below threshold, when

when Pc =

we

signal

AM signal with modulation

shall use a

vides an output

m(t) < 1. To demodulate this network which accepts the modulated carrier and pro\

\

which follows the waveform of the envelope of the

carrier.

The

diode demodulator of Sec. 3.6 is a physical circuit which performs the required operation to a good approximation. As usual, as in Fig. 8.3-1, the demodulator is preceded by a bandpass filter with center frequency fc and bandwidth 2/ M followed by a low-pass baseband filter of bandwidth fM

and

,

is

.

very large. The threshold point

is

It

convenient in the present discussion to use the noise representation

is

given in Eq. (7.11-2): n(t)

=

wc t -

n c (t) cos

n s (t) sin

co c t

(8.7-1)

has a power spectral density n/2 in the range \f — fc
If the

n(t)

\

.

s,(t)

where A

+

n t (t)

=

/i[l

=

{A[\

+

m(t)] cos co c

+

m{t)2

+

t

+

n c (t) cos

n c (t)} cos coc

-

t

coc

-

t

n,(t) sin co c t

n s (t) sin co c

t

(8.7-2a) (8.7-26)

the carrier amplitude and m(f) the modulation. In a phasor diagram, term of Eq. (8.7-26) would be represented by a phasor of amplitude All + m(t)~] + n c (t), while the second term would be represented by a phasor perpendicular to the first and of amplitude n s(f)- The phasor sum of the two terms is the

is

first

then represented by a phasor of amplitude equal to the square root of the the squares of the amplitudes of the

baseband

just prior to

s 2 (t)

+

filtering is the

n 2 (t)

= =

{(Ail

{A

+

We

should

now

like to

make

2

[\

two terms. Thus, the output

sum

of

signal plus noise

envelope (phasor sum):

+

m(ty]

+

m(t)]

2

« c (t)

+

n

+ 2

2

(f)}

n c (t))

+

2

2A[1

+

n

+

2

(t)}

,/2

(8.7-3a)

m(f)]n c (t)

1/2

(8.7-36)

the simplification in Eq. (8.7-36) that

would be and n s (t) were much smaller than the carrier amplitude A. The difficulty is that n and n„ are noise " waveforms " for c which an explicit time function may not be written and which are described only allowed

if

we might assume

that both

|

n c {t)

|

\

|

in terms of the statistical distributions of their instantaneous amplitudes.

matter Figure 8.6-4 Performance of a square-law demodulator illustrating the

phenomenon of threshold.

how

n s (t), there

large

is

A and how

No

small the values of the standard deviation of n {t) or c always a finite probability that n c (t) n s (t) or both, will be com|

\

,

|

,

|

366 PRINCIPLES OF COMMUNICATION SYSTEMS

parable

On

or even larger than, A.

to,

the other hand,

approach or exceed A is

is

is

n

2

exceeds 4 times

it

,

if

Since both square-law demodulation and envelope demodulation exhibit a

A and we assume

(t) -4

that

n c (t)

|

|

«

its

com parison

threshold, a

m 2 (t) ^

lation that

greater than twice the standard devi-

only 0,045 and only 0.00006 that

Hence,

the standard deviations

smaller than A, the likelihood that n c or n„ will rather small. For example, since n c and n s have gaussian

distributions, the probability that n c (t)

ation.

if

much

of n c (t) and n s (t) are

ation

NOISE IN AMPLITUDE-MODULATION SYSTEMS 367

and Eq.

is

of interest.

n 2 (t)

*

{A 2 [l

+

m(t)¥

+ 2AU +

Using now the further approximation that have finally that s 2 (t)

+

n 2 (t)

The output-signa pow er measured l

terms,

is

power

after

S„

= A 2 m 2 (t). baseband

Eq.

(8.5-3),

we

+

as

+

x)

m(f)]

after the

1 '2

+

m(t)>c (t)} 1/2

as

1

power

N = 0

2t\fM

.

+

x/2 for small x

baseband

we

(8.7-5)

filter,

=

>/,

and neglecting dc the output-noise

Again using the symbol N M ( = rjfu ) baseband range fM and using

at the input in the

,

find that

y

=

SJN a =

m 2 (t)

Sa

is

the carrier power,

I

Pc

,

(8.7-8)

'

N which

0

compared with Eq.

to be

is

1.1

\N

(8.6-24) giving

SJN 0

below threshold

for the

square-law demodulator.

The comparison indicates that, below threshold, the square-law demodulator performs better than the envelope detector. Actually this advantage of the square-law demodulator is of dubious value. Generally, when a demodulator is operated below threshold to any appreciable extent, the performance may be so

(8.7-4a)

n c (t)

Since the spectral density of n c (t)

filtering is

to stand for the noise

(1

square-law demodu-

in

3 A 2 /2 = Pc

f

standard devi-

A, the assumption

Assuming then that |n c (t)| « A and |n s (r)|
+

had assumed

becomes

(8.7-7)

usually valid.

s 2 (t)

We

Then, as noted above, S

1.

poor as to be nearly useless. What is of greater significance is that the comparison suggests that the threshold in square-law demodulation is lower than the threshold in envelope demodulation. Therefore a square-law demodulator will operate above threshold on a weaker signal than will an envelope demodulator.

summary, on strong signals all demo dulat ors work equally well except demodulator requires that m 2 (t) < 1 to avoid baseband-signal distortion. On weak signals, synchronous demodulation does best since it exhibits no threshold. When synchronous demodulation is not feasible, squareIn

that the square-law

law demodulation does better than envelope demodulation. It is also interesting to note that voice signals require a 40-dB output signal-to-noise ratio for high

=

mHt)

(8.7-6)

quality. In this case

both the linear-envelope detector and the square-law detector

operate above threshold.

The result is the same as given in Eq. (8.5-13) for synchronous dem odula tion. To make a comparison with the square-law demodulator, we assume m 2 {t) <$ 1. In this case, as before, S, 3 P and Eq. (8.7-6) reduces to Eq. (8.5-6). Hence we have c

REFERENCE

,

the important result that above threshold the synchronous demodulator, the square-la w dem odulator, and the envelope demodulator all perform equally well,

provided

m

2

(f) <|

1.

Davenport, W., and W. Root: York, 1958.

"Random

I

PROBLEMS

Like the square-law demodulator, the envelope demodulator exhibits a threshold. ratio decreases, a point is reached where the signalto-noise ratio at the output decreases more rapidly than at the input. The calcu-

As the input signal-to-noise

lation of signal-to-noise ratio

is

quite complex, an d

to simply state the result 1 that for Si/N M


and

we

shall therefore be content

m 2 (t) «

A superheterodyne receiver using an IF frequency of 455 kHz is tuned to 650 kHz. It is found that the receiver picks up a transmission from a transmitter whose carrier frequency is 1560 kHz. Suggest a reason for this undesired reception and suggest a remedy. (These frequencies, 650 kHz and 1560 kHz, are referred to as image frequencies. Why?) 8.1-1. (a)

Let

8.3-1.

1 I

g{t)

/ > /] Show .

1

8.3-2.

be a waveform characterized by a power spectral density G{f). Assume G(f) that the time-average value of g(t) cos 2nfr t is zero if fc >ft

s,<()

=

in Sec. 3.10,

m(l) cos

2nfe

t

if

+

m(t)

obviously indicates a poorer performance than indicated by

(a) (fc)

(8.7-6),

which applies above threshold.

Show Show

that

m2

if

(r),

is

for

an arbitrary baseband waveform, a received SSB signal may be 2nfc t. Here m(f) is derived from m(t) by shifting by 90° the

in m(t).

have the same power spectral densities and that m 2 (t) = m 2 (t). m(t) has a spectrum which extends from zero frequency to a maximum

that m(r)

quency /„ then ,

0

m(t) sin

phase of every spectral component (8.7-7)

=

.

As noted

written

Eq.

New

McGraw-Hill Book Company,

1.

Threshold

Equation

Signals and Noise,"

and

m2

(r),

m(r)

and

m{t)m(t) all

have spectra which extend from zero frequency to 2f„

fre-

368 PRINCIPLES OF COMMUNICATION SYSTEMS

(c)

Prob.

Show

that

power

the normalized

NOISE IN AMPLITUDE-MODULATION SYSTEMS 369

of s {t)

m 2(f) = m 2 (r).

is

t

(Hint:

Use the

results

of

Calculate the normalized power S„ of the demodulated

SSB

signal,

i.e .,

the signal

s,(t)

by cos 2nft

10~ 3 watt/Hz. The signal plus noise

t

83-3. Prove that Eq. (8.3-1

8_M. A baseband

1) is

signal m(l)

density of m([)

correct by sketching the

is

SSB

transmitted using

power

spectral density of Eq. (8.3-10).

as in Prob. 8.3-2.

Assume

that the

power

The

signal [e

+

m(!)] cos

multiply the incoming signal,

bandwidth spec-

shown

B, as

is

m

c

t

A

synchronously detected. The reference signal cos m,

n(0.G„(/> -

of

|

ij/2 is

added

to the

SSB

o vD (t)

signal, find the i><<)

=

received

lo/

t

SSB

+/„=

signal has a

1.003

!«

+ m(i)l cos uc

t

power spectrum which extends over the frequency range from fc = signal is accompanied by noise with uniform power spectral

Reference

MHz. The

signal, vg

The noise

n(t) is

expressed as

n(f)

=

n c (t) cos 2nfe

components « t (r) and

n,(t)

t

-

n,(t) sin

2nfc

t.

The

signal plus

accompanying noise

its

is

U)

Find the power spectral

of the noise in the spectral range specified as

B«2f

r,s/s /,+/„. fh\

filter

The input signal power, The output signal power.

densities of the quadrature

power

used to

in Fig. P8.5-1.

density 10** watt/Hz. (a)

i

White gaussian noise

.

MHz

is

l/l
(r) If white gaussian noise with power spectral density output SNR. The baseband filter cuts off at/ = /M

I

demodu-

is

obtained by passing the input signal through a narrowband

Find:

8-3-5.

shown. The signal

Comment on

(c)

\f\>f*

(fc)

filter

volt.

Calculate v R (t) due to the input signal alone. Calculate the noise power accompanying rjt\. the effect of the noisy reference on the output SNR.

(fc)

GJf) =

(a)

1

Calculate S,.

(a)

l/l

passed through the

Find the output noise power.

(fc)

8.5-1.

is

is

by multiplication with a local carrier of amplitude (a) Find the output signal power.

multi-

and then passed through a baseband filter. Show that S„ = m 2 (r)/4 and hence that SJSi =i. [Note: This problem establishes more generally the result given in Eq. (8.3-6) which was derived on the basis of the assumption that the modulating waveform is a sinusoid.]

tral

=

lated

(
plied

if/2

8.3-1.)

multiplied by a local carrier cos 2nf c

t.

Plot the

spectral density of the noise at the output of the multiplier.

(<•) The signal plus noise, after multiplication, is passed through a baseband filter and an ampliwhich provides a voltage gain of 10. Plot the power spectral density of the noise at the amplifier output, and calculate the total noise output power.

fier

8.4-1.

Show

that

m 2 (r) in

8.4-i Repeat Prob. 8.3-4 8.4-3.

A

Eq. (8.4-10) if

DSB

carrier of amplitude 10

form of frequency 750 Hz.

It

is

is

bandlimited to 2fM

rather than

mV

.

SSB modulation

is

employed.

50 percent amplitude-modulated by a sinusoidal waveaccompanied by thermal noise of two-sided power spectral density at

fe

is

Figure P8.5-1

IWII 8.5-2. Verify Eqs. (8.5-5) 8.5-3. (o)

Show

and

(8.5-8).

that the output

dent of the IF bandwidth;

i.e.,

SNR

of a

Eq. (8.4-8)

DSB-SC

is

signal which

synchronously detected

is

is

indepen-

independent of the IF bandwidth.

(fc) Show that the output SNR of an SSB signal which is synchronously detected is dependent on the IF bandwidth. To do this, consider an IF bandwidth which extends from/0 — B tof„ +fu where M > B > 0. Calculate the output SNR using this bandwidth. ,

1

amplitude-modulated signal

8.5- 4. In the received

GJf)

spectral density spectral density 8.6- 1. Verify

Given

number -1000

-fc

-£ + 1000

The

s (t)

A[\

+

power accompanied by noise of power

m(()] cos 2nf,t, m(() has the is

Calculate the output signal-to-noise ratio.

rj/2.

2K +

=

t

received signal

Eq. (8.6-4) by showing graphically that

outside the range \f\ 8.6-2.

specified in Prob. 8.3-4.

1


all

the neglected terms have spectra falling

.

spectral

components of

a noise

waveform spaced by intervals A/ Show that the A/ is p = 2K — 1 — p [i.e., verify the dis-

of pairs of components separated by a frequency p

cussion leading to Eq. (8.6-15)].

lOOOf

/-500

fc

4+1000 /+500 f

/,

Hz

1 MHz and of amplitude 2 volts is amplitudeby a sinusoidal baseband signal of frequency 5 kHz. The signal 6 corrupted by white noise of two-sided power spectral density 10" watt/Hz. The demodulator is a

8.6-3. In

a

modulated Figure P8.4-3 is

DSB

transmission, a carrier of frequency

to the extent of 10 percent

370 PRINCIPLES OF COMMUNICATION SYSTEMS whose input/output

device

characteristic

output and input voltage. The IF istic

filter,

of unity gain and 10-kHz bandwidth.

at 1.002

given by

is

v.

=

3d?

where

,

v„

and

are respectively the

v,

before the demodulator, has a rectangular transfer character-

By

error, the

IF

filter is

tuned so that

its

center frequency

CHAPTER

is

NINE

MHz. Calculate the signal waveform at the demodulator output and calculate

(a)

its

normalized

power.

PJN M = 8.6- 5.

NOISE IN FREQUENCY-MODULATION SYSTEMS

Calculate the noise power at the demodulator output and the signal-to-noise ratio.

(fc)

8.6-4. Plot (1/m

2 )

(SJNJ

versus P,/N u in Eq. (8.6-22) and

show

that the 1-dB threshold occurs

when

4.6 dB.

Assume

that

m 2 (t) =

0.1

and that

SJS0 =

30 dB. Find

PJN U

in Eq. (8.6-22).

Are we above

threshold? 8.7- 1.

A

baseband

signal m(I)

is

superimposed as an amplitude modulation on a carrier

transmission from a transmitting station. density which

falls

off linearly

from

a

The instantaneous amplitude of

maximum

at

m=

extends over a frequency range from zero to 10 kHz.

0 to zero at

The

level

m=

in a

DSB

m(r) has a probability

0.1 volt.

The spectrum

of m(t)

of the modulating waveform applied to

is adjusted to provide for maximum allowable modulation. It is required that under condition the total power supplied by the transmitter to its antenna be 10 kW. Assume that the antenna appears to the transmitter as a resistive load of 72 ohms.

the

modulator

this

(a)

Write an expression for the voltage

(ft)

At a

receiver,

diode demodulator lator input

if

is

at the

input to the antenna.

tuned to pick up the transmission, the

3 volts.

What

is

the

maximum

level

of the carrier at the input to the

allowable power spectral density at the

the signal-to-noise ratio at the receiver output

is

to be at least 20

8.7-1 Plot (1/m 2 ) {SJN,) versus Sj/N M for the envelope demodulator. the intersection of the above-threshold

demodu-

dB?

Assume

that

m2

<S

1,

and

find

and the below-threshold asymptotes.

In this chapter

we

discuss the performance of frequency-modulation systems in

the presence of additive noise.

ment

We shall

show how,

output signal-to-noise power ratio can be bandwidth.

9.1

in

in

an

FM

system, an improve-

made through

the sacrifice oi

AN FM DEMODULATOR AM

The

FM

receiving system of Fig. 8.1-1 may be used with an or signal When used to recover a frequency-modulated signal, the demodulator is replaced by an demodulator such as the limiter-discriminator shown in Fig. 9.1-1.

AM

FM

The Limiter In an FM system, the baseband signal varies only the frequency of the carrier. Hence any amplitude variation of the carrier must be due to noise alone. The

used to suppress such amplitude-variation noise. In a limiter, diodes, used to construct a circuit in which the output voltage t>! is related to the input voltage u, in the manner shown in Fig. 9.1 -la. The output follows the input only over a limited range. A cycle of the carrier is limiter is

transistors, or other devices are

shown

in Fig. 9.1 -2b

and the output waveform

in Fig. 9.1 -2c. In limiter operation

371

Find the

10.11-1.

<

which n/2

<

i//

differential

equation of the piecewise-linear second-order

PLL

in the region in

in/2.

CHAPTER

10.11- 2. Find the differential equation of the second-order

PLL

if

the phase comparator has a sinus-

ELEVEN

oidal characteristic.

Show

10.12- 1.

that Fig. 10.12-1 represents the first-order

10.12-2. Plot

and fu

=

10.14-1.

5

Show

bandpass 10.14-2.

N/fu versus SJnfu at threshold. This kHz, find JV to produce threshold. that the ratio SJt\fu

shown

filter

Show

S /nB p with l

is

called the threshold hyperbola. If S,/V w

is

unchanged when measured

=

20

DATA TRANSMISSION

dB

after the carrier filter or the

in Fig. 10.13-1.

FMFB, SNp N r Use + aG„)B p ]/Af= 4 (this corresponds

that for the

[(1

PLL.

.

Eq. (10.14-6) and plot to choosing

1

+ «G 0 =

NJSN 0,

as a function of

which represents rea-

sonably good design). 10.14- 3. Find the threshold extension possible for 10.15- 1. Refer to Fig. 10.15-3.

when

Show

that

if,

due

fl

=

3

and

to noise, t

Q

5, if is

1

+ «G 0 =

ft.

momentarily displaced to

r

=

3T„/4 or

Q will return to t a = TJ2. 10.15- 2. Design a time-to-voltage converter having the characteristics that 20 past samples are used to compute v„ and that all samples receive equal weight. TJ4, then

10.16- 1. will

Show

the disturbance subsides,

that

if

H(s)

=

«(1

+

\/s)

be able to follow jitter occurring

10.16-2.

r

and the

at the rate

PLL is adjusted to have a f4 with peak deviations less

Repeat Prob. 10.16-1 for the Costas loop shown

bandwidth than

2Mft

,

the

PLL

Af.

in Fig. 10.16-2.

A

data transmission system using binary encoding transmits a sequence of binary is, l's and 0's. These digits may be represented in a number of ways. For example, a 1 may be represented by a voltage V held for a time T, while a zero is represented by a voltage - V held for an equal time. In general the binary digits, that

encoded so that a 1 is represented by a signal s,(t) and a 0 by a signal where s,(f) and s 2 (t) each have a duration T. The resulting signal may be transmitted directly or, as is more usually the case, used to modulate a carrier. The received signal is corrupted by noise, and hence there is a finite probability digits are

s 2 (f),

that the receiver will

whether a

1

In this chapter

methods

make an

error in determining, within each time interval,

or a 0 was transmitted.

we make

calculations of such error probabilities

minimize them. The discussion matched filter and correlator. to

will

and discuss

lead us to the concept of the

A BASEBAND SIGNAL RECEIVER

11.1

Consider that a binary-encoded signal consists of a time sequence of voltage + V or — K If there is a guard interval between the bits, the signal forms a sequence of positive and negative pulses. In either case there is no particular levels

waveform of the signal after reception. We are interknowing within each bit interval whether the transmitted voltage V or — V. With noise present, the received signal and noise together will

interest in preserving the

ested only in

was yield shall

+

sample values generally different from ± V. In this case, what deduction the sample value concerning the transmitted bit?

we make from

441

Suppose that the noise probability density which

entirely symmetrical with respect to zero volts.

is

Dump

gaussian and therefore the noise voltage has a

is

Then

sample value is the same as the probability that the noise has decreased the sample value. It then seems entirely reasonable that we can do no better than to assume that if the sample value is positive the transmitted level was + V, and if the sample value is negative the

White gaussian noise, n(t)

the probability that the noise has increased the

—K

transmitted level was

may

noise voltage

It is,

V and

bit.

sustained over an interval

is

noise. If

+V

level

now

T

from

so that the voltage

the sampling should

happen

t

l

to

made

t

2

.

Noise has been superim-

and At, an

to take place at a time

t

=

/

t

4-

can reduce the probability of error by processing the received signal plus manner that we are then able to find a sample time where the sample voltage due to the signal is emphasized relative to the sample voltage due

after

t

= T

is shown in Fig. 11.1-2. The signal input As a matter of convenience we have set t = 0 at The waveform of the signal s(t) before t = 0 and

has not been indicated since, as

and future

The

signal

will

with added white gaussian noise

s(t)

SWj

at time

t

=0—

,

interval, at

this t

=

sampling switch T.

The

SW2

.

0

may

thus relieving

during the previous interval. The sample

by closing

=

t

is

C

Figure 11.1-2

A

receiver for a binary

coded

n(t)

of power spectral density

require that capacitor

C

Peak Signal The

to

RMS Noise Output Voltage Ratio

integrator yields an output which

1/RC. Using t

= RC, we

v„(T)

=

-

f

the integral of

is

[s(0

+

=

n(t)] dt

T Jo

The sample voltage due

-\ T

=-

to the noise

it

may have

is

signal processing indicated in Fig. 11.1-2

is

n(t)

dt

(1

1.1-1)

C

T

V

dt

=

VT —

(11.1-2)

t

is

=

n 0 (T)

bit

-

n(t)dt

(11.1-3)

I

T Jo

acquired

taken at the end of the

f

1 Jo

is

1

s 0 (T)

be ensured by a brief closing of

of any charge

dt.+-

s(t)

Jo

to the signal

The sample voltage due

input multiplied by

its

have

be

taken at the output of the integrator

dump

T)

signal.

appear, the operation of the

+ we

This sample

the phrase integrate and dump, the term

(

Integrator

bit intervals.

uncharged. Such a discharged condition switch

1

t Jo

presented to an integrator. At time

is

r

0

independent of the waveform during past

is

v„

"„(<)

(receiver)

receiver during each bit interval

n/2

oN,

bit interval is indicated.

the beginning of the interval.

Sample switch SW,

represents the received signal

v

We

Such a processer

_

amplifier

V

as

noise in such a

during a

High- gain

represented by the voltage

is

error will have been made.

to the noise.

ij/2

<0

of a polarity opposite to

In this case an error will be

indicated in Fig. 11.1-1. Here the transmitted bit

+ F which

=

SW,

of course, possible that at the sampling time the

be of magnitude larger than

the polarity assigned to the transmitted

posed on the

G„ •

switch

oN>-

This noise-sampling voltage n 0 (T) n(t),

described by

which

is

a

is

a gaussian

random process. n„(T) was found in Sec.

random

variable in contrast with

gaussian

The variance

of

7.9 [see

Eq. (7.9-17)] to be

referring to the abrupt discharge of

the capacitor after each sampling.

and, as noted in Sec. - 1

V

— +

n„(T) has a gaussian probability density. switch, is

is

t>„(()

=

s 0 (t)

a ramp, in each bit

T. At the end of the interval the ramp attains the voltage which is + VT/x or - VT/x, depending on whether the bit is a 1 or a 0. At the end of each interval the switch SWt in Fig. 11.1-2 closes momentarily to dis-

interval, of duration

V

s0 (T)

charge the capacitor so that

s 0 (t) drops to zero. The noise n 0 (t\ shown in each interval with «„(0) = 0 and has the random value n a (T) at the end of each interval. The sampling switch 2 closes briefly just

i

0

7.3,

The output of the integrator, before the sampling n a (t). As shown in Fig. 11.1 -3a, the signal output s„(t)

h

«i

!«

T-

*

Fig. 11.1 -3b, also starts

-

SW

before the closing of Figure 11.1-1 Illustration that noise in the

may

SW

i

and hence reads the voltage

cause an error

determination of a transmitted voltage

level.

v„(T)

=

s n (T)

+

n 0 (T)

(11.1-5)

•„(<)

We ian

have seen that the probability density of the noise sample n„{T)

and hence appears

as in Fig.

1.2-1.

1

The

,

the variance,

is

a „2

==

is

is

gauss-

therefore given by

==-

/[n.(7TJ= where a 2

density

(11.2-1)

2

n (T) given by Eq. (11.1-4). Suppose, then, that

voltage is held at, say, — V. Then, at the sample time, the signal sample voltage is s 0 (T) = - VT/x, while the noise sample is n„(T). If n (T) is positive and larger in magnitude than VT/x, the total sample 0 voltage v„(T) = s 0 {T) + n 0 (T) will be positive. Such a positive sample voltage will result in an error, since as noted earlier, we have instructed the receiver to interpret such a positive sample voltage to mean that the signal voltage was + V during the bit interval. The probability of such a misinterpretation, that is, the

during some

probability that n 0 (T)

(»)

Fig. Figure

1

1.1-3 («)

The

signal output

and

(h)

bit interval the input-signal

1

1.2-1.

The

> VT/x,

We

would naturally

like the

output signal voltage to be as large as possible

comparison with the noise voltage. Hence a

figure of merit of interest

is

Defining x

Jf.

enhances the signal relative to the noise, and

this

enhancement increases with

in Eq. (1 1.1-6).

PROBABILITY OF ERROR

Since the function of a receiver of a data transmission is to distinguish the bit 1 bit 0 in the presence of noise, a most important characteristic is the

from the

probability that an error will be this error probability

P e for

made

in

such a determination.

We now

the integrate-and-dump receiver of Fig.

1

calculate

1.1-2.

in

which E s

dn 0 (T)

,

1

It is instructive to note that the integrator filters the signal and the noise such that the signal voltage increases linearly with time, while the standard deviation (rms value) of the noise increases more slowly, as Thus, the integrator

11.2

using Eq.

n£T)l^/2o 0 and using Eq.

P.

This result is calculated from Eqs. (11.1-2) and (11.1-4). Note that the signal-tonoise ratio increases with increasing bit duration T and that it depends on V 2 T which is the normalized energy of the bit signal. Therefore, a bit represented by a narrow, high amplitude signal and one by a wide, low amplitude signal are equally effective, provided V 2 T is kept constant.

shown

=

the

signal-to-noise ratio

time as

is,

(1 1.2-1).

the noise output of the integrator of Fig. 11.1-2.

/K(T)]

in

given by the area of the shaded region in

is

probability of error

2

is

(11.1-4), Eq. (11.2-2)

dn„(T)

may be

(1 1.2-2)

rewritten as

m C

= 5 -7=

= V2 T

-===-

=

e-*>dx

the signal energy of a

bit.

+ V during some bit interval, then it from the symmetry of the situation that the probability of error would again be given by P e in Eq. (11.2-3). Hence Eq. (11.2-3) gives P r quite generally. If

is

the signal voltage were held instead at

clear

As shown

in Fig. 11.3-1 the input,

addition of noise

0.5

n(t).

The noise

is

which

is s,(t)

or

s 2 (f), is

corrupted by the

gaussian and has a spectral density G(f). [In white, so that G(f) = r\jl. However, we shall

most cases of interest the noise is assume the more general possibility, since it introduces no complication to do so.] The signal and noise are filtered and then sampled at the end of each bit interval. The output sample is either v 0 (T) = s (T) + n„(T) or v„(T) = s„ (T) oi 2 + n„{T). We assume that immediately after each sample, every energy-storing element

has been discharged. have already considered in Sec. 2.22, the matter of signal determination in the presence of noise. Thus, we note that in the absence of noise the output sample would be v B (T) = s oX (T) or s„ 2 (T). When noise is present we have shown that to minimize the probability of error one should assume that s,(f) has been in the filter

We

transmitted

if

v„(T)

is

has been transmitted

midway between

fore

Fig. 11.1-2,

e„(T)

=

0.

where

closer to s 0l (T) than to s n2 (T). Similarly, we assume s (t) 2 v„(T) is closer to s„ 2 (T). The decision boundary is there-

if

In general,

and

s„,(T)

so1 (T)

we

c

Thus, even

if

the receiver

the signal

is

entirely lost in the noise so that

a sheer guess, the receiver cannot be on the average. is

J.

any determination of

wrong more than

half the time

The probability of error

=

We

We

(T)^so2 (T)

is

we

shall discuss

optimum

filters

as

>

larger in

if

Hence the probability of error

S° l{T)

- S°> iT) (11.3-2)

is

»

contemplated in the system of indeed the optimum filter. However, before returning

specifically to the integrator receiver,

may be deduced

was transmitted. If, at magnitude than the voltage difference 2 [_s oi (T) + s„ 2 (TJ] — s„ 2 (T), an error will have been made. That is, an error [we decide that .s,(r) is transmitted that s 2 (t)

g

shall find that for the received signal

Fig. 11.1-2 the integrator

(n3i)

filter (i.e.,

the integrator), so that at the sampling time the signal voltage might be emphasized in comparison with the noise voltage. are naturally led to ask whether the integrator is the optimum filter for the purpose of minimizing the probability

of error.

to be

an extension s o2 (T) and the sampling time, the noise n 0 (T) is positive and

n 0 (T)>

was passed through a

soi

in the baseband system of VT/x, the decision boundary is

boundary

for this general case

THE OPTIMUM FILTER

In the receiver system of Fig. 11.1-2, the signal

= -

of the considerations used in the baseband case. Suppose that s oi (T)

rather than s 2 (r)] will result

11.3

s„ 2 (T)

shall take the decision

Vo(T)

The probability of error P., as given in Eq. (11.2-3), is plotted in Fig. 11.2-2. Note that P e decreases rapidly as EJq increases. The maximum value of P is

For example,

s„ 2 (T).

= VT/x and

J"[i.i(r)- J .2(r)]/2

-n„2(D/2»„2

=^dnlT)

(11.3-3)

7

N/ 27t
more gen-

erally.

We

assume that the received signal is a binary waveform. One binary digit represented by a signal waveform s^t) which persists for time T, while the other bit is represented by the waveform s 2 (t) which also lasts for an interval T.

(bit) is

For example, s i(t)

= + cos

V, while s 2 (t)

o) 0 t;

Spectral

density,

On (f) Sample

in the case of

are transmitted.

—A

Gaussian noise, n(()

- —

transmission at baseband, as shown in Fig. 11.1-2, V; for other modulation systems, different waveforms

For example,

while for

FSK,

for

s^t)

PSK signalling, s,(r) = A cos w 0 1 and s 2 (t) = = A cos (co„ + Q)f and s 2 (t) = A cos (
every

T sec

Figure 113-1

A

receiver for binary

coded

signalling.

f„(T)

-

«„i(r) + n 0 (T) i

or «

M <mi.0 (r>

Tf

we make

Pc =

=

x

the substitution

n„( T)jjl(ta

2

1

n

If

Eq. (11.3-3) becomes

H( f)

is

the transfer function of the

*~"

(U.3-4a)

<**

and

'[s 0 i(r)-s„2(7')]/2y2«„

Po (T)

Note 4),

-r

The complementary its

argument. (See Fig.

difference s o1 (T) smaller.

(ll.3-4/>)

J VT/x and

s„ 2 (T)

= - VT/x,

and, using Eq.

(1 1.1-

The input

optimum

noise to the

-

error function

The optimum

becomes

filter,

then,

larger

is

the

\H(f)\ G„(f)

is

to be anticipated, P„ decreases as the

and

filter

rms noise voltage a„ becomes which maximizes the ratio as the

Using Parseval's theorem (Eq. power,

the noise variance

i.e.,

We now

1.13-5), ,

we

find that the normalized output noise

is

2

-a±-L

22i_!

(11.3-5)

From

Eqs.

(1

1.3-9)

and

calculate the transfer function

H(f) of this optimum filter. As a matter of actually maximize y 2 rather than y.

.

(11.3-11)

we now

Equation

(11.3-11)

find that

If?. H(f)P(f)e^ df\ $™ x \H(f)\ 2 G„(f)df

P o(T)

2

\H(f)\ G„(f)df

00

2

mathematical convenience we shall

(11.3-10)

2 tr

G„.(f)df=\ n„V.

=

The output noise is n 0 (t) which power spectral density of

related to the

is

2

GJf) =

-

y

(11.3-9)

a monotonically decreasing function of

is

Hence, as

11.2-2.)

s o2 (T)

df

GO

filter is n(t).

has a power spectral density G„ c (f) and the input noise G„( f) by

1.3-4b) reduces to Eq. (11.2-3) as expected.

(1

H(f)P(fy 2 >" T

f°°

J—

00

,2(71 1

rfc

=

that for the case s ol (T)

Eq.

(n.3-8)

= f" P0(fy2 " T df= J-

p

filter,

wmn

pjj) =

f"

-^= V

,

2

(11.3-12)

is unaltered by the inclusion or deletion of the absolute value numerator since the quantity within the magnitude sign p„(T) is a positive real number. The sign has been included, however, in order to allow further development of the equation through the use of the Schwarz inequality. The Schwarz inequality states that given arbitrary complex functions X(f) and Y(f) of a common variable /, then

1.3-12)

(1

sign in the

Calculation of the Optimum-Filter Transfer Function H(f)

The fundamental requirement we make of a binary encoded data receiver is that it distinguishes the voltages s (t) + n(t) and s (t) + n(t). We have seen that the 2 ability of the receiver to do so depends on how large a particular receiver can make y. It is important to note that y is proportional not to s^t) nor to s (f), but 2 rather to the difference between them. For example, in the baseband system we represented the signals by voltage levels + V and - V. But clearly, if our only interest was in distinguishing levels, we would do just as well to use +2 volts and 0 volt, or + 8 volts and + 6 volts, etc. (The + V and - V levels, however, have the advantage of requiring the least average power to be transmitted.) Hence, while Sj(t) or s 2 (t) is the received signal, the signal which is to be compared with t

the noise, is

i.e.,

the signal which

is

relevant in

all

=

s,(t)

Thus, for the purpose of calculating the output signal of the

The equal

sign applies

where

K

is

signal to the

filter is

P„(t\

P(f) and

2

df\ J-

\

Y(f)\

2

df

(11.3-13)

00

= KY*(f)

(11.3-14)

an arbitrary constant and Y*(f) is the complex conjugate of Y( f). apply the Schwarz inequality to Eq. (11.3-12) by making the identifi-

We now

-

s 2 (t)

minimum

optimum

filter

(11.3-6)

we shall The corresponding

error probability, is

p(t).

and

X(f)

= jGjf)H(f)

Y(f)

=

(11.3-15)

1

P(f)e J2

*T

'

(1

1.3-16)

V G n(f)

then Po(t)

shall let

\X{f)\

when X(f)

Using Eqs.

We

I"00 J-

cation p(t)

assume that the input

<

0

our error-probability calculations,

the difference signal

X

2

X(f)Y(f) df

P0 (f)

s ol (t)

-

s„ 2 (t)

be the Fourier transforms, respectively, of

(11.3-7)

pit)

and

we may

(11.3-15)

and

(1

1.3-16)

and using the Schwarz

rewrite Eq. (11.3-12) as

pjmji^x {f )Y(f)df\

2

f-

inequality, Eq.(11.3-13),

or,

using Eq.

(1 1.3-16),

Finally, since p(f)

r

O0

The

2

ratio p (T)/(r

Eq. (11.3-18)

may

2

J-

2

will attain its

be employed as 2

ratio p (T)/n

2

s,(r)

(n.3-18)

is

Eq.

s 2 (t) [see

—

=

[s,(T

(1 1.3-6)],

-

t)

-

we have

s 2 (T

-

(1 1-4-4)

t)]

1

maximum

the case

-

h(t)

J- 00

CO

from Eqs. (11.3-15) and (11.3-16) that the

maximum

r ^£df G„(/)

i/=

iy(/)i

=

value when the equal sign in when X(f) = KY*(f). We then find optimum filter which yields such a

has a transfer function

The

significance of these results for the

appreciated

by applying them to a

matched

may

filter

be more readily

example. Consider then, as in

specific

is a triangular waveform of duration T, while s 2 (f), as shown in Fig. 1.4-16, is of identical form except of reversed polarity. Then p(f) is as shown in Fig. 11.4-lc, and p( — t) appears in Fig. 1 1.4-1 d. The waveform p( — t) is the waveform p(t) rotated around the axis t = 0. Finally, the waveform p(T — t)

Fig. 11. 4- la, that s,(f) 1

P*(

f)

mf)^K-f/e-^ T

(11.3-19)

called for as the impulse response of the filter in Eq. (11.4-36)

Correspondingly, the

maximum

ratio

is,

from Eq.

(1 1.3-18),

|P(/)|

~pl(T)\

f°°

waveform

2

we

(1 1.1

3-20) to a

number

translated in the positive

translation ensures that

~F77V df

(11.3-20)

t

=

=

0 for

t

<

0 as

0 and then delayed long enough

direction

t

is

required for a causal

matched

(i.e.,

this rotated

is

by amount T. This

last

filter.

consists of p(t) rotated

filter

a time T) to

make

the

filter realiz-

have occasion to apply Eqs. (11.13-19) and

shall

of cases of interest. •l

(O a

11.4

h(t)

In general, the impulsive response of the

about In succeeding sections

p( — t)

WHITE NOISE: THE MATCHED FILTER

z\

T

An optimum filter which yields a maximum ratio p (T)/
t

2

T

filter

becomes

-a

H(f)

The impulsive response of strength impulse applied at

h(t)

=

=

K^

this filter,

r

=

A

r°°

J-o

(11.4-1)

the response of the

filter

to a unit

=

2K —

f

\ T

00

P^/V^'V *" d/ 2

J-oo

P*(f)e>M-T)

P(-«)

df

(11.4-26)

filter will

2K m=— t1 p(/y 2K = — p(T-t) C

t

(11.4-2a)

have an impulse response which is real, i.e., not complex. Therefore h(t) = h*(t). Replacing the right-hand member of Eq. (11.4-26) by its complex conjugate, an operation which leaves the equation unaltered, we have physically realizable

2a

(e)

n

*l

-»" T

- »j(0-»2<«)

0, is

*[//(/)]

2k

i.e.,

e

p(»)

2 *" r

-» df

(i

An

-r

id)

(

i.4-3a) Figure 11.4-1 The signals (c) p(t)

(11.4-36)

axis

(

-

=

=

The waveform amount T.

0. (e)

the right by

(a) s,((), (b) s 2 (l),

Sj((). (d) p(t)

and

rotated about the in

(rf)

translated to

able.

We may

duce would

in

Rewriting Eq.

note in passing, that any additional delay that a filter might introno way interfere with the performance of the filter, for both signal

and noise would be delayed by the same amount, and (which would need similarly to be delayed) the ratio of

at the

(1

1.3-46) using p„(T)

Combining Eq.

=

i

2

-

maximum

PROBABILITY OF ERROR OF THE MATCHED FILTER

we

results

when employing a matched

maximum signal-to-noise ratio = ij/2, Eq. (11.3-20) becomes

found by evaluating the Fq. (11.3-20). With G„(f)

\P(.f)\

L al

Jra.»

2

2

lp

filter,

(T)/(t*~\ m!lx

find that the

j"

|P(/)| \P(f)\

2

(11.5-6)

J

minimum

error probability

is

2

may

(11.5-7)

be

given by

(11.5-8)

We df

(11.5-1)

1 J-

we have

Parseval's theorem

1/2

P (T)

1

more

note that Eq. (11.5-8) establishes

generally the idea that the error

and not on the signal waveshape. point only for signals which had constant

probability depends only on the signal energy

Previously

From

8ff »

L

value of v\{J)la 02

erfc

The probability of error which

we have

rp„ (T)i

erfc

2

L2V2 aJ

(11.5-6) with (11.5-5),

(P e) min corresponding to a 11.5

s o2 (T), 2

would

remain unaltered.

—

s oi (T)

r p»(^)

erfc

x

sampling time

signal to noise

=

we had

established this

voltage levels.

d/= df--

p

2 (f)

dt

=

p\t) dt

(11.5-2)

J* In the last integral in Eq. (11.5-2), the limits take account of the fact that p(t) per-

We note also that Eq. (11.5-8) gives (P e) min for the case of the matched filter and when s,(f) = — s 2 (t). In Sec. 11.2 we considered the case when s,(r) = + V and s 2 (f) = — V and the filter employed was an integrator. There we found [Eq. (1 1.2-3)] that the result for P e was identical with (P e ) m n given in Eq. (1 1.5-8). This agreement leads us to suspect that for an input signal where s,(f) = + V and s 2 (t) = — V, the integrator is the matched filter. Such is indeed the case. For when ,

only a time T. With

sists for

write Eq.

(1

p(f)

=

s^t)

-

s 2 (t),

and using Eq.

we may

(1 1.5-2),

1.5-1) as

we have .

I'M -'MY* ^Pl a Jmax -IT JO o

(11.5-3fl)

=1

-

s\(t)

j

dt+\

LJo

si(t)

dt

-

Jo

2 f

s,(f)s 2 (f) dt

= V

and

s 2 (t)

= -V

the impulse response of the

matched

(11.5-9o)

< < T

0

(11.5-96)

t

(1 1.5-3fc)

Jo

+ Es2 — 2E sl2 )

~~

0
s,(r)

*l

(11.5-3c) h(t)

=

IK —

filter is,

[s,(r

-

t)

from Eq.

-

-

s 2 (T

(1 1.4-4),

(11.5-10)

if]

1

Here £ sl and £, 2 are the energies,

respectively, in s^t)

energy due to the correlation between s^f) and

and

s 2 (t),

while

£sl2

is

the

s 2 (t).

Suppose that we have selected s^t), and let ^(t) have an energy E sl Then it if s 2 (f) is to have the same energy, the optimum choice of s (f) 2

The quantity s^T — t) — s 2 {T — t) is a pulse of amplitude 2V extending from t = 0 to t = T and may be rewritten, with u{t) the unit step,

.

can be shown that

Ht)

is

is

optimum

in that

given signal energy. Letting

s 2 (t)

£ and Eq.

(1

1.5-3c)

s

i

it

=

yields a

-s^f),

output signal pl(T) for a

The constant the gain of the

find

signal

— Es2 = —Esi2 = £

(2V)tu(t)

-

u{t

-

(11.5-11)

T)l

(115-4)

maximum

we

IK — 1

S2«=-Si(0 The choice

=

filter,

1.

4KV/n

in the expression for h(t) (that

is,

has no effect on the probability of error since the gain affects

alike.

AKV/n =

tion of the

becomes

filter)

and noise

so that

s

factor of proportionality

We may

Then

therefore select the coefficient

the inverse transform of

becomes, with

h(t),

that

K

is,

in

Eq. (11.5-11)

the transfer func-

the Laplace transform variable, '

(11.5-5) .

<*»

Jn,.,

f

H(s)

=

1

s

e

(11.5-12) s

s

The first term in Eq. (11.5-12) represents an integration beginning at t = 0, while the second term represents an integration with reversed polarity beginning

= T. The overall response of the matched filter is an integration from = 0 to t = T and a zero response thereafter. In a physical system, as already described, we achieve the effect of a zero response after t = T by sampling at = T, so that so far as the determination of one bit is concerned we ignore the response after - T. at

t

t

matched

the impulsive response of the

If h(t) is

matched have

filter,

then the output of the

can be found using the convolution integral

filter v„(t)

=

v„(t)

v/(X)h(t

-

k)

dk

=

Vi (k)h(t

|

t

J-

-

k)

(see Sec. 1.12).

dk

We

(1 1.6-3)

I

Jo

oo

t

The in

limits

the

Eq.

(1

have been changed to 0 and T since we are interested bit which extends only over that interval. Using which gives h(t) for the matched filter, we have

on the

1.4-4)

integral

response to a

filter

COHERENT RECEPTION: CORRELATION

11.6

h(t)

We

discuss

now an

identical in

alternative type of receiving system which, as

performance with the matched

Fig. 11.6-1, the input

The

a binary data

is

filter

waveform

we

or

shall see,

shown

is

in

T. The received signal plus noise v,{t) is multiplied by a waveform s,(f) - s 2 (t). The output of the multiplier is passed through an integrator whose output is sampled at t = T. As before, immediately n(t).

h{t-k)

so that

corrupted by noise

s 2 (t)

Substituting Eq.

each sampling, at the beginning of each new bit interval, all energy-storing elements in the integrator are discharged. This type of receiver is called

(1 1.6-5)

=

v„(t)

we

are correlating the received signal

and noise with the waveform

s,(t)

n

Since

v,(k)

=

s,(A)

+

into

2K —

a correla-

since

-

s 2 (T

(H.6-4)

f)]

t

+

X)

—

s 2 (T

—t+

A)]

—

s 2 (T

-t+

A)]

(11.6-5)

1

after

-

-

bit length is

locally generated

tor,

t)

1

receiver. Again, as

s,(t)

2K — 0,(7 2K = — 0,(7 —

=

n(k),

we have

(1 1.6-3),

"T v,{X)ls,(T

—t+

X)

dk

(11 .6-6)

J

and

t>„(0

=

s 0 (t)

+

setting

n„(t),

= T

t

yields

s 2 (t).

The output

signal

and noise of the correlator shown

s.m =

; T

s,{t) is

=-

either s,(t) or s 2 (f),

1.6-1

are

s f (A)[5!(A)

«/(0[s 1 (t)

-

s 2 (t)] it

(1 1.6-1)

n(t)[*i(')

-

s 2 (t)] dt

(H.6-2)

I

P

z Jo

where

1

and where

the constant of the integrator (i.e., 1/t times the integral of its input). now compare these

the integrator output is outputs with the matched

t is

We

filter

where

s,(A) is

equal to s,(A) or

Similarly

s 2 (k).

-

we

s 2 (A)]

dk

(11.6-7)

find that

Jo

1

» 0 (T>

r in Fig.

n 0 (T)

=

2K —

C

1

Jo

T n(A)[ Sl (A)

-

s 2 (A)]

dk

(1 1.6-8)

Thus sJT) and n„(T), as calculated from Eqs. (1 1.6-1) and (1 1.6-2) for the corand as calculated from Eqs. (1 1.6-7) and (1 1.6-8) for the matched filter receiver, are identical. Hence the performances of the two systems are identirelation receiver,

outputs. cal.

The matched

filter

and the correlator are not simply two

distinct,

independent

techniques which happen to yield the same result. In fact they are two techniques Local

MO-

signal «2

of synthesizing the

(0

filter h(t).

"oil)

Input

f«i (')+»(«)

A.

11.7 Integrator

-o Sample

Multiplier

at

i

- T

or

\-

0

(T)

An

coherent system of signal reception.

important application of the coherent reception system of Sec. keying (PSK). Here the input signal is

1

1.6 is its use

in phase-shift

Correlator

A

PHASE-SHIFT KEYING

*oi{T) + n 0 [T)\

f(t)

U2 (0 + n(0

Figure 11.6-1

optimum

or

Sj(t)

=A

s 2 (t)

= -A

cos

w0

(11.7-1)

t

cos
1

(11.7-2)

modulating baseband signal. For SSB modulation the modulating baseband signal and the modulated other cases are doubled sideband systems and hence:

W

is

the bandwidth of both

carrier.

However,

the (two-sided) bandwidth of the modulated carrier.

Pe

in

is

to be designed,

it

place that, in addition to such practical constraints as cost, volume,

is

etc.,

In such circumstances, plots as

on allowable error rate. Here P r is

in Fig. 11. 20-1 are very useful.

fixed at

Pe =

\0

5

As we

.

shall

see in Sec. 13.7, there is an ultimate limit to the performance of a communications system. This limit, deduced by Shannon, is given by the inequality:

For the sake of

every case tabulated the

(11.20-2)

.

and (1

fJW,

communication system which

In connection with a

common

there be a specification (11.20-1)

having a uniform basis of comparison, we have error probability

of the

shown

W=| B being

all

bits/s/Hz

P,

=

10"'

1

1.20-1. In Fig.

P e we

shall

right of

Shannon's limiting

,

always find that

MPSK,

than does

1

all

modulation schemes

(see

width of

Table

MPSK

1

1.20-1

any given error probability. yield plots which lie to the

QAM more closely approaches Shannon's curve QAM a more efficient system for operation

indicating that

is

white gaussian noise environment. Note that

MFSK

have included a plot of Eq.

1.20-1, as well as for

plot.

Figure 11.20-1 shows that in a

We

independant of the error probability.

is

1.20-2) in Fig.

and

Fig.

1

decreases while the

1.20-1)

SNR

we

when we compare

find that as

MPSK

with

M increases the band-

required to obtain

P e = 10" 5

increases.

However, with MFSK the bandwidth increases %hile the SNR decreases. In all cases however we note that the slope of each curve is positive, so that an increase in j'JW requires an increase in E h jt\ if a constant bit error rate is to be maintained,

i.e.,

for a constant error rate

(11.20-3)

W If

the error rate were, for example,

imately

dB

1

r\

P r = 10" 6 each

curve would

to the right, except of course, for Shannon's curve

shift

which

is

approxindepen-

dent of P,

REFERENCES 1.

Stein, S.,

and

J.

Jones:

"Modern Communication

Principles,"

McGraw-Hill Book Company,

New

York, 1967. 2.

W. R. Bennett, and S. Stein: "Communication Systems and Techniques," McGrawBook Company, New York, 1966. Wozencraft, J., and I. Jacobs: "Communication Engineering," John Wiley and Sons, New York. Schwartz, M., Hill

3.

1966.

PROBLEMS 11.1-1. (n)

Asymptotic value Figure 11.20-1 Bits/s/Hz

vs.

for

M—>

Find the power spectral density G„(/) of noise

R n {x) =

oo

EJr\ for Probability of error

=

10

(b)

-5 .

The

noise in (a)

is

ff

applied to an integrator at

the noise output of the integrator at

(

=

T.

n(f)

which has an autocorrelation function

2 e-l"'ol

t

-

0.

Find the mean square value [n„(7TT of

The

(c)

-V

noise in

accompanies a signal which consists of either the voltage

(a)

sustained for a time T. At time

= T

t

find the ratio of the integrator output

+V due

or the voltage

to the signal to

rms noise voltage.

the

A

11.1- 2.

received signal

= ±V

s(t)

11.9-1. Plot Eq. (11.9-1) versus EJt).

M-ary

11.11-1. (a)

held for an interval T.

is

gaussian noise of power spectral density

The

t;/2.

The

received signal

signal

is

accompanied by white

(ft)

to be processed as in Fig. 11.1-2.

is

However, as an approximation to the required integrator we use a low-pass RC circuit of 3-dB bandwidth/, Calculate the value of/, for which the signal-to-noise voltage, at the sampling time, will be a maximum. For this value of/c calculate the signal-to-rms noise ratio and compare with Eq. (11.1-6) which applies when an integrator is used. Show that for the RC network

(c)

PSK

Show how

involves choosing signals of the form (cos ro 0

to

(

+

for

ft,)

choose the 0 so that the probability of error of each

M values of

fl.

/.

the same,

is

{

Find the correlator detector. Obtain an expression for the probability of error.

11.11- 2 Verify the entries in Table 11.11-1.

.

the signal-to-noise ratio

about

dB

1

A

11.2- 1.

smaller than for the integrator.

+V

accompanied by white gaussian

Assume

than zero

+ V or - V is transmitted. Consider that the i while the probability of transmitting - V is i The signal is

is

noise.

for

(ft)

I.13-5a and

(I

and

minimum and

a

ii

(b)

cos

A

When a

received,

T

time

it is

embedded

+

—V

V, 0,

white gaussian noise.

in

A

The

with equal likelihood,

transmitted.

is

receiver integrates the signal

and noise

2

ity

of error

A

11.3- 1.

power

±

V,

so that the probability of error

(ft)

T

during which a signal must be sustained

for P,

=

4

lfr

.

An

is

indepen-

integrator

is

for this

is

= G 0 /[l +

2

(/ft)

Find

Repeat Prob. 11.5-1

Compare

power

r

if

the signal

the outputs of the

function of time

A

its

its

(

for

0

<

t

<

T.

is s(t)

MF

= ±A(l -

and the

Assume white gaussian

.

signal

(a)

Draw

(b)

If

when

noise.

waveform

0

<, t <,

T.

The

is

is

the input signal

is

either

Are the outputs the same

(a)

(<)

and

(1

(11.15-7)

assume

V, as a /,

or

output of a matched filter receiver. to be no larger than 10" 4 find the minimum allowable inter-

Prove

FSK

rr

QAM

versus EJt] and

satisfies

compare with

FSK

EJv\ for

BPSK

and

results.

4/,.

that the arc of a circle

is

MPSK

for

if

is

same length

of the

the above approximation

is

as the

not made.

to be correct to within 50 percent deter-

M=

4, 8,

16, 32,

and 64 are

to be the

to (EJij) KKyi ?

using the Union Bound.

Compare your answer

with that of

Pp

for 16

QAM without using the Union Bound approximation.

Using the Union Bound approximation, calculate P, for 64 and 256 QAM. determine the ratio of each EJtj as comTo obtain the same P t for 16, 64 and 256

QAM

A

NRZ

9.6 kb/s

data stream

11. 15- 8.

A

4 error rate of 10~ is

when

EJt]

Can that

is

(minimum

NRZ

9.6 kb/s is

Due

is

to

be transmitted over a 2.4 kHz bandwidth channel. An and its phase characteristics, intersymbol

to channel nonlinearities

produced which

when

if

EJrj)7

data stream

desired.

to be transmitted over a 2.4 kHz bandwidth channel. What 4 an error rate of 10~ is to be achieved with a minimum

is

modulation system would you choose

results in

the transmitter

is

an "eye pattern" which closes by

3

dB compared

to the

connected directly to the receiver (by-passing the channel)

is 12 dB. 4 a modulation system be found to achieve this error rate of 10~ ?

What

is

the value of EJrf

needed?

SIT

=

11.16- 2. ror, s,(«)

and

s 2 (r)

are orthogonal.

this statement,

Plot the P„ in binary

and the

they differ?

is

,

Calculate P„. Plot

compared

The

probability of a bit being in error

message being

Note how quickly the probability of in

1.13-lOu)

(1

B(BFSK) =

and

2/;

in error,

a

(ft)

is

the results given in Eq.

as a function of CIT. Select EJt\

=

(1 1.8-8).

15.

The

message being

probability of a bit being in error

The

10"

3

is

(a) If a

.

Repeat

1.8-8) versus EJij.

Q

and

1.13-4)

BPSK, BFSK and

P, for 16

the probability of the

frequency offset

are

11.15-6. Verify Eq. (11.15-16).

11.15-7.

11.16- 1.

1.8-2. If the

11.8-3.

±

for all

corrupted by white gaussian noise of

11.7- 1. Verify Eq. (11.7-5). (1



so that

pared to (£,yi/W-

at the

the probabilily of error P,


1.15-5. (n)

response obtained signal

and

—

1.14-3) reduces to Eq. (11.14-4).

the ratio of their EJt]

interference (ISI) is s(t)

the signal

11.8- 1. Plot Eq.

is

signal-to-noise ratio

= T?

= ±2(t/T) for spectral density 10~ 6 volt 2 /Hz.

(ft)

(b).

cos 2nf„t).

correlator,

0

Eq. (11.15-11).

(ft)

P,and compare with

val T.

1

(1

1.15-5)

(1

11.15-2. If the probability of

same, what

1

readability,

is either s,(f) = A cos 2nf t or s (r) = 0 for an interval T = n/f0 with n an integer. 0 2 corrupted by white noise with G„{f) = r,/2. Find the transfer function of the matched signal. Write an expression for the probability of error P

11.6- 1.

and n 2 (r)

w,(t)

find a> 2

Why do

for Prob. 11.13-1.

the approximation is to be used but the resulting P, mine the minimum value of M.

11.15-4. Calculate

].

filter.

ti,

which is/t /W and an abscisssa which

.

the probabil-

with equal probability; the channel noise has the

employed rather than the optimum

11.5- 2.

when

if

signal

signal

11.6- 2.

answer obtained

1(T' and 1(T 6 Use Eqs.

11.14- 1. Verify that Eq.

11.15-3. Calculate

±V

e

just

to the

chord, {a) Derive an expression for P, to replace Eq. (11.15-7)

not to exceed 10"*?

transmitter transmits the signals

A

11.5-1.

time

where

(j>)

(b) If

Find the transfer function H(f) of the matched filter, and comment on Find the average probability of error when using the matched filter.

(<)

filter

minimum

spectral density G„(/)

(a)

The

is

the

is

such that the unit vectors

= ^/2/fh cos [w 2 t + - n < ©, 4> <

u 2 (f)

received signal

what

to,

are orthonormal.

Compare your answer

11.15- 1. Equations

receiver,

-

for

is either + 2V or -2V held for a time T. The signal is corrupted by white gaussian noise of power spectral density lfr" volt J /Hz. If the signal is processed by an integrate and

dump

to 2

calculate the corresponding

s.

Write an expression for the threshold voltages dent of which signal is transmitted. 11.2- 3.

+ 0) and

t

B(BPSK) =

which can take on the voltages

signal,

((«,

11.13- 3. Plot a graph having an ordinate

BFSK is

= ^2/7;

11.13-2. (n) If u,(f)

probability of error. 11.2-2.

Determine

b).

independent random variables uniformly distributed between u,

between the two possible signals is V, rather the probability that an error in decision will be made. An

and dump receiver is used as in Fig. 11.1-2. Find V, such that the probability of error

integrate

11.13- 1. Refer to Eq.

that the threshold voltage for decision

Write an expression

volts.

11.12- 1. Verity Eq. (11.12-2).

are orthonormal.

which can assume one of the voltage

signal

probability of transmitting

(a)

is

(a), if

message consists of 10

the message length

in error increases

10~

3 .

(a)

What

is

toward

is

bits,

calculate

100, 1000, 10,000.

unity.

the probability of a 6-bit

word

grouped so that instead of transmitting 6 bits, bit-by-bit, a 64 phase, 64 PSK signal is sent. (The symbol rate is one sixth of the bit rate.) Calculate the probability of a 6-bit symbol being in error, (r) Repeat (ft) using 64 QAM. (d) Repeat using 64 FSK. containing an error?

(p)

(ft)

Discuss your results.

11.17- 1. Verify Eq. (11.17-2).

bits are

11.17- 2. Consider that bits are

16

will

grouped so

that there are 4 bits/symbol

and then modulated using arrange the Grey code so that an error in a symbol with high probability correspond to an error in only of the 4 bits.

QAM.

Following

Fig. 11.17-1

show how

to

CHAPTER

TWELVE

I

11.18- 1. Prove that

H D(f)

as defined

impulse transmitted through

by Eqs.

1.18-2)

(1

and (11.18-3)

the

is

matched

filter

for

an

H t(/).

11.18-2. Verify Eq. (11.18-6). 11.18-3. Verify Eq. (11.18-9)

NOISE IN PULSE-CODE AND DELTA-MODULATION SYSTEMS

and Eq. (11.18-10).

11.18-4. Verify Eq. (11.18-11).

A„(kTs ) and A,(kTJ can each assume one of

11.18-5.

indicate nine signal points.

A

11.18-6.

probability of error of 1(T

the bit rate for

80 kb/s, 16 PSK, 16

is

5 is

QAM,

each of these systems.

11.19MSK FM detection 11.19- 2. MSK

is

=

0.75 fh

bit, i.e. for

(a)

,

IF

filter

very

little

Th

a time

Using Eqs.

is set

to

required signal

filter is

with a peak frequency deviation offJA. If coherent 1.8) calculate the probability of error.

B=

Why

11.20- 1.

\s

power

lies

MSK

is zero at outside this region and hence ISI can be

often replaced by an integrate-and-dump

(9.2-16), (9.2-21),

1

discriminator. Referring to

filter,

which integrates over

.

f = fJ41 (c) How much

Using Table

FM

1.5/, (since the spectral density of

and

(10.6-1)

wherein =f„/4, I

th)

would

(1 1.18-15).

1

noncoherent manner using an

in a

space to yield Eq.

QPR

FM system MFSK (see Sec.

bandwidth

The baseband

neglected).

each

as in

can be detected

Fig. 10.2-1. the

/-/<•

employed

in signal

desired and a channel bandwidth of 20kHz is available If or can be used. Calculate the value of EJq required

can be viewed as an

1.

three values. Thus, a signal space sketch

Determine the nine signal points

1.20-1

degradation

and Shannon's

is

produced when the

calculate the probability of error.

FM discriminator

limit, plot Fig. 11.20-1 for

P.

=

is

employed?

10""".

12.1

A

PCM TRANSMISSION

binary

m{t)

is

PCM

transmission system

is

shown

in Fig. 12.1-1.

quantized, giving rise to the quantized signal

m q(t) =

m(f)

+

e(t)

m ?(f),

The baseband

signal

where (12.1-1)

The term e(t) is the error signal which results from the process of quantization. The quantized signal is next sampled. Sampling takes place at the Nyquist rate. The sampling interval is Ts = \/2fM where fM is the frequency to which the signal ,

m(f)

is

bandlimited.

Sampling is accomplished by multiplying the signal m q (t) by a waveform which consists of a periodic train of pulses, the pulses being separated by the sampling interval Ts We shall assume that the sampling pulses are narrow enough so that the sampling may be considered as instantaneous. It will be recalled (Sec. 5.1) that with such instantaneous sampling, the sampled signal may be reconstructed exactly by passing the sequence of samples through a low-pass Now, as a matter of mathematical convefilter with cut-off frequency at fM nience, we shall represent each sampling pulse as an impulse. Such an impulse is infinitesimally narrow yet is characterized by having a finite area. The area of an impulse is called its strength, and an impulse of strength / is written I5(t). The .

.

sampling-impulse train

is

therefore S(f), given by

(12.1-2)

487