K. Deergha Rao - Signals And Systems (2018, Birkhäuser)_2

1 t (iii) xðt Þ ¼ : 0 t  1: E¼

ð1 1

jxðt Þj2 dt ¼

¼ 4 ln ½t 1 1 ¼1

ð1 1

4 dt t

16

1 Introduction

1 T!1 T

ð T=2

1 T!1 T

ð T=2

4 dt t T=2 1  

 1T=2  1 1 T 1 ¼ 4 lim ln ½t  ln  ln ½1 ¼ 4 lim T!1 T T!1 T 2 T   1 T ¼ 4 lim ln T!1 T 2 0 1 T ln B 2 C C ¼ 4 lim B @ T!1 T A

P ¼ lim

x2 ðt Þdt ¼ lim

Using L’Hospital’s rule, we see that the power of the signal is zero. That is 2  T  ln 2 P ¼ 4 lim ¼ 4 lim T ¼ 0 T!1 T!1 1 T The energy of the signal is infinite and its average power is zero; x(t) is neither energy signal nor power signal. (iv) x(t) ¼ eat for real value of a ð1

ð1

jeat j dt ¼ 1, jxðt Þj dt ¼ 1 1 ð ð 1 T=2 2 1 T=2 2at P ¼ limT!1 x ðt Þdt ¼ limT!1 e dt T T=2 T T=2  aT   aT   aT  e  eaT e e ¼ lim ¼ lim  lim T!1 T!1 2aT T!1 2aT 2aT  aT  e ¼ lim 0 T!1 2aT

E¼

2

2

Using L’Hospital’s rule, we see that the power of the signal is infinite. That is,  P ¼ lim

T!1

eaT 2aT



 aT  e ¼1 T!1 2

¼ lim

The energy of the signal is infinite and its average power is infinite; x(t) is neither energy signal nor power signal. (v) x(t) ¼ cos(t) E¼

ð1 1

2

jxðt Þj dt ¼

ð1 1

cos 2 ðt Þdt ¼ 1,

1.4 Classification of Signals

17

ð ð 1 T=2 2 1 T=2 x ðt Þdt ¼ limT!1 cos 2 ðt Þ dt T T=2 T T=2

ð 1 T=2 1 þ cos ð2t Þ ¼ limT!1 dt T T=2 2 ð ð 1 T=2 1 1 T=2 cos ð2t Þ dt þ limT!1 dt ¼ limT!1 T T=2 2 T T=2 2 1 ¼ 2

P ¼ limT!1

The energy of the signal is infinite and its average power is finite; x(t) is a power signal. (vi) xðt Þ ¼ ejð2tþ4Þ , jxðt Þj ¼ 1. π

E¼

ð1

ð1

2

dt ¼ 1, jxðt Þj dt ¼ 1 ð T=2 ð 1 1 T=2 2 P ¼ limT!1 x ðt Þdt ¼ limT!1 1 dt ¼ limT!1 1 ¼ 1 T T=2 T T=2 1

The energy of the signal is infinite and its average power is finite; x(t) is a power signal. Example 1.11 Consider the following signals, and determine the energy of each signal shown in Figure 1.11. How does the energy change when transforming a signal by time reversing, sign change, time shifting, or doubling it?

x(t)

(t)

2

2

t

2

(t) 2

t

-2

( )

-2

2

( ) 2

4

t

4

2

Figure 1.11 Signals of example 1.11

t

t

18

1 Introduction

(

Solution xðtÞ ¼

t

0
0

otherwise

2 t 3  8 Ex ¼ jxðtÞj dt ¼ t dt ¼  ¼ 3 0 3 1 0 ( t 2 < t  0 x1 ðtÞ ¼ 0 otherwise 0 ð1 ð0 t3  8 E x1 ¼ jx1 ðtÞj2 dt ¼ t 2 dt ¼  ¼ 3 2 3 1 2 ( t 0 < t  2 x2 ðtÞ ¼ 0 otherwise 2 ð1 ð2 t3  8 E x2 ¼ jx2 ðtÞj2 dt ¼ t 2 dt ¼  ¼ 3 0 3 1 0 ( ðt  2Þ 2 < t  4 x3 ðtÞ ¼ 0 otherwise ð1 ð2 ð4 jx3 ðtÞj2 dt ¼ ðt  2Þ2 dt ¼ ðt 2  4t þ 4Þdt E x3 ¼ ð1

ð2

2

2

1

 ¼ ¼ x4 ðtÞ ¼

4  t  2t 2 þ 4t  3 2

8 3 (

0

2

3

2t

0
0 ð1

otherwise

2 t 3  32 E x4 ¼ jx4 ðtÞj dt ¼ 4t dt ¼ 4  ¼ 3 3 1 0 0 2

ð2

2

The time reversal, sign change, and time shifting do not affect the signal energy. Doubling the signal quadruples its energy. Similarly, it can be shown that the energy of k x(t) is k2Ex. Proposition 1.2 The sum of two sinusoids of different frequencies is the sum of the power of individual sinusoids regardless of phase. Proof Let us consider a sinusoidal signal x(t) ¼ Acos(Ωt + θ). The power of x(t) is given by

1.4 Classification of Signals

P ¼ limT!1

1 T

ð T=2 T=2

19

x2 ðtÞdt ¼ limT!1

1 T

ð T=2 T=2 ð T=2

A2 cos 2 ðΩt þ θÞdt

1 A2 ½1 þ cos 2 ð2Ωt þ 2θÞdt 2T T=2 "ð # ð T=2 A2 T=2 ¼ limT!1 dt þ cos ð2Ωt þ 2θÞdt 2T T=2 T=2

¼ limT!1

¼

A2 A2 ½T þ 0 ¼ 2T 2 ð1:12Þ 2

Thus, a sinusoid signal of amplitude A has a power A2 regardless of the values of its frequency Ω and phase θ. Now, consider the following two sinusoidal signals: x1 ðt Þ ¼ A1 cos ðΩ1 t þ θ1 Þ x2 ðt Þ ¼ A2 cos ðΩ2 t þ θ2 Þ Let xs ðt Þ ¼ x1 ðt Þ þ x2 ðt Þ The power of the sum of the two sinusoidal signals is given by 1 Ps ¼ limT!1 T ¼ limT!1 ¼

1 T

ðT

2

T 2

x2s ðtÞ

ð T=2

½A1 cos ðΩ1 t þ θ1 Þ þ A2 cos ðΩ2 t þ θ2 Þ T=2 ðT 1 2 2 limT!1 A cos 2 ðΩ1 t þ θ1 Þdt T T 1 2 ðT 1 2 2 A cos 2 ðΩ2 t þ θ2 Þdt þ limT!1 T T 2 2 ðT 2A1 A2 2 cos ðΩ1 t þ θ1 Þcos ðΩ2 t þ θ2 Þdt þ limT!1 T T 2

2

dt

The first and second integrals on the right-hand side are the powers of the two sinusoidal signals, respectively, and the third integral becomes zero since cos ðΩ1 t þ θ1 Þ cos ðΩ2 t þ θ2 Þ ¼ cos ½ðΩ1 þ Ω2 Þt þ ðθ1 þ θ2 Þ þ cos ½ðΩ1  Ω2 Þt þ ðθ1  θ2 Þ Hence, Ps ¼

A21 A22 þ 2 2

ð1:13Þ

20

1 Introduction

It can be easily extended to sum of any number of sinusoids with distinct frequencies

1.4.7

Deterministic and Random Signals

For any given time, the values of deterministic signal are completely specified as shown in Figure 1.12(a). Thus, a deterministic signal can be described mathematically as a function of time. A random signal takes random statistically characterized random values as shown in Figure 1.12(b) at any given time. Noise is a common example of random signal.

1.5 1.5.1

Basic Continuous-Time Signals The Unit Step Function

The unit step function is defined as  uð t Þ ¼

1 0

t>0 t<0

ð1:14Þ

Amplitude

which is shown in Figure 1.13. It should be noted that u(t) is discontinuous at t ¼ 0. 1 0.5 0 −0.5 −1 0

1 0.8 0.6 0.4 0.05

0.1 0.15 Time (sec)

0.2 0.2 0 0

(a)

50

100

150

(b)

Figure 1.12 (a) Deterministic signal, (b) random signal

Figure 1.13 Unit step function

u(t) 1

t

1.5 Basic Continuous-Time Signals

1.5.2

21

The Unit Impulse Function

The unit impulse function also known as the Dirac delta function, which is often referred as delta function is defined as δðt Þ ¼ 0, t 6¼ 0 ð1 δðt Þ ¼ 1:

ð1:15aÞ ð1:15bÞ

1

The delta function shown in Figure 1.14(b) can be evolved as the limit of the rectangular pulse as shown in Figure 1.14(a). δðt Þ ¼ lim pΔ ðtÞ

ð1:16Þ

Δ!0

As the width Δ ! 0, the rectangular function converges to the impulse function δ(t) with an infinite height at t ¼ 0, and the total area remains constant at one. Some Special Properties of the Impulse Function • Sampling property If an arbitrary signal x(t) is multiplied by a shifted impulse function, the product is given by xðt Þδðt  t 0 Þ ¼ xðt 0 Þδðt  t 0 Þ

ð1:17aÞ

implying that multiplication of a continuous-time signal and an impulse function produces an impulse function, which has an area equal to the value of the continuous-time function at the location of the impulse. Also, it follows that for t0 ¼ 0, xðt Þδðt Þ ¼ xð0Þδðt Þ

ð1:17bÞ

• Shifting property ð1 1

Figure 1.14 (a) Rectangular pulse, (b) unit impulse

xðt Þδðt  t 0 Þdt ¼ xðt 0 Þ

ð1:18Þ

p Δ (t) 1/Δ

−Δ/2

Δ/2

(a)

d (t)

t

0

(b)

t

22

1 Introduction

• Scaling property   1 b δðat þ bÞ ¼ δ t þ a j aj

ð1:19Þ

• The unit impulse function can be obtained by taking the derivative of the unit step function as follows: δ ðt Þ ¼

duðt Þ dt

ð1:20Þ

• The unit step function is obtained by integrating the unit impulse function as follows: uð t Þ ¼

1.5.3

ðt 1

δðt Þdt

ð1:21Þ

The Ramp Function

The ramp function is defined as  r ðt Þ ¼

t 0

t>0 t<0

ð1:22aÞ

which can also be written as r ðt Þ ¼ tuðt Þ The ramp function is shown in Figure 1.15.

1.5.4

The Rectangular Pulse Function

The continuous-time rectangular pulse function is defined as Figure 1.15 The ramp function

ð1:22bÞ

1.5 Basic Continuous-Time Signals

23

x(t)

Figure 1.16 The rectangular pulse function

1

- 1

0

1

t

x(t)=

Figure 1.17 The signum function

1

0

t

-1

 xð t Þ ¼

1 0

jt j  T 1 jt j > T 1

ð1:23Þ

which is shown in Figure 1.16.

1.5.5

The Signum Function

The signum function also called sign function is defined as 8 t>0 < 1 sgnðt Þ ¼ 0 t¼0 : 1 t<0

ð1:24Þ

which is shown in Figure 1.17.

1.5.6

The Real Exponential Function

A real exponential function is defined as xðt Þ ¼ Aeσt

ð1:25Þ

where both A and σ are real. If σ is positive, x(t) is a growing exponential signal. The signal x(t) is exponentially decaying for negative σ. For σ ¼ 0, the signal x(t) is equal to a constant. Exponentially decaying signal and exponentially growing signal are shown in Figure 1.18(a) and (b), respectively.

24

1 Introduction

x(t)

x(t)

A

A

t

0

0

(a)

t

(b)

Figure 1.18 Real exponential function. (a) Decaying, (b) growing

1.5.7

The Complex Exponential Function

A real exponential function is defined as xðt Þ ¼ AeðσþjΩÞt

ð1:26Þ

xðt Þ ¼ Aeσt ejΩt

ð1:26aÞ

ejΩt ¼ cos ðΩt Þ þ j sin ðΩt Þ

ð1:27Þ

Hence

Using Euler’s identity

Substituting Eq. (1.27) in Eq. (1.26a), we obtain xðt Þ ¼ Aeσt ð cos ðΩt Þ þ j sin ðΩt ÞÞ

ð1:28Þ

Real sine function and real cosine function can be expressed by the trigonometric identities as jΩt jΩt jΩt jΩt cos ðΩt Þ ¼ e þe and sin ðΩt Þ ¼ e e 2 2j

1.5.8

The Sinc Function

The continuous-time sinc function is defined as Sincðt Þ ¼ which is shown in Figure 1.19

sin ðπt Þ πt

ð1:28Þ

1.5 Basic Continuous-Time Signals

25

Figure 1.19 The sinc function

Example 1.12 State whether the following signals are causal, anticausal, or noncausal. (a) x(t) ¼ e2tu(t) (b) x(t) ¼ tu(t)  t(u(t  1) þ e(1t)u(t  1)) (c) x(t) ¼ et cos (2πt)u(1  t) Solution (a)

x(t) 1 0.25 1

t

It is causal since x(t) ¼ 0 for t < 0 (b)

x(t) 1

1 It is causal since x(t) ¼ 0 for t < 0

t

26

1 Introduction

(c)

(c ) u(1-t)

t

1

for t<0

It is non causal since

Example 1.13 Determine and plot the even and odd components of the following continuous-time signal xðt Þ ¼ tuðt þ 2Þ  tuðt  1Þ Solution

xðt Þ ¼ tuðt þ 2Þ  tuðt  1Þ xðt Þ ¼ tuðt þ 2Þ þ tuðt þ 1Þ

x(t)

x(-t)

1

1

-2

1

t

-2

1

-2

-2

xðtÞ þ xðtÞ 2  1
¼ t uðt þ 2Þ  uðt  1Þ  uðt þ 2Þ þ uðt þ 1Þ 2

xe ðtÞ ¼

2

t

1.5 Basic Continuous-Time Signals

27 ( )

1 1

2

-2

t

-0.5 -1

xðt Þ  xðt Þ 2 1 ¼ t ðuðt þ 2Þ  uðt  1Þ þ uðt þ 2Þ  uðt þ 1ÞÞ 2

xo ðt Þ ¼

1 1

2

-2 -0.5 -1

Example 1.14 Simplify the following expressions:
 t (a) tsin 2 þ3 δðt Þ (b) (c)

4þjt

Ð3jt 1

δ ð t  1Þ

1

ð4t  3ÞÞδðt  1Þdt

Solution
 ð0Þ (a) sin 0þ3 δðt Þ ¼ 0 (b) (c)

4þjt 4þj 3jt δðt  1Þ ¼ 3j δðt  1Þ Ð1 1 ð4ð1Þ  3ÞÞδðt  1Þdt

¼

Ð1

1

1δðt  1Þdt ¼ 1:

t

28

1.6

1 Introduction

Generation of Continuous-Time Signals Using MATLAB

An exponentially damped sinusoidal signal can be generated using the following MATLAB command: x(t) ¼ A ∗ sin (2 ∗ pi ∗ f0 ∗ t + θ) ∗ exp (a ∗ t)where a is positive for decaying exponential. Example 1.15 Write a MATLAB program to generate the following exponentially damped sinusoidal signal. x(t) ¼ 5 sin (2πt)e0.4t  10  t  10 Solution The following MATLAB program generates the exponentially damped sinusoidal signal as shown in Figure 1.20. MATLAB program to generate exponentially damped sinusoidal signal clear all;clc; x =inline('5*sin(2*pi*1*t).*exp(-.4*t)','t'); t = (-10:.01:10); plot(t,x(t)); xlabel ('t (seconds)'); ylabel ('Ámplitude');

250 200 150

Ámplitude

100 50 0 -50 -100 -150 -200 -250 -10

-8

-6

-4

-2

0 2 t (seconds)

4

6

8

10

Figure 1.20 Exponentially damped sinusoidal signal with exponential parameter a ¼ 0.4.

1.6 Generation of Continuous-Time Signals Using MATLAB

29

2 1.5 1

Ámplitude

0.5 0 -0.5 -1 -1.5 -2 -5

-4

-3

-2

-1

0

1

2

3

4

5

t (seconds)

Figure 1.21 Unit step function

Example 1.16 Generate unit step function over [5,5] using MATLAB Solution The following MATLAB program generates the unit step function over [5,5] as shown in Figure 1.21. MATLAB program to generate unit step function over [5,5] clear all;clc; u=inline('(t>=0)','t'); t=-5:0.01:5; plot(t,u(t)) xlabel ('t (seconds)'); ylabel ('Ámplitude') axis([-5 5 -2 2])

Example 1.17 Generate the following rectangular pulse function rect(t) using MATLAB:  t 1, 5 < t < 5 rect 10 ¼ 0, elsewhere Solution The following MATLAB program generates the rectangular pulse function as shown in Figure 1.22.

30

1 Introduction 2 1.5 1

Ámplitude

0.5 0 -0.5 -1 -1.5 -2 -10

-8

-6

-4

-2

0

2

4

6

8

10

t (seconds)

Figure 1.22 Rectangular pulse function

MATLAB program to generate rectangular pulse function clear all;clc; u=inline('(t>=-5)& (t<5)','t'); t=-10:0.01:10; plot(t,u(t)) xlabel ('t (seconds)'); ylabel ('Ámplitude') axis([-10 10 -2 2])

1.7 1.7.1

Typical Signal Processing Operations Correlation

Correlation of signals is necessary to compare one reference signal with one or more signals to determine the similarity between them and to determine additional information based on the similarity. Applications of cross correlation include crossspectral analysis, detection of signals buried in noise, pattern matching, and delay measurements.

1.7 Typical Signal Processing Operations

1.7.2

31

Filtering

Filtering is basically a frequency domain operation. Filter is used to pass certain band of frequency components without any distortion and to block other frequency components. The range of frequencies that is allowed to pass through the filter is called the passband, and the range of frequencies that is blocked by the filter is called the stopband. A low-pass filter passes all low-frequency components below a certain specified frequency Ωc, called the cutoff frequency, and blocks all high-frequency components above Ωc. A high-pass filter passes all high-frequency components above a certain cutoff frequency Ωc and blocks all low-frequency components below Ωc. A band-pass filter passes all frequency components between two cutoff frequencies Ωc1 and Ωc2 where Ωc1 < Ωc2 and blocks all frequency components below the frequency Ωc1 and above the frequency Ωc2. A band-stop filter blocks all frequency components between two cutoff frequencies Ωc1 and Ωc2 where Ωc1 < Ωc2 and passes all frequency components below the frequency Ωc1 and above the frequency Ωc2. Notch filter is a narrow band-stop filter used to suppress a particular frequency, called the notch frequency.

1.7.3

Modulation and Demodulation

Transmission media, such as cables and optical fibers, are used for transmission of signals over long distances; each such medium has a bandwidth that is more suitable for the efficient transmission of signals in the high-frequency range. Hence, for transmission over such channels, it is necessary to transform the low-frequency signal to a high-frequency signal by means of a modulation operation. The desired low-frequency signal is extracted by demodulating the modulated high-frequency signal at the receiver end.

1.7.4

Transformation

The transformation is the representation of signals in the frequency domain, and inverse transform converts the signals from the frequency domain back to the time domain. The transformation provides the spectrum analysis of a signal. From the knowledge of the spectrum of a signal, the bandwidth required to transmit the signal can be determined. The transform domain representations provide additional insight into the behavior of the signal and make it easy to design and implement algorithms, such as those for filtering, convolution, and correlation.

32

1.7.5

1 Introduction

Multiplexing and Demultiplexing

Multiplexing is used in situations where the transmitting media is having higher bandwidth, but the signals have lower bandwidth. Thus, multiplexing is the process in which multiple signals, coming from different sources, are combined and transmitted over a single channel. Multiplexing is performed by multiplexer placed at the transmitter end. At the receiving end, the composite signal is separated by demultiplexer performing the reverse process of multiplexing and routes the separated signals to their corresponding receivers or destinations. In electronic communications, the two basic forms of multiplexing are timedivision multiplexing (TDM) and frequency-division multiplexing (FDM). In time-division multiplexing, transmission time on a single channel is divided into non-overlapped time slots. Data streams from different sources are divided into units with same size and interleaved successively into the time slots. In frequency-division multiplexing (FDM), numerous low-frequency narrow bandwidth signals are combined for transmission over a single communication channel. A different frequency is assigned to each signal within the main channel. Code-division multiplexing (CDM) is a communication networking technique in which multiple data signals are combined for simultaneous transmission over a common frequency band.

1.8 1.8.1

Some Examples of Real-World Signals and Systems Audio Recording System

An audio recording system shown in Figure 1.23(a) takes an audio or speech as input and converts the audio signal into an electrical signal, which is recorded on a magnetic tape or a compact disc. An example of recorded voice signal is shown in Figure 1.23(b).

Audio Recording System

(a)

Audio output signal

(b)

Figure 1.23 (a) Audio recording system, (b) the recorded voice signal “don’t fail me again”

1.8 Some Examples of Real-World Signals and Systems

1.8.2

33

Global Positioning System

The satellite-based global positioning system (GPS) consists of a constellation of 24 satellites at high altitudes above the earth. Figure 1.24 shows an example of the GPS used in air, sea, and land navigation. It requires signals at least from four satellites to find the user position (X, Y, and Z) and clock bias from the user receiver. The measurements required in a GPS receiver for position finding are the ranges, i.e., the distances from GPS satellites to the user. The ranges are deduced from measured time or phase differences based on a comparison between the received and receivergenerated signals. To measure the time, the replica sequence generated in the receiver is to be compared to the satellite sequence. The correlator in the user GPS receiver determines which codes are being received, as well as their exact timing. When the received and receiver-generated sequences are in phase, the correlator supplies the time delay. Now, the range can be obtained by multiplying the time delay by the velocity of light. For example, assuming the time delay as 3 ms (equivalent to 3 blocks of the C/A code of satellite 12), the correlation of satellite 12 producing a peak after 3 ms [Rao06] is shown in Figure 1.25.

1.8.3

Location-Based Mobile Emergency Services System

Mobile emergency services (MES) refer to the use of mobile positioning technology to pinpoint mobile users for purposes of providing enhanced wireless emergency dispatch services (including fire, ambulance, and police) to mobile phone users. In this emergency service system, user should have assisted GPS-enabled mobile handset unit. Network service providers will support “Mobile Location Protocol Figure 1.24 A pictorial representation of GPS positioning

34

1 Introduction

Figure 1.25 The correlation of satellite 12 producing a peak

160 140 120 100 80 60 40 20 0 −20

0

1000 2000 3000 4000 5000 6000 7000 8000 9000

(MLP).” The MLP serves as the interface between a location server and a location services (LCS) client. Whenever user requires an emergency service, he will dial the specified number for emergency calling. Dialing of emergency service number will generate an “emergency location immediate service (ELIS).” ELIS is used to retrieve the position of a mobile subscriber that is involved in an emergency call or has initiated an emergency service in some other way. The service consists of the following messages: emergency location immediate request (ELIR) and emergency location immediate answer (ELIA). When user has dialed the emergency number, emergency location immediate request is sent to network service provider. After receiving the emergency location immediate request from the user, network service provider extracts the position information and sends emergency location immediate answer to the mobile user, and service provider asks him to select the service from ambulance, police, and fire services. Mobile user selects the service, which he actually needs. The service provider would find the nearest emergency service center and send an emergency location report to that center. Whenever an emergency location report is received, a mark will appear on the corresponding digital map. This mark will indicate the user’s location. A schematic block diagram of location-based mobile emergency service system and tracking a mobile user are shown in Figure 1.26 (a) and (b), respectively.

1.8.4

Heart Monitoring System

In cardiac cells of the human body, a small electrical current is produced by the movement of sodium (Naþ) and potassium (Kþ) ions. The electrical potential

1.8 Some Examples of Real-World Signals and Systems

35

Figure 1.26 (a) Schematic block diagram (b) tracking a mobile user of location-based mobile emergency service system

Figure 1.27 One cycle of ECG signal

generated by these ions is known as an electrocardiogram (ECG) signal. The ECG signal is used by physicians to analyze heart conditions. The ECG signal is very small (normally 0.0001 to 0.003 volt). These signals are within the frequency range of 0.05 to 100 Hz. A typical one cycle ECG tracing of a normal heartbeat consists of a P wave, a QRS complex, and a T wave as shown in Figure 1.27. A small U wave is normally visible in 50 to 75% of ECGs. The processing of ECG signal yields information, such as amplitude and timing, required for a physician to analyze a patient’s heart condition. Detection of R-peaks and computation of R-R interval of an ECG record are important requirements of comprehensive  arrhythmia analysis systems. Heart rate is computed as Heart rate ¼ RR interval1 in seconds  60. An ECG signal with variations in heart rate is shown in Figure 1.28.

1 Introduction

Amplitude

Amplitude

36

Recorded ECG signal 0.5 0 −0.5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

Filtered ECG signal 1 0 −1

0

0.2

0.4

0.6

0.8

1

1.2

2 ⫻104

1.4

1.6

1.8

2 ⫻10

Amplitude

4

ECG signal R peaks 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

2

1.8

Amplitude

⫻104

Heart rate 100 50 0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

2

1.8 ⫻10

4

Figure 1.28 An ECG signal with variations in heart rate

1.8.5

Human Visual System

The human visual system (HVS) can widely perform a number of image processing operations in a manner superior to anything we are currently able to execute with computers. To perform such signal processing operations, we have to understand the way HVS works. When the reflection from an object (light ray) is observed by the eye, first, it passes through the cornea, eventually through the aqueous humor, the iris, the lens, the vitreous humor, and finally reaching the retina. The retina consists photosensitive cells called cones and rods, which are responsible to convert the incident light energy into neural signals that are carried to human brain by the optic nerve (Figure 1.29).

1.8.6

Magnetic Resonance Imaging

When an oscillating strong magnetic field is applied at a certain frequency on a certain part of the human body, the hydrogen atoms in the body emit radio-frequency waves to form image of the particular part of the body, which is captured by the MRI machine. An MRI imaging system and a MRI image with brain tumor are shown in Figure 1.30(a) and (b), respectively.

1.9 Problems

37

Figure 1.29 Human visual system

Figure 1.30 (a) MRI imaging system, (b) MRI image with brain tumor

1.9

Problems

1. Classify the following continuous-time signals as periodic or aperiodic. If periodic, determine the period.   π  t (i) xðt Þ ¼ cos 2π 3 t þ 2 sinp2ffiffi ffi  (ii) xðt Þ ¼ cos ð2πt Þ þ sin 2πt (iii) xðt Þ ¼ 12  12 cos ð2t Þ   (iv) xðt Þ ¼ 1 þ sin ð4t Þ þ cos 6t þ π3 (v) x(t) ¼ ej(4t + π/5)  (vi) xðt Þ ¼ cos 2t þ π4 (vii) x(t) ¼ cos(2πt)u(t) (viii) x(t) ¼ cos2(t) 2. A periodic signal x1(t) has a period 2, and another periodic signal x2(t) has a period 3. Find the fundamental frequency and period for the signal y (t) ¼ x1(t) þ x2(t).

38

1 Introduction

3. Classify the following continuous-time signals as even or odd signals or neither even nor odd. Determine power and energy for each incase of power or energy signal. (i) (ii) (iii) (iv) (v) (vi)

x(t) ¼ (1þ t2)cos2(5t) x(t) ¼ u(t) x(t) ¼ tu(t) x(t) ¼ tsin(2t) x(t) ¼ t + cos(2t) x(t) ¼ e2tsin(2t))

4. Consider the following continuous-time signal:   2π ðt  T Þ xðt Þ ¼ 2 sin 10 Determine the values of T for which the signal is (i) An even function (ii) An odd function 5. Classify the following continuous-time signals as power or energy signals or neither. Determine power and energy for each incase of power or energy signal. (i) (ii) (iii) (iv)

x(t) ¼ sin(2πt)cos(πt) x(t) ¼ tu(t) x(t) ¼ e3tu(t) x(t) ¼ ej3t

6. Determine energy for each of the following signals and comment on the results.

x(t)

x(t)

x(t)

1

1

0

0

1

2

t

0

(a)

1

2

2

t

3

-1

(b)

(c)

x(t) 2

0

1

1

(d)

2

t

t

1.10

MATLAB Exercises

39

7. What is the energy of the signal x(t) ¼ cx(at  b), where a 6¼ 0? 8. Verify that ect is neither energy nor a power signal for a complex value of c with nonzero real part. 9. Show that the energy of x(t)  y(t) is Ex + Ey, if x(t) and y(t) are orthogonal. 10. Derive an expression for the power of the following continuous-time signal x (t) ¼ A1cos(Ω1t + θ1) þ A2cos(Ω2t + θ2) for Ω1 ¼ Ω2. 11. Determine the power of the signal x(t) ¼ AejΩt. 12. Find power for each of the following signals: (i) x(t) ¼ (5 þ 3sin(2t))cos(5t) (ii) x(t) ¼ 5 cos(5t) cos (10t) (iii) x(t) ¼ 2sin(5t) cos (10t) 13. Find odd and even components for each of the following signals: (a) x(t) ¼ u(t) (b) x(t) ¼ eatu(t) 14. Evaluate the following expressions: ð1 π (i) cos ðt  5ÞÞδð2t  3Þdt 1  2  2t (ii) e cos 50 t þ 1Þ π t δð  ð1 π 50 2t t δðt þ 1Þdt (iii) e cos 2 π ð1 1 (iv) ðt þ cos ð2πt ÞÞδðt  1Þdt ð1 1 dδðt Þ dt (v) et dt ð1 1 (vi) et δðt  1Þdt 1

1.10

MATLAB Exercises

1. Use MATLAB to generate the continuous-time signal shown in Figure 1.p1.1. 2. Generate and plot each of the following continuous-time signals using MATLAB: (i) x(t) ¼ 10 sin(2πt) cos (πt  4) for 10  t  10 (ii) x(t) ¼ 2e0.1t sin(2πt) for 5  t  5

40

1 Introduction 5 4 3

Ámplitude

2 1 0 -1 -2 -3 -4 -5 -10

-8

-6

-4

-2

0 2 t (seconds)

4

6

8

10

Figure p1.1 Signal of MATLAB exercise 1

Further Reading 1. Pierce, J.R., Noll, A.M.: Signals: The Science of Telecommunications. American Library, New Delhi (1960) 2. Lathi, B.P.: Linear Systems and Signals, 2nd edn. Oxford University Press, New York (2005) 3. Mandal, M., Asif, A.: Continuous and Discrete Time Signals and Systems. Cambridge University Press, Cambridge (2007)

Chapter 2

Continuous-Time Signals and Systems

This chapter presents time-domain analysis of continuous-time systems. It develops representation of signals in terms of impulses. The notions of linearity, timeinvariance, causality, stability, memorability, and invertibility are introduced. It has shown that the input-output relationship for linear time-invariant (LTI) continuous systems is described in terms of a convolution integral. The differential equation representation of LTI continuous systems and classical solutions of differential equations are also presented. Next, block-diagram representation of LTI continuous-time systems is introduced. Furthermore, a brief discussion on singularity functions is provided. Finally, the state-space representation of continuous-time LTI systems is described.

2.1

The Representation of Signals in Terms of Impulses

Consider pulse or staircase approximation b x ðt Þ to continuous-time signal x(t) as shown in Figure 2.1. Then, the approximation signal can be expressed as sum of all these pulse signals. Define 8 < 1, 0 < t < Δ ð2:1Þ δΔ ðt Þ ¼ Δ : 0, otherwise Since δΔ(t)Δ ¼ 1, b x ðt Þ can be expressed as b x ðt Þ ¼

X1 k¼1

xðkΔÞ δΔ ðt  kΔÞΔ

© Springer International Publishing AG, part of Springer Nature 2018 K. D. Rao, Signals and Systems, https://doi.org/10.1007/978-3-319-68675-2_2

ð2:2Þ

41

42

2 Continuous-Time Signals and Systems

x(t)

Figure 2.1 Representation of a signal in terms of impulses

........

........

........

0

t

As Δ approaches zero, the above approximation b x ðt Þ can be written as b x ðt Þ ¼ limΔ!0

X1 k¼1

xðkΔÞ δΔ ðt  kΔÞΔ

ð2:3Þ

Also, as Δ ! 0, the summation approaches an integral, and the pulse approaches unit impulse. Therefore, Eq. (2.3) can be rewritten as b x ðt Þ ¼

ð1

xðτÞδðt  τÞ dτ

ð2:4Þ

1

Thus, a continuous-time signal can be represented as weighted superposition of shifted impulses. Here the superposition is integration due to nature of the continuous-time input. The weight x(τ) dτ on the impulse δ(t  τ) is determined from the value of the input signal x(t) at the time of occurrence of each impulse.

2.2 2.2.1

Continuous-Time Systems Linear Systems

Let x1(t) and x2(t) are the inputs applied to a system characterized by the transformation operator ℜ[] and y1(t) and y2(t) are the system outputs. A linear system should satisfy the principles of homogeneity and superposition. Hence, the following equations hold for a linear system y1 ðt Þ ¼ ℜ½x1 ðt Þ,

ð2:5aÞ

y2 ðt Þ ¼ ℜ½x2 ðt Þ,

ð2:5bÞ

ℜ½ax1 ðt Þ ¼ ay1 ðt Þ,

ð2:6aÞ

ℜ½bx2 ðt Þ ¼ by2 ðt Þ

ð2:6bÞ

Principle of homogeneity:

2.2 Continuous-Time Systems

43

Principle of superposition: ℜ½x1 ðt Þ þ ℜ½x2 ðt Þ ¼ y1 ðt Þ þ y2 ðt Þ

ð2:7Þ

ℜ½ax1 ðt Þ þ ℜ½bx2 ðt Þ ¼ ay1 ðt Þ þ by2 ðt Þ

ð2:8Þ

Linearity:

where a and b are arbitrary constants.

2.2.2

Time-Invariant System

A system is time invariant if the behavior and characteristics of the system are fixed over time. A system is time invariant if a time shift in the input signal results in an identical time shift in the output signal. For example, a time-invariant system should produce y(t  t0) as the output when x(t  t0) is the input. Mathematically it can be specified as yð t  t 0 Þ ¼ ℜ½ xð t  t 0 Þ  Example 2.1 Check for linearity and time-invariance of the following system yðt Þ ¼ txðt Þ Solution Linearity: Let x1(t) and x2(t) be two distinct inputs applied to the system, then y1 ðt Þ ¼ ℜ½x1 ðt Þ ¼ tx1 ðt Þ, y2 ðt Þ ¼ ℜ½x2 ðt Þ ¼ tx2 ðt Þ If an input equal to sum of the inputs ax1(t), bx2(t), x(t) ¼ ax1(t) þ bx2(t) is applied, then yðt Þ ¼ tax1 ðt Þ þ tbx2 ðt Þ ¼ ay1 ðt Þ þ by2 ðt Þ Hence, the system is linear. Time-invariance: yðt Þ ¼ ℜ½xðt Þ ¼ txðt Þ The output y(t) of the system delayed by t0 can be written as yðt  t 0 Þ ¼ ðt  t 0 Þxðt  t 0 Þ

ð2:9Þ

44

2 Continuous-Time Signals and Systems

For example, for an input x1(t) ¼ x(t  t0), the output y1(t) can be written as y1 ðt Þ ¼ ℜ½x1 ðt Þ ¼ tx1 ðt Þ ¼ txðt  t 0 Þ yðt  t 0 Þ 6¼ y1 ðt Þ Hence, it is a time variant system. Example 2.2 Check for linearity and time-invariance of the following system: yðt Þ ¼ sin ðxðt ÞÞ Solution Linearity: Let x1(t) and x2(t) be two distinct inputs applied to the system, then y1 ðt Þ ¼ ℜ½x1 ðt Þ ¼ sin ðx1 ðt ÞÞ, y2 ðt Þ ¼ ℜ½x2 ðt Þ ¼ sin ðx2 ðt ÞÞ If an input equal to sum of the inputs ax1(t), bx2(t),x(t) ¼ ax1(t) þ bx2(t) is applied, then yðt Þ ¼ sin ðax1 ðt ÞÞ þ sin ðbx2 ðt ÞÞ 6¼ ay1 ðt Þ þ by2 ðt Þ Hence, the system is nonlinear. Time-invariance: yðt Þ ¼ ℜ½xðt Þ ¼ sin ðxðt ÞÞ The output y(t) of the system delayed by t0 can be written as yðt  t 0 Þ ¼ sin ðxðt  t 0 ÞÞ For example, for an input x1(t) ¼ x(t  t0), the output y1(t) can be written as y1 ðt Þ ¼ ℜ½x1 ðt Þ ¼ sin ðx1 ðt ÞÞ ¼ sin ðxðt  t 0 ÞÞ yð t  t 0 Þ ¼ y1 ð t Þ Hence, it is a time-invariant system. Example 2.3 Determine if the following continuous-time systems are linear or nonlinear: (i)

dyðt Þ dt

þ 2tyðt Þ ¼ t 2 xðt Þ

(ii) 2y(t) þ 3 ¼ x(t) ðt (iii) yðt Þ ¼ xðτÞdτ 1

(iv)

dyðt Þ dt

þ 3yðt Þ ¼ xðt Þ dxdtðtÞ

2.2 Continuous-Time Systems

45

Solution (i) Let x1(t) and x2(t) be two distinct inputs applied to the system, then dy1 ðt Þ þ 2ty1 ðt Þ ¼ t 2 x1 ðt Þ dt dy2 ðt Þ þ 2ty2 ðt Þ ¼ t 2 x2 ðt Þ dt If an input equal to sum of the inputs ax1(t), bx2(t), x(t) ¼ ax1(t) þ bx2(t) is applied, then a

dy1 ðt Þ dy ðt Þ þ 2aty1 ðt Þ þ b 2 þ 2bty2 ðt Þ ¼ at 2 x1 ðt Þ þ bt 2 x2 ðt Þ dt dt

Hence, the system is linear. (ii)

yð t Þ ¼ xð t Þ 

3 2

Let x1(t) and x2(t) be two distinct inputs applied to the system, then y1 ðt Þ ¼ ℜ½x1 ðt Þ ¼ x1 ðt Þ  32 , y2 ðt Þ ¼ ℜ½x2 ðt Þ ¼ x2 ðt Þ  32 If an input equal to sum of the inputs ax1(t), bx2(t),x(t) ¼ ax1(t) þ bx2(t) is applied, then 3 3 yðt Þ ¼ ax1 ðt Þ  þ bx2 ðt Þ  6¼ ay1 ðt Þ þ by2 ðt Þ 2 2 Hence, the system is nonlinear. (iii) Let x1(t) and x2(t) be two distinct inputs applied to the system, then ðt ðt y1 ðt Þ ¼ ℜ½x1 ðt Þ ¼ x1 ðτÞdτ, y2 ðt Þ ¼ ℜ½x2 ðt Þ ¼ x2 ðτÞdτ 1

1

If an input equal to sum of the inputs ax1(t), bx2(t), x(t) ¼ ax1(t) þ bx2(t) is applied, then ðt yð t Þ ¼

ax1 ðτÞdτ þ

1

ðt 1

bx2 ðτÞdτ ¼ ay1 ðt Þ þ by2 ðt Þ

Hence, the system is linear. (iv) Let x1(t) and x2(t) be two distinct inputs applied to the system, then dy1 ðt Þ dx1 ðt Þ þ 3y1 ðt Þ ¼ x1 ðt Þ dt dt dy2 ðt Þ dx2 ðt Þ þ 3y2 ðt Þ ¼ x2 ðt Þ dt dt

46

2 Continuous-Time Signals and Systems

If an input equal to sum of the inputs ax1(t), bx2(t), x(t) ¼ ax1(t) þ bx2(t) is applied, then a

dy1 ðt Þ dy ðt Þ dx1 ðt Þ dx2 ðt Þ þ 3ay1 ðt Þ þ b 2 þ 3by2 ðt Þ 6¼ a2 x1 ðt Þ þ b2 x 2 ð t Þ dt dt dt dt

The system is nonlinear. Example 2.4 Determine if the following continuous-time systems are time invariant or time variant: (i) y(t) ¼ x(t), ðt (ii) yðt Þ ¼ xðτÞdτ, 1

(iii) y(t) ¼ x(4t) (iv) yðt Þ ¼ ð2 þ sin ðt ÞÞxðt ÞðvÞyðt Þ ¼ dxdtðtÞ Solution (i)

y(t) ¼ ℜ[x(t)] ¼ x(t) The output y(t) of the system delayed by t0 can be written as yðt Þ ¼ ℜ½xðt Þ ¼ xðt Þ yðt  t 0 Þ ¼ xððt  t 0 ÞÞ ¼ xðt þ t 0 Þ For example, for an input x1(t) ¼ x(t  t0), the output y1(t) can be written as y1 ðt Þ ¼ ℜ½x1 ðt Þ ¼ x1 ðt Þ ¼ xðt  t 0 Þ yðt  t 0 Þ 6¼ y1 ðt Þ Hence, it is a time-varying system.

(ii) Let x(t) ¼ δ(t), then yðt Þ ¼ ℜ½xðt Þ ¼

ð3

δðt Þdt ¼ 1

3

Now, for an input x1(t) ¼ x(t  6), the output y1(t) can be written as y1 ð t Þ ¼ ℜ½ x 1 ð t Þ  ¼

Ð3 3

δðt  6Þdt ¼ 0

yðt  6Þ 6¼ y1 ðt Þ Hence, it is a time-varying system. (iii) The output y(t) of the system delayed by t0 can be written as yðt Þ ¼ ℜ½xðt Þ ¼ xð4t Þ yðt  t 0 Þ ¼ xð4ðt  t 0 ÞÞ ¼ xð4t  4t 0 Þ

2.2 Continuous-Time Systems

47

For example, for an input x1(t) ¼ x(t  t0), the output y1(t) can be written as y1 ðt Þ ¼ ℜ½x1 ðt Þ ¼ x1 ð4t Þ ¼ xð4t  t 0 Þ yðt  t 0 Þ 6¼ y1 ðt Þ Hence, it is a time-varying system. y(t) ¼ ℜ[x(t)] ¼ (2 þ sin (t))x(t)

(iv)

The output y(t) of the system delayed by t0 can be written as yðt  t 0 Þ ¼ ð2 þ sin ðt  t 0 ÞÞxðt  t 0 Þ For example, for an input x1(t) ¼ x(t  t0), the output y1(t) can be written as y1 ðt Þ ¼ ℜ½x1 ðt Þ ¼ ð2 þ sin ðt ÞÞxðt  t 0 Þ yðt  t 0 Þ 6¼ y1 ðt Þ Hence, it is a time-varying system. (v) yðt Þ ¼ ℜ½xðt Þ ¼

dxðt Þ dt

The output y(t) of the system delayed by t0 can be written as yð t  t 0 Þ ¼

dxðt  t 0 Þ dt

For example, for an input x1(t) ¼ x(t  t0), the output y1(t) can be written as y1 ð t Þ ¼ ℜ½ x 1 ð t Þ  ¼

dxðt  t 0 Þ dt

yð t  t 0 Þ ¼ y1 ð t Þ Hence, it is a time-invariant system. Example 2.5 Consider an LTI system with the response y(t) as shown in Figure 2.2 to the input signal x(t) ¼ u(t)  u(t  2). Figure 2.2 Response y(t) to the input x(t)

y(t) 2

2

4

t

48

2 Continuous-Time Signals and Systems

(t) 2

2

t

8

4

-2

Figure 2.3 Response y1(t) to the input x1(t)

(t)

Figure 2.4 Response y2(t) to the input x2(t)

2

-2

2

4

t

Determine and sketch the response of the system to the following inputs: (i) x1(t) ¼ x(t)  x(t  4) (ii) x2(t) ¼ x(t) þ x(t þ 2) Solution (i) x1(t) ¼ x(t)  x(t  4) Since it is an LTI system, the response y1(t) to the input x1(t) is given by y1(t) ¼ y(t)  y(t  4) as depicted in Figure 2.3. (ii) x2(t) ¼ x(t) þ x(t þ 2) Since it is an LTI system, the response y2(t) to the input x2(t) is given by y2(t) ¼ y (t) þ y(t þ 2) as depicted in Figure 2.4.

2.2.3

Causal System

The causal system generates the output depending upon present and past inputs only. A causal system is non-anticipatory.

2.3 The Convolution Integral

2.2.4

49

Stable System

When the system produces bounded output for bounded input, then the system is called bounded-input and bounded-output stable. If the signal is bounded, then its magnitude will always be finite.

2.2.5

Memory and Memoryless System

The output of a memory system at any specified time depends on the inputs at that specified time and at other times. Such systems have memory or energy storage elements. The system is said to be static or memoryless if its output depends upon the present input only.

2.2.6

Invertible System

A system is said to be invertible if the input can be recovered from its output. Otherwise the system is noninvertible system.

2.2.7

Step and Impulse Responses

If the input to the system is unit impulse input δ(t), the system output is called the impulse response and denoted by h(t): hðt Þ ¼ ℜ½δðt Þ

ð2:10Þ

If the input to the system is a unit step input u(t), then the system output is called the step response s(t);that is, s ð t Þ ¼ ℜ½ uð t Þ 

2.3

ð2:11Þ

The Convolution Integral

The output of a system for an input expressed as weighted superposition as in Eq. (2.4) is given by ð 1  yð t Þ ¼ ℜ½ xð t Þ  ¼ ℜ xðτÞδðt  τÞdτ ð2:12Þ 1

50

2 Continuous-Time Signals and Systems

From the linearity property of the system, Eq. (2.12) can be rewritten as ð1 yð t Þ ¼

xðτÞℜ½δðt  τÞdτ

ð2:13Þ

1

For a time-invariant system, ℜ[δ(t  τ)] ¼ h(t  τ). Hence, we obtain ð1 yð t Þ ¼

xðτÞhðt  τÞdτ

ð2:14Þ

1

Thus, the output y(t) of a linear time-invariant system to an arbitrary input x(t) is obtained in terms of the unit impulse input δ(t). Eq. (2.14) is referred to as the convolutional integral and is denoted by the symbol * as ð1 yðt Þ ¼ xðt Þ∗hðt Þ ¼

xðτÞhðt  τÞdτ

ð2:15Þ

1

2.3.1

Some Properties of the Convolution Integral

2.3.1.1

The Commutative Property x1 ðt Þ∗x2 ðt Þ ¼ x2 ðt Þ∗x1 ðt Þ

Proof This property can be proved by a change of variable. By the definition of the convolution integral Ð1 x1 ðt Þ∗x2 ðt Þ ¼ 1 x1 ðτÞx2 ðt  τÞdτ Let V ¼ t  τ so that τ ¼ t  V, and dτ ¼ dV:

ð2:16Þ

ð2:17Þ

Then Ð 1 x1 ðt Þ∗x2 ðt Þ ¼  1 x1 ðt  V Þx2 ðV ÞdV Ð1 ¼ 1 x1 ðt  V Þx2 ðV ÞdV

ð2:18Þ

¼ x2 ðt Þ∗x1 ðt Þ

2.3.1.2

The Distributive Property x1 ðt Þ∗½x2 ðt Þ þ x3 ðt Þ ¼ x1 ðt Þ∗x2 ðt Þ þ x1 ðt Þ∗x3 ðt Þ

ð2:19Þ

2.3 The Convolution Integral

51

Proof By the definition of the convolution integral ð1  x1 ðt Þ∗½x2 ðt Þ þ x3 ðt Þ ¼ x1 ðτÞ x2 ðt  τÞ þ x3 ðt  τÞdτ 1

ð1 ¼

x1 ðτÞx2 ðt  τÞdτ þ

1

ð1

x1 ðτÞx3 ðt  τÞdτ

ð2:20Þ

1

¼ x1 ðt Þ∗x2 ðt Þ þ x1 ðt Þ∗x3 ðt Þ

2.3.1.3

The Associative Property ½x1 ðt Þ∗x2 ðt Þ∗x3 ðt Þ ¼ x1 ðt Þ∗½x2 ðt Þ∗x3 ðt Þ

Proof The left-hand side of the property can be expressed by ð1 ½x1 ðt Þ∗x2 ðt Þ∗x3 ðt Þ ¼ x1 ðτ1 Þx2 ðt  τ1 Þdτ1 ∗x3 ðt Þ

ð2:21Þ

ð2:22Þ

1

where x1(t) * x2(t) is expressed as a convolution integral. Expanding the second convolution gives ð 1 ð 1 ½x1 ðt Þ∗x2 ðt Þ∗x3 ðt Þ ¼ 1

 x1 ðτ1 Þx2 ðτ2  τ1 Þdτ1 x3 ðt  τ2 Þdτ2

ð2:23Þ

1

Reversing the order of integration gives ð1 ð1 dx1 ðt Þ∗x2 ðt Þe∗x3 ðt Þ ¼ 1

x1 ðτ1 Þx2 ðτ2  τ1 Þ x3 ðt  τ2 Þdτ1 dτ2

ð2:24Þ

1

Similarly, the right-hand side of the property can be written as Ð 1  x1 ðt Þ∗dx2 ðt Þ∗x3 ðt Þe ¼ x1 ðt Þ∗ 1 x2 ðτ2 Þ x3 ðt  τ2 Þdτ2 Ð1 Ð1 ¼ 1 1 x1 ðτ1 Þx2 ðτ2 Þx2 ðt  τ1  τ2 Þ dτ1 dτ2

ð2:25Þ

Now, it is to be shown that ð1 ð1 1

1

x1 ðτ1 Þx2 ðτ2  τ1 Þx3 ðt  τ2 Þ dτ1 dτ2 ¼

ð1 ð1 1

x1 ðτ1 Þx2 ðτ2 Þ

1

x2 ðt  τ1  τ2 Þdτ1 dτ2 In the right hand τ1 integration, let v ¼ τ1 + τ2 and dτ1 ¼ dv.

ð2:26Þ

52

2 Continuous-Time Signals and Systems

Then ð1 ð1 1

x1 ðτ1 Þx2 ðτ2  τ1 Þ∗x3 ðt  τ2 Þdτ1 dτ2 ¼

ð1 ð1

1

1

x1 ðV  τ2 Þx2 ðτ2 Þ

1

x3 ðt  V Þ dV dτ2 ð2:27Þ Next, let u ¼ V  τ2 anddτ2 ¼ du Then ð1 ð1 1

x1 ðτ1 Þx2 ðτ2  τ1 Þx3 ðt  τ2 Þdτ1 dτ2 ¼ 

1

ð 1 ð 1 x1 ðuÞx2 ðV  uÞ 1

1

x3 ðt  V ÞdV d u ð1 ð1 1

x1 ðτ1 Þx2 ðτ2  τ1 Þx3 ðt  τ2 Þdτ1 dτ2 ¼

ð2:28Þ

ð1 ð1 x1 ðuÞx2 ðV  uÞ

1

1

1

ð2:29Þ

x3 ðt  V ÞdV d u The above right-hand side and left-hand side integrals are the same except for change of the variables. Hence, the associative property is proved.

2.3.1.4

Convolution with an Impulse xðt Þ∗δðt Þ ¼ xðt Þ

ð2:30Þ

Proof By definition ð1 xðt Þ∗δðt Þ ¼

xðτÞδðt  τÞ dτ

ð2:31Þ

1

Since δ(t  τ) is an impulse at τ ¼ t and by sampling property of the impulse, Ð1 1

xðτÞδðt  τÞdτ ¼ xðτÞjτ¼t ¼ xð t Þ

Hence xðt Þ∗δðt Þ ¼ xðt Þ

ð2:32Þ

2.3 The Convolution Integral

2.3.1.5

53

Convolution with Delayed Input and Delayed Impulse Response

If y(t) ¼ x(t) * h(t), then xðt  t 1 Þ∗hðt  t 2 Þ ¼ yðt  t 1  t 2 Þ

ð2:33Þ

Proof By the convolution integral, we have ð1 yðt Þ ¼ xðt Þ∗hðt Þ ¼ xðτÞhðt  τÞ dτ

ð2:34Þ

1

and

ð1 xðt  t 1 Þ∗hðt  t 2 Þ ¼

xðτ  t 1 Þhðt  τ  t 2 Þ dτ

ð2:35Þ

1

Let τ  t1 ¼ υ. Then τ ¼ υ + t1, and Eq. (2.35) becomes ð1 xðt  t 1 Þ∗hðt  t 2 Þ ¼

xðυÞhðt  t 1  t 2  υÞ dυ

ð2:36Þ

1

It is observed that replacing t by t  t1  t2 in Eq. (2.34), we obtain Eq. (2.36). Thus, it is proved that xðt  t 1 Þ∗hðt  t 2 Þ ¼ yðt  t 1  t 2 Þ Example 2.6 Determine the continuous-time convolution of x(t) and h(t) for the following: (i) x(t) ¼ u(t) h(t) ¼ u(t) (ii) x(t) ¼ u(t  a) h(t) ¼ u(t  b) (iii) x(t) ¼ u(t þ 1)  u(t  1) h(t) ¼ u(t þ 1)  u(t  1) (iv) x(t) ¼ e(t2)u(t  2) h(t) ¼ u(t þ 2)

x(t)

h(t)

1

(v)

(t-1)

1

t

2

4

t

54

2 Continuous-Time Signals and Systems

(vi) x(t) ¼ u(t) h(t) ¼ etu(t) (vii) x(t) ¼ 2(u(t)  u(t  2)) h(t) ¼ et/2u(t) Solution

Ð1

xðτÞhðt  τÞ ¼ xðt( 1Þ Ðt Ð1 0 1dτ, ¼ 1 uðτÞuðt  τÞdτ ¼ 0, ( t, t > 0 ¼ ¼ tuðt Þ 0, t < 0

(i) yðt Þ ¼

1

t>0 t<0

(ii) uðt  aÞ∗uðt  bÞ ¼ ðuðt Þ∗δðt  aÞÞ∗ðuðt Þ∗δðt  bÞÞ ¼ ðuðt Þ∗uðt ÞÞ∗ðδðt  aÞ∗δðt  bÞÞ ¼ ðuðt Þ∗uðt ÞÞ∗δðt  a  bÞ Since u(t)*u(t) ¼ tu(t), uðt  aÞ∗uðt  bÞ ¼ ðtuðt ÞÞ∗δðt  a  bÞ ¼ ðt  a  bÞuðt  a  bÞ (iii) xðt Þ∗hðt Þ ¼ ½uðt þ 1Þ  uðt  1Þ∗½uðt þ 1Þ  uðt  1Þ ¼ uðt þ 1Þ∗uðt þ 1Þ  uðt þ 1Þ∗uðt  1Þ  uðt  1Þ∗uðt þ 1Þ þ uðt  1Þ∗uðt  1Þ ¼ uðt þ 1Þ∗uðt þ 1Þ  2uðt þ 1Þ∗uðt  1Þ þuðt  1Þ∗uðt  1Þ ¼ ðt þ 2Þuðt þ 2Þ  2tuðt Þ þ ðt  2Þuðt  2Þ as shown in Figure 2.5

x(t)*h(t)

Figure 2.5 The convolution of x(t) and h(t)

2

-2

0

2

tt

2.3 The Convolution Integral

(iv)

yð t Þ ¼ ¼ ¼

Ð1 1

Ð1

55

xðτÞδðt  τ  1Þ ¼ xðt  1Þ

ðτ2Þ uðτ  2Þuðt  τ þ 2Þdτ 1 e ( Ð tþ2 eðτ2Þ dτ, t > 0 2

t<0

0, Letting τ 1 ¼ τ – 2, ( Ð tþ2 2

yðt Þ ¼

eτ1 dτ1

( ¼

2  et ,

t>0

0,

t<0

0 ð1 yð t Þ ¼

(v)

xðτÞδðt  τ  1Þ ¼ xðt  1Þ

1

Hence, y(t) is a shifted version of x(t) as shown in Figure 2.6. (vi)

yð t Þ ¼

Ð1 1

Ð1

xðτÞhðt  τÞdτ

ðtτÞ uðt  τÞdτ 1 uðτ Þe Ð t ðtτÞ dτ, t > 0, ¼ 0e t ¼ eðtτÞ=2 0 ¼ ð1  et Þ, t

¼

yð t Þ ¼ 0 (vii)

yð t Þ ¼ ¼ yð t Þ ¼ ¼

> 0,

t<0 Ð t ðtτÞ=2 dτ, 2  t  0, 0 2e

 t=2 , 2  t  0, 4 1e Ð 2 ðtτÞ=2 dτ, t  2, 0 2e 

 2 4eðtτÞ=2 0 ¼ 4 eðt2Þ=2  et=2 ,

¼ 4et=2 ð e

1 Þ,

t  2,

t  2,

yð t Þ ¼ 0 t  0 Example 2.7 Consider LTI system with the impulse response h(t); for an input x(t), the output y(t) is as shown in Figure 2.7. Figure 2.6 The shifted version of x(t)

y(t) 1

1

2

4

5

t

56

2 Continuous-Time Signals and Systems

y(t)

Figure 2.7 Response y(t) to the input x(t)

1

1

2

4

5

t

y(t-2) 1

1

2

4

t

5

Figure 2.8 Response y(t  2) to the input x(t  2)

(t)

Figure 2.9 Response y1(t) to the input x1(t)

1

1

2

4

5

t

Determine the output of the system for an input x1(t) ¼ x(t)  x(t  2)) Solution Since the system is LTI, for an input x(t  2), the output is y(t  2) as shown in Figure 2.8. The output y1(t) for the input x1(t) ¼ x(t)  x(t  2) is given by y1(t) ¼ y(t)  y(t  2), which is shown in Figure 2.9. Example 2.8 Consider a LTI system with input and output related through the equation ðt eðtτÞ xðτ  3Þ dτ

yðt Þ ¼ 1

2.3 The Convolution Integral

57

(i) Determine the impulse response h(t) of the system. (ii) Determine the output y(t) of the system for the input x(t) ¼ u(t þ 1)  u(t  3). Solution ðt (i)

eðtτÞ xðτ  3Þ dτ

yðt Þ ¼ 1

Let τ 1 ¼ τ  3, then

ðt

eðt3τ 1 Þ xðτ 1 Þ dτ 1

yð t Þ ¼ 1

Hence, h(t) ¼ e(t  3)u(t  3) ðt (ii) yð t Þ ¼ eðt3τ 1 Þ ½uðt  τ 1 þ 1Þ  uðt  τ 1  3Þ dτ 1 1

ðt ¼

eðt3τ 1 Þ ½uðt  τ 1 þ 1Þ  uðt  τ 1  3Þ dτ 1

3

x(t  τ) and h(τ) are shown in Figure 2.10. Using Figure 2.10, y(t) can be written as 8 0, > > >ð > tþ1 > > < eðτ 1 3Þ dτ1 ¼ 1  eðt2Þ , yð t Þ ¼ 3 > > ð tþ1 > >  > > : eðτ 1 3Þ dτ ¼ eðt6Þ 1  e4 , t23

1

t-3

1

1

0

t+1

Figure 2.10 x(t  τ) and h(τ)

0

t  2, 2 < t  6, t > 6:

58

2 Continuous-Time Signals and Systems

2.3.2

Graphical Convolution

An understanding of graphical interpretation of convolution is very useful in computing the convolution of more complex signals. The stepwise procedure for graphical convolution is as follows: Step 1: Make x(τ) fixed. Step 2: Invert h(τ) about the vertical axis (t ¼ 0) to obtain h(τ). Step 3: Shift the h(τ) along the τ axis by t0 seconds so that the shifted h(τ) is representing h(t0  τ). Step 4: The area under the product of x(τ) and h(t0  τ) is y(t0), the value of convolution at t ¼ t0. Step 5: Repeat steps 3 and 4 for different values of positive and negative to obtain y (t) for all values of t. Example 2.9 Graphically determine the continuous-time convolution of h(t) and x(t) for the following: ( 1, 0  t  4 (i) xðt Þ ¼ 0, otherwise ( 1, 0  t  4 hð t Þ ¼ 0, otherwise Solution

1

1

0

4

0

4

To compute y(t) ¼ x(t) * h(t), first h(τ) is to be obtained by inverting h(τ) about the vertical axis. Then, the product of x(τ) and h(t  τ) is formed, point by point, and this product is integrated to compute y(t). Thus, the overlap area between the rectangles forming x(τ) and h(t  τ) is y(t). Clearly, y(0) ¼ 0 because there is no overlap between the rectangles forming x(τ) and h(t  τ) at t ¼ 0. For 0 < t < 8, there is overlap between the rectangles forming x (τ) and h(t  τ). For t  8, there is no overlap, and hence, y(8) ¼ 0. These are illustrated in Figure 2.11 with the final result for y(t). The shaded portion represents the overlap area of the product x(τ) and h(t  τ).

2.3 The Convolution Integral

59 (0)= ( )h(0 − ) = 0

h( − )|

( )

=0 1

-4

0

4

h(1 − )

8

(1)= ( )h(1 − ) = 1

( ) 1

-4

0

-3

4

h(2 − )

8

(2)= ( )h(2 − ) = 2

( ) 1

-4

0

-2

h(3 − )

4

8

(3)= ( )h(3 − ) = 3

( ) 1

-4

0

-1

4

8

h(4 − ) (4)= ( )h(4 − ) = 4

( ) 1

-4

-1

0

4

Figure 2.11 Steps in the convolution and the final result

8

60

2 Continuous-Time Signals and Systems

h(5 − ) (5)= ( )h(5 − ) = 3

( ) 1

-4

0

-1

8

4

h(2 − )

(2)= ( )h(2 − ) = 2

( ) 1

-4

0

-2

h(3 − )

4

8

(3)= ( )h(3 − ) = 3

( ) 1

-4

0

-1

4

8

h(4 − ) (4)= ( )h(4 − ) = 4

( ) 1

-4

-1

0

4

8

h(5 − ) (5)= ( )h(5 − ) = 3

( ) 1

-4

Figure 2.11 (continued)

-1

0

4

8

2.3 The Convolution Integral

61

h(6 − ) ( )

(6)= ( )h(6 − ) = 2

1

-4

-1

0

4

8

h(7 − ) ( )

(7)= ( )h(7 − ) = 1

1

-4

-1

0

4

8

h(8 − ) (8)= ( )h(8 − ) = 0

( ) 1

-4

-1

0

8

4

y(t) 4

2

Figure 2.11 (continued)

4

6

8

t

62

2 Continuous-Time Signals and Systems

Example 2.10 Determine graphically y(t) ¼ x(t) * h(t) for the following x(t) and h(t) shown.

1 1

0

1

-1

3

0

1

3

0

1

3

-1

Solution

1

-3

-1

1

0

1

3

-1

There is no overlap area between x(τ) and h(τ) at t ¼ 0, y(0) ¼ 0. For 0 < t < 6, there is overlap between the rectangles forming x(τ) and h(τ). For t  6, there is no overlap, and hence, y(6) ¼ 0. These are illustrated in Figure 2.12 with the final result for y(t). The shaded portion represents the overlap area of the products x(τ) and h(t  τ). Example 2.11 Consider the RC circuit shown in Figure 2.13. Determine the Vout(t) for Vin(t) ¼ u(t  1)  u(t  2), and assume the time constant RC ¼ 1 sec. Assume the capacitor is initially discharged. Solution The impulse response of the RC low-pass filter is t Ð 1 hðtÞ ¼ e uðtÞ V out ðt Þ ¼ 1 V in ðτÞhðt  τÞ ¼ V in ðt Þ∗hðtÞ

2.3 The Convolution Integral

63

h(1 − )

( ) (1)= ( )h(1 − ) = 1

1

-2

0

-1

3

h(2 − )

( )

-1

(2)= ( )h(2 − ) = 2

1

h(1 − ) -1

0

3

-1

h(2 − ) h(3 − )

( )

(3)= ( )h(3 − ) = 1 − 1=0

1

-1

0

3

-1

h(3 − ) h(4 − )

( )

(4)= ( )h(4 − ) = − 2 1

-1

0

3

5

-1

h(4 − ) Figure 2.12 Steps in the convolution and the final result

64

2 Continuous-Time Signals and Systems

( )

h( 5 − ) ( 5) = ( ) h ( 5 − ) = − 1

1

4

0

6

-1

h( 6 − )

( )

( 6) = ( ) h ( 6 − ) = 0 1

-1

4

0

6

7

-1

h( 6 − )

y(t) 2

2

4

6

8

t

-2

Figure 2.12 (continued)

The steps involved in the convolution are illustrated in Figure 2.14. V out ðt Þ ¼ 0,

t<1

2.3 The Convolution Integral

65

Figure 2.13 RC circuit

Figure 2.14 Illustration of steps in the convolution

Ðt

eðtτ Þ dτ, 1  t  2, t

 ¼ eðtτÞ 1 ¼ 1  eðt1Þ , 1  t  2, Ð V out ðt Þ ¼ 12 eðtτ Þ dτ, 2  t, 2

 ¼ eðtτÞ 1 ¼ eðt2Þ  eðt1Þ , 2  t: V out ðt Þ ¼

1

which is shown in Figure 2.15. Example 2.12 If (t) ¼ x(t) * h(t), then show that y(2t) ¼ 2x(2t) * h(2t) ð1

Solution yð2t Þ ¼

xð2t  τÞhðτÞdτ

1

τ Letting τ1 ¼ , we have 2 Ð1 Ð1 yð2t Þ ¼ 1 xð2t  2τ1 Þhð2τ1 Þ2dτ1 ¼ 2 1 xð2t  2τ1 Þhð2τ1 Þdτ1 ¼ 2xð2t Þ∗hð2t Þ Example 2.13 If x(t) and h(t) are odd signals, then show that y(t) ¼ x(t) * h(t) is an even signal.

66

2 Continuous-Time Signals and Systems 0.7 0.6

Amplitude

0.5 0.4 0.3 0.2 0.1 0

0

2

1.5

1

0.5

2.5

3

3.5

4

Time

Figure 2.15 Times versus Vout(t)

Solution y(t) ¼ x(t) * h(t) yðt Þ ¼ xðt Þ∗hðt Þ Ð1 ¼ 1 xððt  τÞÞhðτÞdτ Ð1 ¼ 1 xðt þ τÞhðτÞdτ Since x(t) and h(t) are odd signals, yðtÞ ¼

Ð1 1

xðt  τÞhðτÞdτ

¼ yð t Þ Hence, y(t) is even because y(t) ¼ y(t). Example 2.14 Consider an LTI system with the impulse response h(t) ¼ etu(t). Find the system response for the input x(t) ¼ sin2tu(t). Solution

yð t Þ ¼ ¼

Ð1 1

Ð1 0 Ð1 0

xðτÞhðt  τÞdτ

sin ð2τÞeðtτÞ dτ

sin ð2τÞeðtτÞ dτ i  t Ð1 ¼ sin ð2τÞeðtτÞ τ¼0  0 2 cos ð2τÞeðtτÞ dτ uðtÞ Ð1 ¼ sin ð2t Þuðt Þ  uðt Þ 0 2 cos ð2τÞeðtτÞ dτ ¼

h

2.3 The Convolution Integral

67

Hence, Ð1 sin ð2τÞeðtτÞ dτ ¼ sin ð2t Þuðt Þ  uðt Þ 0 2 cos ð2τÞeðtτÞ dτ

 t Ð1 ¼ sin ð2t Þuðt Þ  uðt Þ 2 cos ð2τÞeðtτÞ τ¼0  0 4 sin ð2τÞeðtτÞ dτ Ð1 ¼ sin ð2t Þuðt Þ  ð2 cos ð2t Þ þ et Þuðt Þ  0 4 sin ð2τÞeðtτÞ dτ

Ð1 0

The above equation can be rewritten as ð1 5

sin ð2τÞeðtτÞ dτ ¼ ½ sin ð2t Þ  2 cos ð2t Þ þ et uðt Þ

0

Therefore, ð1 yð t Þ ¼ 0

1 sin ð2τÞeðtτÞ dτ ¼ ½ sin ð2t Þ  2 cos ð2t Þ þ 2et uðt Þ 5

Example 2.15 If the response of an LTI system to input x(t) is the output y(t), dy show that the response of the system to dx dt is dt , and using this result determines the impulse response of an LTI system having the response y(t) ¼ sin2t for an input x(t) ¼ e4tu(t). Solution

yðtÞ ¼ xðt Þ∗hðt Þ Ð1 ¼ 1 hðτÞxðt  τÞdτ

Differentiating both sides with respect to t, ð1 dy dx ¼ hðτÞ ðt  τÞdτ dt dt 1 dy dx ¼ hðtÞ∗ dt dt dx 4t For given y(t), dy þ e4t δðt Þ: dt ¼ 2 sin 2t, and for given xðt Þ, dt ¼ 4e From sampling property impulse function, it is known that x(t) δ (t) ¼ x(0)δ(t). Since e4tδ(t) ¼ e0δ(t) ¼ δ(t), dx dt can be written as

dx ¼ 4e4t þ δðt Þ dt Let x1(t) ¼ 4e4tu(t), then by homogeneity, the corresponding output y1 ðt Þ ¼ 4yðt Þ ¼ 4 sin 2t

68

2 Continuous-Time Signals and Systems 4t Let x2 ðt Þ ¼ dx þ δðt Þ the corresponding output dt ¼ 4e

y2 ðt Þ ¼ 2 sin 2t As it is LTI system, if (x1(t) þ x2(t)) is the input to the system, the corresponding output is(y1(t) þ y2(t)) since x1(t) þ x2(t) ¼ 4e4t  4e4t + δ(t) ¼ δ(t), the impulse response h (t) ¼ y1(t) þ y2(t) ¼ 4 sin 2t þ 2 sin 2t Example 2.16 Consider a continuous-time LTI system with the unit step response s(t): (i) Deduce that the response y(t) of the system to the input x(t) is ð1 yð t Þ ¼ 1

and also show that

ð1 xð t Þ ¼ 1

dxðτÞ sðt  τÞdτ dt dxðτÞ uðt  τÞdτ: dt

(ii) Determine the response of an LTI system with step response

 sðt Þ ¼ e2t  et þ 1 uðt Þ to an input x(t) ¼ etu(t). Solution (i) The step response s(t) is sðt Þ ¼ hðt Þ∗uðt Þ Ð1 ¼ 1 hðτÞuðt  τÞdτ Ðt ¼ 1 hðτÞdτ Consider the following equivalence.

x(t)

h(t)

y(t)

=

x(t)

From the above equivalence, we obtain

h(t)

y(t)

2.3 The Convolution Integral

69

h(t)

Thus,

y(t)

ð t  dxðt Þ ∗ yð t Þ ¼ hðτÞdτ dt 1 dxðt Þ ¼ ∗sðt Þ dt

Since y(t) ¼ x(t) ∗ h(t) and if h(t) ¼ δ(t) and y(t) ¼ ð tx(t) as x(t)∗ δ(t) ¼ x(t),thus, dxðt Þ ∗ putting h(t) ¼ δ(t) in yð t Þ ¼ hðτÞdτ , it becomes dt 1 Ð  t xðt Þ ¼ dxdtðtÞ ∗ 1 δðτÞdτ Ð  t δ ð τ Þdτ ¼ uð t Þ 1 Since dxðt Þ ∗uðt Þ xðt Þ ¼ dt (ii)

xðt Þ ¼ et uðt Þ dxðt Þ ¼ et uðt Þ þ δðtÞet dt δðtÞet ¼ δðtÞe0 ¼ δðtÞ Since dxðt Þ ¼ et uðt Þ þ δðtÞ dt

 sðt Þ ¼ e2t  et þ 1 uðt Þ dxðtÞ ∗sðtÞ yðtÞ ¼ dt Ð 1 dxðτÞ ¼ 1 sðt  τÞdτ Ð 1 dt τ ¼ 1 fe uðτÞ þ δðτÞgfe2ðtτÞ  eðtτÞ þ 1guðt  τÞdτ Ðt ¼ 0 et fe2ðtτÞ  eðtτÞ þ 1gdτ þ fe2t  et þ 1guðtÞ Ðt ¼ 0 fe2tþ3τ  etþ2τ þ eτ gdτ þ sðtÞ 1 1 ¼ e2t ðe3t  1Þ  et ðe2t  1Þ þ ðet  1Þ þ sðtÞ 3 2 1 2t 1 t 5 t ¼  e þ e þ e  1 þ sðtÞ 3 2 6

Example 2.17 Consider h(t) be the triangular pulse and x(t) be the unit impulse train as shown in Figure 2.16. Determine y(t) ¼ x(t) * h(t) for T ¼ 2.

70

2 Continuous-Time Signals and Systems

x (t)

h (t)

1

1

t

1

-1

-2T

-T

0

T

2T

t

Figure 2.16 x(t) and h(t) of Example 2.17

y(t) 1

-3

-2

-1

0

1

3

2

t

Figure 2.17 Time versus y(t)

Solution xðt Þ ¼ yðt Þ ¼ xðt Þ∗hðt Þ ¼

1 X n¼1

1 X

δðt  nT Þ

n¼1

hðt Þ∗δðt  nT Þ ¼

1 X

hðt  nT Þ

n¼1

which is shown in Figure 2.17.

2.3.3

Computation of Convolution Integral Using MATLAB

MATLAB provides a function conv() that performs a discrete-time convolution of two discrete-time sequences. A new function convint() that uses conv() to numerically integrate the continuous-time convolution is as follows.

2.3 The Convolution Integral

71

function[y,ty]=convint(x,tx,h,th) %Inputs: %x is the input signal vector %tx is the times of the samples in x %h is the impulse response vector %th is times of the samples in h %outputs: %y is the output signal vector, %length(y)=length(x)+length(h)-1 %ty is the time of the samples in y dt=tx(2)-tx(1); y=conv(x,h)*dt; ty=(tx(1)+th(1))+[0:(length(y)-1)]*dt;

The computation of convolution of continuous-time signals using MATLAB is illustrated through the following numerical examples. Example 2.18 (i) Verify the result of Example 2.9 using MATLAB. (ii) Verify the result of Example 2.10 using MATLAB. Solution (i) The following MATLAB program 2.1 is used to compute the convolution of x(t) and h(t) of Example 2.9. Program 2.1 clc; clear all; close all; tx=[0:0.01:4]; x=ones(1,length(tx)); th=[0:0.01:4]; h=ones(1,length(th)); [y ty]=convint(x,tx,h,th); figure; plot(ty,y); xlabel('Time') ylabel('Amplitude'); axis([0 8 0 4]);

The output y(t) ¼ x(t) * h(t) of the above program is shown in Figure 2.18. It is observed to be the same as that shown in Figure 2.11. Thus, it is verified. (ii) The following MATLAB program 2.2 is used to compute the convolution of x(t) and h(t) of Example 2.10.

72

2 Continuous-Time Signals and Systems 4 3.5

Amplitude

3 2.5 2 1.5 1 0.5 0

0

1

2

3

4 Time

5

6

7

8

Figure 2.18 Time versus y(t)

2 1.5

Amplitude

1 0.5 0 -0.5 -1 -1.5 -2

0

Figure 2.19 Time versus y(t)

1

2

3

4

5

6

2.3 The Convolution Integral

73

Program 2.2 clc; clear all; close all; tx=[0:0.01:3]; tx1=[0:0.01:1]; x=[zeros(1,length(tx1)) ones(1,(length(tx)-length(tx1)))]; th1=[-1:0.01:1]; th2=[1.01:0.01:3]; h=[ones(1,length(th1)) -1*ones(1,length(th2))]; th=[-1:0.01:3]; [y ty]=convint(x,tx,h,th); figure; plot(ty,y); xlabel('Time') ylabel('Amplitude'); axis([0 6 -2 2]);

The output y(t) ¼ x(t) * h(t) of tie above program is shown in Figure 2.19 It is observed to be the same as that shown in Figure 2.12. Thus, it is verified Example 2.19 Consider the RC circuit of Example 2.11 with time constant RC ¼ 13 sec . Determine the Vout(t) using MATLAB for Vin(t) ¼ (u(t  3)  u (t  5)). Assume the capacitor is initially discharged. Solution The impulse response of the RC circuit is given by hð t Þ ¼

1  t=RC e uðtÞ ¼ 3e3t uðtÞV out ðt Þ ¼ RC

ð1

V in ðτÞhðt  τÞ ¼ V in ðt Þ∗hðtÞ

1

The following MATLAB program 2.3 is used to compute the convolution of vin(t) and h(t). Program 2.3 clc; clear all; close all; tx=[0:0.01:5]; tx1=[0:0.01:3]; x=[zeros(1,length(tx1)) ones(1,(length(tx)- length(tx1)))]; th=[0:0.01:5]; h =(3)* exp(-3*th); [y ty]=convint(x,tx,h,th); figure; plot(ty,y); xlabel('Time') ylabel('Amplitude');

74

2 Continuous-Time Signals and Systems 0.12 0.1

Amplitude

0.08

0.06 0.04 0.02 0

0

1

2

3

4

5

6

7

8

9

10

Time

Figure 2.20 Time versus Vout(t)

(a)

(b)

Figure 2.21 (a) Cascade connection of two systems. (b) Equivalent system

The output Vout(t) ¼ Vin(t) * h(t) of the above program is shown in Figure 2.20.

2.3.4

Interconnected Systems

2.3.4.1

Cascade Connection of Systems

The system shown in Fig. 2.21 is formed by connecting two systems in cascade. The impulse responses of the systems are given by h1(t) and h2(t), respectively. Let y(t) be the output of the first system. By the definition of convolution y1 ðt Þ ¼ xðt Þ∗h1 ðt Þ

ð2:37Þ

Then, the output of the overall system y(t) is given by yðt Þ ¼ y1 ðt Þ∗h2 ðt Þ ¼ ½xðt Þ∗h1 ðt Þ∗h2 ðt Þ

ð2:38Þ

2.3 The Convolution Integral

75

By the associativity property of convolution, Eq. (2.38) can be rewritten as yðt Þ ¼ y1 ðt Þ∗h2 ðt Þ ¼ xðt Þ∗½h1 ðt Þ∗h2 ðt Þ

ð2:39Þ

Hence, the impulse response of the overall system is given by hðt Þ ¼ h1 ðt Þ∗h2 ðt Þ ¼ ¼

2.3.4.2

Ð1 1

h1 ðτÞ h2 ðt  τÞdτ

1

h2 ðτÞ h1 ðt  τÞdτ

Ð1

ð2:40Þ

Parallel Connection of Two LTI Systems

The system shown in Fig. 2.22 is formed by connecting two systems in parallel. The impulse responses of the systems are given by h1(t) and h2(t), respectively. Let y1(t) and y2(t) be the outputs of the first system and second system, respectively. By the definition of convolution y1 ðt Þ ¼ xðt Þ∗h1 ðt Þ

ð2:41Þ

y2 ðt Þ ¼ xðt Þ∗h2 ðt Þ

ð2:42Þ

Then, the output of the overall system y(t) is given by yðt Þ ¼ y1 ðt Þ þ y2 ðt Þ ¼ xðt Þ∗h1 ðt Þ þ xðt Þ∗h2 ðt Þ

ð2:43Þ

By the distributive property of convolution, Eq. (2.43) can be rewritten as yðt Þ ¼ y1 ðt Þ þ y2 ðt Þ ¼ xðt Þ∗½h1 ðt Þ þ h2 ðt Þ Hence, the impulse response of the overall system is given by hð t Þ ¼ h1 ð t Þ þ h2 ð t Þ

(a)

ð2:44Þ

(b)

Figure 2.22 (a) Parallel connection of two systems. (b) Equivalent system

76

2 Continuous-Time Signals and Systems

Example 2.20 An LTI system consists of two subsystems in cascade. The impulse responses of the subsystems are, respectively, given by h1 ðt Þ ¼ e3t uðt Þ; h2 ðt Þ ¼ et uðt Þ Find the impulse response of the overall system. Solution The overall impulse response of the system is given by hðt Þ ¼ h1 ðt Þ∗h2 ðt Þ Ð1 ¼ 1 e3τ uðτÞeðtτÞ uðt  τÞdτ Ðt ¼ 0 e3τ eðtτÞ dτ Ðt ¼ et 0 e2τ dτ  1

¼ et  e2t uðtÞ 2

2.3.5

Periodic Convolution

If the signals x1(t) and x2(t) are periodic with common period T, it can be easily shown that the convolution of x1(t) and x2(t) does not converge. In such a case, the periodic convolution of x1(t) and x2(t) is defined as yðt Þ ¼ x1 ðt Þ

O

ðT x2 ð t Þ ¼

x1 ðτÞx2 ðt  τÞdτ

ð2:45Þ

0

Example 2.21 Let y(t) be the periodic convolution of x1(t) and x2(t). Show that yð t Þ ¼ yð t þ T Þ ðT Solution (i) yðt þ T Þ ¼

x1 ðτÞx2 ðt þ T  τÞdτ

0

Since x2(t) is periodic with period T, x2(t þ T  τ) ¼ x2(t  τ) ÐT 0

x1 ðτÞx2 ðt þ T  τÞdτ ¼

ÐT 0

x1 ðτÞx2 ðt  τÞdτ

yðt þ T Þ ¼ yðt Þ:

2.4 Properties of Linear Time-Invariant Continuous-Time System

2.4 2.4.1

77

Properties of Linear Time-Invariant Continuous-Time System LTI Systems With and Without Memory

The output y(t) of a memoryless system depends only on the present input x(t). If the system is LTI, then the relationship between the y(t) and x(t) for a memoryless system is yðt Þ ¼ cxðt Þ

ð2:46Þ

where c is an arbitrary constant. Since the output of a continuous-time system can be written as ð1

hðτÞxðt  τÞdτ

yð t Þ ¼ 1

the corresponding impulse response is h(t) ¼ cδ(t). Thus, a continuous-time system is memoryless if and only if hðt Þ ¼ cδðtÞ

2.4.2

ð2:47Þ

Causality for LTI Systems

The output of a continuous-time system can be written as ð1

hðτÞxðt  τÞdτ

yð t Þ ¼ 1

Since the impulse response h(τ) ¼ 0 for τ < 0 for a causal continuous-time system, the output of a causal system can be expressed by the following convolution integral: ð1 yð t Þ ¼

hðτÞxðt  τÞdτ

ð2:48Þ

0

2.4.3

Stability for LTI Systems

A continuous-time system is BIBO stable if and only if the impulse response is absolutely integrable, that is,

78

2 Continuous-Time Signals and Systems

ð1

hðτÞdτ < 1

ð2:49Þ

1

Example 2.22 Check stability of continuous-time system having the following impulse responses: (i) h(t) ¼ etu(t) (ii) h(t) ¼ et cos (2t)u(t) (iii) h(t) is periodic and nonzero Solution ð1

jhðτÞjdτ ¼

(i) 1

ð1

eτ dτ ¼ 1

0

Indicating that h(t) is absolutely integrable and, hence, h(t) is the impulse response of a stable system, ð1 ð1 (ii) eτ j cos ð2τÞjdτ jhðτÞjdτ ¼ 1

0 τ

Since e |cos (2τ)| is exponentially decaying for 0  t  1 , h(t) is absolutely summable, and hence, h(t) is the impulse response of a stable system. (iii) If h(t) is periodic with period T, then ð1 1

ð T=2 jhðτÞjdτ ¼ N

jhðτÞjdτ T=2

where N ! 1

ð1 Since h(t) is nonzero jhðτÞjdτ ! 1, hence, h(t) is absolutely summable and, 1

hence, h(t) is the impulse response of an unstable system. Example 2.23 Determine if each of the following system is causal or stable: (i) (ii) (iii) (iv) (v) (vi)

h(t) ¼ etu(t  1) h(t) ¼ etu(t þ 1) h(t) ¼ e2tu(t þ 10) h(t) ¼ tetu(t) h(t) ¼ e+tu(t  1) h(t) ¼ e2|t|

ð1 Solution (i) Causal because h(t) ¼ 0 for t < 0. Stable because jhððτÞjdτ < 1. 1 1 (ii) Not causal because h(t) 6¼ 0 for t < 0. Unstable because jhðτÞjdτ ¼ 1. ð 1 1 (iii) Not causal because h(t) 6¼ 0 for t < 0. Stable because jhðτÞjdτ < 1. 1

2.4 Properties of Linear Time-Invariant Continuous-Time System

ð1 (iv) Causal because h(t) ¼ 0 for t < 0. Stable because

79

jhðτÞjdτ < 1.

1 ð1

(v) Not causal because h(t) 6¼ 0 for t < 0. Stable because

jhðτÞjdτ < 1.

1ð

1

(vi) Not causal because h(t) 6¼ 0 for t < 0. Unstable because

jhðτÞjdτ ¼ 1.

1

2.4.4

Invertible LTI System

A system is invertible if its input x(t) can be recovered from its output y(t) ¼ x(t) * h (t). The cascade of an LTI system having impulse response h(t) with a LTI inverse system having impulse response g(t) ¼ h1(t) is shown in Figure 2.23. The process of recovering x(t) from x(t) * h(t) is called deconvolution as it corresponds to reverse of the convolution operation. The overall impulse response of the invertible system shown in Figure 2.23 is the convolution of h(t) and g(t). It is required that the output of the invertible system is equivalent to the input:

 xðt Þ∗ hðt Þ∗h1 ðt Þ ¼ xðt Þ

ð2:50Þ

hðt Þ∗h1 ðt Þ ¼ δðt Þ

ð2:51Þ

implying that

As an example, it is verified that the inverse system for a continuous-time integrator is a differentiator as follows: d dt

ð t

 xðτÞdτ ¼ xðtÞ

ð2:52Þ

1

Hence, the input-output relation for the inverse system shown in Figure 2.24 is xðtÞ ¼

x(t)

h(t)

dyðt Þ dt

y(t)

Figure 2.23 Cascade connection of an LTI system and its inverse

ð2:53Þ

(t)

x(t)

80

2 Continuous-Time Signals and Systems

x(t)

x(t)

y(t)

ò

Figure 2.24 Input-output relation for the inverse system

Example 2.24 (i) An echo of an auditorium can be modeled as a LTI system with an impulse response consisting of a train of impulses: hð t Þ ¼

1 X

hk δðt  kT Þ

k¼0

The inverse LTI system with impulse response g(t) is yðt Þ∗gðt Þ ¼ xðt Þ where g(t) is also an impulse train that is modeled as gð t Þ ¼

1 X

gk δðt  kT Þ

k¼0

Obtain the relationship between hkand gk. Solution y(t) ¼ x(t) * h(t) and x(t) ¼ g(t) *y(t), then gðt Þ∗hðt Þ ¼ δðt Þ: However, gðt Þ∗hðt Þ ¼

1 1 X Ð1 X gk δðt  τ  kT Þ hm δðτ  mT Þdτ 1 m¼0

k¼0

1 X 1 X gk hm δðt  τ  kT Þδðt  ðm þ kÞT Þ ¼ k¼0 m¼0

Let n ¼ m þ k, then m ¼ n-k, and g(t) * h(t) can be rewritten as gðt Þ∗hðt Þ ¼

1 1 X X n¼0

k¼0

! gk hnk δðt  nT Þ

2.4 Properties of Linear Time-Invariant Continuous-Time System

81

Hence, 1 X

( gk hnk ¼

k¼0

1,

n ¼ 0,

0,

n 6¼ 0:

Implying that g0 h0 ¼ 1, g0 h1 þ g1 h0 ¼ 0, g0 h2 þ g1 h1 þ g2 h0 ¼ 0, and so on, solution of the above equations leads to 1 , h0 g0 h1 h1 h1 ¼ ¼ 2 , g1 ¼ h0 h0 h0 h0 ! ! 1 1 h1 1 h2 h21  h2  2 h1 ¼   g0 h2 þ g1 h1 þ g2 h0 ¼  h0 h0 h0 h0 h20 h0 g0 ¼

(ii) Consider the following echo generation model characterized by yðt Þ ¼ xðt Þ þ ayðt  T Þ where 0 < a < 1 and T is delay. Construct the corresponding inverse system and obtain its impulse response. Solution Assuming y(t) ¼ 0 for t < 0 and x(t) ¼ 0 for t < 0, the impulse response h(t) of the echo generation system is given by hðtÞ ¼

1 X

ak δðt  kTÞ

k¼0

Thus, h0 ¼ 1, h1 ¼ a, hi ¼ 0 for i > 2. The inverse system has to obtain x(t) from the output y(t). Hence, the inverse system is characterized by xðt Þ ¼ yðt Þ  ayðt  T Þ and represented as depicted in Figure 2.25.

82

2 Continuous-Time Signals and Systems

y(t)

Figure 2.25 An inverse system

x(t) Delay T

-a

The impulse response g(t) of the inverse system is given by gð t Þ ¼

1 X

ðaÞk δðt  kT Þ

k¼0

Hence, g0 ¼ 1, g1 ¼ a. Example 2.25 Check y(t) ¼ x(2t) for causality and invertibility. Solution

yðt Þ ¼ xð2t Þ

At time t ¼ 1 yð1Þ ¼ xð2Þ indicating that the value of y(t) at time t ¼ 1 depends on x(t) at a time t ¼ 2. Therefore, y(t) ¼ x(2t) is not causal. y(t) is invertible; xðt Þ ¼ yðt=2Þ

2.5 2.5.1

Systems Described by Differential Equations Linear Constant-Coefficient Differential Equations

A general Nth-order linear constant-coefficient differential equation is given by XN

a n¼0 n

dn yðt Þ X M d n xð t Þ b n n ¼ k¼0 dt dt n

ð2:54Þ

where coefficients an and bn are real constants. The order N refers to the highest derivative of y(t) in Eq. (2.54). For example, consider the RC circuit considered in Example 2.11, the input x(t) and the output y(t) ¼ vo (t). If the current flowing through the RC circuit is i(t), using Kirchhoff’s voltage law, we write

2.5 Systems Described by Differential Equations

xðt Þ þ Riðt Þ þ

83

1 c

ð iðt Þdt ¼ 0

ð2:55Þ

which can be rewritten as Riðt Þ þ

1 c

ð iðt Þdt ¼ xðt Þ

ð2:56Þ

Since iðt Þ ¼ c dvdt0 ðtÞ ¼ c dydtðtÞ , substituting iðt Þ ¼ c dydtðtÞ in Eq. (2.56), we obtain the following first-order constant-coefficient differential equation dyðt Þ 1 1 þ yð t Þ ¼ xð t Þ dt RC RC

ð2:57Þ

relating the voltage across the capacitor y(t) and the input x(t). Example 2.26 Find the differential equation relating the current y(t) and the input voltage x(t) for the RLC circuit shown in Figure 2.26 assuming R ¼ 3 Ohms, L ¼ 1 Henry, and C ¼ 12 Farad: Solution Using Kirchhoff’s voltage law, we write the following loop equation for the given RLC circuit: ð dyðt Þ 1 þ yðt Þdt ¼ 0 xðt Þ þ Ryðt Þ þ L dt c For R ¼ 3, L ¼ 1, and C ¼ 12 , the above equation becomes ð dyðt Þ þ 3yðtÞ þ 2 yðtÞdt ¼ xðtÞ dt Differentiating this equation, we obtain d 2 yð t Þ dyðt Þ dxðt Þ þ 2yðtÞ ¼ þ3 2 dt dt dt

Figure 2.26 RLC circuit

84

2 Continuous-Time Signals and Systems

Figure 2.27 Operational amplifier circuit

Example 2.27 Find the differential equation relating the input voltage V i ðt Þ and the output voltage V o ðt Þ for the operational amplifier circuit shown in Figure 2.27. Solution ‘I r1 ¼ I r2 þ I c1 ; I r2 ¼ I c2 ; V 2 ¼ V 0 Rewriting the current node equations, we get Vi  V1 V1  V0 d ¼ þ c1 ð V 1  V 0 Þ dt r1 r2 which can be rewritten as r 1 r 2 c1

dV 1 dV 0  r 1 r 2 c1 þ ðr 1 þ r 2 ÞV 1  r 1 V 0 ¼ V i r 2 dt dt V1  V0 dV 0 ; ¼ c2 r2 dt dV 0 þ V0 V 1 ¼ r 2 c2 dt

Substituting the above equation for V 1 , the input-output relation can be written as c2 c1 r 2 r 1

d2 V 0 dV 0 þ V0 ¼ Vi þ c2 ðr 1 þ r 2 Þ 2 dt dt

which is rewritten as d2 V 0 r 1 þ r 2 dV 0 V0 Vi þ þ ¼ 2 r r c dt r r c c r r dt 1 2 1 1 2 1 2 1 2 c1 c2

2.5 Systems Described by Differential Equations

2.5.2

85

The General Solution of Differential Equation

The general solution of Eq. (2.54) for a particular input x(t) is given by yð t Þ ¼ yc ð t Þ þ yp ð t Þ

ð2:58Þ

where yc(t) is called the complementary solution and yp(t) is called the particular solution. The complementary solution yc(t) is obtained by setting x(t) ¼ 0 in Eq. (2.54). Thus yc(t) is the solution of the following homogeneous differential equation XN

a n¼0 n

dn yðt Þ ¼0 dt n

ð2:59Þ

Example 2.28 Consider the RC circuit of Example 2.11 with time constant RC ¼ 1 sec. Determine the voltage across the capacitor for an input x(t) ¼ e2tu(t). Assume the capacitor is initially discharged. Solution As the time constant RC ¼ 1, the input x(t) and the output yðt Þ ¼ V 0 ðt Þ of the RC circuit are related by dyðt Þ þ yðt Þ ¼ e2t uðtÞ yð0Þ ¼ 0 dt The particular solution for the exponential input is of the form yp ðt Þ ¼ Ae2t

t>0

Substituting yp(t) in the above differential equation, we get 2Ae2t þ Ae2t ¼ e2t Solving for A, we obtain A ¼ 1 and yp ðt Þ ¼ e2t To obtain complementary solution, let us assume yc ðt Þ ¼ Bekt Substituting this into dyc ðt Þ þ yc ð t Þ ¼ 0 dt

t>0

86

2 Continuous-Time Signals and Systems

yields Bkekt þ Bekt ¼ 0 ðk þ 1ÞBekt ¼ 0 Thus, k ¼ 1 and yc ðt Þ ¼ Bet Now, yðt Þ ¼ yc ðt Þ þ yp ðt Þ ¼ Bet  e2t at t ¼ 0 y(0) ¼ B – 1. Since the capacitor is initially discharged, y(0) ¼ 0, and we obtain B ¼ 1. Hence, the voltage across the capacitor is given by

 yðt Þ ¼ et  e2t uðt Þ

2.5.3

Linearity

The system specified by Eq. (2.54) is linear only if all of the initial conditions are zero. For instance, in the Example 2.11, if the capacitor is not assumed to be discharged initially, then yð0Þ ¼ V 0 ð0Þ 6¼ 0 A linear system has the property that zero input produces zero output. However, if we let x(t) ¼ 0, then yðt Þ ¼ yc ðt Þ ¼ yð0Þet

ð2:60Þ

Thus, this system is nonlinear if y(0) 6¼ 0. If the capacitor is assumed to be discharged initially, then yð0Þ ¼ V 0 ð0Þ ¼ 0: Then for x(t) ¼ 0, yðt Þ ¼ yc ðt Þ ¼ 0

ð2:61Þ

the system is linear

2.5.4

Causality

A linear system described by Eq. (2.54) is causal when it is initially relaxed. It implies that if x(t) ¼ 0 for t  t0, then y(t) ¼ 0 for t  t0, thus, the response for t > to with the initial conditions

2.5 Systems Described by Differential Equations

yð t 0 Þ ¼

87

dyðt 0 Þ dN1 yðt 0 Þ ... ¼ ¼0 dt dt N1

dn yðt 0 Þ dn yðt Þ ¼ dt n dt n t¼t0

2.5.5

ð2:62aÞ

ð2:62bÞ

Time-Invariance

For a linear causal system, initial rest also implies time-invariance. For example, consider the system described by dyðt Þ þ yðt Þ ¼ xðt Þ yð0Þ ¼ 0 dt

ð2:63Þ

Let y1(t) be the response to an input x1(t) and x1 ðt Þ ¼ 0 t  0

ð2:64Þ

dy1 ðt Þ þ y1 ðt Þ ¼ x1 ðt Þ dt

ð2:65Þ

y 1 ð 0Þ ¼ 0

ð2:66Þ

so that

and

Now, let x2(t) ¼ x1(t  τ) and y2 (t) be the corresponding response. From Eq. (2.64), we get x2 ð t Þ ¼ 0

tτ

ð2:67Þ

Then y2(t) should satisfy dy2 ðt Þ þ y2 ðt Þ ¼ x2 ðt Þ dt

ð2:68Þ

y2 ð τ Þ ¼ 0

ð2:69Þ

and

From Eq. (2.65), we write dy1 ðt  τÞ þ y1 ð t  τ Þ ¼ x1 ð t  τ Þ ¼ x2 ð t Þ dt

88

2 Continuous-Time Signals and Systems

By letting y2(t) ¼ y1(t  τ), we obtain from Eq. (2.66) y2 ðτÞ ¼ y1 ðτ  τÞ ¼ y1 ð0Þ ¼ 0: Eqs. (2.68) and (2.69) are satisfied and thus the system is time invariant.

2.5.6

Impulse Response

The impulse response h(t) of the continuous-time LTI system described by Eq. (2.54) satisfies the differential equation XN

a n¼0 n

d n hð t Þ X M dn δðt Þ þ b n n¼0 dt n dt n

ð2:70Þ

with the initial rest condition. For example, let us consider the RC circuit of Example 2.11 with the time constant RC ¼ 1 sec described by the following differential equation: dyðt Þ þ yðt Þ ¼ xðt Þ dt The impulse response h(t) should satisfy the differential equation dhðt Þ þ hðt Þ ¼ δðt Þ dt Then, the complimentary solution hc(t) satisfies  dhc ðt Þ þ hc ð t Þ ¼ 0 dt To obtain complementary solution, let us assume yc ðt Þ ¼ Bekt Substituting this into  dhc ðt Þ þ hc ð t Þ ¼ 0 dt yields Bkekt þ Bekt ¼ 0 ðk þ 1ÞBekt ¼ 0

2.5 Systems Described by Differential Equations

89

Thus, k ¼ 1 and hc ðt Þ ¼ Bet uðtÞ The particular solution hp(t) is zero since hp(t) cannot contain δ(t). Thus, hðt Þ ¼ Bet uðtÞ To find the constant B, substituting h(t) ¼ Betu(t)into dhðt Þ þ hðt Þ ¼ δðt Þ dt yields Bet uðtÞ þ Bet Bet

duðtÞ þ Bet uðtÞ ¼ δðtÞ dt

duðtÞ ¼ Bet δðtÞ ¼ δðtÞ dt

such that B ¼ 1 and hence hðt Þ ¼ et uðtÞ: Example 2.29 Consider the RL circuit shown in Figure 2.28: (i) Determine the impulse response. (ii) Determine the step response. Solution Writing the loop equation using Kirchhoff’s voltage law assuming i(t) is the current flowing through the circuit, we obtain xðtÞ þ RiðtÞ þ L But iðtÞ ¼ yRðtÞ :

Figure 2.28 RL circuit

diðt Þ ¼0 dt

90

2 Continuous-Time Signals and Systems

Substituting iðtÞ ¼ yRðtÞ in the above equation, we get L dyðtÞ þ yðtÞ ¼ xðtÞ R dt which can be rewritten as dyðt Þ R R þ yðtÞ ¼ xðt Þ dt L L (i) The impulse response h(t) should satisfy the differential equation dhðt Þ R R þ hðt Þ ¼ δðt Þ dt L L Then, the complimentary solution hc(t) satisfies dhc ðtÞ R þ hc ðtÞ ¼ 0 dt L To obtain complementary solution, let us assume yc ðt Þ ¼ Bekt Substituting this into dhc ðtÞ R þ hc ðtÞ ¼ 0 dt L yields R Bkekt þ Bekt ¼ 0

L R kþ Bekt ¼ 0 L Thus, k ¼ RL and R

hc ðt Þ ¼ BeLt uðtÞ The particular solution hp(t) is zero since hp(t)) cannot contain δ(t). Thus, R

hðt Þ ¼ BeLt uðtÞ R

To find the constant B, substituting hðt Þ ¼ BeLt uðtÞinto dhðt Þ R R þ hðt Þ ¼ δðt Þ dt L L

2.5 Systems Described by Differential Equations

91

yields R duðtÞ R R R R R þ B eLt uðtÞ ¼ δðtÞ B eLt uðtÞ þ BeLt L dt L L R R R  L t duðtÞ Lt R Be ¼ Be δðtÞ ¼ δðtÞ dt L L

such that B ¼ RL and hence hð t Þ ¼

R R t e L uðtÞ: L

(ii) The step response s(t) is given by Ðt

ÐtR R hðτÞdτ ¼ 0 eLτ dτ

L  R R Lτ t Lt ¼ e j0 ¼ 1  e uðtÞ

sðtÞ ¼

2.5.7

0

Solution of Differential Equations Using MATLAB

The response y(t) of a system described by Eq. (2.54) for an input x(t) can be determined by using the MATLAB command dsolve(‘eqn1’,‘eqn2’, ...) which accepts symbolic equations representing ordinary differential equations and initial conditions. Several equations or initial conditions may be grouped together, separated by commas, in a single input argument. Example 2.30 Verify the result of Example 2.28 using MATLAB. Solution The relation between the output y(t) and the input x(t) is related by dyðt Þ þ yðt Þ ¼ e2t uðtÞ yð0Þ ¼ 0 dt The output response y(t) is determined and displayed by the MATLAB commands y = dsolve('Dy+y=exp(-2*t)', ,'y(0)=0', 't'); disp (['y(t) = (',char(y), ')u(t) ' ]);

The displayed output is

 yðt Þ ¼ et  e2t uðt Þ

92

2 Continuous-Time Signals and Systems

Example 2.31 Consider the RLC circuit of Example 2.26 and determine the current y(t) for the input voltage x(t) ¼ 10e3t u(t) where the initial inductor current is zero and the initial capacitor voltage is equal to 5 volts. Solution The relation between the output y(t) and the input x(t) is related by d2 y dy dx þ 3 þ 2y ¼ 2 dt dt dt dyðt Þ yð0Þ ¼ 0, ¼ 5: dt t¼0 The current y(t) is determined and displayed by the MATLAB commands

y = dsolve('D2y+3*Dy+2*y=-30*exp(-3*t)', 'y(0)=0', 'Dy(0)=5', 't'); disp (['y(t) = (', char(y), ')u(t) ' ]);

The displayed y(t) is

 yðt Þ ¼ 25e2t  10et  15e3t uðt Þ

2.5.8

Determining Impulse Response and Step Response for a Linear System Described by a Differential Equation Using MATLAB

In general, for a system described by Eq. (2.54), the impulse response can be determined by using the following MATLAB command: h ¼ impluseðb; a; t Þ For example, for the differential equation given by d2 y dy þ 3 þ 2y ¼ xðt Þ dt dt 2 we use the following MATLAB program 2.4 to determine impulse response and step response.

2.6 Block-Diagram Representations of LTI Systems Described by Differential Equations

93

Program 2.4 clc; clear all; close all; th=0:.01:4; b = [1]; a = [1 3 2]; h=impulse(b,a,th); tx=[0:0.01:4]; x=[ones(1,length(tx))]; [y ty]=convint(x,tx,h,th); figure,plot(th,h) xlabel('Time') ylabel('Amplitude'); figure, plot(ty,y); xlabel('Time') ylabel('Amplitude');

The impulse response and step response obtained from the above program are shown in Figure 2.29(a) and (b), respectively.

2.6

Block-Diagram Representations of LTI Systems Described by Differential Equations

The block diagram of a continuous system describes how the internal operations are ordered, whereas the differential equation description gives only the input and output relation. Hence, the block diagram is more detailed representation of continuoustime systems than the differential equation description. Integrators are preferred to differentiators in the block-diagram representation of continuous-time systems as the integrators can be easily built from analog components and noise in a system will be smoothed out. Let us define the following three basic elements adder, scalar multiplier, and integrator used in the block-diagram representation of continuous-time systems (Figure 2.30). Consider the system described by Eq. (2.54), which is repeated here for convenience assuming M ¼ N: XN

a n¼0 n

dk yðtÞ X N d n xðtÞ ¼ b n n¼0 dt k dt n

ð2:71Þ

94

2 Continuous-Time Signals and Systems 0.25

Amplitude

0.2

0.15

0.1

0.05

0

0

0.5

1

1.5

2

2.5

3

3.5

4

5

6

7

8

Time

(a) 0.5 0.45 0.4

Amplitude

0.35 0.3 0.25 0.2 0.15 0.1 0.05 0

0

1

2

3

4 Time

(b) Figure 2.29 (a) impulse response, (b) step response

n

If it is assumed that the system is at rest, then the Nth integral of d dtyðntÞ is yðNnÞ ðt Þ, n and the Nth integral of is d dtxðntÞ is xðNnÞ ðt Þ. Hence, taking the Nth integral of Eq. (2.71), we obtain the integral description of the system as

2.7 Singularity Functions

95

Figure 2.30 Blockdiagram representation of basic elements (a) adder, (b) multiplier, (c) integrator

(a)

(b)

XN n¼0

an yðNkÞ ðt Þ ¼

XN n¼0

bn xðNnÞ ðt Þ

ð2:72Þ

Since y(0)(t) ¼ y(t), Eq. (2.72) can be rewritten as yð t Þ ¼

i XN1 1 hX N b x ð t Þ  a y ð t Þ n n ð Nn Þ ð Nn Þ n¼0 n¼0 aN

ð2:73Þ

The direct form I and the direct form II implementations of Eq. (2.73) are shown in Figure 2.31(a) and (b), respectively. Example 2.32 Obtain the block-diagram representation of a system described by d2 y dy d 2 x dx þ 2 þ y ¼ 2   6xðt Þ 2 dt dt dt dt Solution

2.7

Singularity Functions

The unit impulse δ(t) is one of a class of signals known as singularity functions. Consider a LTI system for which the input and the output are related by yðt Þ ¼

dxðt Þ dt

ð2:74Þ

The unit impulse response of the considered system is the derivative of the unit impulse which is referred as the unit doublet u1(t).

96

2 Continuous-Time Signals and Systems

(a)

(b) Figure 2.31 Block-diagram representation for continuous-time system described by integral Eq. (2.73) (a) direct form I, (b) direct form II

2.7 Singularity Functions

97

Figure 2.32 Block diagram representation in direct form II for a second order continuous time system described by the above differential equation is shown in Figure 2.32

By the convolution representation of LTI systems, we represent dxðt Þ ¼ xðt Þ∗u1 ðt Þ dt

ð2:75Þ

for any input signal x(t). Similarly, for an LTI system described by yð t Þ ¼

d 2 xð t Þ , dt 2

ð2:76Þ

we obtain

 d 2 xðt Þ d dxðt Þ ¼ ¼ xðt Þ∗u1 ðt Þ∗u1 ðt Þ dt dt dt 2

ð2:77Þ

¼ xðt Þ∗u2 ðt Þ where u2(t) ¼ u1(t) ∗ u1(t) is the second derivative of unit impulse. Thus, uk(t) for k > 0 is the kth derivative of the unit impulse and is the impulse response of a LTI system that takes the kth derivative of the input: uk ðt Þ ¼ u1 ðt Þ∗u1 ðt Þ . . . . . . k times

ð2:78Þ

If we consider a system described by Eq. (2.75) with input x(t) ¼ 1,then dxðt Þ ¼ xðt Þ∗u1 ðt Þ ¼ dt

ð1

u1 ðτÞxðt  τÞdτ ¼

1

ð1

u1 ðτÞdτ ¼ 0

ð2:79Þ

1

Hence, the unit doublet has zero area. In addition to the singularity functions, the successive integrals of the unit impulse function, it is known that ðt uð t Þ ¼ 1

δðτÞdτ,

ð2:80Þ

98

2 Continuous-Time Signals and Systems

and hence the unit step function is the impulse response of an integrator, and we have ðt xðt Þ∗uðt Þ ¼

xðτÞdτ

ð2:81Þ

1

Similarly, the impulse response of a system consisting of two integrators in cascade can be denoted by u1(t) which can be expressed as the convolution of u(t) with itself: ðt u1 ðt Þ ¼ uðt Þ∗uðt Þ ¼ uðτÞdτ ð2:82Þ 1

Since u(t) ¼ 0 for t < 0 and u(t) ¼ l, u1(t) can be expressed as u1 ðt Þ ¼ tuðt Þ

ð2:83Þ

Example 2.33 For a system if y(t) ¼ x(t) * h(t), show that

ð t  dhðt Þ (i) yðtÞ ¼ xðτÞdτ ∗ dt 1 (ii) yðtÞ ¼ Solution (i)

dxðt Þ ∗ dt

ðt

hðτÞdτ

1

yðtÞ ¼ xðt Þ∗hðt Þ, ¼ xðt Þ∗u1 ðt Þ∗u1 ðt Þ∗hðt Þ Ðt dhðt Þ ¼ 1 xðτÞdτ∗ dt

(ii)

yðtÞ ¼ xðt Þ∗hðt Þ, ¼ xðt Þ∗u1 ðt Þ∗hðt Þ∗u1 ðt Þ ðt dxðtÞ ∗ ¼ hðτÞdτ dt 1

2.8 2.8.1

State-Space Representation of Continuous-Time LTI Systems State and State Variables

The state of a system at time t0 is the minimal information required that is sufficient to determine the state and the output of the system for all times t  t0 for the known system input at all times t  t0.The variables that contain this information are called state variables.

2.8 State-Space Representation of Continuous-Time LTI Systems

2.8.2

99

State-Space Representation of Single-Input SingleOutput Continuous-Time LTI Systems

Consider a single-input single-output continuous-time LTI system described by the following Nth-order differential equation: d N yð t Þ dN1 yðt Þ dyðt Þ þ a 0 y ð t Þ ¼ ℧ð t Þ þ a þ . . . þ a1 N1 N N1 dt dt dt

ð2:84Þ

where y(t)) is the system output and ʊ(t) is the system input. Define the following useful set of state variables x1 ðt Þ ¼ yðt Þ, x2 ðt Þ ¼

dyðt Þ d 2 yð t Þ dN1 yðt Þ , x3 ðt Þ ¼ , . . . , x ð t Þ ¼ N dt dt 2 dt N1

ð2:85Þ

Taking derivatives of the first N  1 state variables of the above, we get dx1 ðt Þ dx2 ðt Þ dxN1 ðt Þ ¼ x2 ðt Þ, ¼ x3 ðt Þ, . . . , ¼ xN : dt dt dt

ð2:86Þ

Rearranging Eq. (2.84) and using Eq. (2.86), we obtain dxN ðt Þ ¼ a0 x1 ðt Þ  a1 x2 ðt Þ      aN1 xN1 ðt Þ þ ℧ðt Þ dt

ð2:87Þ

yð t Þ ¼ x1 ð t Þ

ð2:88Þ

_, Denoting dxðtÞ dt by x Eqs. (2.86), (2.87), and (2.88) can be written in matrix form as 3 2 0 x_ 1 ðtÞ 7 6 6 _ x ðtÞ 7 6 0 6 2 7 6 6 6 ⋮ 7¼6 ⋮ 7 6 6 7 6 6 4 x_ N1 ðtÞ 5 4 0 a0 x_ N ðtÞ 2

1

0

...

0

0 ⋮

1 ⋮

... ⋱

0 ⋮

0

0

a1

a2

32

x1 ðtÞ

3

2

0

3

7 6 7 76 76 x2 ðtÞ 7 6 0 7 7 6 7 76 76 ⋮ 7 þ 6 0 7℧ðtÞ 7 6 7 76 7 6 7 76 ... 1 54 xN1 ðtÞ 5 4 ⋮ 5 1 xN ðtÞ . . . aN1 ð2:89aÞ 2

yð t Þ ¼ ½ 1

0 0

...

x1 ð t Þ

3

7 6 6 x2 ð t Þ 7 7 6 7 6 0 6 ⋮ 7 7 6 6 x ðt Þ 7 4 N1 5 xN ð t Þ

ð2:89bÞ

100

2 Continuous-Time Signals and Systems

Define a Nx1 dimensional vector called state vector as 2

x1 ð t Þ

3

7 6 6 x2 ð t Þ 7 7 6 7 6 X ðt Þ ¼ 6 ⋮ 7 7 6 6 x ðt Þ 7 4 N1 5

ð2:90Þ

xN ð t Þ The derivative of X(t) becomes 2

x_ 1 ðtÞ

3

7 6 6 x_ 2 ðtÞ 7 7 6 7 6 dXðtÞ 6 _ ¼ X ðtÞ ¼ 6 ⋮ 7 7 dt 7 6 6 x_ N1 ðtÞ 7 5 4 x_ N ðtÞ

ð2:91Þ

More compactly Eqs. (2.89a) and (2.89b) can be written as X_ ðt Þ ¼ AXðt Þ þ b℧ðt Þ

ð2:92Þ

yðt Þ ¼ cX ðt Þ

ð2:93Þ

where 2

0

1

0

...

0

6 6 0 6 6 A¼6 ⋮ 6 6 0 4

0

1

...

0

⋮

⋮

⋱

⋮

0

0

...

1

a0

a1

a2

. . . aN1

3

2

0

3

7 6 7 7 6 0 7 7 6 7 7 6 7 7; b ¼ 6 0 7; c ¼ ½ 1 7 6 7 7 6⋮7 5 4 5

0

0 ...

0

1

Eqs. (2.92) and (2.93) are called N-dimensional state-space representation or state equations of the system. In general state equations of a system are described by X_ ðt Þ ¼ AXðt Þ þ b℧ðt Þ

ð2:94Þ

yðt Þ ¼ cXðt Þ þ d℧ðt Þ

ð2:95Þ

2.8 State-Space Representation of Continuous-Time LTI Systems

101

Example 2.34 Obtain the state-space representation of a system described by the following differential equation: d3 yðt Þ d 2 yð t Þ dyðt Þ þ 4yðt Þ ¼ ℧ðt Þ þ2 þ3 3 dt dt dt 2 Solution: The order of the differential equation is three. Hence, the three-state variables are x1 ðt Þ ¼ yðt Þ, x2 ðt Þ ¼

dyðt Þ d 2 yð t Þ , x3 ð t Þ ¼ dt dt 2

The first derivatives of the state variables are x_ 1 ðt Þ ¼ x2 ðt Þ x_ 2 ðt Þ ¼ x3 ðt Þ x_ 3 ðt Þ ¼ 4x1 ðt Þ  3x2 ðt Þ  2x3 ðt Þ þ ℧ðt Þ The state-space representation in matrix form is given by 2

x_ 1 ðt Þ

3

2

0

1

6 7 6 4 x_ 2 ðt Þ 5 ¼ 4 0

0

x_ 3 ðt Þ

-4

-3

yð t Þ ¼ ½ 1

2 3 0 76 7 6 7 1 54 x2 ðt Þ 5 þ 4 0 5℧ðt Þ 0

32

x1 ð t Þ

3

x3 ð t Þ 2 3 x1 ð t Þ 6 7 0 0 4 x2 ðt Þ 5 -2

1

x3 ð t Þ Example 2.35 Obtain the state-space representation for the electrical circuit shown in Figure 2.33 considering V c1 , i1, and V c2 as state variables and vc1 as output y(t). i1

+ vc1 – 1F vs(t)

+ –

Figure 2.33 Third-order electrical circuit

1H

i2

i3 1W

1F

+ Vc2 -

102

2 Continuous-Time Signals and Systems

Solution The state variables for the circuit are x1 ¼ V c1 x2 ¼ i1 x3 ¼ V c2 From the relationship between the voltage V c1 and current i1, we obtain dx1 ðt Þ ¼ x2 dt Kirchhoff’s voltage equation around the closed loop gives V s þ x1 þ

dx2 ðt Þ þ x3 ¼ 0 dt

This equation can be rewritten as dx2 ðt Þ ¼ x1  x3 þ V s dt The current i3 ¼ x3 and the current i2 ¼ dxdt3 ðtÞ By Kirchhoff’s current law i1 ¼ i2 þ i3 implying that x2 ¼

dx3 ðt Þ þ x3 dt

Hence, dx3 ðt Þ ¼ x2  x3 dt The voltage V c1 is taken as the output y(t): yð t Þ ¼ x1 The state-space representation of the circuit is given by

2.8 State-Space Representation of Continuous-Time LTI Systems

103

1Ω i1(t )

iL(t ) 1H

+

+

vs (t )

1F

−

vc (t )

1Ω

Figure 2.34 Electrical circuit of Example 2.36

2

x_ 1 ðt Þ

3

2

0

1

6 7 6 6 x_ 2 ðt Þ 7 ¼ 6 1 4 5 4 x_ 3 ðt Þ

0

0

yð t Þ ¼ ½ 1

2 3 0 76 7 6 7 6 7 6 7 1 7 54 x2 ðt Þ 5 þ 4 0 5V s 0

32

x1 ð t Þ

1 x3 ð t Þ 2 3 x1 ð t Þ 6 7 7 0 6 4 x2 ð t Þ 5

1

0

3

1

x3 ð t Þ Example 2.36 Obtain state-space representation of the circuit shown in Figure 2.34 considering the current through the inductor and voltage across the capacitor as state variables and voltage across the capacitor as the output y(t). Solution The state variables for the circuit are x1 ð t Þ ¼ i L ð t Þ x2 ð t Þ ¼ V c ð t Þ Since the voltage across the capacitor is equal to the voltage across the series inductor and resistor branch, we obtain dx1 ðt Þ þ x1 ðt Þ ¼ x2 ðt Þ dt This equation can be rewritten as

104

2 Continuous-Time Signals and Systems

dx1 ðt Þ ¼ x1 ðt Þ þ x2 ðt Þ dt Kirchhoff’s voltage equation around the closed loop gives V s ðt Þ þ i1 ðt Þ þ x2 ðt Þ ¼ 0 Hence i1 ðt Þ ¼ x2 ðt Þ þ V s ðt Þ By Kirchhoff’s current law i1 ðt Þ ¼ x1 ðt Þ þ

dx2 ðt Þ dt

Therefore x2 ðt Þ þ V s ðt Þ ¼ x1 ðt Þ þ

dx2 ðt Þ dt

This equation can be rewritten as dx2 ðt Þ ¼ x1 ðt Þ  x2 ðt Þ þ V s ðt Þ dt The voltage V c ðt Þ is taken as the output y(t): yð t Þ ¼ x2 ð t Þ The state-space representation of the circuit is given by "

x_ 1 ðt Þ x_ 2 ðt Þ

#

" ¼

1 1

yð t Þ ¼ ½ 0

2.8.3

1

1

#"

x1 ðt Þ

1 x2 ðt Þ " # x1 ð t Þ

# þ

" # 0 1

vs ð t Þ

x2 ð t Þ

State-Space Representation of Multi-input Multi-output Continuous-Time LTI Systems

The state-space representation of continuous-time system with m inputs and l output and N state variables can be expressed as

2.9 Problems

105

X_ ðt Þ ¼ AX ðt Þ þ B℧ðt Þ

ð2:96Þ

yðt Þ ¼ CX ðt Þ þ D℧ðt Þ

ð2:97Þ

where 2

...

a22

...

⋮

⋱

7 a2N 7 7 ⋮ 7 5

aN1 aN2 2 c11 c12 6 6 c21 c22 C¼6 6⋮ ⋮ 4

...

aNN

cl2

...

6 6 a21 A¼6 6⋮ 4

cl1

2.9

3

a12

a11

a1N

. . . c1N

3

2

6 6 b21 B¼6 6⋮ 4 NN

7 . . . c2N 7 7 ⋱ ⋮7 5 clN

b11

lN

b12

...

b22

...

⋮

⋱

b1m

3

7 b2m 7 7 ⋮ 7 5

bN1 bN2 . . . bNm Nm 3 2 d11 d12 . . . d 1m 7 6 6 d21 d22 . . . d 2m 7 7 6 D¼6 7 4⋮ ⋮ ⋱ ⋮5 dl1

dl2

...

d lm

lm

Problems

1. Check the following for linearity and time-invariance: (i) (ii) (iii) (iv) (v) (vi)

yðt Þ ¼ dxdtðtÞ y(t) ¼ tx(t) y(t) ¼ t2x2(t) dyðt Þ dt þ tyðt Þ ¼ xðt Þ y(t) ¼ ln (x(t)) y(t) ¼ x(t) þ cons tan t

2. Determine which of the following systems are linear and which are nonlinear: (i) (ii) (iii) (iv)

dyðt Þ dt dyðt Þ dt

þ 3yðt Þ ¼ x2 ðt Þ

dyðt Þ dt

þ sin ðt Þyðt Þ ¼ dxdtðtÞ þ 3xðt Þ



þ y2 ð t Þ ¼ xð t Þ 2 dyðt Þ þ 3yðt Þ ¼ xðt Þ dt

3. An amplifier has an output y(t) ¼ cos (ωt). If y(t) is limited by clipping resulting in the clipped output yc(t), check for linearity and time-invariance of yc(t). 4. An input signal x(t) and two possible outputs of a linear time-invariant system are shown in Figure P2.1. Which outputs are possible given that the system is linear and time invariant?

106

2 Continuous-Time Signals and Systems x (t )

y (t )

(i)

(ii)

Figure P2.1 Input signal and two possible outputs of problem 4

5. If x(t) ¼ u(t + 1)  u(t  1), compute (x *x *x)(t) and sketch. 6. Determine graphically the convolution y(t) ¼ x(t) * h(t) for the following: (i) (ii)

xðt Þ ¼ uðt Þ-uðt-4Þ hðt Þ ¼ uðt-4Þ-uðt-6Þ xðt Þ ¼ et uðt Þ

hðt Þ ¼ e2t uðt Þ xðtÞ ¼ 2uðt  1Þ  2uðt  2Þ (iii) hðtÞ ¼ uðt þ 1Þ  2uðt  1Þ þ uðt  2Þ xðt Þ ¼ uðt Þ ( et , t  0, (iv) hð t Þ ¼ et , t  0 xðt Þ ¼ uðt þ 1Þ  uðt  1Þ (v) 1 hðt Þ ¼ t ðuðt Þ  uðt  3ÞÞ 3 7. Consider a continuous-time LTI system with the step response sðtÞ ¼ et uðtÞ Determine and sketch the output of this system to the input xðtÞ ¼ uðt  1Þ  uðt  3Þ 8. Consider h(t) be the triangular pulse and x(t) be the unit impulse train as shown in Figure 2.16 of Example 2.17. Determine y(t) ¼ x(t) * h(t) and sketch it. (i) for T ¼ 3 (ii) for T ¼ 32 9. An LTI system consists of two subsystems in cascade. The impulse responses of the subsystems are, respectively, given by h1 ðt Þ ¼ δðt Þ  2e2t uðt Þ;

h2 ð t Þ ¼ e t uð t Þ

Find the impulse response of the overall system.

2.9 Problems

107

10. If y(t) ¼ x(t) * h(t) shows that the area of the convolution y(t) is the product of the areas of the signals that are being convolved x(t) and h(t), that is, ð1 1

yðτÞdτ ¼

ð 1

xðτÞdτ

 ð 1

1

 hðτÞdτ

1

11. Compute and sketch the periodic convolution of the square-wave signal x(t) shown in Fig. P2.2 with itself.

x(t)

Figure P2.2 Square wave signal of problem 11

2

-2

-1

0

1

2

t

12. Consider an LTI system with impulse response h(t): (i) Check for its stability, if h(t) is periodic and nonzero. (ii) Check for causality of the inverse of the LTI system if h(t) is causal. (iii) Check for its stability if h(t) is causal. 13. Consider the system described by dyðt Þ dxðt Þ þ 3yðt Þ ¼ xðt Þ þ dt dt Determine the impulse response h(t) of the system. 14. Consider the system described by dyðt Þ þ yðt Þ ¼ xðt Þ yð0Þ ¼ 0 dt (i) Determine the step response of the system. (ii) Determine the impulse response from the step response. 15. Determine the response y(t) of the OP-Amp circuit shown in Figure P2.3 for an input x(t) ¼ u(t)

108

2 Continuous-Time Signals and Systems

Figure P2.3 OP-Amp circuit of problem 15

16. Determine the impulse response y(t) of the OP-Amp circuit shown in Figure P2.4.

Figure P2.4 OP-Amp circuit of problem 16

17. Determine the impulse response h(t) for a system described by d2 y dy dx þ 3 þ 2y ¼ dt dt dt 2 18. Draw block diagrams for direct form II implementation of the corresponding systems (i)

d2 y dt 2

dx þ 5dy dt þ 4y ¼ dt þ x

2.10

MATLAB Exercises

109

dx (ii) dy dt þ 3y ¼ 2 dt þ xðt Þ 2 2 (iii) d y/dt  ady/dt ¼ a dx/dt + abx(t)

19. For a given signal x(t)

(i) Show that xðt Þu1 ðt Þ ¼ xð0Þu1 ðt Þ  dxdtðtÞ

t¼0

δðt Þ:

(ii) Determine the value of ð1

xðτÞu2 ðτÞdτ:

1

(iii) Find an expression for x(t) u2(t) similar to (i). 20. Obtain the state-space representation for the electrical circuit shown In Figure P2.5 considering i1, V 1, and V 2 as state variables and current through the inductor as the output y(t) − v1 +

Figure P2.5 Electrical circuit of prblem 20

1Ω i1

i2

i3

1F

+

1F

1H 1Ω

2.10

+ v2 −

MATLAB Exercises

1. Verify the solution of problem 2 using MATLAB. 2. Write a MATLAB program to compute the convolution of the input x(t) and the impulse response h(t) shown in Figure M2.1.

1

1

0

5

0

Figure M2.1 Input and Impulse response of MaTLAB exercise 2

0.5

1.5

110

2 Continuous-Time Signals and Systems

3. If x(t) ¼ u(t  1)  u(t  2), write a MATLAB program to compute the result y10(t) convolving ten x(t) functions together, that is y10 ðt Þ ¼ xðt Þ∗xðt Þ∗xðt Þ∗xðt Þ∗xðt Þ∗xðt Þ∗xðt Þ∗xðt Þ∗xðt Þ∗xðt Þ and comment on the result. 4. Write a MATLAB program to find the impulse response h(t) and step response for a system described by d 2 yð t Þ dyðt Þ dxðt Þ þ 6yðt Þ ¼ þ xðt Þ: þ5 dt dt dt 2

Further Reading 1. Oppenheim, A.V., Willsky, A.S.: Signals and Systems. Prentice-Hall, Englewood Cliffs (1983) 2. Hsu, H.: Signals and Systems Schaum’s Outlines, 2nd edn. McGraw-Hill, New York (2011) 3. Kailath, T.: Linear Systems. Prentice-Hall, Englewood Cliffs (1980) 4. Zadeh, L., Desoer, C.: Linear System Theory. McGraw-Hill, New York (1963)

Chapter 3

Frequency Domain Analysis of ContinuousTime Signals and Systems

The continuous-time signals and systems are often characterized conveniently in a transform domain. This chapter describes the transformations known as Fourier series and Fourier transform which convert time-domain signals into frequencydomain (or spectral) representations. The frequency domain representation of continuous-time signals are described along with the conditions for the existence of Fourier series for periodic signals and Fourier transform for nonperiodic signals and their properties. Finally, the frequency response of continuous-time systems is discussed.

3.1

Complex Exponential Fourier Series Representation of the Continuous-Time Periodic Signals

It is recalled from Chapter 1 that a signal x(t) is periodic if xðt Þ ¼ xðt þ T Þ for all t

ð3:1Þ

The fundamental period T0 is small minimum, positive nonzero value of T for which Eq. (3.1) is satisfied, and Ω0 ¼ T2π0 is referred to as the fundamental angular frequency. A sinusoidal signal x(t) ¼ cos (ω0t) and the complex exponential signal xðt Þ ¼ ejΩ0 t are the two examples of periodic signals. The complex exponentials related harmonically are expressed by xn ðt Þ ¼ ejnΩ0 t ,

n ¼ 0,  1,  2,   

ð3:2Þ

The fundamental frequency of each of these signals is a multiple of Ω0, and hence each is periodic with period T0.

© Springer International Publishing AG, part of Springer Nature 2018 K. D. Rao, Signals and Systems, https://doi.org/10.1007/978-3-319-68675-2_3

111

112

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Thus, a linear combination of complex exponentials related harmonically can be written as xðt Þ ¼

X1 n¼1

ð3:3Þ

an ejnΩ0 t

Eq. (3.3) is the Fourier series representation of a periodic signal x(t). The Fourier coefficients ak can be determined as follows: Multiplying Eq. (3.3) both sides by ejmΩ0 t , we obtain xðt ÞejmΩ0 t ¼

X1 n¼1

an ejnΩ0 t ejmΩ0 t

ð3:4Þ

Integrating Eq. (3.4) both sides from 0 to T0, we have ðT0

xðt Þe

jmΩ0 t

¼

ðT0 X 1

0

n¼1

0

an ejnΩ0 t ejmΩ0 t dt

ð3:5Þ

Interchanging the integration and summation, Eq. (3.5) can be rewritten as ðT0

xðt ÞejmΩ0 t dt ¼

a n¼1 k

0

ðT0

ejðnmÞΩ0 t dt ¼

0

ðT0

ð T 0

X1

cos ððn  mÞΩ0 t Þ dt þ j

0

 ejðnmÞΩ0 t dt

ð3:6Þ

0

ðT0

sin ððn  mÞΩ0 t Þ dt

ð3:7Þ

0

Hence, ðT0

ejðnmÞΩ0 t dt ¼

0



T0 0

m¼n m 6¼ n

ð3:8Þ

Thus, Eq. (3.6) becomes ðT0

xðt ÞejmΩ0 t dt ¼ an T 0

ð3:9Þ

0

Therefore, the Fourier coefficients an are given by an ¼

1 T0

ðT0

xðt ÞejnΩ0 t dt

ð3:10Þ

0

Thus, the Fourier series of a periodic signal is defined by Eq. (3.3) referred to as the synthesis equation and Eq. (3.10) as the analysis equation.

3.1 Complex Exponential Fourier Series Representation of the Continuous. . .

3.1.1

113

Convergence of Fourier Series

The sufficient conditions for guaranteed convergence of Fourier series are the following Dirichlet conditions: 1. x(t) must be absolutely integral over any period, that is, ð j xðt Þ j dt < 1:

ð3:11Þ

T0

which guarantees that each Fourier coefficient ak has finite value. 2. x(t) must have finite number of maxima and minima during any single period of it. 3. x(t) must have finite number of discontinuities in any finite interval of time and each of these discontinuities being finite.

3.1.2

Properties of Fourier Series

Linearity Property If x1(t) and x2(t) are two continuous-time signals with Fourier series coefficients an and bn, then the Fourier series coefficients of a linear combination of x1(t) and x2(t), that is, c1x1(t) þ c2x2(t), are given by c 1 an þ c 2 bn where c1 and c2 are arbitrary constants. Time Shifting Property If x(t) is a continuous-time periodic signal with period T0 and the Fourier coefficients an, then the Fourier series coefficients of the x(t  t0) are given by e Proof by

jnT2π t 0 0

an .

Since Ω0 ¼ T2π0 , the Fourier coefficients an of a periodic signal x(t) are given

an ¼

1 T0

ðT0

xðt Þe

jnT2π t 0

dt

0

Let the Fourier series coefficients of x(t  t0) be b an 1 b an ¼ T0 Letting τ ¼ t  t0

ðT0 0

xðt  t 0 Þe

jnT2π t 0

dt

114

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

b an ¼

1 T0

b an ¼ e

ðT0

xðτÞe

0

2π jn T t 0

b an ¼ e

0

1 T0

2π jn T t 0 0

2π 2π jn T τ jn T t 0

e

0

ðT0

xðτÞe

0

2π jn T τ 0

dτ dτ

0

an

Conjugate Property If x(t) is a continuous-time periodic signal with the Fourier coefficients an, then the Fourier series coefficients of the x*(t) are given by a∗ n . ðT0 1 jn 2π t xðt Þe T 0 dt, then replacing n by n, we get Proof an ¼ T0 0 an ¼

1 T0

ðT0

xðt Þe

jnT2π t 0

dt

0

Taking both sides conjugate of this equation, we have a∗ n

1 ¼ T0

ðT0

x∗ ðtÞe

jnT2π t 0

dt

0

Symmetry for Real Valued Signal If x(t) is a continuous-time real valued signal with the Fourier coefficients an, then an ¼ a∗ n where * stands for the complex conjugate. Proof From conjugate property, we know that a∗ n

1 ¼ T0

ðT0

x∗ ðtÞe

jnT2π t 0

dt

0

Since xðt Þ is real x∗ ðt Þ ¼ xðt Þ, we get ð 2π 1 T0 jn t a∗ ¼ xðt Þe T 0 dt n T0 0 ¼ an implies that Re(an) ¼ Re(a–n), i.e., the real part of an is even, andIm(an) ¼ (Im (an)), i.e., the imaginary part of an is odd. If x(t) is a continuous-time real and even signal, i.e., x*(t) ¼ x(t) x(t) ¼ x(t), it can be easily shown that an¼ an and an ¼ a∗ n. Similarly, if x(t) is a continuous-time real and odd signal, i.e., x*(t) ¼ x(t) x(t) ¼ x(t), it can be easily shown that an ¼ an and an ¼  a∗ n .

3.1 Complex Exponential Fourier Series Representation of the Continuous. . .

115

Frequency Shifting Property If x(t) is a continuous-time periodic signal with period T0 and the Fourier coefficients an, then the Fourier series coefficients of the e

jK T2π t 0

xðt Þ are given by anK.

Proof The Fourier coefficients an of a periodic signal x(t) are given by 1 an ¼ T0

ðT0

1 T0

ðT0

jnT2π t 0

dt

0

Let dn be the Fourier coefficients of e dn ¼

xðt Þe

jK T2π t 0

xðt Þe

xðt Þ, then

2π 2π jK T t jn T t 0 e 0 dt

0

ð 2π 1 T0 jðnK Þ T t 0 dt ¼ xðt Þe T0 0 ¼ anK Thus, it is proved. Time Reversal Property If x(t) is a continuous-time periodic signal with the Fourier coefficients an, then the Fourier series coefficients of the x(t) are given by an. Time Scaling Property If x(t) is a continuous-time periodic signal with period T0 and the Fourier coefficients an, then the Fourier series coefficients of the x(αt) α > 0 are given by an with period Tα0 . Proof Since Ω0 ¼ T2π0 , the Fourier coefficients an of a periodic signal x(t) are given by an ¼

1 T0

ðT0

xðt Þe

jnT2π t 0

dt

0

Let the Fourier series coefficients of x(αt) be b a n , then α b an ¼ T0

ðT0

xðαt Þe

jn2παt T 0

dt

0

Letting τ ¼ αt 1 b an ¼ T0 ¼ an b 0 ¼ T0. Hence, T α

ðT0 0

xðτÞe

2π jn T τ 0

dτ

116

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Differentiation in Time If x(t) is a continuous-time periodic signal with period T0 and the Fourier coefficients an, then the Fourier series coefficients of the dtd xðt Þ are given by jn T2π0 an Proof The Fourier series representation of of x(t) is xð t Þ ¼

X1 n¼1

an ejnΩ0 t

Differentiating this equation both sides with respect to t, we obtain 1 X d xð t Þ ¼ jnΩ0 an ejnΩ0 t dt n¼1

Since Ω0 ¼ T2π0 1 X d 2π xð t Þ ¼ jn an ejnΩ0 t dt T0 n¼1

giving the Fourier series representation of dtd xðt Þ. Comparing this with the Fourier series representation of coefficients an, it is clear that the Fourier coefficients of d 2π dt xðt Þ are jn T 0 an Integration Property If x(t) is a continuous-time periodic signal withÐ period T0 and the Fourier coefficients an, then the Fourier series coefficients of the x(t) are given T0 by jn2π an Proof The Fourier series representation of x(t) is xðt Þ ¼

X1 n¼1

an ejnΩ0 t

Integrating this equation both sides with respect to t, we obtain ð

ð xðt Þdt ¼

1 X

! an e

jnΩ0 t

dt

n¼1

¼

1 X an jnΩ0 t e jnΩ0 n¼1

Since Ω0 ¼ T2π0 ð xðt Þdt ¼

1 X T0 an ejnΩ0 t jn2π n¼1

3.1 Complex Exponential Fourier Series Representation of the Continuous. . .

117

Ð giving the Fourier series representation of x(t)dt. Comparing this with the Fourier Ð series representation of coefficients an, it is clear that the Fourier coefficients of x(t) T0 dt are jn2π an Periodic Convolution If x1(t) and x2(t) are two continuous-time signals with common period T0 and Fourier coefficients an and bn, respectively, then the Fourier series coefficients of the convolution integral of x1(t) and x2(t) are given by T0anbn. Proof The periodic convolution integral of two signals with common period T0 is defined by ð yðt Þ ¼ x1 ðτÞx2 ðt  τÞdτ ð3:12Þ T0

Let cn be the Fourier series coefficients of y(t), then cn ¼

1 T0

ðT0

1 T0

yðt ÞejnΩ0 t dt ¼

0

ð T 0 ð 0

 x1 ðτÞx2 ðt  τÞdτ ejnΩ0 t dt

ð3:13Þ

T0

Letting t  τ ¼ t1, Eq. (3.13) becomes ð T 0 ð

1 cn ¼ T0

0

 x1 ðτÞx2 ðt 1 Þdτ ejnΩ0 ðτþt1 Þ dt 1 T0

Interchanging the order of integration, the above equation can be rewritten as 1 cn ¼ T0

ð x1 ðτÞe

jnΩ0 τ

 ð x2 ðt 1 Þe

dτ

T0

jnΩ0 t 1

 dt 1

ð3:14Þ

T0

By definition of Fourier series 1 T0 hÐ

ð

 x1 ðτÞejnΩ0 τ dτ ¼ an T0

i jnΩ0 t 1 x ð t Þe dt ¼ T 0 bn 2 1 1 T0

Hence cn¼ T0anbn.. Thus, it is proved. Multiplication Property If x1(t) and x2(t) are two continuous-time signals with Fourier coefficients an and bn, respectively, then the Fourier series coefficients of a new signal x1(t)x2(t) are given X1 by a b implying that signal multiplication in the time domain is equivl¼1 l nl alent to discrete-time convolution in the frequency domain. Proof Let dn be the Fourier coefficients of the new signal x1(t)x2 (t). By definition of Fourier series representation of a signal, we have

118

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

dn ¼ ¼ ¼ ¼

1 T0

ðT0

x1 ðτÞx2 ðt ÞejnΩ0 t dt

0

ðT0 X 1

1 T0

al ejlΩ0 t x2 ðt ÞejnΩ0 t dt

l¼1

0

P1

1 l¼1 al T0

P1

l¼1

ðT0

x2 ðt ÞejðnlÞΩ0 t dt

0

al bnl

Parseval’s Theorem If x(t) is a continuous-time signal with period T0 and Fourier coefficients an, then the average power P of x(t) is given by ð X1 1 P¼ ð3:15Þ ja j2 jxðt Þj2 dt ¼ n¼1 n T 0 T0 Proof The average power P of a periodic signal x(t) is defined as ð 1 P¼ jxðt Þj2 dt T0 T0

ð3:16Þ

Assuming that x(t) is complex valued x(t)x*(t) ¼ |x(t)|2 and x*(t) can be expressed in terms of its Fourier series as x∗ ð t Þ ¼

X1 n¼1

jnΩ0 t a∗ n e

ð3:17Þ

Eq. (3.16) can be rewritten as P¼

1 T0

ð

xðt Þx∗ ðt Þ dt

ð3:18Þ

T0

Substituting Eq. (3.17) in Eq. (3.18), we obtain ð hX1 i 1 ∗ jnΩ0 t xð t Þ a e P¼ dt n¼1 n T0 T0

ð3:19Þ

Interchanging the order of integration, Eq. (3.19) can be rewritten as P¼

X1

1 a∗ n¼1 n T 0

ðT0

xðt Þe

jnT2π t

0

By definition of the Fourier series an ¼ Thus,

1 T0

ðT0 0

xðt Þe

jnT2π t 0

dt

0

dt

ð3:20Þ

3.1 Complex Exponential Fourier Series Representation of the Continuous. . .

119

Table 3.1 Some properties of continuous-time Fourier series Property Linearity Time shifting

Periodic signal c1x1(t) þ c2x2(t) x(t  t0)

Time reversal Conjugate Symmetry

x(t) x*(t) x(t) real xe(t) (x(t) real) xo(t) (x(t) real)

Frequency shifting Time scaling

e

jK T2π t

Differentiation in time Integration Periodic convolution property Multiplication property

0

Fourier series coefficients c1an + c2bn jn 2π t

e T 0 0 an an a∗ 8n an ¼ a∗ > n > > > < Re½an  ¼ Re½an  Im½an  ¼ Im½an  > > j a j¼j an j > > : n arg½an  ¼ arg½an  Re[an] jIm[an] anK

xðt Þ

x(αt), α > 0 (periodic with period Tα0 )

an

d xðt Þ Ðdt x(t)dt Ð T 0 x1 ðτ Þx2 ðt  τ Þdτ

jn T2π0 an T0 jn2π an

T0anbn. 1 X al bnl ¼

x1(t)x2(t)

l¼1

P¼

1 T0

ð jxðt Þj2 dt ¼ T0

1 X

jan j2

n¼1

The properties of continuous-time Fourier series are summarized in Table 3.1. Parseval’ s Theorem

1 T0

ð1 1

jxðt Þj2 dt ¼

X1 n¼1

jan j2

Half-Wave Symmetry If the two halves of one period of a periodic signal are of identical shape, except that one is the negation of the other, the periodic signal is said to have a half-wave symmetry. Formally, if x(t) is a periodic signal with period T0, then x(t) has half  wave symmetry if x t  T20 ¼ xðt Þ Example 3.1 Prove that Fourier series representation of a periodic signal with halfwave symmetry has no even-numbered harmonics. Proof A periodic signal x(t) with half-wave symmetry is given by

120

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

( xð t Þ ¼

xð t Þ

0  t < T 0 =2

xðt Þ T 0 =2  t < T 0 ð 1 an ¼ xðt ÞejnΩ0 t dt T 0 T0 ð ð 1 T 0 =2 1 T0 ¼ xðt ÞejnΩ0 t dt þ xðt ÞejnΩ0 t dt T0 0 T 0 T 0 =2   T0 For T 0 =2  t < T 0 xðt Þ ¼ x t  2   Substituting xðt Þ ¼ x t  T20 in the above equation, we get an ¼

1 T0

ð T 0 =2

xðt ÞejnΩ0 t dt 

0

1 T0

  T 0 jnΩ0 t x t dt e 2 T 0 =2

ðT0

By using time shifting property, we obtain an ¼

1 T0

ð T 0 =2

T0

xðt ÞejnΩ0 t dt  ejnΩ0 2

0

1 T0

ð T 0 =2

xðt ÞejnΩ0 t dt

0

Since Ω0 ¼ T2π0 an ¼

1 T0

ð T 0 =2

xðt ÞejnΩ0 t dt  ejnπ

0

ð1  ejnπ Þ ¼ T0

ð T 0 =2

1 T0

ð T 0 =2

xðt ÞejnΩ0 t dt

0

xðt ÞejnΩ0 t dt

0

ð ð1  ð1Þn Þ T 0 =2 ¼ xðt ÞejnΩ0 t dt T0 0 8 for even n < 0 ¼ 2 Ð T 0 =2 : xðt ÞejnΩ0 t dt for odd n T0 0 Example 3.2 Find Fourier series of the following periodic signal with half-wave symmetry as shown in Figure 3.1. Figure 3.1 Periodic signal with half-wave symmetry

3.1 Complex Exponential Fourier Series Representation of the Continuous. . . π Solution The period T0 ¼ 6 and Ω0 ¼ 2π 6 ¼ 3 Fourier coefficients are given by 8 > < 0 ð T 0 =2 an ¼ 2 > xðt ÞejnΩ0 t dt : T0 0

121

for even n for odd n

For odd n ð 2 T 0 =2 xðt ÞejnΩ0 t dt T0 0 ð 2 3 xðt ÞejnΩ0 t dt ¼ 6 0 ð 1 2 jnπt e 3 dt ¼ 3 1

1 j2nπ=3 ¼ e  ejnπ=3 jnπ

an ¼

Example 3.3 Find Fourier series of the following periodic signal with half-wave symmetry (Figure 3.2) π Solution The period T0 ¼ 8 and Ω0 ¼ 2π 8 ¼ 4 Fourier coefficients are given by

8 < 0 an ¼ 2 : T0

for even n Ð T 0 =2 0

xðt ÞejnΩ0 t dt

for odd n

For odd n

x(t) 1

2

4

-1

Figure 3.2 Periodic signal with half-wave symmetry

6

8

t

122

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

an ¼ ¼

2 T0 2 8

ð T 0 =2

xðt ÞejnΩ0 t dt

0

ð4

xðt ÞejnΩ0 t dt

0

ð2

t jnπ t e 4 dt 02 ð 1 2 jnπ t te 4 dt ¼ 8 0 " #  ð2 π 1 t jnπ t 2 1 jn t e 4 þ ¼ e 4 dt 8 jnπ=4 jnπ=4 0 0 " 2 # 1 8j k 16 jnπt  j þ 2 2e 4  ¼ 8 nπ nπ 0 1 ¼ 4

¼

jðkþ1Þ 2 þ 2 2 jðkÞ  1 n π nπ

Example 3.4 Consider the periodic signal x(t) given by xðt Þ ¼ ð2 þ j2Þej3t  j3ej2t þ 6 þ j3ej2t þ ð2  j2Þej3t (i) Determine the fundamental period and frequency of x(t) (ii) Show that x(t) is a real signal (iii) Find energy of the signal Solution (i) x(t) is the sum of two periodic signals with periods T 1 ¼ 2π 3 and T 2 T1 2 ¼ 2π The ratio ¼ is a rational number. 2 3 T2 The fundamental period of the signal x(t) is 3T1 ¼ 2T2 ¼ 2π. The fundamental frequency Ω0 ¼ 2π 2π ¼ 1. (ii) x(t) is exponential Fourier series representation of the form xð t Þ ¼

1 X

an ejnΩ0 t

n¼1

where a3 ¼ 2 þ j2; a2 ¼ j3; a0 ¼ 6; a3 ¼ 2  j2; a2 ¼ j3 and an¼ 0 for all other. It is noticed that an ¼ a∗ n for all n. Hence, x(t) is a real signal.

3.1 Complex Exponential Fourier Series Representation of the Continuous. . .

123

(iii) The average power of the signal x(t) is E¼

1 T0

ð jxðt Þj2 dt ¼ T0

1 X

j an j 2

n¼1

By Parseval’s theorem, E¼

1 X

j an j 2

n¼1

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ¼ 22 þ 22 þ 32 þ 62 þ 32 þ 22 þ 22 ¼ 8 þ 9 þ 36 þ 9 þ 8 ¼ 70 Example 3.5 Find the Fourier series coefficients for each of the following signals: (i) x(t) ¼ cos(Ω0t) þ sin (2Ω0t) (ii) x(t) ¼ 2 cos(Ω0t) þ sin2 (2Ω0t) X1 Solution (i) xðt Þ ¼ a ejnΩ0 t n¼1 n 1 1 1 1 xðt Þ ¼ ejΩ0 t þ ejΩ0 t þ ej2Ω0 t  ej2Ω0 t 2 2 2j 2j 1 1 1 1 a1 ¼ ; a1 ¼ ; a2 ¼ ; a2 ¼  2 2 2j 2j an ¼ 0 for all other n. (ii) xðt Þ ¼ 2 cos ðΩ0 t Þ þ sin 2 ð2Ω0 t Þ ¼ 2 cos ðΩ0 t Þ þ 12 ½1  cos ð4Ω0 t Þ xð t Þ ¼

1 X

an ejnΩ0 t

n¼1



   1 jΩ0 t 1 jΩ0 t 1 1 1 j4Ω0 t 1 j4Ω0 t e xð t Þ ¼ 2 e þ e þ e þ  2 2 2 2 2 2   1 1 1 j4Ω0 t 1 j4Ω0 t e ¼ ejΩ0 t þ ejΩ0 t þ  þ e 2 2 2 2 1 1 1 a1 ¼ 1; a0 ¼ ; a1 ¼ 1; a4 ¼  ; a4 ¼  2 4 4 an ¼ 0 for all other n.

124

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Example 3.6 Find Fourier series coefficients of the following continuous-time periodic signal and plot the magnitude and phase spectrum of it: xðt Þ ¼ 2s inð2πt  3Þ þ s inð6πt Þ Solution xðt Þ ¼

2 jð2πt3Þ 2 jð2πt3Þ 1 jð6πtÞ 1 jð6πtÞ e  e þ e  e 2j 2j 2j 2j

1 1 1 1 ¼ ej3 ejð2πtÞ  ej3 ejð2πtÞ þ ejð6πtÞ  ejð6πtÞ j j 2j 2j 1 1 1 1 ¼  ejð6πtÞ  ej3 ejð2πtÞ þ ej3 ejð2πtÞ þ ejð6πtÞ 2j j j 2j Since Ω0¼ 2π 1 1 1 1 xðt Þ ¼  ejð3Ω0 tÞ  ej3 ejðΩ0 tÞ þ ej3 ejðΩ0 tÞ þ ejð3Ω0 tÞ 2j j j 2j 1 X xð t Þ ¼ an ejnΩ0 t n¼1

1 j 1 π a3 ¼  ¼ ¼ ejð2Þ 2j 2 2 ej3 a1 ¼  ¼ jej3 ¼ ej1:7124 j ej3 ¼ jej3 ¼ ej1:7124 a1 ¼ j π 1 j 1 a3 ¼ ¼  ¼ e j ð  2 Þ 2j 2 2 an¼ 0 for all other n. The magnitudes of Fourier coefficients are ja3 j ¼ ja3 j ¼ 0:5 ja1 j ¼ ja1 j ¼ 1:0 The magnitude spectrum and phase spectrum are shown in Figures 3.3 and 3.4, respectively. Since x(t) is a real valued, its magnitude spectrum is even and the phase spectrum is odd. Example 3.7 (i) Obtain x(t) for the following non-zero Fourier series coefficients of a continuoustime real valued periodic signal x(t) with fundamental period of 8. a1 ¼ a∗ 1 ¼ j, a5 ¼ a5 ¼ 1

3.1 Complex Exponential Fourier Series Representation of the Continuous. . .

125

Figure 3.3 Magnitude spectrum of x(t)

Figure 3.4 Phase spectrum of x(t) Figure 3.5 Magnitude spectrum of a signal

(ii) Consider a continuous periodic signal with the following magnitude spectra shown in Figure 3.5. Find the DC component and average power of the signal. Solution (i) xð t Þ ¼

1 X

an ejnΩ0 t

n¼1

xðt Þ ¼ ej5Ω0 t þ ej5Ω0 t þ jðejΩ0 t  ejΩ0 t Þ ¼ 2 cos ð5Ω0 t Þ  2 s inðΩ0 t Þ π

¼ 2 cos ð5Ω0 t Þ þ 2 cos Ω0 t þ 2

126

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

(ii) The DC component is given by a0 ¼ 1: By using Parseval’s relation, the average power is computed as X 1 ja j2 ¼ 12 þ 22 þ 12 þ 22 þ 12 ¼ 11 n¼1 n Example 3.8 Which of the following signals cannot be represented by the Fourier series? (i) x(t) ¼ 4 cos(t) þ 6 cos(t) (ii) x(t) ¼ 3 cos(πt) þ 6 cos(t) (iii) x(t) ¼ cos(t) þ 0.75 (iv) x(t) ¼ 2 cos(3πt) þ 3 cos(7πt) (v) x(t) ¼ e|t| sin (5πt) Solution (i) x(t) ¼ 4 cos(t) þ 6 cos(t) is periodic with period 2π. (ii) x(t) ¼ 3 cos(πt) þ 6 cos(t) The first term has period T1 ¼

2π ¼2 π

T2 ¼

2π ¼ 2π 1

The second term has period

The ratio TT 12 ¼ π2 is not a rational number. Hence, x(t) is not a periodic signal. (iii) x(t) ¼ cos(t) þ 0.75 is periodic with period 2π. (iv) x(t) ¼ 2 cos(3πt) þ 3 cos(7πt) The first term has period T1 ¼

2π 2 ¼ 3π 3

T2 ¼

2π 2 ¼ 7π 7

The second term has period

The ratio TT 12 ¼ 73 is a rational number. Hence, x(t) is a periodic signal. (v) Due to decaying exponential function, it is not periodic. So Fourier series cannot be defined for it. Hence, (ii) and (v) cannot be represented by Fourier series. Since the remaining three are periodic; they can be represented by Fourier series.

3.1 Complex Exponential Fourier Series Representation of the Continuous. . .

127

Example 3.9 Find the Fourier series of a periodic square wave with period T0 defined over one period by  xðt Þ ¼

1 0

j t j< T 0 =4 T 0 =4 <j t j T 0 =2

Solution For n¼0, a0 ¼

1 T0

ð T 0 =4 T 0 =4

dt ¼

2T 0 =4 1 ¼ T0 2

For n 6¼ o, an ¼ T10

T 0 =4  1  ejnΩ0 t dt ¼  ejnΩ0 t   jnΩ0 T 0 T 0 =4

ð T 0 =4



jnΩ0 T 0 =4

¼

2 e nΩ0 T 0

¼

2 sin ðnΩ0 T 0 =4Þ nΩ0 T 0

e 2j

jnΩ0 T 0 =4



T 0 =4

Since Ω0 ¼ T2π0 ,   2π T 0 =4 T0 an ¼ 0 nπ 1 2π BT 0 T 0 C C sin B @ 4 A sin

n 6¼ 0

¼

nπ

nπ sin 1 2 ¼ 2 nπ=2 Example 3.10 Find the Fourier series of the periodic signal shown in Figure 3.6. Solution The period T0 ¼ 2. Ω0 ¼ T2π0 ¼ π: For n¼0, 1 a0 ¼ 2 For n 6¼ o,

ð1 1

tdt ¼ 0

128

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure 3.6 Periodic signal

an ¼

1 2

ð1 "

tejnπt dt

1

# 1 ð  1 t 1 1 jnπt jnπt  e ¼ e dt  þ jnπ 2 jnπ 1 1 " #1 3 1 jnπt  1 t e 5 ¼ ejnπt   2 jnπ ðjnπ Þ2 1 1 " # 1  1 t ejnπt  þ 0 ¼ 2 jnπ 1

¼

1 ðe 2

jnπ

þe Þ jnπ jnπ

Since ejnπ + e jnπ ¼ 2(1)n an ¼

3.2

ð1Þnþ1 jnπ

Trigonometric Fourier Series Representation

The trigonometric Fourier series representation of a periodic signal x(t) is expressed by xð t Þ ¼

a0 Xþ1 þ ðan cos ðnΩ0 t Þ þ bn sin ðnΩ0 t ÞÞ n¼1 2

ð3:21Þ

The Fourier coefficients an and bn are given by an ¼ bn ¼

2 T0 2 T0

ðT0

xðt Þ cos ðnΩ0 t Þdt

ð3:22aÞ

xðt Þ sin ðnΩ0 t Þdt

ð3:22bÞ

0

ðT0 0

3.2 Trigonometric Fourier Series Representation

3.2.1

129

Symmetry Conditions in Trigonometric Fourier Series

If x(t) is an even periodic signal, then ð 2 T 0=2 a0 ¼ xðt Þdt T0 0 ð 4 T 0=2 an ¼ xðt Þ cos ðnΩ0 t Þdt T0 0

ð3:23aÞ ð3:23bÞ

bn¼ 0 for all n. If x(t) is an odd periodic signal, thena0¼ 0;an ¼ 0 for all n bn ¼

4 T0

ð T 0=2

xðt Þ sin ðnΩ0 t Þdt

ð3:24Þ

0

Therefore, for every even signal bn ¼ 0. Hence, Fourier series of an even signal contains DC term and cosine terms only. Fourier series of an odd signal contains sine terms only. Example 3.11 Find the trigonometric Fourier series representation of the periodic signal shown in Figure 3.7 with A ¼ 3 and period T0¼ 2π. Solution The period T0¼2π. Ω0 ¼ T2π0 ¼ 1. The periodic signal x(t) defined over one period is  xð t Þ ¼

3 3

π  t < 0 0t<π

Since x(t) has odd symmetry, a0 ¼ 0 and an ¼ 0 ð 4 0 3 sin ðnt Þ dt 2π π  6 cos ðnt Þ 0 ¼ π π n 6 1  cos ððnπ ÞÞ ¼ nπ ( 0 for n even ¼ 12 for n odd nπ

bn ¼

Figure 3.7 Periodic signal

130

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure 3.8 Periodic triangle wave

The trigonometric Fourier series representation of x(t) is given by 1 X

xð t Þ ¼

bn sin ðnt Þ ¼

n¼1

1 12 X sin ðð2n  1Þt Þ π n¼1 ð2n  1Þ

Example 3.12 Find the trigonometric Fourier series representation of the periodic triangle wave shown in Figure 3.8 with A ¼ 2 period T0 ¼ 2. Solution The period T0¼2. Ω0 ¼ T2π0 ¼ π. The periodic triangle wave x(t) defined over one period with A¼2 is  xð t Þ ¼

4t j t j< 1=2 4 ð1  t Þ 1=2 < t < 3=2

Since x(t) has odd symmetry, a0 ¼ 0 and an ¼ 0 bn ¼

4 2

ð 1=2 1=2

"

4t sin ðnπt Þ dt

1=2 1 ¼8 cos ðnπt Þ1=2 þ nπ h i  1=2 ¼ 8 0 þ n21π 2 sin ðnπt Þ1=2  nπ  2 ¼ 8 0 þ 2 2 sin nπ 2

16 nπ ¼ 2 2 sin n π 2 t nπ

ð 1=2 1=2

# cos ðnπt Þ dt

The trigonometric Fourier series representation of x(t) is xð t Þ ¼

  16 1 1 1 sin ð 3πt Þ þ sin ð 5πt Þ  sin ð 7πt Þ þ    sin ð πt Þ  π2 9 25 49

Example 3.13 Find the trigonometric Fourier series representation of the following periodic signal with period T0 ¼ 2.

3.2 Trigonometric Fourier Series Representation

xð t Þ ¼

8 > > > <0 cos ð3πt Þ > > > : 0

131

1 1  t   2 1 1  t< 2 2 1=2  t < 1

Solution The period T0¼2. Ω0 ¼ T2π0 ¼ π. 2 Since x(t) has even symmetry, bn ¼ 0 a0 ¼ 2 ð1 þ22 0 dt

ð 1=2 1

0 dt þ

1=2

¼

1=2 sin ð3πt Þ 3π 1 2

¼

ð 1=2

2 an ¼ 2

1=2

2 3π

cos ð3πt Þ cos ðnπt Þ dt

For n ¼ 1, a1 ¼

ð 1=2 1=2

cos ð3πt Þ cos ðπt Þ dt ¼ 0

For n ¼ 2, a2 ¼

ð 1=2 1=2

cos ð3πt Þ cos ð2πt Þ dt ¼

6 5π

cos ð3πt Þ cos ð3πt Þ dt ¼

1 2

For n ¼ 3 a3 ¼

ð 1=2 1=2

For n ¼ 4,5,6,. . . . an ¼ ¼

ð 1=2

cos ð3πt Þ cos ðnπt Þ dt nπ

1=2

6 cos

2 n2 π  9π

The trigonometric Fourier series representation of x(t) is

2 2

ð 1=2 1=2

cos ð3πt Þ dt

132

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

xð t Þ ¼ ¼

þ1 a0 X þ an cos ðnπt Þ 2 n¼1 1 6 cos X

nπ

1 6 1 2 cos ðnπt Þ þ cos ð2πt Þ þ cos ð3πt Þ 2 π  9π n 3π 5π 2 n¼4

Example 3.14 If the input to the half-wave rectifier is an AC signal x(t) ¼ cos(2πt), find trigonometric Fourier series representation of output signal of the half-wave rectifier. Solution The output y(t) of half-wave rectifier is  yð t Þ ¼

xð t Þ 0

for xðt Þ  0 for xðt Þ < 0

The sinusoidal input and output of half-wave rectifier are shown in Figure 3.9. Since y(t) is a real and even function, its Fourier coefficients are real and even. The period T0¼1. Ω0 ¼ T2π0 ¼ 2π

Input x(t)=cos(2p t) 1 0.5 0 −0.5 −1 −1.5 −1.25 −1 −0.75 −0.5 −0.25

0

0.25

0.5 0.75

1

1.25

1.5

0.5 0.75

1

1.25

1.5

Output y (t ) 1 0.5 0 −0.5 −1 −1.5 −1.25 −1 −0.75 −0.5 −0.25

0 0.25 t (sec)

Figure 3.9 Input and output of half-wave rectifier

3.3 The Continuous Fourier Transform for Nonperiodic Signals

133

For n ¼ 0, a0 ¼

1 1

ð 1=4 1=4

cos ð2πt Þdt ¼

1 π

Hence, the DC component is π1 For n6¼0 an ¼

2 1

ð 1=4 1=4

cos ð2πt Þ cos ðnΩ0 t Þdt

"ð # ð 1=4 2 1=4 ¼ cos ð2π ðn þ 1Þt Þdt þ cos ð2π ðn  1Þt Þdt 1 1=4 1=4 " #   sin ð2π ðn þ 1Þt Þ1=4 sin ð2π ðn  1Þt Þ1=4 ¼ þ 2π ðn þ 1Þ 1=4 2π ðn þ 1Þ 1=4 π

π

3 2 2 sin ðn þ 1Þ 2 sin ðn þ 1Þt 2 2 5 þ ¼4 2π ðn þ 1Þ 2π ðn þ 1Þ π 

π  3 π 

2 n n n cos cos 2 cos 1 2 2 2 5¼  ¼ 4 nþ1 n1 π ð 1  n2 Þ π The trigonometric Fourier series representation of y(t) is þ1 a0 X þ an cos ðnπt Þ 2 n¼1 π 

1 2 cos n X   1 2 þ cos nπt ¼ 2π n¼4 π ð1  n2 Þ

xð t Þ ¼

3.3

The Continuous Fourier Transform for Nonperiodic Signals

Consider a nonperiodic signal x(t) as shown in Figure 3.10 (a) with finite duration, i.e., x(t) ¼ 0 for |t| >T1. From this nonperiodic signal, a periodic signal ~x ðt Þ can be constructed as shown in Figure 3.10(b). The Fourier series representation of ~x ðt Þ is ~x ðt Þ ¼

X1 n¼1

an ejnΩ0 t

ð3:25aÞ

134

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

x(t)

0

− 1

t

1

(a) ~ (t)

−2

0

−

0

− 1

0

2

0

1

0

t

(b) Figure 3.10 (a) Nonperiodic signal (b) periodic signal obtained from (a)

1 an ¼ T0

ð T 0 =2 T 0 =2

~x ðt ÞejnΩ0 t dt

ð3:25bÞ

Since ~x ðt Þ ¼ xðt Þ for j t j< T20 and also since x(t) ¼ 0 outside this interval, so we have 1 an ¼ T0

ð T 0 =2 T 0 =2

xðt Þe

jnΩ0 t

1 dt ¼ T0

ð1 1

xðt ÞejnΩ0 t dt

ð3:26Þ

Define X ðjΩÞ ¼

ð1 1

xðt ÞejΩt dt

ð3:27Þ

Then an ¼

1 X ðjnΩ0 Þ T0

ð3:28Þ

and ~x ðt Þ can be expressed in terms of X( jΩ), that is, ~x ðt Þ ¼ ¼

P1 n¼1

1 X ðjnΩ0 ÞejnΩ0 t T0

1 X1 X ðjnΩ0 ÞejnΩ0 t Ω0 n¼1 2π

ð3:29Þ

3.3 The Continuous Fourier Transform for Nonperiodic Signals

135

As T0 tends to infinity, ~x ðt Þ ¼ xðt Þ and summation becomes integration, Eq. (3.29) becomes 1 xð t Þ ¼ 2π

ð1 1

X ðjΩÞejΩt dΩ

ð3:30Þ

Eq. (3.27) is referred to as the Fourier transform of x(t), and Eq. (3.30) is called the inverse Fourier transform.

3.3.1

Convergence of Fourier Transforms

The sufficient conditions referred to as the Dirichlet conditions for the convergence of Fourier transform are: 1. x(t) must be absolutely integrable, that is, ð1 1

j xðt Þ j dt < 1

ð3:31Þ

2. x(t) must have a finite number of maxima and minima within any finite interval. 3. x(t) must have a finite number of discontinuities within any finite interval, and each of these discontinuities is finite. Although the above Dirichlet conditions guarantee the existence of the Fourier transform for a signal, if impulse functions are permitted in the transform, signals which do not satisfy these conditions can have Fourier transforms. Example 3.15 Determine x(0) and X(0) using the definitions of the Fourier transform and the inverse Fourier transform Solution By the definition of Fourier transform, we have X ðjΩÞ ¼ F½xðt Þ ¼

ð1 1

xðt ÞejΩt dt

Substituting Ω ¼ 0 in this equation, we obtain X ð 0Þ ¼

ð1 1

xðt Þ dt

By the definition of the inverse Fourier transform, we have 1 xð t Þ ¼ 2π

ð1 1

X ðjΩÞ ejΩt dΩ

136

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Substituting t¼0, it follows that x ð 0Þ ¼

3.3.2

ð1

1 2π

1

X ðjΩÞ dΩ

Fourier Transforms of Some Commonly Used Continuous-Time Signals

The unit impulse The Fourier transform of the unit impulse function is given by F½δðt Þ ¼

ð1 1

δðt Þ ejΩt dt ¼ 1

ð3:32Þ

implying that the Fourier transform of the unit impulse contribute equally at all frequencies Example 3.16 Find the Fourier transform of x(t) ¼ ebtu(t) for b > 0. Solution F½xðt Þ ¼ ¼

ð1

ð1 1 0

1 ¼ b þ jΩ

uðt Þe

jΩt

ð1

ebt ejΩt dt  1 ðjΩþbÞt 1 ðjΩþbÞt e e dt ¼  jΩ þ b e

bt

dt ¼

0

0

b>0

Example 3.17 Find the Fourier transform of x(t) ¼ 1. Solution By definition of the inverse Fourier transform and sampling property of the impulse function, we have F1 ½δðΩÞ ¼

1 2π

ð1 1

δðΩÞejΩt dΩ ¼

1 2π

1 ¼ δðΩÞ and thus F½1 ¼ 2πδðΩÞ Hence, F 2π Example 3.18 Find the Fourier transforms of the following (i) sin (Ω0t) (ii) cos (Ω0t) Solution (i)

sin ðΩ0 tÞ ¼ e

jΩ0 t

0ejΩ0 t 2j

By the sampling property of the impulse function, we have

3.3 The Continuous Fourier Transform for Nonperiodic Signals

137

ð 1 1 1 F ½δðΩ  Ω0 Þ ¼ δðΩ  Ω0 ÞejΩt dΩ ¼ ejΩ0 t 2π 1 2π

1 jΩ t  Hence, F 2π e 0 ¼ δ ð Ω  Ω0 Þ

jΩ t 

 0 ̀ ¼ 2πδðΩ  Ω0 Þ, F ejΩ0 t ̀ ¼ 2πδðΩ þ Ω0 Þ Thus F e Therefore, F½ sin ðΩ0 tÞ ¼ πj ½δðΩ  Ω0 Þ  δðΩ þ Ω0 Þ 1

(ii) cos ðΩ0 tÞ ¼ e

jΩ0 t

þejΩ0 t 2

 jΩ0 t  e þ ejΩ0 t F½ cos ðΩ0 tÞ ¼ F ¼ π½δðΩ  Ω0 Þ þ δðΩ þ Ω0 Þ 2 Example 3.19 Find the Fourier transform of the rectangular pulse signal shown in Figure 3.11 

Solution xð t Þ ¼

1 0

XðjΩÞ ¼ F½xðtÞ ¼ ¼

jt j  T 1 jt j > T 1 ð1 ð1 T1

xðtÞejΩt dt ejΩt dt

T 1

1 jΩt T 1 e jT 1 jΩ ejΩT1  ejΩT1 ¼2 2jΩ sin ðΩT 1 Þ ¼2 Ω ¼

Since sinc ðt Þ ¼ function as

sin ðπt Þ , πt

2

sin ðΩT 1 Þ can be written in terms of the sinc Ω

  sin ðΩT 1 Þ ΩT 1 2 ¼ 2T 1 sin c Ω π

x(t)

Figure 3.11 Rectangular pulse signal

1

− 1

0

1

t

138

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure 3.12 Fourier transform of a signal

(

Ω)

1

Ω

Ω

-Ω Hence,  X ðjΩÞ ¼ F½xðt Þ ¼ 2T 1 sinc

ΩT 1 π



Example 3.20 Consider the Fourier transform X( jΩ) of a signal shown in Figure 3.12. Find the inverse Fourier transform of it. 

Solution X ðjΩÞ ¼

1 0

jΩj  Ω jΩj > Ω

By the inverse Fourier transform definition, we have xðt Þ ¼

1 2π

¼

1 2π

ð1 1

ðΩ

X ðjΩÞejΩt dΩ ejΩt dΩ

Ω

  1 1 jΩt  Ω e Ω 2π jΩ   1 ejΩt þ ejΩt ¼ 2π jt

¼

  sin ðΩt Þ Ω Ωt ¼ sinc ¼ πt π π Example 3.21 Determine the Fourier transform of Gaussian signal xðt Þ ¼ et2σ2 2

Solution XðjΩÞ ¼ F½xðt Þ ¼ Letting b ¼

1 2σ 2

ð1 1

t 2 jΩt e dt e2σ2

3.3 The Continuous Fourier Transform for Nonperiodic Signals

XðjΩÞ ¼ ¼ ¼

Ð1

ebt ejΩt dt 2

1

Ð1

139

bðt 2 þðjΩ=bÞÞ dt 1 e Ð 1 cðtþðjΩ=2bÞÞ2 Ω2 =4b dt 1 e Ð 2 2 2 1 eΩ =4b 1 ebðtðjΩ=2bÞÞ Ω =4b dt

¼ pffiffiffi pffiffiffiffiffi Letting τ ¼ t b, dτ ¼ b dt XðjΩÞ ¼ eΩ

2

Ð =4b 1

1 e

b tðjΩ=2bÞ

2 Ω2 =4b



2 ð pffiffi eΩ =4b 1  τðjΩ=2 bÞ e dτ ¼ pffiffiffi b 1 pffiffi 2 pffiffiffi eðτðjΩ=2 bÞÞ dτ ¼ π

dt

2

ð1 Since 1

eΩ =4b pffiffiffi π XðjΩÞ ¼ pffiffiffi b 2

Substituting b ¼ 2σ1 2 eΩ =4b pffiffiffi π XðjΩÞ ¼ pffiffiffi pffiffiffiffiffi σb2 Ω2 =2 ¼ σ 2π e 2

3.3.3

Properties of the Continuous-Time Fourier Transform

Linearity If x1(t) and x2(t) are two continuous-time signals with Fourier transforms X1( jΩ) and X2( jΩ), then the Fourier transform of a linear combination of x1(t) and x2(t) is given by F½a1 x1 ðt Þ þ a2 x2 ðt Þ ¼ a1 X 1 ðjΩÞ þ a2 X 2 ðjΩÞ

ð3:33Þ

where a1 and a2 are arbitrary constants. Example 3.22 Find the Fourier transform of an impulse train with period T as given by xð t Þ ¼

1 X k¼1

δðt  kT Þ

140

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Solution A periodic signal x(t) with period T is expressed by xð t Þ ¼

P1 k¼1

ak ejkΩ0 t

Ω0 ¼

2π T

Taking Fourier transform both sides, we obtain "

1 X

F½xðt Þ ¼ F

# ak e

jkΩ0 t

k¼1

 Using F ejΩ0 t ̀ ¼ 2πδðΩ  Ω0 Þand the linearity property, we have " F

1 X

# ak e

jkΩ0 t

¼ 2π

k¼1

1 X

ak δðΩ  kΩ0 Þ

k¼1

If x(t) is an impulse train with period T as given by xð t Þ ¼

1 X

δðt  kT Þ

k¼1

Since

X1

1 X1 ejkΩ0 t k¼1 T " # 1 1 X 2π X F δðt  kT Þ ¼ δðΩ  kΩ0 Þ T k¼1 k¼1

δðt  TÞ ¼ k¼1

Symmetry for Real Valued Signal If x(t) is a continuous-time real valued signal with Fourier transform X( jΩ), then X ðjΩÞ ¼ X ∗ ðjΩÞ where * stands for the complex conjugate.

Ð 1 ∗ X ∗ ðjΩÞ ¼ Ð 1 xðt ÞejΩt dt 1 ¼ 1 x∗ ðt ÞejΩt dt

Proof

Since x(t) is real x*(t) ¼ x(t), we get ∗

X ðjΩÞ ¼

ð1 1

xðt ÞejΩt dt ¼ X ðjΩÞ

The X( jΩ) can be expressed in rectangular form as

ð3:34Þ

3.3 The Continuous Fourier Transform for Nonperiodic Signals

141

X ðjΩÞ ¼ Re½X ðjΩÞ þ jIm X ðjΩÞ If x(t) is real, then Re ½X ðjΩÞ ¼ Re½X ðjΩÞ Im ½X ðjΩÞ ¼ Im ½X ðjΩÞ implying that the real part is an even function of Ω and the imaginary part is an odd function of Ω. For real x(t) in polar form jX ðjΩÞj ¼ jX ðjΩÞj arg½X ðjΩÞ ¼ arg½X ðjΩÞ indicating that the magnitude is an even function of Ω and the phase is an odd function of Ω. Symmetry for Imaginary Valued Signal If x(t) is a continuous-time imaginary valued signal with Fourier transform X( jΩ), then X ∗ ðjΩÞ ¼ X ðjΩÞ Proof

X ∗ ðjΩÞ ¼ ¼

ð3:35Þ

Ð 1

∗ jΩt dt 1 xðt Þe Ð1 ∗ jΩt 1 x ðt Þe dt

Since x(t) is purely imaginary, we get x(t) ¼ x*(t), and we get X ∗ ðjΩÞ ¼ 

ð1 1

xðt ÞejΩt dt ¼ X ðjΩÞ

Re ½X ðjΩÞ ¼ Re ½X ðjΩÞ Im ½X ðjΩÞ ¼ Im ½X ðjΩÞ Symmetry for Even and Odd Signals (i) If x(t) is a continuous-time real valued and has even symmetry, then X ∗ ðjΩÞ ¼ X ðjΩÞ

ð3:36aÞ

(ii) If x(t) is a continuous-time real valued and has odd symmetry, then X ∗ ðjΩÞ ¼ X ðjΩÞ

ð3:36bÞ

Proof Since x(t) is real x(t) ¼ x*(t) and x(t) has even symmetry x(t) ¼ x(t), we get

142

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

∗

X ðjΩÞ ¼ ¼ ¼

ð 1 1 ð1 1 ð1

xðt Þe

∗ dt

x∗ ðt Þ ejΩt dt x ðt Þ ejΩt dt

1 ð1

¼

jΩt

1

xðt Þ ejΩðtÞ dt

Letting τ ¼t, we obtain Ð1 X ∗ ðjΩÞ ¼ 1 xðτÞejΩτ dτ ¼ X ðjΩÞ The condition X*( jΩ) ¼ X( jΩ) holds for the imaginary part of X( jΩ) to be zero. Therefore, if x(t) is real valued and has even symmetry, then X( jΩ) is real. Similarly, for real valued x(t) having odd symmetry, it can be shown that X* ( jΩ)¼X( jΩ) and X( jΩ) is imaginary. Time Shifting If x(t) is a continuous-time signal with Fourier transform X( jΩ), then the Fourier transform of x(t-t0) the delayed version of x(t) is given by F½xðt  t 0 Þ ¼ ejΩt0 X ðjΩÞ Proof F½xðt  t 0 Þ ¼

ð1 1

ð3:37Þ

xðt  t 0 ÞejΩt dt

Letting τ ¼ tt0, we obtain F½xðt  t 0 Þ ¼ ¼

Ð1

jΩðτt 0 Þ dτ 1 xðτÞe Ð 1 jΩt 0 jΩτ e dτ 1 xðτÞe jΩt 0

¼e

X ðjΩÞ

Therefore, time shifting results in unchanged magnitude spectrum but introduces a phase shift in its transform, which is a linear function of Ω. Example 3.23 Find the Fourier transform of δ(tt0) Solution

δðjΩÞ ¼ F½δðt Þ ¼ 1

Hence, F½δðt  t 0 Þ ¼ ejΩt0 δðjΩÞ ¼ ejΩt0

3.3 The Continuous Fourier Transform for Nonperiodic Signals

143

Frequency Shifting If x(t) is a continuous-time signal with Fourier transform X ( jΩ), then the Fourier transform of the signal ejΩ0 t xðt Þ is given by

 F ejΩ0 t xðt Þ ¼ X ðjðΩ  Ω0 ÞÞ Proof F½ejΩ0 t xðt Þ ¼ ¼

ð1 1 ð1 1

ð3:38Þ

ejΩ0 t xðt ÞejΩt dt xðt ÞejðΩΩ0 Þt dt

¼ X ð j ð Ω  Ω0 Þ Þ Thus, multiplying a sequence x(t) by a complex exponential ejΩ0 t in the time domain corresponds to a shift in the frequency domain. Time and Frequency Scaling If x(t) is a continuous-time signal with Fourier transform X( jΩ), then the Fourier transform of the signal x(at) is given by F½xðat Þ ¼

   1 Ω X j a j aj

ð3:39Þ

where a is a real constant. Proof F½xðat Þ ¼

ð1 1

xðat Þ ejΩt dt

Letting τ ¼ at, we obtain ð  Ω 1 1 F½xðatÞ ¼ xðτÞej a τ dτ a 1   ð Ω 1 1 ¼ xðτÞej a τ dτ a 1

for a > 0 f or a < 0

Thus,    1 Ω F½xðat Þ ¼ X j a j aj Differentiation in Time If x(t) is a continuous-time signal with Fourier transform X ( jΩ), then the Fourier transform of the dtd xðt Þ is given by 

 d F xðt Þ ¼ jΩX ðjΩÞ dt

ð3:40Þ

144

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Proof By the definition of the inverse Fourier transform, it is known that xð t Þ ¼

1 2π

ð1 1

X ðjΩÞ ejΩt dΩ

Differentiating this equation both sides with respect to t, we obtain ð d 1 1 xð t Þ ¼ X ðjΩÞjΩejΩt dΩ dt 2π 1 d xðt Þ ¼ jΩxðt Þ dt Taking the Fourier transform of this equation both sides, we get  F

 d xðt Þ ¼ jΩX ðjΩÞ dt

Thus, differentiation in the time domain corresponds to multiplication by jΩ in the frequency domain. By repeated application of this property, we obtain  F

 dn x ð t Þ ¼ ðjΩÞn X ðjΩÞ dt n

Example 3.24 Determine the Fourier transform of x(t) ¼ u(t) Solution Decomposing the unit step function into even and odd components, it is written as uðt Þ ¼ xe ðt Þ þ xo ðt Þ where the even component xe ðt Þ ¼ 12 and the odd component xo ðt Þ ¼ uðt Þ  12 Hence, F½uðtÞ ¼ F½xe ðt Þ þ F½xe ðt Þ ¼ X e ðjΩÞ þ X o ðjΩÞ 1 1 X e ðjΩÞ ¼ F½xe ðt Þ ¼ F½1 ¼ 2πδðΩÞ ¼ πδðΩÞ 2 2 d d xo ðt Þ ¼ uðt Þ ¼ δðt Þ dt dt

d  Thus, F dtxo ðt Þ ¼ jΩX o ðjΩÞ jΩX o ðjΩÞ ¼ F½δðt Þ ¼ 1 1 X o ðjΩÞ ¼ jΩ Therefore, U ðjΩÞ ¼ F½uðtÞ ¼ X e ðjΩÞ þ X o ðjΩÞ 1 ¼ πδðΩÞ þ jΩ

3.3 The Continuous Fourier Transform for Nonperiodic Signals

145

Example 3.25 Determine the Fourier transform of (i) x(t) ¼ sin(Ω0t)u(t) (ii) x(t) ¼ cos(Ω0t)u(t) Solution (i) 

 ejΩ0 t  ejΩ0 t uðtÞ 2j  1  1 F½ sin ðΩ0 tÞuðtÞ ¼ F ejΩ0 t uðtÞ  F ejΩ0 t uðtÞ 2j 2j sin ðΩ0 tÞuðtÞ ¼

1 since F½uðtÞ ¼ πδðΩÞ þ jΩ By frequency shifting property, we get

  π 1 1 1  F½ sin ðΩ0 tÞuðtÞ ¼ ½δðΩ  Ω0 Þ  δðΩ þ Ω0 Þ þ 2j 2j jðΩ  Ω0 Þ jðΩ þ Ω0 Þ π Ω0  ¼ ½ δ ð Ω  Ω0 Þ  δ ð Ω þ Ω0 Þ  þ  2 2j Ω0  Ω 2 (ii) 

 ejΩ0 t þ ejΩ0 t uð t Þ 2  1  1 F½ cos ðΩ0 tÞuðtÞ ¼ F ejΩ0 t uðtÞ þ F ejΩ0 t uðtÞ 2 2 cos ðΩ0 tÞuðtÞ ¼

1 Since F½uðtÞ ¼ πδðΩÞ þ jΩ By frequency shifting property, we get

  π 1 1 1 þ F½ cos ðΩ0 tÞuðtÞ ¼ ½δðΩ  Ω0 Þ þ δðΩ þ Ω0 Þ þ 2 2 jðΩ  Ω0 Þ jðΩ þ Ω0 Þ π jΩ  ¼ ½δðΩ  Ω0 Þ þ δðΩ þ Ω0 Þ þ  2 2 Ω0  Ω2 Example 3.26 Determine the Fourier transform of (i) xðt Þ ¼ ebt sin ðΩ0 tÞuðt Þ b > 0 (ii) xðt Þ ¼ ebt cos ðΩ0 tÞuðt Þ b > 0 Solution (i)

  jΩ0 t  

 e  ejΩ0 t F ebt sin ðΩ0 tÞuðt Þ ¼ F ebt uð t Þ 2j     ejΩ0 t ejΩ0 t uðt Þ  F ebt uð t Þ ¼ F ebt 2j 2j

Let x1(t) ¼ ebtu(t)

146

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Ðt

Ðt

0

0

ebt ejΩt dt ¼ 1 b > 0: ¼ b þ jΩ

F½x1 ðtÞ ¼

eðbþjΩÞt dt

By frequency shifting property

 F ejΩ0 t x1 ðt Þ ¼ X 1 ðjðΩ  Ω0 ÞÞ     ejΩ0 t 1 1 F ebt uð t Þ ¼ 2j b þ jðΩ  Ω0 Þ 2j Similarly,     ejΩ0 t 1 1 uð t Þ ¼ F ebt 2j b þ jðΩ þ Ω0 Þ 2j Hence,    

bt  1 1 1 1 F e sin ðΩ0 tÞuðt Þ ¼  2j b þ jðΩ  Ω0 Þ 2j b þ jðΩ þ Ω0 Þ ¼

Ω0 ðb þ jΩÞ2 þ Ω0 2

b>0

(ii)

 F ebt cos ðΩ0 tÞuðt Þ ¼ ¼ ¼ ¼

  jΩ0 t jΩ0 t   e þe F ebt uð t Þ 2     ejΩ0 t ejΩ0 t uðt Þ þ F ebt uð t Þ F ebt 2 2     1 1 1 1 þ 2 b þ j ð Ω  Ω0 Þ 2 b þ jðΩ þ Ω0 Þ b þ jΩ ðb þ jΩÞ2 þ Ω0 2

b>0

Differentiation in Frequency If x(t) is a continuous-time signal with Fourier transform X( jΩ), then the Fourier transform of the -jtx(t) is given by F½jtxðt Þ ¼

d X ðjΩÞ dΩ

Proof By the definition of the Fourier transform, it is known that

ð3:41Þ

3.3 The Continuous Fourier Transform for Nonperiodic Signals

X ðjΩÞ ¼

ð1

147

xðt ÞejΩt dt

1

Differentiating this equation with respect to Ω, we have d X ðjΩÞ ¼ dΩ

ð1 1

jtxðt ÞejΩt dt

implying that F½jtxðt Þ ¼ Thus, differentiation in the frequency domain corresponds to multiplication by jt in the time domain. d F½jtxðt Þ ¼ dΩ X ðjΩÞ can also be expressed as d dΩ X ðjΩÞ

d F½txðt Þ ¼ j X ðjΩÞ dΩ Example 3.27 Find the Fourier transform of the following continuous-time signal: xð t Þ ¼

t n1 bt e uð t Þ ðn  1Þ!

Solution For n ¼ 1, xðt Þ ¼ ebt uðt Þ,

b>0

X ðjΩÞ ¼

1 b þ jΩ

For n ¼ 2, x(t) ¼ tebtu(t) By differentiation in frequency property,  

 d 1 X ðjΩÞ ¼ F tebt uðt Þ ¼ j dΩ ðb þ jΩÞ d ¼ j ðb þ jΩÞ1 dΩ j ¼ ð1Þðb þ jΩÞ2 1 1 ¼ ðb þ jΩÞ2 t bt For n ¼ 3, xðt Þ ¼ 2! e uð t Þ 2

148

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

" #  t 2 bt j d 1 X ðjΩÞ ¼ F e uðt Þ ¼ 2dΩ ðb þ jΩÞ2 2! 

j d ðb þ jΩÞ2 2dΩ j ¼ ð2Þðb þ jΩÞ3 j 2 1 ¼ ðb þ jΩÞ3 ¼

t bt For n ¼ 4, xðt Þ ¼ 3! e uð t Þ 3

" #  t 3 bt j d 1 X ðjΩÞ ¼ F e uðt Þ ¼ 3dΩ ðb þ jΩÞ3 3! 

j d ðb þ jΩÞ3 3dΩ j ¼ ð3Þðb þ jΩÞ4 j 3 1 ¼ ðb þ jΩÞ4 ¼

Thus t n1 bt uð t Þ for n, xðt Þ ¼ ðn1 Þ! e " #  n1  t j d 1 bt e uð t Þ ¼ X ðjΩÞ ¼ F ðn  1ÞdΩ ðb þ jΩÞn1 ðn  1Þ! j d ðb þ jΩÞnþ1 ðn  1ÞdΩ j   ðn  1Þðb þ jΩÞnþ11 j ¼ ð n  1Þ 1 ¼ ðb þ jΩÞn

¼

Integration If x(t) is a continuous-time signal with Fourier transform X( jΩ), then ðt xðτÞ dτ is given by the Fourier transform of the 1

F

ð t

 1 xðτÞ dτ ¼ X ðjΩÞ jΩ 1

ð3:42Þ

3.3 The Continuous Fourier Transform for Nonperiodic Signals

Proof Letting yðt Þ ¼

ðt 1

149

xðτÞ dτ and differentiating both sides, we obtain d yð t Þ ¼ xð t Þ dt

Now taking the Fourier transform of both sides, it yields   d F yðt Þ ¼ F½xðt Þ ¼ X ðjΩÞ dt jΩY ðjΩÞ ¼ X ðjΩÞ Hence

Ð t  F½yðt Þ ¼ F 1 xðτÞ dτ 1 ¼ X ðjΩÞ jΩ Parseval’s theorem If x(t) is a continuous-time signal with Fourier transform X ( jΩ), then the energy E of x(t) is given by ð1

1 E¼ jxðt Þj dt ¼ 2π 1 2

ð1 1

jX ðjΩÞj2 dΩ

ð3:43Þ

where |X( jΩ)|2 is called the energy density spectrum. Proof The energy E of x(t) is defined as E¼

ð1 1

jxðt Þj2 dt

ð3:44Þ

Assuming that x(t) is complex value x(t)x∗(t) ¼ |x(t)|2 and x*(t) can be expressed in terms of its Fourier transform as 1 x ðt Þ ¼ 2π ∗

ð1 1

X ∗ ðjΩÞejΩt d Ω

ð3:45Þ

Eq. (3.44) can be rewritten as E¼

ð1 1

xðt Þx∗ ðt Þ dt

Substituting Eq (3.45) in Eq. (3.46), we obtain

ð3:46Þ

150

3 Frequency Domain Analysis of Continuous-Time Signals and Systems



ð1

1 E¼ xð t Þ 2π 1

ð1 1

∗

X ðjΩÞe

jΩt

 d Ω dt

ð3:47Þ

Interchanging the order of integration, Eq. (3.47) can be rewritten as 1 E¼ 2π

ð1 1

∗

X ðjΩÞ

ð 1 1

xðt Þe

jΩt

 ð3:48Þ

d t dΩ

By definition of the Fourier transform X ððjΩÞÞ ¼

ð1 1

xðt ÞejΩt dt

Thus, ð1

1 E¼ jxðt Þj dt ¼ 2π 1 2

ð1 1

jX ðjΩÞj2 dΩ

Example 3.28 Consider a signal x(t) with its Fourier transform given by 8 < 2 jΩj  1 X ðjΩÞ ¼ 1 1 < j Ωj  2 : 0 otherwise (i) Determine the energy of the signal x(t) (ii) Find x(t) Solution (i) By Parseval’s theorem, the energy E of x(t) is given by Ð1

1 E ¼ 1 jxðt Þj dt ¼ 2π 2

ð1 1 ð1

jX ðjΩÞj2 dΩ

1 1 4dΩ þ 2π 1 2π 8 1 1 þ þ ¼ 2π 2π 2π 5 ¼ π

¼

ð2

1dΩ þ

1

(ii) 8 <2 X ðjΩÞ ¼ 1 : 0 can be written as

jΩj  1 1 < j Ωj  2 otherwise

1 2π

ð 1 1dΩ 2

3.3 The Continuous Fourier Transform for Nonperiodic Signals

151

X ðjΩÞ ¼ X 1 ðjΩÞ þ X 2 ðjΩÞ where  X 1 ðjΩÞ ¼  X 2 ðjΩÞ ¼

1 0

jΩj  1 jΩj > 1

1 0

jΩj  2 jΩj > 2

which are depicted as 1(

W)

2(

1

1

0

Since F

h

sin ðΩ0 t Þ πt

W)

W

1

i

 ¼

-2

0

2

W

j Ωj  Ω0 j Ωj > Ω0

1 0

By linearity property, xðt Þ ¼ F1 ðX 1 ðjΩÞÞ þ F1 ðX 2 ðjΩÞÞ ¼

sin ðt Þ sin ð2t Þ þ πt πt

The Convolution Property If x1(t) and x2(t) are two continuous-time signals with Fourier transforms X1( jΩ) and X2( jΩ), then the Fourier transform of the convolution integral of x1(t) and x2(t) is given by F½x1 ðt Þ∗x2 ðt Þ ¼ X 1 ðjΩÞX 2 ðjΩÞ

ð3:49Þ

Hence, convolution of two sequences x1(t) and x2(t) in the time domain is equal to the product of their frequency spectra. Proof By the definition of convolution integral, yðt Þ ¼ x1 ðt Þ∗x2 ðt Þ ¼

ð1 1

x1 ðτÞx2 ðt  τÞ dτ

Taking the Fourier transform of Eq. (3.50), we obtain

ð3:50Þ

152

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Y ðjΩÞ ¼ F½yðt Þ ¼

ð1 ð1 1

1

½x1 ðτÞx2 ðt  τÞ dτejΩt dt

ð3:51Þ

Interchanging the order of integration, Eq. (3.51) can be rewritten as Y ðjΩÞ ¼

ð1 1

ð1 By the shifting property,

1

x1 ð τ Þ

ð1

1

 x2 ðt  τÞejΩt dt dτ

ð3:52Þ

 x2 ðt  τÞejΩt dt ¼ ejΩτ X 2 ðjΩÞ.

Hence, Eq. (3.52) becomes YðjΩÞ ¼ By X 1 ðjΩÞ ¼

ðthe 1 1

ð1 1

x1 ðτÞejΩt X 2 ðjΩÞdτ ¼ X 2 ðjΩÞ

definition x1 ðτÞe

jΩτ

of

ð1 1

continuous-time

x1 ðτÞejΩτ dτ

ð3:53Þ

Fourier

transform,

dτ

Thus Y ðjΩÞ ¼ X 1 ðjΩÞX 2 ðjΩÞ Example 3.29 Determine the Fourier transform of the triangular output signal y(t) of an LTI system as shown in Figure 3.13. Solution A triangular signal can be represented as the convolution of two rectangular pulse signals x1(t) and x2(t) defined by  x1 ð t Þ ¼ x2 ð t Þ ¼

1 0

jt j < 1 jt j > 1

yðt Þ ¼ x1 ðt Þ∗x2 ðt Þ By definition,

y(t)

Figure 3.13 Triangular output signal

2

-2

2

t

3.3 The Continuous Fourier Transform for Nonperiodic Signals

Ð1 dt ¼ 1 ejΩt dt  1  jΩ sin Ω e  ejΩ ¼ 2 ¼ jΩ Ω

X 1 ðjΩÞ ¼

Ð1

153

1 x1 ðt Þe

jΩt

By the convolution property, Y( jΩ) the Fourier transform of y(t) is given by Y ðjΩÞ ¼ X 1 ðjΩÞX 2 ðjΩÞ sin ðΩÞ sin ðΩÞ 2 ¼2 Ω Ω 2 sin ðΩÞ ¼4 Ω2 Example 3.30 Consider an LTI continuous-time system with the impulse response hðt Þ ¼ sin ðtΩ0 tÞ. Find the output y(t) of the system for an input xðt Þ ¼

sin ð2Ω0 t Þ : t

h i  1 jΩj  Ω0 0 tÞ Solution Since F sin ðΩ ¼ πt 0 jΩj > Ω0   sin ðΩ0 t Þ HðjΩÞ ¼ F½hðt Þ ¼ F t  π j Ωj  Ω 0 ¼ 0 j Ωj > Ω0   sin ð2Ω0 t Þ XðjΩÞ ¼ F½xðt Þ ¼ F t  π jΩj  2Ω0 ¼ 0 jΩj > 2Ω0 shown as (

-2W0

0

(

W)

2W 0

W

-W 0

0

W)

W0

W

154

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

The output y(t) of the system is given by yðt Þ ¼ xðt Þ∗hðt Þ By convolution property, we have YðjΩÞ ¼ F½yðt Þ ¼ XðjΩÞHðjΩÞ ( π 2 j Ωj  Ω 0 ¼ 0 jΩj > Ω0 which is depicted as (

W) 2

-W 0

0

W

W0

Therefore, yðt Þ ¼ F1 ðYðjΩÞÞ ¼ π

sin ðΩ0 t Þ t

Duality property For a given Fourier transform pair F

xðt Þ $ X ðjΩÞ By interchanging the roles of time and frequency, a new Fourier transform pair is obtained as F

X ðjt Þ $ 2πxðΩÞ For example, the duality exists between the Fourier transform pairs of Examples 3.19 and 3.20 as given by  xð t Þ ¼

1 0

jt j  T 1 jt j > T 1

F

$ X ðjΩÞ ¼ 2T 1 sin c

  ΩT 1 π

3.3 The Continuous Fourier Transform for Nonperiodic Signals

  Ω Ωt xðt Þ ¼ sin c π π

F

$

 X ðjΩÞ ¼

155

1 0

jt j  Ω jt j > Ω

The Modulation Property Due to duality between the time domain and frequency domain, the multiplication in the time domain corresponds to convolution in the frequency domain. If x1(t) and x2(t) are two continuous-time signals with Fourier transforms X1( jΩ) and X2( jΩ), then the Fourier transform of the product of x1(t) and x2(t) is given by F½x1 ðt Þx2 ðt Þ ¼

1 ½X 1 ðjΩÞ∗X 2 ðjΩÞ 2π

ð3:54Þ

This can be easily proved by dual property. Eq. (3.54) is called the modulation property since the multiplication of two signals often implies amplitude modulation. Example 3.31 Find the Fourier transform of ejΩ0 t xðt Þ Solution Let x1 ðt Þ ¼ ejΩ0 t and x2(t) ¼ x(t) X 1 ðjΩÞ ¼ F½ejΩ0 t  ¼ ` 2πδðΩ  Ω0 Þ X 2 ðjΩÞ ¼ F½xðtÞ ¼ ` XðjΩÞ F½x1 ðtÞx2 ðtÞ ¼ F½ejΩ0 t xðtÞ ¼

1 ½2πδðΩ  Ω0 Þ∗XðjΩÞ 2π

¼ XðjðΩ  Ω0 ÞÞ Example 3.32 Let y(t) be the convolution of two signals x1(t) and x2(t) defined by x1 ðt Þ ¼ sin cð2t Þ x2 ðt Þ ¼ sin cðt Þ cos ð3πt Þ Determine the Fourier transform of y(t). Solution By dual property, the Fourier transform of sinc(t) is given by 

Ω F½sincðt Þ ¼ rect 2π



The Fourier transform of x1(t) is given by F½x1 ðt Þ ¼ X 1 ðjΩÞ ¼ F½ sin cð2t Þ   1 Ω=2 ¼ rect 2 2π   1 Ω ¼ rect 2 4π

156

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

The Fourier transform of x2(t) is given by F½x2 ðt Þ ¼ X 2 ðjΩÞ ¼ F½sin c ðt Þ cos ð3πt Þ  j3πt  e þ ej3πt ¼ F sin c ðt Þ 2 By modulation property        j3πt  e þ ej3πt 1 ðΩ  3π Þ ðΩ þ 3π Þ X 2 ðjΩÞ ¼ F sin c ðt Þ þ rect ¼ rect 2 2π 2π 2 By convolution property

F yðt Þ ¼ F½x1 ðt Þ∗x2 ðt Þ ¼ X 1 ðjΩÞX 2 ðjΩÞ       Ω þ 3π 1 Ω ðΩ  3π Þ ¼ rect rect þ rect 4 4π 2π 2π

1

1

4p

-2p

0

-2p

4p

2(

0

1(

W) =

2p

2p

1 W) = rect W 2

4

4p

(W + 3 ) (W − 3 ) 1 + rect rect 2 2 2

4p

W

There is no overlap between the two transforms X1( jΩ) and X2( jΩ), and hence X 1 ðjΩÞX 2 ðjΩÞ ¼ 0 Therefore, YðjΩÞ ¼ F½x1 ðtÞ∗x2 ðtÞ ¼ X 1 ðjΩÞX 2 ðjΩÞ ¼0

3.3 The Continuous Fourier Transform for Nonperiodic Signals

157

Example 3.33 Determine the value of ð1 1

sin c2 ð2t Þ dt

Solution Since the Fourier transform of sinc (2t) is theorem, ð1 1

 

1 Ω 2 rect 4π

, using Parseval’s

  ð 1  2 1 Ω rect2 dΩ 4π 1 2 ð 1 2π 1dΩ ¼ 8π 2π

sin c2 ð2t Þdt ¼

1 2π

4π 8π 1 ¼ 2

¼

Example 3.34 Consider a signal x(t) with its Fourier transform given by  X ðjΩÞ ¼

π 0

j Ωj  Ω0 j Ωj > Ω0

Find the Fourier transform the system output y(t) given by yðt Þ ¼ xðt Þ cos ðΩc t Þ where

Ωc > Ω0

Solution By the definition of the inverse Fourier transform, we have ð 1 Ω0 πejΩt dΩ xð t Þ ¼ 2π Ω0    1 1  jΩ0 t ¼ e  ejΩ0 t 2 jt ¼

sin ðΩ0 t Þ t

Hence, yðt Þ ¼ xðt Þ cos ðΩc t Þ ¼ sin ðtΩ0 tÞ cos ðΩc t Þ By modulation property, we obtain YðjΩÞ ¼ XðjΩÞ∗F½cos ðΩc tÞ ¼ XðjΩÞ∗½πδðΩ  Ωc Þ þ πδðΩ þ Ωc Þ

158

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

2

−W − W 0 −W

−W + W 0

0 W −W0

W

W + W0

W

The Fourier transform properties of continuous-time signals are summarized in Table 3.2 and 3.3. Duality : for given Fourier transform pair xðt Þ $F X ðjΩÞ X ðjt Þ $F 2π xðΩÞ ð Ð1 1 1 2 Parseval’ s theorem x ð t Þ dt ¼ j j jX ðjΩÞj2 dΩ 1 2π 1

Table 3.2 Some properties of continuous-time Fourier transforms Property Linearity Time shifting Symmetry

Aperiodic signal a1x1(t)+a2x2(t) x(t  t0) x∗ ðtÞ xðtÞ real xe(t) (x(t) real) xo(t) (x(t) real)

Time reversal Frequency shifting Time and frequency scaling

x(n) ejΩ0 t xðt Þ x(at)

Differentiation in time

d dt xðt Þ

jΩX( jΩ)

Differentiation in frequency

jtx(t) ðt xðτÞ dτ

d dΩ X ðjΩÞ 1 jΩ X ðjΩÞ

x1(t) ∗ x2(t)

Xl( jΩ)X2( jΩ)

x1(t) x2(t)

1 2π ½X 1 ðjΩÞ∗X 2 ðjΩÞ

Integration

Fourier transform a1X1( jΩ)+a2X2( jΩ) ejΩt0 X ðjΩÞ X ∗ ðjΩÞ 8 ∗ X ðjΩÞ ¼ X ðjΩÞ > > > > < Re ½X ðjΩÞ ¼ Re ½X ðjΩÞ Im ½X ðjΩÞ ¼ Im X ðjΩÞ > > jX ðjΩÞj ¼ jX ðjΩÞj > > : arg½X ðjΩÞ ¼ arg½X ðjΩÞ Re[X( jΩ) jIm[X( jΩ)] X(ejω) X( j(ΩΩ0))  Ω 1 jaj X j a

1

Convolution property Modulation property

3.4 The Frequency Response of Continuous-Time Systems

159

Table 3.3 Basic Fourier transform pairs Signal δ(t) ebt uðt Þ 1 ejΩ0 t X1 k¼1

Fourier transform 1 b>0

1 bþjΩ

2πδ(Ω) 2πδ(ΩΩ0) 1 X 2π ak δðΩ  kΩ0 Þ

ak ejkΩ0 t

k¼1 1 X

1 2π X δðΩ  kΩ0 Þ T k¼1

δðt  kTÞ

k¼1

sin(Ω0t) cos(Ω0t)  1 jt j < T 1 xðt Þ ¼ 0 jt j > T 1 Periodicsquare wave with period T0 1 jt j < T 1 xðt Þ ¼ 0 T 1 jt j  T 0 =2 and x(t) ¼ x(t+T0) Ωt Ω π sin c π

π[δ(ΩΩ0)δ(ΩΩ0)] π[δ(ΩΩ0)+δ(ΩΩ0)]/j   2T 1 sin c ΩTπ 1

δ(tt0) u(t)

ejΩt0 1 πδðΩÞ þ jΩ

Sgn(t)

2 jΩ

tebt uðt Þ

1 X 2 sin ðkΩ0 T 1 Þ δðΩ  kΩ0 Þ k k¼1

 X ðjΩÞ ¼

b>0

Ω 6¼ 0

1 ðbþjΩÞ2 1 ðbþjΩÞn

t bt uðt Þ ðn1Þ! e t2 e2σ 2 n1

3.4

1 jΩj  Ω 0 jΩj > Ω

pffiffiffiffiffi 2 2 σ 2π e σ2Ω

The Frequency Response of Continuous-Time Systems

As in chapter 2, the input output relation of a useful class of continuous-time LTI systems satisfies the linear constant coefficient differential equation XN

a n¼0 n

dn yðtÞ X M d n xðtÞ ¼ b n n¼0 dt n dt n

ð3:55Þ

where coefficients an and bn are real constants. From convolution property, it is known that Y ðjΩÞ ¼ H ðjΩÞX ðjΩÞ which can be rewritten as

ð3:56Þ

160

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

H ðjΩÞ ¼

Y ðjΩÞ X ðjΩÞ

ð3:57Þ

Applying Fourier transform to both sides of Eq. (3.55), we obtain F

X N

 X  d n yð t Þ d n xð t Þ M a b ¼F n¼0 n dt n n¼0 n dt n

ð3:58Þ

By linearity property, Eq. (3.58) becomes  n  X  n  d yð t Þ M d xð t Þ a F bF ¼ n¼0 n n¼0 n dt n dt n

XN

ð3:59Þ

From the differentiation property, Eq. (3.59) can be rewritten as XN n¼0

an ðjΩÞn Y ðjΩÞ ¼

XM n¼0

bn ðjΩÞn X ðjΩÞ

ð3:60Þ

which can be rewritten as Y ðjΩÞ

hX N

i XM n a ð jΩ Þ b ðjΩÞn ¼ X ð jΩ Þ n n¼0 n¼0 n

ð3:61Þ

Thus, the frequency response of a continuous-time LTI system is given by HðjΩÞ ¼

PM YðjΩÞ bn ðjΩÞn ¼ Pn¼0 N n XðjΩÞ k¼0 an ðjΩÞ

ð3:62Þ

The function H(jΩ) is a rational function being a ratio of polynomials in (jΩ).

3.4.1

Distortion During Transmission

Eq. (3.56) implies that the transmission of an input signal x(t) through the system is changed into an output signal y(t). The X( jΩ) and Y( jΩ) are the spectra of the input and output signals, and H( jΩ) is the frequency response of the system. During the transmission, the input signal amplitude spectrum |X( jΩ)| is changed to|X( jΩ)||H(jΩ)|. Similarly, the input signal phase spectrum ∠X(jΩ) is changed to ∠X( jΩ) þ ∠ H(jΩ). An input signal spectral component of frequency Ω is modified in amplitude by a |H(jΩ)| factor and is shifted in phase by an angle ∠H(jΩ). Thus, the output waveform will be different from the input waveform during transmission through the system introducing distortion.

3.4 The Frequency Response of Continuous-Time Systems

161

Example 3.35 Consider an LTI system described by the following differential equation d2 yðt Þ dyðt Þ dxðt Þ þ 2yðt Þ ¼ 4  xð t Þ þ3 dt dt dt 2 Find the Fourier transform of the impulse response of the system. Solution Apply the Fourier transform on both sides of the differential equation, then we obtain  2    d yð t Þ dyðt Þ dxðt Þ F þ 2yðt Þ ¼ F 4  xð t Þ þ3 dt dt dt 2      2  d yð t Þ dyðt Þ dxðt Þ F þ 2F ½ y ð t Þ  ¼ 4F  F½xðt Þ þ 3F dt dt dt 2 ðjΩÞ2 YðjΩÞ þ 3jΩY ðjΩÞ þ 2YðjΩÞ ¼ 4jΩX ðjΩÞ  XðjΩÞ

 Ω2 þ j3Ω þ 2 YðjΩÞ ¼ ½j4Ω  1XðjΩÞ The Fourier transform of the impulse response H(jΩ) is given by HðjΩÞ ¼

YðjΩÞ 1  j4Ω ¼ XðjΩÞ Ω2  j3Ω  2

Example 3.36 Find the frequency response H( jΩ) of the following circuit

Solution

dv0 ðt Þ v0 ðt Þ þ dt R diðt Þ vi ðt Þ þ L þ v0 ð t Þ ¼ 0 dt diðt Þ ¼ vi ð t Þ  v0 ð t Þ L dt i ðt Þ ¼ C

LC

d2 v0 ðt Þ L dv0 ðt Þ þ v0 ð t Þ ¼ vi ð t Þ þ R dt dt 2

162

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Taking Fourier transform both sides of this equation, we obtain L LCðjΩÞ2 v0 ðjΩÞ þ jΩ v0 ðjΩÞ þ v0 ðjΩÞ ¼ vi ðjΩÞ R   L 2 1  LCΩ þ jΩ v0 ðjΩÞ ¼ vi ðjΩÞ R H ðjΩÞ ¼

3.5 3.5.1

v0 ðjΩÞ ¼ vi ðjΩÞ

1 1  LCΩ2 þ jΩ

L R

Some Communication Application Examples Amplitude Modulation (AM) and Demodulation Amplitude Modulation

In amplitude modulation, the amplitude of the carrier signal c(t) is varied in some manner with the baseband signal (message signal) m(t) also known as the modulating signal. The AM signal is given by s(t)

m(t)

c(t)

sðt Þ ¼ mðt Þ:cðt Þ

ð3:63Þ

cðt Þ ¼ cos ðΩc t þ θc Þ

ð3:63aÞ

For convenience, if it is assumed that θc ¼ 0 cðt Þ ¼ cos ðΩc t Þ

ð3:64Þ

CðjΩÞ ¼ F½cðtÞ ¼ π½δðΩ  Ωc Þ þ δðΩ þ Ωc Þ

ð3:65Þ

SðjΩÞ ¼ F½mðtÞcðtÞ ¼

1 ½M ðjΩÞ∗CðjΩÞ 2π

ð3:66Þ

M ðjΩÞ∗δðΩ  Ωc Þ ¼ M ðjðΩ  Ωc ÞÞ

ð3:67Þ

1 SðjΩÞ ¼ ½M ðjðΩ  Ωc ÞÞ þ M ðjðΩ þ Ωc ÞÞ 2

ð3:68Þ

Eq. (3.61) implies that the AM shifts the message signal so that it is centered at Ωc. The message signal m(t) can be recovered if Ωc >Ωm so that the replica spectra

3.5 Some Communication Application Examples

163

Ω

Ω

Ω

0

−Ω

Ω

Ω −Ω

Ω Ω

0.5

Ω −Ω − Ω

−Ω −Ω

Ω

Ω − Ω

Ω

Ω

Ω

Figure 3.14 Amplitude modulation

w(t)

s(t)

Low pass filter

ˆ (t)

( Ω)

c(t) Figure 3.15 Amplitude demodulation

do not overlap. The AM modulation in frequency domain is illustrated in Figure 3.14. Amplitude Demodulation The message signal m(t) can be extracted by multiplying the AM signal s(t) by the same carrier c(t) and passing the resulting signal through a low-pass filter as shown in Figure 3.15.

164

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

wðt Þ ¼ sðt Þcðt Þ ¼ mðt Þc2 ðt Þ ¼ mðt Þcos 2 ðΩc t Þ  1 þ cos ð2Ωc t Þ ¼ m ðt Þ 2

ð3:69Þ

The amplitude demodulation in frequency domain is illustrated in Figure 3.16.

3.5.2

Single-Sideband (SSB) AM

The double-sideband modulation is used in Section 3.5.1. By removing the upper sideband by using a low-pass filter with cutoff frequency Ωc or a lower sideband by high-pass filter with cutoff frequency Ωc, single-sideband modulation that requires half the bandwidth can be used. The frequency domain representation of singlesideband modulation is shown in Figure 3.17. However, the single-sideband modulation requires nearly ideal filters and increases the transmitter cost.

3.5.3

Frequency Division Multiplexing (FDM)

In frequency division multiplexing, multiple signals are transmitted over a single wideband channel using a single transmitting antenna. Different carriers with adequate separation are used to modulate for each of these signals with no overlap between the spectra of the modulated signals. The different modulated signals are summed before sending to the antenna. At the receiver, to recover a specific signal, the corresponding frequency is extracted through a band-pass filter. The FDM spectra for three modulated signals are shown in Figure 3.18.

3.6

Problems

1. Find the exponential Fourier series representation for each of the following signals: (i) (ii) (iii) (iv) (v) (vi)

x(t) ¼ cos(Ω0t) x(t) ¼ sin(Ω0t)  xðt Þ ¼ cos 2t þ π6 x(t) ¼ sin2(t) x(t) ¼ cos(6t) þ sin (4t)   xðt Þ ¼ ½1 þ cos ð2πt Þ sin 5πt þ π4

3.6 Problems

165 Ω

0.5

−Ω − Ω

−Ω −Ω

Ω

Ω − Ω

Ω

Ω

Ω

Ω

Ω

Ω −Ω

Ω

Ω

0.5

− Ω

−Ω

Ω

Ω

Ω

Ω

Ω

Ω

−Ω

0

Ω

Figure 3.16 Illustration of amplitude demodulation in frequency domain

Ω

166

3 Frequency Domain Analysis of Continuous-Time Signals and Systems Ω

0.5

Ω −Ω −Ω

Ω

Ω

Ω

Ω

Figure 3.17 Single-sideband AM Ω Ω

Ω

1

1

1

Ω

Ω

Ω

Ω

0.5

Ω Figure 3.18 Illustration of FDM for three signals

2. What signal will have the following Fourier series coefficients an ¼

1 sin 2 ðnπ=2Þ 4 ðnπ=2Þ2

3. If x1 (t) and x2(t) are periodic signals with fundamental period T0, find the Fourier series representation of x(t) ¼ x1(t)x2(t). 4. Consider the periodic signal x(t) given by xðt Þ ¼ ð2 þ j2Þej3t  j3ej2t þ 6 þ j3ej2t þ ð2  j2Þej3t Determine the trigonometric Fourier series representation of the signal x(t). 5. Find the Fourier series of a periodic signal x(t) with period 3 defined over one period by  xð t Þ ¼

t þ 2 2  t  0 2  2t 0  t  1

6. Find the Fourier series of a periodic signal x(t) with period 6 defined over one period by

3.6 Problems

167

8 0 > > > > < tþ2 xð t Þ ¼ 1 > > t þ2 > > : 0

3  t  2 2  t  1 1  t  1 1t2 2t3

7. Find Fourier series of the periodic signal shown in Figure P3.1.

Figure P3.1 Periodic signal of problem 7

8. Determine the exponential Fourier series representation of the periodic signal depicted in Figure P3.2 Figure P3.2 Periodic signal of problem 8

9. Determine trigonometric Fourier series representation of the signal shown in Figure P3.3

Figure P3.3 Periodic signal of problem 9

10. Plot the magnitude and phase spectrum of the periodic signal shown in Figure P3.4

168

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure P3.4 Periodic signal of problem 10

11. Find the Fourier transform of xðt Þ ¼  eat uðt Þ a > 0 t jtj  1 12. Find the Fourier transform of xðtÞ ¼ 0 jtj > 1 13. Find the Fourier transform of rectangular pulse given by  xð t Þ ¼

1 0

jt j  T jt j > T

14. Find the Fourier transform of xðtÞ ¼ π24t2 sin 2 ð2tÞ 15. Find the Fourier transform of the complex sinusoidal pulse given by  xð t Þ ¼

ej5t jt j  π 0 otherwise

16. Find Fourier transform of the following signal shown in Figure P3.5 Figure P3.5 Signal x(t)

17. Consider the following two signals x(t) and y(t) as shown in Figure P3.6. Determine Fourier transform y(t) using the Fourier transform of x(t), time shifting property, and differentiation property

3.6 Problems

169

Figure P3.6 Signals x(t) and y(t)

18. Find the inverse Fourier transform of  XðjΩÞ ¼

2cos ðΩÞ jΩj  π 0 jΩj > π

19. Find the inverse Fourier transform of X ðjΩÞ ¼

jΩ 2

ðjΩÞ þ 3jΩ þ 2

20. Consider the following communication system shown in Figure P3.7 to transmit two signals simultaneously over the same channel.

Figure P3.7 Communication system

Plot the spectra of x(t), y(t), and z(t) for given the following spectra of the two input signal shown in Figure P3.8.

170

3 Frequency Domain Analysis of Continuous-Time Signals and Systems

Figure P3.8 Spectra of x1(t) and x2(t)

21. Determine y(t) of an LTI system with input xðt Þ ¼ an ejnΩ0 t and the following H( jΩ) depicted in Figure P3.9 where H( jΩ) is H ( jnΩ0) evaluated at frequency nΩ0.

Figure P3.9 H( jΩ) of LTI system of problem 21

22. Sketch amplitude single-sideband modulation and demodulation if the message signal m(t) ¼ cos(Ωmt).

Further Reading 1. Lanczos, C.: Discourse on Fourier Series. Oliver Boyd, London (1966) 2. Körner, T.W.: Fourier Analysis. Cambridge University Press, Cambridge (1989) 3. Walker, P.L.: The Theory of Fourier Series and Integrals. Wiley, New York (1986) 4. Churchill, R.V., Brown, J.W.: Fourier Series and Boundary Value Problems, 3rd edn. McGrawHill, New York (1978) 5. Papoulis, A.: The Fourier Integral and Its Applications. McGraw-Hill, New York (1962) 6. Bracewell, R.N.: Fourier Transform and Its Applications, rev, 2nd edn. McGraw-Hill, New York (1986) 7. Morrison, N.: Introduction to Fourier Analysis. Wiley, New York (1994) 8. Lathi, B.P.: Linear Systems and Signals, 2nd edn. Oxford University Press, New York (2005) 9. Oppenheim, A.V., Willsky, A.S.: Signals and Systems. Englewood Cliffs, NJ, Prentice- Hall (1983)

Chapter 4

Laplace Transforms

The Laplace transform is a generalization of the Fourier transform of a continuous time signal. The Laplace transform converges for signals for which the Fourier transform does not. Hence, the Laplace transform is a useful tool in the analysis and design of continuous time systems. This chapter introduces the bilateral Laplace transform, the unilateral Laplace transform, the inverse Laplace transform, and properties of the Laplace transform. Also, in this chapter, the LTI systems, including the systems represented by the linear constant coefficient differential equations, are characterized and analyzed using the Laplace transform. Further, the solution of state-space equations of continuous time LTI systems using Laplace transform is discussed.

4.1 4.1.1

The Laplace Transform Definition of Laplace Transform

The Laplace transform of a signal x(t) is defined as ð1 X ðsÞ ¼ Lfxðt Þg ¼ xðt Þest dt 1

ð4:1Þ

The complex variable s is of the forms ¼ σ + jΩ, with a real part σ and an imaginary part Ω. The Laplace transform defined by Eq. (4.1) is called as the bilateral Laplace transform.

© Springer International Publishing AG, part of Springer Nature 2018 K. D. Rao, Signals and Systems, https://doi.org/10.1007/978-3-319-68675-2_4

171

172

4.1.2

4 Laplace Transforms

The Unilateral Laplace Transform

The unilateral Laplace transform plays an important role in the analysis of causal systems described by constant coefficient linear differential equations with initial conditions. The unilateral Laplace transform is mathematically defined as X ðsÞ ¼ Lfxðt Þg ¼

ð1 0þ

xðt Þest dt

ð4:2Þ

The difference between Eqs. (4.1) and (4.2) is on the lower limit of the integration. It indicates that the bilateral Laplace transform depends on the entire signal, whereas the unilateral Laplace transform depends on the right-sided signal, i.e., x(t) ¼ 0 for t < 0.

4.1.3

Existence of Laplace Transforms

The Laplace transform is said to exist if the magnitude of the transform is finite, that is, |X(s)| < 1. Piecewise continuous A function x(t) is piecewise continuous on a finite interval a  t  b, if x is continuous on [a,b], except possibly at finitely many points at each of which x has a finite left and right limit. Sufficient Condition The sufficient condition for existence of Laplace transforms is that if x(t) is piecewise continuous on (0, 1) and there exist some constants k and M such that |x(t)|  Mekt, then X(s) exists for s > k. Proof As x(t) is piecewise continuous on (0, 1), x(t)est is integrable on (0, 1). ð 1  ð1 ð1   st   Mekt est dt jxðt Þjest dt  jLfxðt Þgj ¼  xðt Þe dt   0

0

0

M  ðskÞt 1  M M e ð 0  1Þ ¼ ¼ ¼ 0 ks ks sk

ð4:3Þ

For s > k, |ℒ{x(t)}| < 1 .

4.1.4

Relationship Between Laplace Transform and Fourier Transform

When the complex variable s is purely imaginary, i.e., s ¼ jΩ, Eq. (4.1) becomes

4.1 The Laplace Transform

173

X ðjΩÞ ¼

ð1 1

xðt ÞejΩt dt

Eq. (4.4) is the Fourier transform of x(t), that is,  X ðsÞs¼jΩ ¼ Ffxðt Þg

ð4:4Þ

ð4:5Þ

If s is not purely imaginary, Eq. (4.1) can be written as X ðσ þ jΩÞ ¼

ð1 1

xðt Þ eðσþjΩÞt dt

ð4:6Þ

xðt Þeσt ejΩt dt

ð4:7Þ

Eq. (4.6) can be rewritten as X ðσ þ jΩÞ ¼

ð1 1

The right hand side of Eq. (4.7) is the Fourier transform of x(t)eσt. Thus, the Laplace transform can be interpreted as the Fourier transform of x(t) after multiplication by a real exponential signal.

4.1.5

Representation of Laplace Transform in the S-Plane

The Laplace transform is a ratio of polynomials in the complex variable, which can be represented by X ðsÞ ¼

N ðsÞ DðsÞ

ð4:8Þ

where N(s) is the numerator polynomial and D(s) represents the denominator polynomial. The Eq. (4.8) is referred to as rational. The roots of the numerator polynomial are referred to as zeros of X(s) ¼ 0 because for those values of s, X(s) becomes zero. The roots of the denominator polynomial are called the poles of X(s), as for those values of s, X(s) ¼ 1. A rational Laplace transform can be specified by marking the locations of poles and zeros by x and o in the s-plane, which is called as pole-zero plot of the Laplace transform. For a signal, the Laplace transform converges for a range of values of s. This range is referred to as the region of convergence (ROC), which is indicated as shaded region in the pole-zero plot.

174

4.2

4 Laplace Transforms

Properties of the Region of Convergence

Property 1

ROC of X(s) consists of strips parallel to the jΩ axis.

The ROC of X(s) contains the values of s ¼ σ + jΩ for which the Fourier transform of x(t)eσt converges. Thus, the ROC of X(s) is on the real part of s not on the frequency Ω. Hence, ROC of X(s) contains strips parallel to the jΩ axis Property 2

ROC of a rational Laplace transform should not contain poles.

In the ROC, X(s) should be finite for all s since X(s) is infinite at a pole and Eq. (4.1) does not converge at a pole. Hence, the ROC should not contain poles Property 3

ROC is the entire s-plane for a finite duration x(t), if there is at least one value of s for which the Laplace transform converges.

Proof A finite duration signal is zero outside a finite interval as shown in Figure 4.1. Let us assume that x(t)eσt is absolutely integrable for some value of σ ¼ σ 1 such that ð t2

jxðt Þjeσ1 t < 1

ð4:9Þ

t1

Then, the line ℜe(s) ¼ σ 1 is in the ROC. For ℜe(s) ¼ σ 2 also to be in the ROC, it is required that ð t2

jxðt Þje

σ 2 t

¼

t1

ð t2

jxðt Þjeσ 1 t eðσ2 σ1 Þt < 1

ð4:10Þ

t1

If σ 2 > σ 1 such that eðσ 2 σ 1 Þt is decaying, then the maximum value of eðσ 2 σ 1 Þt becomes eðσ 2 σ1 Þt1 for nonzero x(t) over the interval. Hence, ð t2 t1

jxðt Þje

σ 2 t

<e

ðσ 2 σ 1 Þt 1

ð t2

jxðt Þjeσ 1 t

ð4:11Þ

t1

The RHS of Eq. (4.11) is bounded and hence the LHS. Thus, the ℜe(s) > σ 1 must also be in the ROC. Similarly, if σ 2 < σ 1, it can be shown that xðt Þeσ 2 t is absolutely integrable. Hence, the ROC is the entire s-plane.

Figure 4.1 Finite duration signal

t

4.2 Properties of the Region of Convergence

Property 4:

175

If ROC of a right-sided signal contains the line ℜe(s) ¼ σ 1, then ℜe(s) > σ 1will also be in the ROC for all values of s.

Proof For a right-sided signal, x(t) ¼ 0 prior to some finite time t1 as shown in Figure 4.2 If the Laplace transform converges for some value of σ ¼ σ 1, then ð1

jxðt Þjeσ 1 t < 1

ð4:12Þ

jxðt Þjeσ 1 t < 1

ð4:13Þ

1

If x(t) is right sided, then ð1 t1

For σ 2 > σ 1 , xðt Þ eσ 2 t is absolutely integrable as eσ 2 t decays faster than eσ1 t as t ! 1. Thus, ℜe(s) > σ 1 will also be in the ROC for all values of s. Property 5:

If ROC of left-sided signal contains the line ℜe(s) ¼ σ1, then ℜe(s) < σ1 will also be in the ROC for all values of s.

Proof For left-sided signal, x(t) ¼ 0 after some finite time t2 as shown in Figure 4.3. This can be proved easily with the same argument and intuition for the property 4. Figure 4.2 Right-sided signal

x(t)

t

Figure 4.3 Left-sided signal

x(t)

t

176

4 Laplace Transforms

Property 6:

If ROC of a two-sided signal contains the line ℜe(s) ¼ σ0, then ROC will contain a strip, which includes the line.

Proof A two-sided signal is of infinite duration for both t > 0 and t < 0 as shown in Figure 4.4(a) Let us choose an arbitrary time t0 that divides the signal into as sum of right-sided signal and left-sided signal as shown Figure 4.4(b) and (c). The Laplace transform of x(t) converges for the values of s for which both the right-handed signal and lefthanded signal converge. It is known from property 4 that the ROC of Laplace transform of right-handed signal Xr(s) consists of a half plane ℜe(s) > σ r for some value σ r; and from property 5, it is known that Xr(s) consists of a half plane ℜe (s) > σ l for some value σ l. Then the overlap of these two half planes is the ROC of the two-sided signal x(t) as shown in Figure 4.4(d) with the assumption that σ r < σ l. If σ r is not less than σ l, then there is no overlap. In this case, X(s) does not exist even Xr(s) and Xl(s) individually exist.

4.3

The Inverse Laplace Transform

From Eq. (4.6), it is known that the Laplace transform X(σ þ jΩ) of a signal x(t) is given by X ðσ þ jΩÞ ¼

ð1 1

xðt Þ eσt ejΩt dt

ð4:14Þ

Applying the inverse Fourier transform on the above relationship, we obtain xðtÞeσt ¼ F 1 fXðσ þ jΩÞg ¼

1 2π

ð1 1

ðXðσ þ jΩÞÞ ejΩt dΩ

ð4:15Þ

Multiplying both sides of Eq. (4.15) by eσt, it follows that xðtÞ ¼

1 2π

ð1 1

ðXðσ þ jΩÞÞ eðσþjΩÞt dΩ

ð4:16Þ

As s ¼ σ + jΩ and σ is a constant, ds ¼ j dΩ. Substituting s ¼ σ + jΩ ds ¼ j dΩ in Eq. (4.16) changing the variable of integration from s to Ω, we arrive at the following inverse Laplace transform xð t Þ ¼

1 2πj

ð σþj1 σj1

X ðsÞ est ds

ð4:17Þ

x(t)

t

t

(d)

(b)

Im

x(t)

Figure 4.4 (a) Two-sided signal (b) Right-sided signal. (c) Left-sided signal. (d) ROC of the two-sided signal

(c)

(a)

Re

s-plane

t

4.3 The Inverse Laplace Transform 177

178

4.4

4 Laplace Transforms

Properties of the Laplace Transform

Linearity If x1(t) and x2(t) are two signals with Laplace transforms X1(s) and X2(s) and ROCs R1 and R2, respectively, then the Laplace transform of a linear combination of x1(t) and x2(t) is given by Lfa1 x1 ðtÞ þ a2 x2 ðtÞg ¼ a1 X 1 ðsÞ þ a2 X 2 ðsÞ

ð4:18Þ

whose ROC is at least (R1 \ R2), a1 and a2 being arbitrary constants. Proof Lfa1 x1 ðtÞ þ a2 x2 ðtÞg ¼

ð1

fa1 x1 ðtÞ þ a2 x2 ðtÞg est dt

1 ð1

¼ a1

1

x1 ðtÞest dt þ a2

ð1 1

x2 ðtÞest dt

¼ a1 X 1 ðsÞ þ a2 X 2 ðsÞ

ð4:19Þ

ð4:20Þ

The result concerning the ROC follows directly from the theory of complex variables concerning the convergence of a sum of two convergent series. Time Shifting If x(t) is a signal with Laplace transform X(s) and ROC R, then for any constant t0  0, the Laplace transform of x(t – t0) is given by Lfxðt  t 0 Þg ¼ est0 X ðsÞ whose ROC is the same as that of X(s). Proof

ð1

xðt  t 0 Þ est dt

ð4:22Þ

xðτÞ esðτþt0 Þ dτ ð1 st0 xðτÞ esτ dτ ¼e

ð4:23Þ

Lfxðt  t 0 Þg ¼

1

Substituting τ ¼ t – t0, L fx ð t  t 0 Þ g ¼

ð1

1

1

¼e

st0

X ðsÞ

Shifting in the s-domain. If x(t) is a signal with Laplace transform X(s) and ROC R, then the Laplace transform of the signal es0 t xðt Þ is given by Lfes0 t xðt Þg ¼ X ðs  s0 Þ whose ROC is the R þ ℜe(s)

ð4:24Þ

4.4 Properties of the Laplace Transform

179

ð1

Proof Lfes0 t xðt Þg ¼

1 ð1

¼

1

es0 t xðt Þest dt xðt Þeðss0 Þt dt

ð4:25Þ

¼ X ðs  s0 Þ Time Scaling. If x(t) is a signal with Laplace transform X(s) and ROC R, then the Laplace transform of the x(at) for any constant a, real or complex, is given by 1 s Lfxðat Þg ¼ X a a

ð4:26Þ

whose ROC is the Ra Proof Lfxðat Þg ¼

ð1 1

xðat Þ est dt

ð4:27Þ

Letting τ ¼ at; dt ¼ dτ/a Then, Lfxðat Þg ¼

Ð1 1

τ1 xðτÞ esa dτ a

ð sτ 1 1 ¼ xðτÞe a dτ a 1 1 s ¼ X a a

ð4:28Þ

Differentiation in the Time Domain. If x(t) is a signal with the Laplace transform X (s) and ROC R, then   dx ¼ sX ðsÞ L dt

ð4:29Þ

with ROC containing R. Proof This property can be proved by differentiating both sides of the inverse Laplace transform expression xð t Þ ¼

1 2πj

ð σþj1 σj1

X ðsÞ est ds

Then, dx 1 ¼ dt 2πj

ð σþj1 σj1

sX ðsÞest ds

ð4:30Þ

180

4 Laplace Transforms

From the above expression, it can be stated that the inverse Laplace transform of sX(s) is dx dt . Differentiation in the s-Domain. If x(t) is a signal with the Laplace transform X(s), then dX ðsÞ ¼ Lftxðt Þg ds

ð4:31Þ

Proof From the definition of the Laplace transform, X ð s Þ ¼ L fx ð t Þ g ¼

ð1 1

xðt Þ est dt

Differentiating both sides of the above equation, we get ð1 dX ðsÞ ¼ txðt Þ est dt ds 1 ¼ Lftxðt Þg

ð4:32Þ

with ROC ¼ R Division by t If x(t) is a signal with Laplace transform X(s), then  L h

xð t Þ t t!0

provided that lim

xðt Þ t

 ¼

i

ð1

X ðuÞ du

ð4:33Þ

s

exists.

Proof Let x1 ðt Þ ¼ xðttÞ, then (t) ¼ t x1(t). By using the differentiation in the s-domain property, d L fx 1 ð t Þ g ds

ð4:34Þ

dLfx1 ðt Þg ¼ X ðsÞds

ð4:35Þ

X ðsÞ ¼  which can be rewritten as

Integrating both sides of Eq. (4.35) yields ð Ð dLfx1 ðt Þg ¼  X ðsÞ ds ðs L fx 1 ð t Þ g ¼  X ðuÞ du 1 ð1 X ðuÞ du L fx 1 ð t Þ g ¼ s

ð4:36Þ

4.4 Properties of the Laplace Transform

181

Integration. If x(t) is a signal with Laplace transform X(s) and ROC R, then L

ð t



1

1 ¼ X ðsÞ s

xðt Þ dt

ð4:37Þ

with ROC containing R \ ℜe(s) > 0. Proof This property can be proved by integrating both sides of the inverse Laplace transform expression xð t Þ ¼

1 2πj

ð σþj1 σj1

X ðsÞest ds

Then, ðτ 1

xðτÞ dτ ¼

1 2πj

ð σþj1

1 X ðsÞ est ds σj1 s

Consequently, the inverse Laplace transform of 1s X ðsÞ is

Ðτ

1

xðτÞ dτ:

Convolution in the Time Domain. If x1(t) and x2(t) are two signals with Laplace transforms X1(s) and X2(s) with ROCs R1 and R2, respectively, then Lfx1 ðt Þ ∗ x2 ðt Þg ¼ X 1 ðsÞX 2 ðsÞ

ð4:38Þ

with ROC containing R1 \ R2 Proof Lfx1 ðt Þ ∗ x2 ðt Þg ¼

ð 1 ð 1 1

1

 x1 ðτÞx2 ðt  τÞ dτ est dt

ð4:39Þ

Changing the order of integration, Eq. (4.39) can be rewritten as L fx 1 ð t Þ ∗ x 2 ð t Þ g ¼

ð1 1

x1 ð τ Þ

ð 1 1

x2 ðt  τÞe

st

 dt dτ

ð4:40Þ

Let t1 ¼ t – τ; dt1 ¼ dt; L fx 1 ð t Þ ∗ x 2 ð t Þ g ¼ ¼

ð1 1

ð1

1

x1 ð τ Þ e

sτ

ð 1

st 1



x2 ðt 1 Þe dt 1 dτ ð 1  sτ sτ e x1 ðτÞ X 2 ðsÞdτ ¼ e x1 ðτÞdτ X 2 ðsÞ 1

ð4:41Þ

1

¼ X 1 ðsÞX 2 ðsÞ The ROC includes R1 \ R2 and is large if pole-zero cancellation occurs. Convolution in the Frequency Domain If x1(t) and x2(t) are two signals with Laplace transforms X1(s) and X2(s), then

182

4 Laplace Transforms

Lfx1 ðt Þx2 ðt Þg ¼

ð cþj1

1 2πj

X 1 ðpÞX 2 ðs  pÞdp

ð4:42Þ

cj1

Proof Let x(t) ¼ x1(t)x2(t) with Laplace transforms X1(s) and X2(s) and areas of convergence ℜe(s) > σ 1 and ℜe(s) > σ 2, respectively, then L fx ð t Þ g ¼

ð1

x1 ðt Þ x2 ðt Þ est dt

ð4:43Þ

X 1 ðpÞ ept dp, c > σ 1

ð4:44Þ

0

According to the inverse integral, 1 x1 ð t Þ ¼ 2πj

ð cþj1 cj1

Substituting this relationship in Eq. (4.43), it follows that L fx ð t Þ g ¼

ð1

x2 ðt Þest



0

1 2πj

ð cþj1

X 1 ðpÞept dp dt

ð4:45Þ

cj1

Permuting the sequence of integration, we obtain X ðsÞ ¼

1 2πj

ð cþj1

X 1 ðpÞdp

cj1

ð1

x2 ðt ÞeðspÞt dt

ð4:46Þ

0

where X 2 ðs  p Þ ¼

ð1

x2 ðt ÞeðspÞt dt

ð4:47Þ

0

This integral converges for ℜe(s  p) > σ 2. By substituting Eq. (4.47) in Eq. (4.46), yields the following proving the property X ðsÞ ¼

1 2πj

ð cþj1

X 1 ðpÞX 2 ðs  pÞdp

ð4:48Þ

cj1

The above properties of the Laplace transform are summarized in Table 4.1.

4.4.1

Laplace Transform Properties of Even and Odd Functions

Even Property If x(t) is an even function such that x(t) ¼ x(–t), then X(s) ¼ X(s).

4.4 Properties of the Laplace Transform

183

Proof X ðsÞ ¼

ð1 1

xðt Þ est dt

ð4:49Þ

xðt Þ est dt

ð4:50Þ

Consider X 1 ðsÞ ¼ Let t1 ¼ t; then, X 1 ðsÞ ¼

ð1 1

ð1 1

xðt 1 Þ est1 dt 1

ð4:51Þ

¼ X ðsÞ Since x(t) ¼ x(t) then L{x(t)} ¼ L{x(t)} Thus, X(s) ¼ X(s). Odd Property If x(t) is an odd function such that x(t) ¼ x(t), then X(s) ¼ X(s). Proof X ðsÞ ¼

ð1

xðt Þ est dt

ð4:52Þ

xðt Þ est dt

ð4:53Þ

1

Consider X 1 ðsÞ ¼ Let t1 ¼ t; then X 1 ðsÞ ¼

ð1

ð1 1

xðt 1 Þ e dt 1 ¼ 

ð1

st 1

1

1

xðt 1 Þ est1 dt 1

ð4:54Þ

¼ X ðsÞ Since x(t) ¼ x(t) then L{x(t)} ¼ L{x(t)} Thus, X(s) ¼ X(s).

4.4.2

Differentiation Property of the Unilateral Laplace Transform

Most of the properties of the bilateral transform tabulated in Table 4.1 are the same for the unilateral transform. In particular, the differential property of unilateral Laplace transform is different as it requires that x(t) ¼ 0 for t < 0 and contains no impulses and higher-order singularities. If x(t) is a signal with unilateral Laplace transform X (s), then the unilateral Laplace transform of dx dt can be found by using integrating by parts as

184

4 Laplace Transforms

Table 4.1 Some properties of the Laplace transform Property Linearity Time shifting Shifting in the s-domain Time scaling

Signal a1x1(t) + a2x2(t) x(t – t0) es0 t xðt Þ x(at)

Differentiation in the time domain

dx dt

Differentiation in the s-domain

tx(t) ðt xðt Þdt

dX ðsÞ ds 1 s X ðsÞ

R \ ℜe(s) > 0.

x1(t) * x2(t)

X1(s)X2(s)

R1 \ R2

Integration

Laplace transform a1X1(s) + a2X2(s) est0 X ðsÞ X(s – s0) s 1 aX a sX(s)

ROC At least R1 \ R2 Same as R Shifted version of R R a

At least R R

1

Convolution

ð1 0

 dx st e dt ¼ xðt Þest 1 0þ þ s dt

ð1

xðt Þest dt

ð4:55Þ

0

þ

¼ sX ðsÞ  xð0 Þ 2

Applying this second time yields the unilateral Laplace transform of ddt2x as given by ð1 0

d 2 x st e dt ¼ s2 X ðsÞ  sxð0þ Þ  x_ ð0þ Þ dt 2

ð4:56Þ

+ where x_ ð0þ Þ is the dx dt evaluated at t ¼ 0 . Similarly, applying this for third time yields the unilateral Laplace transform

ð1 0

d 3 x st e dt ¼ s3 XðsÞ  s2 xð0þ Þ  s_x ð0þ Þ  €xð0þ Þ dt 3

ð4:57Þ

where €xð0þ Þ is the ddt2x evaluated at t ¼ 0+. n Continuing this procedure for the nth time, the unilateral Laplace transform of ddtnx is given by 2

ð1 0

d n x st e dt ¼ sn X ðsÞ  sn1 xð0þ Þ  sn2 x_ ð0þ Þ  sn3€xð0þ Þ: . . . dt n

ð4:58Þ

Example 4.1 Determine whether the following Laplace transforms correspond to the even time function or odd time function and comment on the ROCs. (a) X ðsÞ ¼ ðsþ2Ks Þðs2Þ sþj2Þðsj2Þ (b) X ðsÞ ¼ Kððsþ2 Þðs2Þ (c) Comment on the ROCs

4.4 Properties of the Laplace Transform

185

Solution Ks (a) X ðsÞ ¼ ðsþ2Ks Þðs2Þ ; X ðsÞ ¼ ðsþ2Þðs2Þ ¼ X ðsÞ; Hence, the corresponding x(t) is an odd function. sþj2Þðsj2Þ (b) X ðsÞ ¼ Kððsþ2 Þðs2Þ ¼ X ðsÞ Hence, the corresponding x(t) is an even function. (c) The ROCs for (a) and (b) are shown in Figure 4.5(a) and (b), respectively. From Figure 4.5(a) and (b), it can be stated that for the time function to be even or odd, the ROC must be two sided. Example 4.2 A real and even signal x(t) with its Laplace transform X(s) has four poles with one pole located at 12 ejπ=4 , with no zeros in the finite s-plane and X (0) ¼ 16. Find X(s). Solution Since X(s) has four poles with no zeros in the finite s-plane, it is of the form X ðsÞ ¼

K ð s  p1 Þ ð s  p2 Þ ð s  p3 Þ ð s  p 4 Þ

As x(t) is real, the poles of X(s) must occur as conjugate reciprocal pairs. Hence, p2 ¼ p1 ∗ ; p4 ¼ p3 ∗ and X(s) becomes X ðsÞ ¼

K ð s  p1 Þ ð s  p1 ∗ Þ ð s  p3 Þ ð s  p 3 ∗ Þ

(a)

(b)

Im

Im j2

s plane

s plane

Re -2

Re

2

2

-2

-j2

K ðsþj2Þðsj2Þ Figure 4.5 (a) ROC of X ðsÞ ¼ ðsþ2Ks Þðs1Þ. (b) ROC of X ðsÞ ¼ ðsþ2Þðs2Þ

186

4 Laplace Transforms

Since x(t) is even, the X(s) also must be even, and hence the poles must be symmetric about the jΩ axis. Therefore, p3 ¼ p1 ∗ . Thus, XðsÞ ¼

K ðs  p1 Þðs  p∗ Þðs þ p∗ 1 1 Þðs þ p1 Þ

Assuming that the given pole location is that of p1, that is, p1 ¼ 12 ejπ=4 , We obtain K     XðsÞ ¼  1 jπ=4 1 jπ=4 1 1 s e s e s þ ejπ=4 s þ ejπ=4 2 2 2 2     1 jπ=4 1 π π 1 1 1 1 1 pffiffiffi þ jpffiffiffi ¼ pffiffiffi þ j pffiffiffi e ¼ cos þ jsin ¼ 2 2 4 4 2 2 2 2 2 2 2 K   XðsÞ ¼  1 1 1 1 2 2 p ffiffi ffi p ffiffi ffi S  sþ S þ sþ 4 4 2 2 K when s ¼ 0, X ð0Þ ¼ 1=16 ¼ 16, and therefore K ¼ 1. Hence,

1   X ðsÞ ¼  s2  p1ffiffi2s þ 14 s2 þ p1ffiffi2s þ 14

4.4.3

Initial Value Theorem

For a signal x(t) with Laplace transform X(s) and x(t) ¼ 0 for t < 0, then xð0þ Þ ¼ lims!1 sX ðsÞ

ð4:59Þ

Proof To prove the theorem, let the following integral first be evaluated by using integration by parts  Ð 1 dx st e dt ¼ xðt Þ est 1 0þ þ 0þ dt

ð1 0þ

xðt Þs est dt

¼ xð0þ Þ þ sX ðsÞ

ð4:60Þ

As s tends to 1, the Eq. (4.60) can be expressed as ð1 lim

s!1

0

þ

dx st e dt ¼ lim ½sXðsÞ  xð0þ Þ s!1 dt

ð4:61Þ

4.5 Laplace Transforms of Elementary Functions

187

As the integration is independent of s, the calculation of the limit and the integration can be permuted provided that the integral converges uniformly. If L{x(t)} exists, then lim

s!1

dx st e ¼0 dt

ð4:62Þ

is valid. Hence, we get xð0þ Þ ¼ lims!1 sX ðsÞ

4.4.4

ð4:63Þ

Final Value Theorem

For a signal x(t) with Laplace transform X(s) and x(t) ¼ 0 for t < 0, then xð1Þ ¼ lims!0 sX ðsÞ

ð4:64Þ

Proof To prove this, the following integration is to be evaluated ð1 lim

s!0

0

þ

dx st e dt ¼ sX ðsÞ  xð0þ Þ dt

ð4:65Þ

Again one can permute the sequence of determining the limit and the integration provided the integral converges. The result is ð1 0þ

dx dt ¼ lim ½sX ðsÞ  xð0þ Þ, s!0 dt

ð4:66Þ

and after integration it follows that xð1Þ  xð0þ Þ ¼ lim ½sX ðsÞ  xð0þ Þ s!0

¼ lim ½sX ðsÞ  xð0þ Þ

ð4:67Þ

s!0

Therefore, xð1Þ ¼ lim sX ðsÞ s!0

4.5

ð4:68Þ

Laplace Transforms of Elementary Functions

Unit Impulse Function The unit impulse function is defined by  δðt Þ ¼

1 0

for t ¼ 0 elsewhere

ð4:69Þ

188

4 Laplace Transforms

By definition, the Laplace transform of δ (t) can be written as X ðsÞ ¼ Lfδðt Þg ¼

ð1

δðt Þest dt

0

ð4:70Þ

¼1 The ROC is the entire s-plane. Unit Step Function The unit step function is defined by 

for t  0 elsewhere

1 0

uð t Þ ¼

ð4:71Þ

The Laplace transform of u(t) by definition can be written as X ðsÞ ¼ Lfuðt Þg ¼ ¼

ð1 ð01 0

¼

uðt Þ est dt  1 1est dt ¼  est 1 0 s

1 s

Hence, the ROC for X(s) is ℜe(s) > 0. Example 4.3 Find the Laplace transform of x(t) ¼ δ(t – t0). Solution By using the time shifting property, we get Lfδðt  t 0 Þg ¼ est0 Lfδðt Þg ¼ est0 The ROC is the entire s-plane. Example 4.4 Find the Laplace transforms of the following: (i) x(t) ¼ eαtu(t) (iv) x(t) ¼ eαtu(t)

(ii) x(t) ¼ eαtu(t)

Solution (i) X ðsÞ ¼ Lfeαt uðt Þg ¼ 

(iii) x(t) ¼ eαtu(t) ð1

eαt uðt Þest dt

0

Because u(t) ¼ 1 for t < 0 and u(t) ¼ 0 for t > 0, X ðsÞ ¼  ¼

ð 0 1

1 sþα

eðsþαÞt dt

ð4:72Þ

4.5 Laplace Transforms of Elementary Functions

189

The ROC for X(s) is ℜe(s) <  α ð1 αt (ii) X ðsÞ ¼ Lfe uðt Þg ¼ eαt uðt Þest dt 0

Because u(t) ¼ 1 for t < 0 and u(t) ¼ 0 for t > 0, X ðsÞ ¼  ¼

ð 0

eðsαÞt dt

1

1 sα

The ROC for X (s) is ℜe(s) < α (iii) Let x1(t) ¼ u(t), then X 1 ðsÞ ¼ Lfuðt Þg ¼

1 s

By using the shifting in the s-domain property, we get X ðsÞ ¼ Lfeαt uðt Þg ¼ X 1 ðs þ αÞ ¼

1 sþα

The ROC for X(s) is ℜe(s) >  α (iv) Let x1(t) ¼ u(t), then X 1 ðsÞ ¼ Lfuðt Þg ¼

1 s

By using the shifting in the s-domain property, we get X ðsÞ ¼ Lfeαt uðt Þg ¼ X 1 ðs  αÞ ¼ The ROC for X(s) is ℜe(s) > α. Example 4.5 Find the Laplace transform of xð t Þ ¼

t ðn1Þ uð t Þ ðn  1Þ!

Solution Let x1(t) ¼ u(t), then X 1 ðsÞ ¼ Lfuðt Þg ¼ and for n ¼ 2, x2(t) ¼ tu(t).

1 s

1 sα

190

4 Laplace Transforms

By using the differentiation in the s-Domain property, we get X 2 ðsÞ ¼ Lftuðt Þg ¼ 

dX 1 1 ¼ 2 s ds

2

t Similarly, for n ¼ 3, x3 ðt Þ ¼ ð31 Þ! uðt Þ. Again, by using the differentiation in the s-domain property, we get



t2 uð t Þ ðn  1Þ! n ðn1Þ o 1 In general, X ðsÞ ¼ L ðtn1Þ!uðt Þ ¼ n . s X 3 ðsÞ ¼ L

 ¼

dX 2 1 ¼ 3 s ds

Example 4.6 Find the Laplace transform of xð t Þ ¼

t ðn1Þ αt e uð t Þ ðn  1Þ!

ðn1Þ

Solution Let x1 ðt Þ ¼ ðtn1Þ! uðt Þ, then  X 1 ðsÞ ¼ L

t ðn1Þ uð t Þ ðn  1Þ!

 ¼

1 Sn

By using the shifting in the s-domain property, we get 

 t ðn1Þ αt 1 X ðsÞ ¼ L e uðt Þ ¼ X 1 ðs þ αÞ ¼ for ℜeðsÞ > α ðs þ αÞn ðn  1Þ! Example 4.7 Find the Laplace transform of x(t) ¼ sin ωt u(t) 

 ejωt  ejωt uð t Þ 2j    1   jωt ¼ L e uðt Þ  L ejωt uðt Þ 2j

Solution

Lf sin ωt uðt Þ ¼ L

Using the shifting in the s-domain property, we get

  jωt  1 1   jωt 1 1 L e uð t Þ  L e uð t Þ ¼  2j 2j s  jω s þ jω ω ¼ 2 for ℜeðsÞ > 0 s þ ω2 Therefore, Lf sin ωt uðt Þ ¼

s2

ω for ℜeðsÞ > 0 þ ω2

4.5 Laplace Transforms of Elementary Functions

191

Example 4.8 Find the Laplace transform of x(t) ¼ cos ωt u(t). 

Solution

ejωt þ ejωt Lf cos ωt uðt Þ ¼ L uð t Þ 2    1  ¼ L ejωt uðt Þ þL ejωt uðt Þ 2



Using the shifting in the s-domain property, we get

   1 1  jωt 1 1 L e uðt Þ þ L ejωt uðt Þ ¼ þ 2 2 s  jω s þ jω s for ℜeðsÞ > 0 ¼ 2 s þ ω2 Therefore, Lf cos ωt uðt Þ ¼

s for ℜeðsÞ > 0 s 2 þ ω2

Example 4.9 Find the Laplace transform of x(t) ¼ eαtsin ωt u(t). Solution Let x1(t) ¼ sin ωt u(t), then X 1 ðsÞ ¼ Lf sin ωt uðt Þg ¼

ω s 2 þ ω2

By using the shifting in the s-domain property, we get ω

X ðsÞ ¼ Lfeαt sin ωt uðt Þg ¼ X 1 ðs þ αÞ ¼

ðs þ αÞ2 þ ω2

for ℜeðsÞ > α

Example 4.10 Find the Laplace transform of x(t) ¼ eαtcos ωt u(t). Solution Let x1(t) ¼ cos ωt u(t), then X 1 ðsÞ ¼ Lf cos ωt uðt Þg ¼

s2

s þ ω2

By using the shifting in the s-domain property, we get X ðsÞ ¼ Lfeαt cos ωt uðt Þg ¼ X 1 ðs þ αÞ ¼ Example 4.11 Find the Laplace transform of Solution Lf sin t g ¼

sþα ðs þ αÞ2 þ ω2

sin t t

1 s2 þ 1

for ℜeðsÞ > α

192

4 Laplace Transforms



 ð1 sin t 1 du L ¼ 2þ1 t u s  ¼ tan 1 u1 s π ¼  tan 1 s 2 Example 4.12 Find the Laplace transform of e Solution

4t

e3t t

  L e4t ¼

  1 1 ; L e3t ¼ s4 sþ3  4t  ð1 ð1 3t e e 1 1 du  du L ¼ t s u4 s uþ3  1  ¼ ln ðu  4Þ1 s  ln ðu þ 3Þ s ¼ ln ¼ ln

ðs  4Þ ðs þ 3Þ

ð s þ 3Þ ð s  4Þ

Example 4.13 Consider the signal x(t) ¼ etu(t) + 2e2tu(t). (a) Does the Fourier transform of this signal converge? (b) For which of the following values of a does the Fourier transform of x(t) eαt converge? (i) α ¼ 1 (ii) α ¼ 2.5 (c) Determine the Laplace transform X(s) of x(t). Sketch the location of the poles and zeros of X(s) and the ROC. Solution (a) The Fourier transform of the signal does not converge as x(t) is not absolutely integrable due to the rising exponentials. (b) (i) For α ¼ 1, x(t) eαt ¼ u(t) þ 2etu(t). Although the growth rate has been slowed, the Fourier transform still does not converge. (ii) For α ¼ 2.5, x(t) eαt ¼ e1.5tu(t) + 2e0.5tu(t), the Fourier transform converges (c) The Laplace transform of x(t) is  2 s  32 1 2 2s  3 þ ¼ ¼ X ðsÞ ¼ s  1 s  2 ðs  1Þðs  2Þ ðs  1Þðs  2Þ and its pole-zero plot and ROC are as shown in Figure 4.6.

4.5 Laplace Transforms of Elementary Functions Figure 4.6 Pole-zero plot and ROC

193

Im

Re 1.5

1

2

It is noted that if α > 2, s ¼ α þ jΩ is in the region of convergence, as it is shown in part (b) (ii), the Fourier transform converges. Example 4.14 The Laplace transform H(s) of the impulse response h(t) for an LTI system is given by H ðsÞ ¼

1 ðs þ 2Þ

ℜeðsÞ > 2

Determine the system output y(t) for all t if the input x(t) is given by x(t) ¼ e3t/2 + 2et for all t. Solution From the convolution integral, yð t Þ ¼

ð1 1

hðτÞxðt  τÞdτ

Let x(t) ¼ eαt, then yð t Þ ¼ ð1 1

ð1 1

hðτÞe

αðtτÞ

dτ ¼ e

αt

ð1 1

hðτÞeατ dτ

hðτÞ eατ dτ can be recognized as H(s)|s ¼ α.

Hence, if x(t) ¼ eαt, then yðt Þ ¼ eαt ½H ðsÞjs¼α  Using linearity and superposition, it can be recognized that if x(t) ¼ e3t/2 þ 2et, then  yðt Þ ¼ e3t=2 H ðsÞs¼3=2 þ 2et H ðsÞjs¼1 So that yðt Þ ¼ 2e3t=2 þ 2et

for all t

194

4 Laplace Transforms

Example 4.15 The output y(t) of a LTI system is  yðt Þ ¼ 2  3et þ e3t uðt Þ for an input  xðt Þ ¼ 2 þ 4e3t uðt Þ Determine the corresponding input for an output y1 ðt Þ ¼ ð1  et  tet Þuðt Þ Solution For the input x(t) ¼ (2 þ 4e3t)u(t), the Laplace transform is X ðsÞ ¼

2 4 6ð s þ 1Þ þ ¼ s s þ 3 s ð s þ 3Þ

The corresponding output has the Laplace transform Y ðsÞ ¼

2 3 1 6  þ ¼ s s þ 1 s þ 3 sðs þ 1Þðs þ 3Þ

Y ðsÞ 1 ¼ ℜeðsÞ > 0 X ðsÞ ðs þ 1Þ2 Now, the output y1(t) ¼ (1 – et– tet)u(t) has the Laplace transform

Hence, H ðsÞ ¼

1 1 1 1 þ Y 1 ðsÞ ¼  ¼ s s þ 1 ð s þ 1Þ 2 s ð s þ 1Þ 2

ℜeðsÞ > 0

Hence, the Laplace transform of the corresponding input is X 1 ðsÞ ¼

Y 1 ðsÞ 1 ¼ H ðsÞ s

ℜeðsÞ > 0

The inverse Laplace transform of X1 (s) gives x1 ðt Þ ¼ uðt Þ:

4.6

Computation of Inverse Laplace Transform Using Partial Fraction Expansion

The inverse Laplace transform of a rational function X(s) can be easily computed by using the partial fraction expansion.

4.6 Computation of Inverse Laplace Transform Using Partial Fraction Expansion

4.6.1

195

Partial Fraction Expansion of X(s) with Simple Poles

Consider a rational Laplace transform X(s) of the form X ðsÞ ¼

N ðsÞ ðs  p1 Þðs  p2 Þ: . . . . . . ðs  pn Þ

ð4:73Þ

with the order of N(s) is less than the order of the denominator polynomial. The poles p1, p2, . . . .., pn are distinct. The rational Laplace transform X(s) can be expanded using partial fraction expansion as X ðsÞ ¼

k1 k2 kn þ þ  ðs  p1 Þ ðs  p2 Þ ð s  pn Þ

ð4:74Þ

The coefficients k1, k2, . . . ., kn are called the residues of the partial fraction expansion. The residues are computed as  ki ¼ ðs-pi Þ XðsÞs¼pi i ¼ 1, 2, . . . , n ð4:75Þ With the known values of the coefficients k1, k1, . . . ., kn inverse transform of each term can be determined depending on the location of each pole relative to the ROC.

4.6.2

Partial Fraction Expansion of X(s) with Multiple Poles

Consider a rational Laplace transform X(s) with repeated poles of the form X ðsÞ ¼

N ðsÞ ðs  p1 Þr ðs  p2 Þ: . . . . . . ðs  pn Þ

ð4:76Þ

with multiplicity r poles at s ¼ p1. The X(s) with multiple poles can be expanded as YðsÞ ¼

k11 k12 k1r k2 kn þ þ  þ  þ ðs  p1 Þ ðs  p1 Þ2 ðs  pn Þ ð s  p1 Þ r ð s  p2 Þ

ð4:77Þ

The coefficients k2, . . . ., kn can be computed using the residue formula used in Section 4.6.1. The residues k11, k12, . . ., klr are computed as  k1r ¼ ðs  p1 Þr XðsÞs¼pi ð4:78Þ k1ðr1Þ ¼

 1 d ½ðs  p1 Þr XðsÞs¼pi 1! ds

ð4:79Þ

196

4 Laplace Transforms

k1ðr2Þ ¼

 1 d2 ½ðs  p1 Þr XðsÞs¼pi 2 2! ds

ð4:80Þ

and so on. Example 4.16 Find the time function x(t) for each of the following Laplace transform X(s) sþ2 ℜeðsÞ > 3 s2 þ 7s þ 12 2 s þsþ1 (b) X ðsÞ ¼ 2 0 < ℜeðsÞ < 1 s ð s  1Þ s2  s þ 1 1 < ℜeðsÞ (c) X ðsÞ ¼ ð s þ 1Þ 2 sþ1 (d) X ðsÞ ¼ 2 ℜeðsÞ < 3 s þ 5s þ 6 (a) X ðsÞ ¼

sþ2 Solution (a) X ðsÞ ¼ s2 þ7sþ12

Using partial fraction expansion,

s2

sþ2 k1 k2 ¼ þ þ 7s þ 12 s þ 3 s þ 4 k1 þ k2 ¼ 1 4k1 þ 3k2 ¼ 2

Solving for k1 and k2, k1 ¼  1; k2¼2. sþ2 1 2 ¼ sþ3 þ sþ4 Thus, X ðsÞ ¼ s2 þ7sþ12 Using Table 4.2, we obtain x(t) ¼ e3tu(t) þ 2e4tu(t) (b) X ðsÞ ¼

s2  s þ 1 1 1 1  þ ¼ s 2 ð s  1Þ s  1 sðs  1Þ s2 ðs  1Þ

Using partial fraction expansion, 1 k1 k2 ¼ þ sðs  1Þ s s1 Solving for k1 and k2, k1 ¼ 1; k2 ¼ 1. 1 1 1 ¼ þ s ð s  1Þ s s1 Using partial fraction expansion, 1 k1 k11 k 12 ¼ þ þ 2 s 2 ð s  1Þ s  1 s s

4.6 Computation of Inverse Laplace Transform Using Partial Fraction Expansion Table 4.2 Elementary functions and their Laplace transforms

Signal δ(t) u(t)

Laplace transform 1 1 s 1 s

u(t)

est0

δ(t  t0) eαtu(t)

1 sþα 1 sþα 1 sn

eatu(t) ðn1Þ

t uð t Þ ðn  1Þ! t ðn1Þ αt e uð t Þ ðn  1Þ! sin ωt u(t) cos ωt u(t) αt

e

sin ωt u(t)

eαt cos ωt u(t)

197 ROC All s ℛe(s) > 0 ℛe(s) > 0 All s ℛe(s) >  α ℛe(s) <  α ℛe(s) > 0

1 ðsþαÞn

ℛe(s) >  α

ω s2 þω2 s s2 þω2 ω ðsþαÞ2 þω2 sþα ðsþαÞ2 þω2

ℜe(s) > 0 ℜe(s) > 0 ℜe(s) >  α ℜe(s) >  α

Solving for k1, k11, and k12 k1 ¼ 1; k11 ¼ 1; k12 ¼ 1: s2 ðs

1 1 1 1   2 ¼  1Þ s  1 s s

Thus, X ðsÞ ¼ ¼

s2  s þ 1 s 2 ð s  1Þ 1 1 1 1 1 1 1 1 1  þ þ  þ   ¼ s  1 sðs  1Þ s2 ðs  1Þ s  1 s s  1 s  1 s s2 1 1  ¼ s  1 s2

Using Table 4.2, we obtain xðt Þ ¼ et uðt Þ  tuðt Þ sþ1 3s (c) X ðsÞ ¼ sðsþ1 ¼ 1  ðsþ1 Þ2 Þ2 2

Using partial fraction expansion, 3s ð s þ 1Þ

2

¼

k11 k 12 þ s þ 1 ð s þ 1Þ 2

198

4 Laplace Transforms

Solving for k11 and k12, we get k11 ¼ 3; k12 ¼ 3 Hence, X ðsÞ ¼

s2  s þ 1 ð s þ 1Þ

2

¼1

3 3 þ s þ 1 ð s þ 1Þ 2

Using Table 4.2, we obtain xðt Þ ¼ δðt Þ  3et uðt Þ þ 3tet uðt Þ sþ1 (d) X ðsÞ ¼ s2 þ5sþ6

Using partial fraction expansion,

s2

sþ1 k1 k2 ¼ þ þ 5s þ 6 s þ 3 s þ 2 k1 þ k2 ¼ 1 2k1 þ 3k 2 ¼ 1

Solving for k1 and k2, k1 ¼ 2; k2 ¼ 1. sþ1 2 1 ¼ sþ3  sþ2 Thus, X ðsÞ ¼ s2 þ5sþ6

Using Table 4.2, we obtain xðt Þ ¼ 2e3t uðt Þ þ e2t uðt Þ Example 4.17 Find the inverse Laplace transform of the following X ðsÞ ¼ Solution X ðsÞ ¼ 1 þ

s2 þ s þ 1 s2  s þ 1

2s s2  s þ 1

¼1þ

2s pffiffi2 2 s  12 þ 23

1 s 1 2 ¼ 1 þ 2 pffiffi2 þ  pffiffi2 2 2 s  12 þ 23 s  12 þ 23

4.6 Computation of Inverse Laplace Transform Using Partial Fraction Expansion

199

Using Table 4.2, we obtain xðt Þ ¼ δðt Þ þ 2et=2 cos

pffiffiffi  pffiffiffi  3 3 2 t uðt Þ þ pffiffiffi et=2 sin t uð t Þ 2 2 3

Example 4.18 Determine x(t) for the following conditions if X(s) is given by X ðsÞ ¼

1 ð s þ 2Þ ð s þ 3Þ

(a) x(t) is right sided (b) x(t) is left sided (c) x(t) is both sided Solution Using partial fraction, X(s) can be written as (a) X ðsÞ ¼

1 1 1 ¼  ð s þ 2Þ ð s þ 3Þ ð s þ 2Þ ð s þ 3Þ

If x(t) is right sided, xðt Þ ¼ L

1



   1 1 1 L ¼ e2t uðt Þ  e3t uðt Þ ð s þ 2Þ ð s þ 3Þ

The ROC is to the right of the rightmost pole as shown in Figure 4.7(a). (b) If x(t) is left sided, xð t Þ ¼ L

1



    1 1 1 L ¼ e2t uðt Þ  e3t uðt Þ ð s þ 2Þ ð s þ 3Þ

The ROC is to the left of the leftmost pole as shown in Figure 4.7(b)

(a)

(b)

Im

Im

(c)

s plane

s plane s plane

-3

-2

Re

-3

-2

0

Re

-3

-2

0

Figure 4.7 (a) ROC of right-sided x(t) (b) ROC of left-sided x(t) (c) ROC of both-sided x(t)

Re

200

4 Laplace Transforms

(c) If x(t) is two sided,  ( e2t uðt Þ for right sided 1 L ¼ ð s þ 2Þ e2t uðt Þ for left sided (   e3t uðt Þ for right sided 1 L1 ¼ ð s þ 3Þ e3t uðt Þ for left sided 1



Hence, if x(t) is chosen as, xðt Þ ¼ e2t uðt Þ  e3t uðt Þ The ROC is as shown in Figure 4.7(c) Example 4.19 Determine x(t) first for the following and verify the initial and final value theorems. 1 (i) XðsÞ ¼ sþ4 (ii) XðsÞ ¼ ðsþ3sþ5 Þðsþ4Þ

Solution (i) xðtÞ ¼ L1

n

1 sþ4

o

¼ e4t uðt Þ

xð0þ Þ ¼ lim e4t ¼ 1: t!0

s ¼ lim xð0 Þ ¼ lim sX ðsÞ ¼ lim s!1 s!1 s þ 4 s!1

1

þ

4 1þ s 4t xð1Þ ¼ lim e ¼ 0:

¼

1 4 1þ 1

¼

1 ¼ 1: 1þ0

t!1

s ¼ lim xð1Þ ¼ lim sX ðsÞ ¼ lim s!0 s!0 s þ 4 s!0

(ii) xðtÞ ¼ L1

n

o

sþ5 ðsþ3Þðsþ4Þ

¼ L1

n

o

2 sþ3

1 4 1þ 0

 L1

n

¼

1 sþ4

o

1 4 1þ 0

¼

1 ¼ 0: 1þ1

¼ 2e3t uðt Þ  e4t uðt Þ

4.7 Inverse Laplace Transform by Partial Fraction Expansion Using MATLAB

201

 xð0þ Þ ¼ lim 2e3t  e4t ¼ 1:: t!0       sþ5 2s s þ xð0 Þ ¼ lim sX ðsÞ ¼ lim ¼ lim  lim s!1 s!1 ðs þ 3Þðs þ 4Þ s!1 s þ 3 s!1 s þ 4 2 1 ¼ lim  lim ¼ 2  1 ¼ 1: 3 s!1 4 s!1 1þ 1þ s s  xð1Þ ¼ lim 2e3t  e4t ¼ 0: t!1       sþ5 2s s ¼ lim  lim xð1Þ ¼ lim sX ðsÞ ¼ lim s!0 s!0 s!0 s þ 3 s!0 s þ 4 ð s þ 3Þ ð s þ 4Þ 2 1 ¼ lim  lim ¼ 0  0 ¼ 0: 3 s!0 4 s!0 1þ 1þ s s

4.7

Inverse Laplace Transform by Partial Fraction Expansion Using MATLAB

The MATLAB command residue can be used to find the inverse transform using the power series expansion. To find the partial fraction decomposition of Y(s), we must first enter the numerator polynomial coefficients and the denominator polynomial coefficients as vectors. The following MATLAB statement determines the residue (r), poles (p), and direct terms (k) of the partial fraction expansion of H(s). ½k; p; const ¼ residue ðN; DÞ; where N is the vector of the numerator polynomial coefficients in decreasing order and the vector D contains the denominator polynomial coefficients in decreasing order. Example 4.20 Find the inverse Laplace transform of the following using MATLAB X ðsÞ ¼

sþ1 s2 þ 5s þ 6

Solution The following MATLAB statements are used to find the Laplace transform of given X(s) N ¼ ½ 1 1 ; % coefficients of the numerator polynomial in decreasing order D ¼ ½ 1 5 6 ; % coefficients of the denominator polynomial in decreasing order [k, p, const] ¼ residue (N and D); % computes residues, poles, and constants Execution of the above statements gives the following output

202

4 Laplace Transforms

k¼ 2.0000 1.0000 p¼ 3.0000 2.0000 Thus, the partial fraction decomposition of X(s) is X ðsÞ ¼

2 1  sþ3 sþ2

After getting the partial fraction expansion of X(s), the following MATLAB statements are used to obtain x(t), that is, the inverse Laplace transform of X(s). syms s t X=2/(s+3)-1/(s+2); ilaplace(X)

Execution of the above three MATLAB statements gives xðt Þ ¼ 2e3t  e2t

4.8 4.8.1

Analysis of Continuous-Time LTI Systems Using the Laplace Transform Transfer Function

It was stated in Chapter 2 that a continuous time LTI system can be completely characterized by its impulse response h(t). The output signal y(t) of a LTI system and the input signal x(t) are related by convolution as y ð t Þ ¼ hð t Þ ∗ x ð t Þ

ð4:81Þ

By using the convolution property, we get Y ðsÞ ¼ H ðsÞX ðsÞ

ð4:82Þ

indicating the Laplace transform of the output signal y(t) is the product of the Laplace transforms of the impulse response h(t) and the input signal x(t). The transform H(s) is called the transfer function or the system function and expressed as

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

203

Figure 4.8 Sallen-Key low-pass filter circuit

H ðsÞ ¼

Y ðsÞ X ðsÞ

ð4:83Þ

The roots of the denominator of the transfer function are called poles. The roots of the numerator are called zeros. The places where the transfer function is infinite (the poles) determine the region of convergence. Example 4.21 Obtain the system function of the following Sallen-Key low-pass filter circuit. Solution For the circuit shown in Figure 4.8, the following relations can be established: V i  V 1 ¼ r 1 I r1 ; V 1  V 2 ¼ r 2 I r2 ; I c2 ¼ sc2 V 2 ; I c1 ¼ sc1 ðV 1  V 0 Þ; I r 1 ¼ I r 2 þ I c1 ; I r 2 ¼ I c2 ; V 2 ¼ V 0 Rewriting the current node equations, we get Vi  V1 V1  V0 ¼ þ sc1 ðV 1  V 0 Þ r1 r2 Which can be rewritten as V i r 2 ¼ ðr 1 þ r 2 þ r 1 r 2 sc1 ÞV 1  ðr 1 þ r 1 r 2 sc1 ÞV 0 V1  V0 ¼ sc2 V 0 ; r2 V 1 ¼ ð1 þ r 2 sc2 ÞV 0 Substituting the above Eq. for V 1 , the input-output relation can be written as V i r 2 ¼ ½ðr 1 þ r 2 þ r 1 r 2 sc1 Þð1 þ r 2 sc2 Þ  r 1  r 1 r 2 sc1 V 0  

r1 r1 Vi ¼ 1 þ þ r 1 sc1 ð1 þ r 2 sc2 Þ   r 1 sc1 V 0 r2 r2

204

4 Laplace Transforms

Thus, V 0 ðsÞ 1 

¼  r1 r1 V i ðsÞ 1 þ þ r 1 sc1 ð1 þ r 2 sc2 Þ   r 1 sc1 r2 r2 1 ¼ 1 þ c2 ðr 1 þ r 2 Þs þ s2 c2 c1 r 2 r 1 Hence, the system function is given by 1 r 1 r 2 c1 c2 H ðsÞ ¼ r þ r2 1 1 s2 þ sþ r 1 r 2 c1 c2 r 1 r 2 c1

4.8.2

Stability and Causality

Stabile LTI System A continuous-time LTI system is stable if and only if the impulse response is absolutely integrable, that is, ð1 1

j hðt Þ j dt < 1:

ð4:84Þ

The Laplace transform of the impulse response is known as the system function, which can be written as H ðsÞ ¼

ð1 1

hðt Þest dt

ð4:85Þ

A continuous-time LTI system is stable if and only if the transfer function has ROC that includes the imaginary axis (the line in complex where the real part is zero). Causal LTI System A continuous time LTI system is causal if its output y(t) depends only on the current and past input x(t) but not the future input. Hence, h(t) ¼ 0 for t < 0. For causal system, the system function can be written as HðsÞ ¼

ð1 0

hðtÞest dt

ð4:86Þ

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

205

Im s plane

Re

Figure 4.9 ROC of a causal LTI system

If s ¼ σ + jΩ and h(t)eσt being absolutely integrable for convergence of H(s) leads to the following convergence condition, ð1

j hðt Þeσt j dt < 1

ð4:87Þ

0

Any large value of σ satisfies the above equation. Thus, the ROC is the region to the right of a vertical line that passes through the point ℜe(s) ¼ σ as shown in Figure 4.9. In particular, if H(s) is rational, H ðsÞ ¼ NDððssÞÞ, then the system is causal if and only if its ROC is the right-sided half plane to the right of the rightmost pole and the order of numerator N(s) is no greater than that of the denominator D(s), so that the ROC is a right-sided plane without any poles (even at s ! 1). Stable and Causal LTI System As the ROC of a causal system is to the right of the rightmost pole and for a stable system, the rightmost pole should be in the left half of the s-plane and should include the jΩ axis; all the poles of a system should lie in the left half of the s-plane (the real parts of all poles are negative, ℜe(sp) < 0 for all sp) for a system to be causal and stable as shown in Figure 4.10. Stable and Causal Inverse LTI System For a causal stable system, the poles must lie in the left half of the s-plane. But it is known that the poles of the inverse system are zeros of the original system. Therefore, the zeros of the original system should be in the left half of the s-plane.

206

4 Laplace Transforms

Im s plane

Re

Figure 4.10 ROC of a stable and causal LTI system

Example 4.22 Consider a LTI system with the system function H ðsÞ ¼

s1 s2  s  6

Show the pole-zero locations of the system and ROCs for the following: (a) the system is causal (b) the system is stable, noncausal (c) the system is neither causal nor stable Solution (a) The system function can be rewritten as H ðsÞ ¼

s1 ð s þ 2Þ ð s  3Þ

Since the ROC of a causal system is to the right of the rightmost pole, the pole-zero plot and ROC of the system are shown in Figure 4.11. (b) For a stable system, the ROC should include the imaginary axis. The pole-zero plot and the ROC are shown in Figure 4.12. (c) Since the system is neither causal nor stable, the ROC should not include the imaginary axis and not to the right of the rightmost pole. Hence, the pole-zero plot and the ROC are shown in Figure 4.13.

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

207

Im s-plane

Re -2

1

3

Figure 4.11 Pole-zero plot and ROC of the causal system

Im s-plane

Re -2

1

3

Figure 4.12 Pole-zero plot and ROC of the noncausal, stable system

4.8.3

LTI Systems Characterized by Linear Constant Coefficient Differential Equations

The system function for a system characterized by a linear constant coefficient equation can be obtained by exploiting the properties of the Laplace transforms. The Laplace transform transforms a differential equation into an algebraic equation in the s-domain making it easy to find the time-domain solution of the differential equation.

208

4 Laplace Transforms

Im

s-plane

Re -2

1

3

Figure 4.13 Pole-zero plot and ROC of the system neither causal nor stable

Consider a general linear constant coefficient differential equation of the form an

dn y dðn1Þ y d2 y dy þ a þ   : þ a þ a1 þ a0 y n1 2 n 2 ð n1 Þ dt dt dt dt dm x dðm1Þ x d2 x dx ¼ bm m þ bm1 ðm1Þ þ   : þ b2 2 þ b1 þ b0 x dt dt dt dt

ð4:88Þ

Taking the Laplace transform of both sides of equation repeated use of the differentiation property and linearity property, we obtain ðan sn þ an1 sðn1Þ þ   : þ a2 s2 þ a1 s þ a0 ÞYðsÞ ¼ ðbm sm þ bm1 sðm1Þ þ   : þ b2 s2 þ b1 s þ b0 ÞXðsÞ

ð4:89Þ

Thus, the system function is given by HðsÞ ¼

YðsÞ bm sm þ bm1 sðm1Þ þ   : þ b2 s2 þ b1 s þ b0 ¼ XðsÞ an sn þ an1 sðn1Þ þ   : þ a2 s2 þ a1 s þ a0

ð4:90Þ

The system function is rational for a system characterized by a differential equation with the roots of the numerator polynomial as zeros and the roots of the denominator polynomial as poles. Eq. (4.90) does not specify any ROC since a differential equation by itself does not constrain any region of convergence. However, with the additional knowledge of stability and causality, the ROC can be specified.

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

209

Example 4.23 Consider a continuous LTI system described by the following dif2

ferential equation ddt2y  dy dt  6y ¼ x. (a) Determine the system function (b) Determine h(t) for each of the following: (i) the system is stable, noncausal (ii) the system is causal (iii) the system is neither causal nor stable Solution Taking the Laplace transform of both sides of the given differential equation, we obtain s2 YðsÞ  sYðsÞ  6YðsÞ ¼ XðsÞ The above relation can be rewritten as 

s2  s  6 YðsÞ ¼ XðsÞ

Now, the system function is given by H ðsÞ ¼

Y ðsÞ 1 ¼ X ðsÞ s2  s  6

The pole-zero plot for the system function is shown in Figure 4.14. (b) The partial fraction expansion of H(s) yields H ðsÞ ¼

1 1  5ð s  3Þ 5ð s þ 2Þ

Im

Re -2

Figure 4.14 Pole-zero plot of the system function

3

210

4 Laplace Transforms

(i) For H(s) to be stable, noncausal, the ROC is 2 < ℜe(s) < 3. Hence, hðt Þ ¼ 15 e3t uðt Þ  15 e2t uðt Þ (ii) For the system to be causal, the ROC is ℜe(s) > 3. Therefore, hðt Þ ¼ 15 e3t uðt Þ  15 e2t uðt Þ. (iii) For the system to be neither causal nor stable, the ROC is ℜe(s) <  2. Hence, hðt Þ ¼ 15 e3t uðt Þ þ 15 e2t uðt Þ

4.8.4

Solution of linear Differential Equations Using Laplace Transform

The stepwise procedure to solve a linear differential equation is as follows: Step 1: Take Laplace transform both sides of the equation. Step 2: Simplify the algebraic equation obtained for Y(s) in the s-domain. Step 3: Find the inverse transform of Y(s) to obtain y(t), the solution of the differential equation. Example 4.24 Find the solution of the following differential equation using Laplace transform: d2 y dy  5 þ 6y ¼ 0 2 dt dt

yð0Þ ¼ 2, y_ ð0Þ ¼ 1:

Solution Step 1: Laplace transform both sides of given differential equation is s2  2s  1  5ðsYðsÞ  2Þ þ 6YðsÞ ¼ 0 Step 2: Simplifying the expression for Y(s), ðs2  5s þ 6ÞYðsÞ  2s  1 þ 10 ¼ 0 Y ðsÞ ¼ YðsÞ ¼

ðs2

2s  9  5s þ 6Þ

2s  9 ð s  3Þ ð s  2Þ

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

211

Step 3: Expanding Y(S) using partial fraction expansion, YðsÞ ¼ ¼

k1 k2 þ ð s  3Þ ð s  2Þ k1 ð s  2Þ þ k2 ð s  3Þ ð s  3Þ ð s  2Þ

Comparing the numerator polynomial with the numerator polynomial of Y(s) of step 2, we get k1 þ k2 ¼ 2 2k1  3k2 ¼ 9 Solving for k1 and k2, k1 ¼ 3, k2 ¼ 5: Thus, YðsÞ ¼

3 5 þ ð s  3Þ ð s  2Þ

Inverse transform of Y(s) gives the solution of the differential equation as yðt Þ ¼ 3e3t þ 5e2t Example 4.25 Find the solution of the following differential equation using Laplace transform dy þ y ¼ 2tet dt

yð0Þ ¼ 2

Solution Step 1: Laplace transform both sides of given differential equation is sYðsÞ  ð2Þ þ YðsÞ ¼ Step 2: Simplifying the expression for Y(s),

2 ð s þ 1Þ 2

212

4 Laplace Transforms

ðs þ 1ÞYðsÞ þ 2 ¼ ðs þ 1ÞYðsÞ ¼ Y ðsÞ ¼ Y ðsÞ ¼ ¼

2 ð s þ 1Þ 2 2 ð s þ 1Þ 2

2

2 ð s þ 1Þ ð s þ 1Þ 2 ð s þ 1Þ

3



2



2 ð s þ 1Þ

2 ðs þ 1 Þ

2s2  4s ð s þ 1Þ 3

Step 3: Expanding Y(S) using partial fraction expansion, YðsÞ ¼ ¼

k11 k12 k13 þ þ 2 ð s þ 1Þ ð s þ 1Þ ð s þ 1Þ 3 k11 ðs2 þ 2s þ 1Þ þ k12 ðs þ 1Þ þ k13 ðs þ 1Þ3

Comparing the numerator polynomial with the numerator polynomial of Y(s) of step 2, we get k11 þ k12 þ k13 ¼ 0 2k11 þ k12 ¼ 4 k11 ¼ 2 Solving for k11, k12, and k13, k11 ¼ 2, k12 ¼ 0, and k13 ¼ 2. Thus, YðsÞ ¼

2 2 þ ð s þ 1Þ ð s þ 1Þ 3

Inverse transform of Y(s) gives the solution of the differential equation as yðt Þ ¼ 2et þ t 2 et

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

213

Figure 4.15 Series RLC circuit

Example 4.26 Consider the following RLC circuit with R ¼ 5 ohms, L ¼ 1h, and C ¼ 0.25 f. (a) Determine the differential equation relating V in and V c . (b) Obtain V c ðt Þ using Laplace transform for V in ðt Þ ¼ et uðt Þ with V c ð0Þ ¼ 1, V_ c ð0Þ ¼ 2 (Figure 4.15). Solution (a) By applying Kirchhoff’s voltage law, we can arrive at the following differential equation: di 1 L þ Ri þ dt C

ð idt ¼ vin

It is known that i ¼ C dVdtc , hence the above differential equation can be rewritten as LC

d2 vc dvc þ vc ¼ vin þ RC dt dt2

For given values of R, L, and C, the differential equation becomes d2 vc dvc þ 4vc ¼ 4vin þ5 dt dt2 2

(b) For given vin, ddtv2c þ 5dvdtc þ 4vc ¼ 4et uðt Þ Laplace transform both sides of given differential equation is 

s2 Vc ðsÞ  s  2 þ 5ðsVc ðsÞ  1Þ þ 4Vc ðsÞ ¼

Simplifying the expression for vc(s)

4 sþ1

214

4 Laplace Transforms

2 s Vc ðsÞ  s  2 þ 5ðsVc ðsÞ  1Þ þ 4vc ðsÞ ¼ ðs2 þ 5s þ 4ÞVc ðsÞ  s  2  5 ¼ ðs2 þ 5s þ 4ÞVc ðsÞ ¼ Vc ðsÞ ¼

4 sþ1

4 sþ1

4 þsþ7 sþ1

s2 þ 8s þ 11 ðs2 þ 5s þ 4Þ ðs þ 1Þ

Vc ðsÞ ¼

s2 þ 8s þ 11 ð s þ 1Þ 2 ð s þ 4Þ

Expanding vc(s) using partial fraction expansion Vc ðsÞ ¼

k11 k12 k2 þ þ 2 ð s þ 1Þ ð s þ 1Þ ð s þ 4Þ

Solving for k11, k12, and k2, we obtain 4 5 k11 ¼ 14 9 , k12 ¼ 3, k2 ¼ 9. Thus, Vc ðsÞ ¼

14 4 5 þ  9ð s þ 1Þ 3ð s þ 1Þ 2 9ð s þ 4Þ

Inverse transform of Vc(s) gives V c ðtÞ as vc ðtÞ ¼

14 t 4 t 5 4t e þ te  e 9 3 9

Example 4.27 Consider the following parallel RLC circuit with R¼1 ohm, L¼1 h. (a) Determine the differential equation relating Is and IL. (b) Obtain zero-state response for IL(t) using Laplace transform for Is(t) ¼ e3tu(t). (c) Obtain zero-input response for IL(t) using Laplace transform with IL(0) ¼ 1 (Figure 4.16).

Figure 4.16 Parallel RLC circuit

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

215

Solution (a) The Is and IL are related by the following differential equation dIL þ IL ¼ IS dt (b) For the given Is, dIL þ IL ¼ e3t uðtÞ dt Applying the unilateral Laplace transform to the above differential equation, we obtain sIL ðsÞ  IL ð0Þ þ IL ðsÞ ¼

1 sþ3

1 Since IL(0) ¼ 0 for zero state, ðs þ 1ÞIL ðsÞ ¼ sþ3

IL ðsÞ ¼

1 ðs þ 1Þðs þ 3Þ

By using partial fraction expansion, IL(s) can be expanded as IL ðsÞ ¼

1 1  2ð s þ 1Þ 2ð s þ 3Þ

The inverse unilateral Laplace transform gives 1 1 IL ðtÞ ¼ et uðt Þ  e3t uðt Þ 2 2 (c) For the zero-input response, Is(t) ¼ 0 and given that IL(0) ¼ 1, we have to find the solution of the following differential equation for zero-input response dIL þ IL ¼ 0 dt

IL ð0Þ ¼ 1

The unilateral Laplace transform of the above differential equation is sIL ðsÞ  1 þ IL ðsÞ ¼ 0 Thus, IL ðsÞ ¼

1 sþ1

216

4 Laplace Transforms

The inverse transform of IL(s) is zero-input response given by IL ðtÞ ¼ et uðt Þ

4.8.5

Solution of Linear Differential Equations Using Laplace Transform and MATLAB

Example 4.28 Find the solution of linear differential equation considered in Example 4.24 using MATLAB Solution The following MATLAB statements are used to find the solution of the differential equation considered in Example 4.24 syms s t Y Y1 = s*Y - 2;%Laplace transform

of

with y(0)=2

Y2 = s*Y1 - 1; %Laplace transform of =1 with Sol = solve(Y2 - 5*Y1 + 6*Y, Y);%Y(s) y = ilaplace(Sol,s,t);%inverse Laplace transform of Y(s)

Execution of the above MATLAB statements gives the solution of the differential equation as yðt Þ ¼ 3e3t þ 5e2t Example 4.29 Find the solution of linear differential equation considered in Example 4.25 using MATLAB Solution The following MATLAB statements are used to find the solution of the differential equation considered in Example 4.25

syms s t Y f = 2*t*exp(-t);%input signal F = laplace(f,t,s);% finds Laplace transform of input Y1 = s*Y + 2;% Laplace transform of with y(0)=-2 Sol = solve(Y1 + Y-F, Y);%Y(s) y = ilaplace(Sol,s,t);% inverse of Y(s)

Execution of the above MATLAB statements gives the solution of the differential equation as yðt Þ ¼ 2et þ t 2 et

4.8 Analysis of Continuous-Time LTI Systems Using the Laplace Transform

4.8.6

217

System Function for Interconnections of LTI Systems

Series Combination of Two LTI Systems Impulse response of the series combination of two LTI systems is hðt Þ ¼ h1 ðt Þ ∗ h2 ðt Þ

ð4:91Þ

and from convolution property of the Laplace transform, the associated system function is (Figure 4.17) H ðsÞ ¼ H 1 ðsÞH 2 ðsÞ

ð4:92Þ

Parallel Combination of Two LTI Systems Impulse response of the parallel combination of two LTI systems is hð t Þ ¼ h1 ð t Þ þ h2 ð t Þ

ð4:93Þ

and from linearity property of the Laplace transform, the associated system function is (Figure 4.18) H ðsÞ ¼ H 1 ðsÞ þ H 2 ðsÞ

Figure 4.17 Series combination of two LTI systems

Figure 4.18 Parallel combination of two LTI systems

ð4:94Þ

218

4.9

4 Laplace Transforms

Block-Diagram Representation of System Functions in the S-Domain

Consider the system function HðsÞ ¼

YðsÞ bm sn þ bm1 sðn1Þ þ   : þ b2 s2 þ b1 s þ b0 ¼ X ðsÞ an sn þ an1 sðn1Þ þ   : þ a2 s2 þ a1 s þ a0

ð4:95Þ

Let us define the following basic elements for addition, multiplication, differentiation, and integration in the s-domain (Figure 4.19) The block-diagram representation of the above system function can be obtained as the interconnection of these basic elements similar to the block-diagram representation of differential equations in the time domain carried out in Section 2.6 of Chapter 2. The block-diagram representation of the system function is shown in Figure 4.20. Example 4.30 Is the system represented by the following block-diagram stable? (Figure 4.21) Solution Let the signal at the bottom node of the block diagram be denoted by E(s). Then we have the following relations

Figure 4.19 Blockdiagram representation basic elements (a) adder, (b) multiplier, (c) differentiator, (d) integrator

(a)

(b)

Differentiation (c)

Integration (d)

4.9 Block-Diagram Representation of System Functions in the S-Domain

X(s)

bm

1/an

-an-1

1/s

219

Y(s)

bm-1

1/s -a1

b1

1/s -a0

b0

Figure 4.20 Block-diagram representation of the system function

Figure 4.21 Block-diagram representation of a second-order system function

X ðsÞ  2sEðsÞ  EðsÞ ¼ s2 EðsÞ s2 E ðsÞ  sE ðsÞ  6E ðsÞ ¼ Y ðsÞ Eliminating the auxiliary signal E(s), we get H ðsÞ ¼

Y ðsÞ s2  s  6 ¼ 2 X ðsÞ s þ 2s þ 1

From the system function H(s) found in the previous part, we see that the poles of the system are at s ¼ 1. Since the system is given to be causal, and the rightmost pole of the system is left of the imaginary axis, the system is stable.

220

4.10

4 Laplace Transforms

Solution of State-Space Equations Using Laplace Transform

For convenience, the state-space equations from Chapter 2 are repeated here: X_ ðt Þ ¼ A X ðt Þ þ b℧ðt Þ

ð4:96Þ

yðt Þ ¼ cX ðt Þ

ð4:97Þ

Taking Laplace transform both sides of Eq. (4.96), we obtain SX ðSÞ  X ð0Þ ¼ AX ðsÞ þ b℧ðSÞ



ð4:98Þ

Eq. (4.98) can be rewritten as ½SI  AX ðsÞ ¼ X ð0Þ þ b℧ðSÞ

ð4:99Þ

where I is the identity matrix. From Eq. (4.99), we get XðsÞ ¼ ½SI  A1 ½X ð0Þ þ b℧ðSÞ

ð4:100aÞ

X ðsÞ ¼ ½SI  A1 X ð0Þ þ ½SI  A1 b℧ðSÞ

ð4:100bÞ

Taking inverse Laplace transform both sides of Eq. (4.100b) yields h i h i L1 ½X ðsÞ ¼ L1 ½SI  A1 Xð0Þ þ L1 ½SI  A1 b℧ðSÞ h i L1 ½SI  A1 Xð0Þ ¼ eAt X ð0Þ

ð4:101Þ ð4:102Þ

By using convolution theorem, we obtain L

1

h

i

1

½SI  A b℧ðSÞ ¼

ðt

eAðtτÞ b℧ðτÞ dτ

ð4:103Þ

0

Thus, 1

L ½ X ð s Þ  ¼ X ð t Þ ¼ e X ð 0Þ þ

ðt

At

eAðtτÞ b℧ðτÞ dτ

ð4:104Þ

0

Example 4.31 Consider the electrical circuit given in Example 2.36. Find V c ðt Þ if V s ðt Þ ¼ uðtÞ under an initially relaxed condition.

4.10

Solution of State-Space Equations Using Laplace Transform

""

Solution ½sI  A ¼

s

0

0

s

#

" 

½sI  A1 ¼

1 1

##

"

¼ 1 1 " sþ1 1 2

ðs þ 1Þ þ 1

eAt ¼ L1 ½½sI  A1  ¼ et

221

sþ1 1 1

1 #

#

sþ1

1 s þ 1 " # cost sint

sint cost " # 0 Ðt XðtÞ ¼ eAt Xð0Þ þ 0 eAðtτÞ V s ðτÞdτ 1

Since the circuit is initially relaxed, eAtX(0) ¼ 0. Therefore, ðt

X ðt Þ ¼

e

AðtτÞ

0



0 V ðτÞdτ 1 s

Since V s ðt Þ ¼ uðt Þ X ðt Þ ¼

ðt "

eðtτÞ cos ðt  τÞ

#" # 0

eðtτÞ sin ðt  τÞ

dτ eðtτÞ sin ðt  τÞ eðtτÞ cos ðt  τÞ 1 # ð t " ðtτÞ e sin ðt  τÞ ¼ dτ 0 eðtτÞ cos ðt  τÞ Ðt V c ðt Þ ¼ x2 ðt Þ ¼ 0 eðtτÞ cos ðt  τÞ dτ ðt ðt eðtτÞ cos ðt  τÞdτ ¼ eðtτÞ ðcost cos τ þ sint sin τÞ dτ 0

0

0

¼

ðt

e

ðtτÞ

cost cos τ dτ þ

0

¼

ðt

et cost eτ cos τ dτ þ

0

ðt

ðt

eðtτÞ sint sin τ dτ

0

et sint eτ sin τ dτ

0

integration by parts gives ðt 0



ðt et cost eτ cos τ dτ ¼ et cost eτ cos τj0t þ eτ sin τ dτ 0

  Ðt ¼ et cost et cost  1 þ eτ sin τj0t  0 eτ cos τ dτ

This equation can be written as

222

4 Laplace Transforms

ðt

et cost eτ cos τ dτ ¼ cos 2 t þ sint cost  et cost ð ðt ð cos 2 t þ sint cost  et cost Þ t t e sint eτ sin τ dτ et cost eτ cos τ dτ ¼ 2 0 0

ðt t t τ τ ¼ e sint e sin τj0  e cos τ dτ 2

0

0



¼ et sint et sint  eτ cos τj0  t

ðt

eτ sin τ dτ



0

which can be written as 2

Ðt

Ðt 0

0

et sint eτ sin τ dτ ¼ sin 2 t  sint cost þ et sint

et sint eτ sin τ dτ ¼

ð sin 2 t  sint cost þ et sint Þ 2

Hence, ð cos 2 t þ sint cost  et cost Þ 2 2 ð sin t  sint cost þ et sint Þ þ 2 1 ¼ ð1 þ et sint  et cost Þ, t > 0 2

V c ðt Þ ¼ x2 ðt Þ ¼

4.11

Problems

1. Find the Laplace transforms of the following: (i) x(t) ¼ e2tu(t) þ e3tu(t) (ii) x(t) ¼ etu(t) þ e3tu(t) 2. Find the Laplace transform of



2 sin 2 t hint : sint t t cos 4t cos 5t . t

1

¼2



ð1 cos 2t Þ t

.

3. Find the Laplace transform of 4. Show that the ROC for the Laplace transform of a noncausal signal is the region to the left of a vertical line in the s-plane. 5. Find Laplace transform of a periodic signal with period T, that is, x(t+T) ¼ x(t). 6. By first determining x(t), verify the final value theorem for the following with comment XðsÞ ¼

s2

1 þ1

4.11

Problems

223

7. The transfer function of an LTI system is HðsÞ ¼

1 sþα

Find the impulse response and region of convergence and the value of α for the system to be causal and stable. 8. Consider a continuous LTI system described by the following differential equation d3 y d2 y dy þ 6 þ 11 þ 6y ¼ x 3 2 dt dt dt (a) Determine the system function (b) Determine h(t) for each of the following: (i) The system is causal (ii) The system is stable (iii) The system is neither causal nor stable 9. Find the solution of the following differential equation using Laplace transform d2 y dy  2 þ 2y ¼ cos t dt dt 2

yð0Þ ¼ 1, y_ ð0Þ ¼ 0:

10. Consider the following RC circuit with R¼2 ohms, C¼0.5 f. (a) Determine the differential equation relating V i and V c . (b) Obtain V c ðt Þ using Laplace transform for V i ðt Þ ¼ e3t uðt Þ with V c ð0Þ ¼ 1.

11. Consider the following RLC circuit. Obtain y(t) using Laplace transform for x(t)¼u(t).

224

4 Laplace Transforms

12. Determine the differential equation characterizing the system represented by the following block diagram

Y(s) -5 X(s)

1/s

6 1/s -12

-7

13. Consider the following state-space representation of a system. 2 3Determine the 3 system output y(t) with the initial state condition X ½0 ¼ 4 3 5. 47 2

2 3 0 6 7 6 76 7 6 7 t 6 x_ 2 ðt Þ 7 ¼ 6 0 6 7 6 7 0 1 7 4 5 4 54 x2 ðt Þ 5 þ 4 0 5e x_ 3 ðt Þ x 3 ðt Þ 1 3 3 1 2 3 x1 ð t Þ 6 7 7 yðt Þ ¼ ½ 1 0 0 6 4 x2 ð t Þ 5 x3 ð t Þ x_ 1 ðt Þ

3

2

0

1

0

32

x 1 ðt Þ

3

Further Reading

4.12

225

MATLAB Exercises

1. Write a MATLAB program for magnitude response of Sallen-Key low-pass filter. 2. Verify the solution of problem 8 using MATLAB. 3. Verify the solution of problem 9 using MATLAB.

Further Reading 1. Doetsch, G.: Introduction to the theory and applications of the Laplace transformation with a table of Laplace transformations. Springer, New York (1974) 2. LePage, W.R.: Complex variables and the Laplace transforms for engineers. McGraw-Hill, New York (1961) 3. Oppenheim, A.V., Willsky, A.S.: Signals and systems. Prentice-Hall, Englewood Cliffs (1983) 4. Hsu, H.: Signals and systems, 2nd edn. Schaum’s Outlines, Mc Graw Hill (2011) 5. Kailath, T.: Linear systems. Prentice-Hall, Englewood Cliffs (1980) 6. Zadeh, L., Desoer, C.: Linear system theory. McGraw-Hill, New York (1963)

Chapter 5

Analog Filters

Filtering is an important aspect of signal processing. It allows desired frequency components of a signal to pass through the system without distortion and suppresses the undesired frequency components. One of the most important steps in the design of a filter is to obtain a realizable transfer function H(s), satisfying the given frequency response specifications. In this chapter, the design of analog low-pass filters is first described. Second, frequency transformations for transforming analog low-pass filter into band-pass, band-stop, or high-pass analog filters are considered. The design of analog filters is illustrated with numerical examples. Further, the design of analog filters using MATLAB is demonstrated with a number of examples. Also, the design of special filters by pole and zero placement is illustrated with examples.

5.1

Ideal Analog Filters

An ideal filter passes a signal for one set of frequencies and completely rejects for the rest of the frequencies. Low-Pass Filter The frequency response of an ideal analog low-pass filter HLP (Ω) that passes a signal for Ω in the range –Ωc
Ωc can be expressed by
H LP ðΩÞ ¼

1, 0,

jΩj
Ωc jΩj > Ωc

ð5:1Þ

The frequency Ωc is called the cutoff frequency. The impulse response of the ideal low-pass filter corresponds to the inverse Fourier transform of the frequency response shown in Figure 5.1. © Springer International Publishing AG, part of Springer Nature 2018 K. D. Rao, Signals and Systems, https://doi.org/10.1007/978-3-319-68675-2_5

227

228

5 Analog Filters

1

Stop band

Pass band

Stop band

Figure 5.1 Frequency response of ideal low-pass filter

Hence, hð t Þ ¼

1 2π

ð Ωc Ωc

ejΩc t dΩ ¼

sin Ωc t πt

ð5:2Þ

sinc function can be defined as sincðxÞ ¼

sin πx πx

ð5:3Þ

therefore from sinc function we can express Eq. (5.2) as   sin Ωc t Ωc Ωc t ¼ sinc πt π π

ð5:4Þ

Thus, hlp ðt Þ ¼

  Ωc Ωc t sinc π π

ð5:5Þ

The impulse response for Ωc ¼ 200 Hz is shown in Figure 5.2. The filter bandwidth is proportional to Ωc and the width of the main lobe is proportional to Ω1c . The impulse response becomes narrow with increase in the bandwidth. High-Pass Filter The following system (Figure 5.3) is generally used to obtain high-pass filter from a low-pass filter. The frequency response of an ideal analog high-pass filter HHP (Ω) that passes a signal for |Ω| > Ωc can be expressed by
H HP ðΩÞ ¼

0, 1,

j Ωj
Ωc j Ωj > Ωc

and is shown in Figure 5.4. The frequency Ωc is called the cutoff frequency

ð5:6Þ

5.1 Ideal Analog Filters Figure 5.2 Impulse response for Ωc ¼ 200 Hz

Figure 5.3 System to obtain a high-pass filter from low-pass filter

Figure 5.4 Frequency response of ideal high-pass filter

229

230

5 Analog Filters

From Figure 5.3, the frequency response of the ideal high-pass filter can also be expressed as H HP ðΩÞ ¼ 1  H LP ðΩÞ

ð5:7Þ

Therefore, the impulse response of an ideal high-pass filter is given by the inverse Fourier transform of Eq. (5.7). Hence, the impulse response of the ideal high pass filter is given by   Ωc Ωc t hhp ðt Þ ¼ δðt Þ  sinc π π

ð5:8Þ

Band-Pass Filter The frequency response of band-pass filter can be expressed by
H BP ðΩÞ ¼

1, Ωc1
jΩj
Ωc2 0, jΩj < Ωc1 and jΩj > Ωc2

ð5:9Þ

which is shown in Figure 5.5. From Figure 5.5, the frequency response of the ideal band-pass filter can be expressed as 1 hBP ðt Þ ¼ 2π

ð Ωc1 Ωc2

1 e dΩ þ 2π jΩt

ð Ωc2

ejΩt dΩ

Ωc1

  1 ejΩt  Ωc1 1 ejΩt  Ωc2 hBP ðtÞ ¼ þ   2π jt  Ωc2 2π jt  Ωc1 ¼

  1  jΩc1 t 1  jΩc2 t e e  ejΩc2 t þ  ejΩc1 t j2πt j2πt

¼

½sin Ωc2 t  sin Ωc1 t tπ

Thus, the impulse response of an ideal band-pass filter is

Figure 5.5 Frequency response of ideal band-pass filter

ð5:10Þ

5.1 Ideal Analog Filters

231

    Ωc2 Ωc2 t Ωc1 Ωc1 t sinc sinc hBP ðt Þ ¼  π π π π

ð5:11Þ

Band-Stop Filter The band-stop filter can be realized as a parallel combination of low-pass filter with cutoff frequency Ωc1 and high-pass filter cutoff frequency Ωc2. The frequency response of band-stop filter can be expressed by ( H BS ðΩÞ ¼

1, jΩj
Ωc1 and jΩj  Ωc2 0, Ωc1 < jΩj < Ωc2

ð5:12Þ

which is shown in Figure 5.7. From Figure 5.6, the frequency response of the ideal band-stop filter can also be expressed as H BS ðΩÞ ¼ H LP ðΩÞ þ H HP ðΩÞ

ð5:13Þ

Therefore, the impulse response of an ideal band-stop filter is given by the inverse Fourier transform of Eq. (5.13). Hence, the impulse response of the ideal band-stop filter is given by hBS ðt Þ ¼ δðt Þ þ

    Ωc1 Ωc1 t Ωc2 Ωc2 t sinc sinc  : π π π π

Figure 5.6 System with summation of high-pass filter and low-pass filter

Figure 5.7 Frequency response of ideal band

ð5:14Þ

232

5.2

5 Analog Filters

Practical Analog Low-Pass Filter Design

A number of approximation techniques for the design of analog low-pass filters are well established in the literature. The design of analog low-pass filter using Butterworth, Chebyshev I, Chebyshev II (inverse Chebyshev), and elliptic approximations is discussed in this section.

5.2.1

Filter Specifications

The specifications for an analog low-pass filter with tolerances are depicted in Figure 5.8, where Ωp - Passband edge frequency Ωs - Stopband edge frequency δp- Peak ripple value in the passband δs - Peak ripple value in the stopband Peak passband ripple in dB ¼ αp ¼ 20 log10(1 – δp) dB Minimum stopband ripple in dB ¼ αs ¼ 20 log10 (δs) dB Peak ripple value in passband δp ¼ 1  10αp =20 Peak ripple value in stopband δs ¼ 10αs =20

|H (jW)|

Transition band

1+ 1-

Pass band

Figure 5.8 Specifications of a low-pass analog filter

Stop band

5.2 Practical Analog Low-Pass Filter Design

5.2.2

233

Butterworth Analog Low-Pass Filter

The magnitude-square response of an Nth-order analog low-pass Butterworth filter is given by 1

jH a ðjΩÞj2 ¼

1 þ ðΩ=Ωc Þ2N

ð5:15Þ

Two parameters completely characterizing a Butterworth low-pass filter are Ωs and N. These are determined from the specified band edges Ωp and Ωc, peak passband ripple αp, and minimum stopband attenuation αs. The first (2N  1) derivatives of |Ha( jΩ|2 at Ω ¼ 0 are equal to zero. Thus, the Butterworth low-pass filter is said to have a maximally flat magnitude at Ω ¼ 0. The gain in dB is given by 10log10|Ha( jΩ|2. At Ω ¼ Ωc, the gain is 10log10(0.5) ¼  3 dB; therefore, Ωc is called the 3 dB cutoff frequency. The loss in dB in a Butterworth filter is given by  α ¼ 10log 1 þ ðΩ=Ωc Þ2N

ð5:16Þ

For Ω ¼ Ωp, the passband attenuation is given by 

2N αp ¼ 10log 1 þ Ωp =Ωc

ð5:17Þ

For Ω ¼ Ωs, the stopband attenuation is  αs ¼ 10log 1 þ ðΩs =Ωc Þ2N

ð5:18Þ

Eqs.(5.17) and (5.18) can be rewritten as

Ωp =Ωc

2N

¼ 100:1αp  1

ð5:19Þ

ðΩs =Ωc Þ2N ¼ 100:1αs  1

ð5:20Þ

From Eqs.(5.19) and (5.20), we obtain



Ωs =Ωp ¼



100:1αs  1 100:1αp  1

1=2N ð5:21Þ

Eq. (5.21) can be rewritten as  0:1αs 

 1 10 1 log log Ωs =Ωp ¼ 2N 100:1αp  1 From Eq. (5.22), solving for N we get

ð5:22Þ

234

5 Analog Filters

 N

log

100:1αs 1 100:1αp 1

ð5:23Þ

2logðΩs =ΩpÞ

Since the order N must be an integer, the value obtained is rounded to the next higher integer. This value of N is used in either Eq. (5.19) or Eq. (5.20) to determine the 3 dB cutoff frequency Ωc. In practice, Ωc is determined by Eq. (5.20) that exactly satisfies stopband specification at Ωc, while the passband specification is exceeded with a safe margin at Ωp. We know that |H( jΩ)|2 may be evaluated by letting s ¼ jΩ in H(s)H(s), which may be expressed as H ðsÞH ðsÞ ¼

1



1 þ s2 =Ω2c

ð5:24Þ

N

If Ωc ¼ 1, the magnitude response |HN( jΩ)| is called the normalized magnitude response. Now, we have 2N

N Y 1 þ s2 ¼ ðs  sk Þ

ð5:25Þ

k¼1

where ( sk ¼

ejð2k1Þπ=2N

for n even

ejðk1Þπ=N

for n odd

ð5:26Þ

Since |sk| ¼ 1, we can conclude that there are 2N poles placed on the unit circle in the s-plane. The normalized transfer function can be formed as H N ðsÞ ¼

1 N Q

ð5:27Þ

ð s  pl Þ

l¼1

where pl for l ¼ 1, 2,.., N are the left half s-plane poles. The complex poles occur in conjugate pairs. For example, in the case of N ¼ 2, from Eq. (5.26), we have     ð2k  1Þπ ð2k  1Þπ sk ¼ cos þ j sin 4 4

k ¼ 1, . . . ::, 2N

The poles in the left half of the s-plane are 1 j 1 j s2 ¼ pffiffiffi þ pffiffiffi ; s3 ¼ pffiffiffi  pffiffiffi 2 2 2 2 Hence,

5.2 Practical Analog Low-Pass Filter Design

235

1 j 1 j p1 ¼ pffiffiffi þ pffiffiffi ; p2 ¼ pffiffiffi  pffiffiffi 2 2 2 2 and H N ðsÞ ¼

s2 þ

1 pffiffiffi 2s þ 1

In the case of N ¼ 3, sk ¼ cos

    ðk  1Þπ ðk  1Þπ þ j sin k ¼ 1, . . . ::, 2N 3 3

The left half of s-plane poles are pffiffiffi 1 j 3 s3 ¼  þ ; 2 2

s4 ¼ 1;

pffiffiffi 1 j 3 s5 ¼   2 2

p2 ¼ 1;

pffiffiffi 1 j 3 p3 ¼   ; 2 2

Hence pffiffiffi 1 j 3 p1 ¼  þ ; 2 2 and H N ðsÞ ¼

1 ð s þ 1Þ ð s 2 þ s þ 1Þ

The following MATLAB Program 5.1 can be used to obtain the Butterworth normalized transfer function for various values of N. Program 5.1 Analog Butterworth Low-Pass Filter Normalized Transfer Function N=input(‘enter order of the filter’); [z,p,k] = buttap(N)% determines poles and zeros disp(‘Poles are at’);disp(p); [num,den]=zp2tf(z,p,k); %Print coefficients in powers of s disp(‘Numerator polynomial’);disp(num); disp(‘Denominator polynomial’);disp(den); sos=zp2sos(z,p,k);%determines coefficients of second order sections

The normalized Butterworth polynomials generated from the above program for typical values of N are tabulated in Table 5.1. The magnitude response of the normalized Butterworth low-pass filter for some typical values of N is shown in Figure 5.9. From this figure, it can be seen that the response monotonically decreases both in the passband and the stopband as Ω

236

5 Analog Filters

Table 5.1 List of normalized Butterworth polynomials N 1 2 3 4 5 6 7

Denominator of HN(s) Sþ1 pffiffiffi s2 þ 2s þ 1 (s þ 1)(s2 + s þ 1) (s2 þ 0.76537s þ 1)(s2 þ 1.8477s þ 1) (s þ 1) (s2 þ 0.61803s þ 1)(s2 þ 1.61803s þ 1) pffiffiffi

 ðs2 þ 1:931855s þ 1Þ s2 þ 2s þ 1 ðs2 þ 0:51764s þ 1Þ (s þ 1) (s2 þ 1.80194s þ 1)(s2 þ 1.247s þ 1)(s2 þ 0.445s þ 1)

Figure 5.9 Magnitude response of typical Butterworth low-pass filter

increases. As the filter order N increases, the magnitude responses both in the passband and the stopband are improved with a corresponding decrease in the transition width. Since the normalized transfer function corresponds to Ωc ¼ 1, the transfer function of the low-pass filter corresponding to the actual Ωc can be obtained by replacing s by (s/Ωc) in the normalized transfer function. Example 5.1 Design a Butterworth analog low-pass filter with 1 dB passband ripple, passband edge frequency Ωp ¼ 2000π rad/sec, stopband edge frequency Ωs ¼ 10,000π rad/sec, and a minimum stopband ripple of 40 dB. Solution Since αs ¼ 40 dB, αp ¼ 1 dB, Ωp ¼ 2000π, and Ωs ¼ 10,000π,

5.2 Practical Analog Low-Pass Filter Design



100:1αs  1 log 100:1αp  1



237



104  1 ¼ log 100:1  1

 ¼ 4:5868:

Hence from (5.23),  N

log

104 1 100:1 1



2logð5=1Þ

¼

4:5868 ¼ 3:2811 1:3979

Since the order must be an integer, we choose N ¼ 4. The normalized low-pass Butterworth filter for N ¼ 4 can be formulated as H N ðsÞ ¼

1 ðs2 þ 0:76537s þ 1Þðs2 þ 1:8477s þ 1Þ

From Eq. (5.20), we have Ωc ¼

Ωs 10  1 4

1=2N ¼

10000π 1=8 ¼ 9935 104  1

The transfer function for Ωc ¼ 9935 can be obtained by replacing s by (s/Ωc) ¼ (s/9935)in HN (s). 1  s  s  2 s þ 1 9935 þ 1:8477 þ1 9935 9935 9:7425  1015 

 ¼ 2 s þ 7:604  103 s þ 9:8704225  107 s2 þ 1:8357  104 s þ 9:8704225  107 H a ðsÞ ¼

5.2.3

1

 s 2 9935 þ 0:76537

Chebyshev Analog Low-Pass Filter

Type 1 Chebyshev Low-Pass Filter The magnitude-square response of an Nth-order analog low-pass Type 1 Chebyshev filter is given by jH ðΩÞj2 ¼

1

 1 þ ε2 T 2N Ω=Ωp

where TN(Ω) is the Chebyshev polynomial of order N

ð5:28Þ

238

5 Analog Filters

( T N ð ΩÞ ¼

cos ðN cos 1 ΩÞ,

 cosh Ncosh1 Ω ,

j Ωj
1 j Ωj > 1

ð5:29Þ

The loss in dB in a Type 1 Chebyshev filter is given by



 α ¼ 10log 1 þ ε2 T 2N Ω=Ωp

ð5:30Þ

For Ω ¼ Ωp, TN(Ω) ¼ 1, and the passband attenuation is given by

 αp ¼ 10log 1 þ ε2

ð5:31Þ

From Eq. (5.31), ε can be obtained as ε¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 100:1αp  1

ð5:32Þ

For Ω ¼ Ωs the stopband attenuation is αs ¼ 10logð1 þ22 T 2n ðΩs =Ωp ÞÞ

ð5:33Þ

Since (Ωs/Ωp) > 1, the above equation can be written as 



 αs ¼ 10log 1 þ22 cosh2 Ncosh1 Ωs =Ωp

ð5:34Þ

Substituting Eq. (5.32) for ε in the above equation and solving for N, we get N

cosh1 cosh

qffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1

100:1αs 1 100:1αp 1

Ωs =Ωp



ð5:35Þ

We choose N to be the lowest integer satisfying (5.35). In determining N using the above equation, it is convenient to evaluate cosh1(x) by applying the identity pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 cosh ðxÞ ¼ ln x þ x2  1 . The poles of the normalized Type 1 Chebyshev filter transfer function lie on an ellipse in the s-plane and are given by


  1 ð2k  1Þπ 1 1 sinh sin xk ¼ sinh N 2 2N
  1 1 ð2k  1Þπ yk ¼ cosh sinh1 cos N 2 2N

for k ¼ 1, 2, ::::, N

ð5:36Þ

for k ¼ 1, 2, :::, N

ð5:37Þ

Also, the normalized transfer function is given by H N ðsÞ ¼ where

H0 Π k ðs  p k Þ

ð5:38Þ

5.2 Practical Analog Low-Pass Filter Design

239




 
  1 ð2k  1Þπ 1 ð2k  1Þπ 1 1 1 1 þ j cosh sinh pk ¼ sinh sinh sin cos N 2 2N N 2 2N ð5:39aÞ and H0 ¼

1 1 2N1 ε

ð5:39bÞ

As an illustration, consider the case of N ¼ 2 with a passband ripple of 1 dB. From Eq. (5.32), we have 1 1 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1:965227 0:1αp ε 1 10 Hence   1 ¼ sinh1 ð1:965227Þ ¼ 1:428 sinh ε 1

Therefore, from (5.39a), the poles of the normalized Chebyshev transfer function are given by pk ¼ sinhð0:714Þ sin

ð2k  1Þπ ð2k  1Þπ þ j coshð0:714Þ cos , 4 4

k ¼ 1, 2

Hence p1 ¼ 0:54887 þ j0:89513, p2 ¼ 0:54887  j0:89513 Also, from (5.39b), we have 1 H 0 ¼ ð1:965227Þ ¼ 0:98261 2 Thus for N ¼ 2, with a passband ripple of 1 dB, the normalized Chebyshev transfer function is H N ðsÞ ¼

0:98261 0:98261 ¼ 2 ðs  p1 Þðs  p2 Þ ðs þ 1:098s þ 1:103Þ

Similarly for N ¼ 3, for a passband ripple of 1 dB, we have pk ¼ sinhð1:428=3Þ sin

Thus,

ð2k  1Þπ ð2k  1Þπ þ j coshð1:428=3Þ cos , 6 6

k ¼ 1, 2, 3

240

5 Analog Filters

p1 ¼ 0:24709 þ j0:96600; p2 ¼ 0:49417; p3 ¼ 0:24709  j0:966: Also, from (5.39b), 1 H 0 ¼ ð1:965227Þ ¼ 0:49131 4 Hence, the normalized transfer function of Type 1 Chebyshev low-pass filter for N¼3 is given by H N ðsÞ ¼

0:49131 0:49131 ¼ ðs  p1 Þðs  p2 Þðs  p3 Þ ðs3 þ 0:988s2 þ 1:238s þ 0:49131Þ

The following MATLAB Program 5.2 can be used to form the Type1 Chebyshev normalized transfer function for a given order and passband ripple. Program 5.2 Analog Type 1 Chebyshev Low-Pass Filter Normalized Transfer Function N=input(‘enter order of the filter’); Rp=input(‘enter passband ripple in dB’); [z,p,k] = cheb1ap(N,Rp)% determines poles and zeros disp(‘Poles are at’);disp(p); [num,den]=zp2tf(z,p,k); %Print coefficients in powers of s disp(‘Numerator polynomial’);disp(num); disp(‘Denominator polynomial’);disp(den);

The normalized Type 1 Chebyshev polynomials generated from the above program for typical values of N and passband ripple of 1 dB are tabulated in Table 5.2. The typical magnitude responses of a Type 1 Chebyshev low-pass filter for N ¼ 3, 5, and 8 with 1 dB passband ripple are shown in Figure 5.10. From this figure, it is seen that Type 1 Chebyshev low-pass filter exhibits equiripple in the passband with a monotonic decrease in the stopband. Example 5.2 Design a Type 1 Chebyshev analog low-pass filter for the specifications given in Example 5.1. Table 5.2 List of normalized Type 1 Chebyshev transfer functions for passband ripple ¼ 1 dB N 1 2 3 4 5

Denominator of HN(s) S þ 1.9652 s2 þ 1.0977s þ 1.1025 s3 þ 0.98834s2 þ 1.2384s þ 0.49131 s4 þ 0.95281s3 þ 1.4539s2 þ 0.74262s þ 0.27563 s5 þ 0.93682s4 þ 1.6888s3 þ 0.9744s2 þ 0.58053s þ 0.12283

H0 1.9652 0.98261 0.49131 0.24565 0.12283

5.2 Practical Analog Low-Pass Filter Design

241

Figure 5.10 Magnitude response of typical Type 1 Chebyshev low-pass filter with 1 dB passband ripple

Solution Since αs ¼ 40 dB, αp ¼ 1 dB, Ωp ¼ 2000π, and Ωs ¼ 10,000π, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 100:1αs  1 104  1 1 cosh ¼ cosh ¼ cosh1 ð196:52Þ 0:1αp 10 100:1  1 1

 cosh1 Ωs =Ωp ¼ cosh1 ð5Þ ¼ 2:2924 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 104  1 1 cosh 100:1  1 N ¼ 2:6059 cosh1 ð5Þ 1

Since the order of the filter must be an integer, we choose the next higher integer value 3 for N. The normalized Type 1 Chebyshev low-pass filter for N ¼ 3 with a passband ripple of 1 dB is given from Table 5.2 as H N ðsÞ ¼

s3

þ

0:988s2

0:49131 þ 1:238s þ 0:49131

The transfer function for Ωp ¼ 2000π is obtained by substituting s ¼ (s/Ωp) ¼ (s/2000π) in HN(s): 0:49131  s 2  s s 3 þ 0:988 þ 1:238 þ 0:49131 2000π 2000π 2000π 1:2187  1011 ¼ 3 3 2 s þ 6:2099  10 s þ 4:889  107 s þ 1:2187  1011

H a ðsÞ ¼ 

242

5 Analog Filters

Type 2 Chebyshev Filter The squared-magnitude response of Type 2 Chebyshev low-pass filter, which is also known as the inverse Chebyshev filter, is given by 

jH ðΩÞj2 ¼ 1 þ ε2

1 T 2N ðΩs =Ωp Þ T 2N ðΩs =ΩÞ



ð5:40Þ

The order N can be determined using Eq. (5.35). The Type 2 Chebyshev filter has both poles and zeros, and the zeros are on the jΩ axis. The normalized Type 2 Chebyshev low-pass filter, or the normalized inverse Chebyshev filter (normalized to Ωs ¼ 1), may be formed as H N ðsÞ ¼ H 0

Πk ðs  zk Þ , Πk ðs  pk Þ

k ¼ 1, 2, ::, N

ð5:41Þ

where zk ¼ j

1 cos

ð2k1Þπ N

for

k ¼ 1, 2, ::, N

σk ωk þj 2 for k ¼ 1, 2, ::, N σ 2k þ Ω2k σ k þ Ω2k
  1 1 ð2k  1Þπ sinh1 for k ¼ 1, 2, ::, N σ k ¼ sinh sin N δs 2N
  1 ð2k  1Þπ 1 1 Ωk ¼ cosh sinh for k ¼ 1, 2, ::, N cos N δs 2N pk ¼

1 δs ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:1αs 1 10 H0 ¼

Πk ðpk Þ Πk ðzk Þ

ð5:42aÞ ð5:42bÞ ð5:42cÞ ð5:42dÞ ð5:42eÞ ð5:42fÞ

For example, if we consider N ¼ 3 with a stopband ripple of 40 dB, then from (5.42e), pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 100:1αs  1 ¼ 104  1 ¼ 99:995 δs Hence, sinh1

  1 ¼ 5:28829 δs

Using (5.42c) and (5.42d), we have

5.2 Practical Analog Low-Pass Filter Design

ð2k  1Þπ 6 ð2k  1Þπ Ωk ¼ coshð5:28829=3Þ cos 6

σ k ¼ sinhð5:28829=3Þ sin

243

for

k ¼ 1, 2, 3

for

k ¼ 1, 2, 3

Hence σ 1 ¼ 1:41927, σ 2 ¼ 2:83854, σ 3 ¼ 1:41927 Ω1 ¼ 2:60387, Ω2 ¼ 2:83854, Ω3 ¼ 2:60387 Thus, from (5.42b), the poles are p1 ¼ 0:16115 þ j0:29593, p2 ¼ 0:3523, p3 ¼ 0:16115 þ j0:29593 Also, using (5.42a), the zeros are given by  pffiffiffi  pffiffiffi z1 ¼ j 2= 3 , z2 ¼ j 2= 3 Finally, from (5.42f), H 0 ¼ 0:03 Therefore, the normalized Type 2 Chebyshev low-pass filter for N ¼ 3 with a stopband ripple of 40 dB is given by H N ðsÞ ¼

0:03ðs  z1 Þðs  z2 Þ 0:03ðs2 þ 1:3333Þ ¼ 3 ðs  p1 Þðs  p2 Þðs  p3 Þ ðs þ 0:6746s2 þ 0:22709s þ 0:04Þ

The following MATLAB Program 5.3 can be used to form the Type2 Chebyshev normalized transfer function for a given order and stopband ripple. Program 5.3 Analog Type 2 Chebyshev Low-Pass Filter Normalized Transfer Function N=input(‘enter order of the filter’); Rs=input(‘enter stopband attenuation in dB’); [z,p,k] = cheb2ap(N,Rs);% determines poles and zeros disp(‘Poles are at’);disp(p); [num,den]=zp2tf(z,p,k); %Print coefficients in powers of s disp(‘Numerator polynomial’);disp(num); disp(‘Denominator polynomial’);disp(den);

The normalized Type 2 Chebyshev transfer functions generated from the above program for typical values of N with a stopband ripple of 40 dB are tabulated in Table 5.3.

244

5 Analog Filters

Table 5.3 List of normalized Type 2 Chebyshev transfer functions for stopband ripple ¼ 40 dB Order N 1 2 3 4 5 6

HN (s) 0:01 s þ 0:01 0:01s2 þ 0:02 2 s þ 0:199s þ 0:02 0:03s2 þ 0:04 s3 þ 0:6746s2 þ 0:2271s þ 0:04 0:01s4 þ 0:08s2 þ 0:08 4 s þ 1:35s3 þ 0:9139s2 þ 0:3653s þ 0:08 0:05s4 þ 0:2s2 þ 0:16 5 4 s þ 2:1492s þ 2:3083s3 þ 1:5501s2 þ 0:6573s þ 0:16 0:01s6 þ 0:18s4 þ 0:48s2 þ 0:32 s6 þ 3:0166s5 þ 4:5519s4 þ 4:3819s3 þ 2:8798s2 þ 1:2393s þ 0:32

Figure 5.11 Magnitude response of typical Type 2 Chebyshev low-pass filter with 20 dB stopband ripple

The typical magnitude response of a Type 2 Chebyshev low-pass filter for N ¼ 4 and 7 with 20 dB stopband ripple is shown in Figure 5.11. From this figure, it is seen that Type 2 Chebyshev low-pass filter exhibits monotonicity in the passband and equiripple in the stopband. Example 5.3 Design a Type 2 Chebyshev low-pass filter for the specifications given in Example 5.1. Solution The order N is chosen as 3, as in Example 5.2, since the equation for order finding is the same for both Type 1 and Type 2 Chebyshev filters. The normalized

5.2 Practical Analog Low-Pass Filter Design

245

Type 2 Chebyshev low-pass filter for N ¼ 3 with a stopband ripple of 40 dB has already been found earlier and is given by H N ðsÞ ¼

ðs3

0:03ðs2 þ 1:3333Þ þ 0:6746s2 þ 0:2271s þ 0:04Þ

For Ωs ¼ 10,000π, the corresponding transfer function can be obtained by substituting s ¼ (s/Ωs) ¼ (s/10000π) in the above expression for HN(s). Thus, the required filter transfer function is

s 2 0:03 10000π þ 0:04  H a ðsÞ ¼



 3 2 s s þ 0:6746 10000π þ 0:22709 10000π

s þ 0:04 10000π 9:4252  102 s2 þ 1:2403  1012 ¼ 3 s þ 2:1193  104 s2 þ 2:2413  108 s þ 1:2403  1012

5.2.4

Elliptic Analog Low-Pass Filter

The square-magnitude response of an elliptic low-pass filter is given by jH a ðjΩÞj2 ¼

1 1þ

ε2 U

N

 Ω=Ωp

ð5:43Þ

where UN(x) is the Jacobian elliptic function of order N and ε is a parameter related to the passband ripple. In an elliptic filter, a constant k, called the selectivity factor, representing the sharpness of the transition region is defined as k¼

Ωp Ωs

ð5:44Þ

A large value of k represents a wide transition band, while a small value indicates a narrow transition band. For a given set of Ωp, Ωs, αp, and αs, the filter order can be estimated using the formula Nffi

 100:1αs 1 log 16  10 0:1αp 1 log10 ð1=ρÞ

ð5:45Þ

where ρ can be computed using pffiffiffiffi 1  k0  ρ0 ¼ pffiffiffiffi 2 1 þ k0

ð5:46Þ

246

5 Analog Filters

k0 ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi 1  k2

ρ ¼ ρ0 þ 2ðρ0 Þ5 þ 15ðρ0 Þ9 þ 150ðρ0 Þ13

ð5:47Þ ð5:48Þ

The following MATLAB Program 5.4 can be used to form the elliptic normalized transfer function for given filter order and passband ripple and stopband attenuation. The normalized passband edge frequency is set to 1. Program 5.4 Analog Elliptic Low-Pass Filter Normalized Transfer Function N=input(‘enter order of the filter’); Rp=input(‘enter passband ripple in dB’); Rs=input(‘enter stopband attenuation in dB’); [z,p,k] = ellipap(N,Rp,Rs)% determines poles and zeros disp(‘Poles are at’);disp(p); [num,den] =zp2tf(z,p,k); %Print coefficients in powers of s disp(‘Numerator polynomial’);disp(num); disp(‘Denominator polynomial’);disp(den);

The normalized elliptic transfer functions generated from the above program for typical values of N and stopband ripple of 40 dB are tabulated in Table 5.4. The magnitude response of a typical elliptic low-pass filter is shown in Figure 5.12, from which it can be seen that it exhibits equiripple in both the passband and the stopband. Example 5.4 Design an elliptic analog low-pass filter for the specifications given in the Example 5.1.

Table 5.4 List of normalized elliptic transfer functions for passband ripple ¼ 1 dB and stopband ripple ¼ 40 dB Order N 1 2 3 4 5 6

HN(s) 1:9652 s þ 1:9652 0:01s2 þ 0:9876 s2 þ 1:0915s þ 1:1081 0:0692s2 þ 0:5265 3 s þ 0:9782s2 þ 1:2434s þ 0:5265 0:01s4 þ 0:1502s2 þ 0:3220 4 s þ 0:9391s3 þ 1:5137s2 þ 0:8037s þ 0:3612 0:0470s4 þ 0:2201s2 þ 0:2299 s5 þ 0:9234s4 þ 1:8471s3 þ 1:1292s2 þ 0:7881s þ 0:2299 0:01s6 þ 0:1172s4 þ 0:28s2 þ 0:186 6 5 s þ 0:9154s þ 2:2378s4 þ 1:4799s3 þ 1:4316s2 þ 0:5652s þ 0:2087

5.2 Practical Analog Low-Pass Filter Design

247

Figure 5.12 Magnitude response of typical elliptic low-pass filter with 1 dB passband ripple and 30 dB stopband ripple

Solution k¼

Ωp 2000π ¼ 0:2 ¼ Ωs 10000π

and k0 ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1  k2 ¼ 1  0:04 ¼ 0:979796:

Substituting these values in Eq. (5.46) and Eq. (5.47), we get ρ0 ¼ 0:00255135, ρ ¼ 0:0025513525 and hence N¼

 104 1 log 16  10 0:1 1 1 Þ ¼ 2:2331: log10 ð0:0025513525

Choose N ¼ 3. Then, for N ¼ 3, a passband ripple of 1 dB, and a stopband ripple of 40dB, the normalized elliptic transfer function is as given in Table 5.4. For Ωp ¼ 2000π, the corresponding transfer function can be obtained by substituting s ¼ (s/Ωp) ¼ (s/2000π) in the expression for HN(s). Thus, the required filter transfer function is

248

5 Analog Filters

s 2 0:0692 2000π þ 0:5265 H a ðsÞ ¼   s 2  s s 3 þ 0:97825 þ 1:2434 þ 0:5265 2000π 2000π 2000π 4:348  102 s2 þ 1:306  1011 ¼ 3 s þ 6:1465  103 s2 þ 4:9087  107 s þ 1:306  1011

5.2.5

Bessel Filter

Bessel filter is a class of all-pole filters that provide linear phase response in the passband and characterized by the transfer function H a ðsÞ ¼

a0 þ a1 s þ a2

s2

1 þ    þ aN1 sN1 þ aN sN

ð5:49Þ

where the coefficients an are given by an ¼

ð2N  nÞ! 2 n!ðN  nÞ! Nn

ð5:50Þ

The magnitude responses of a third-order Bessel filter and Butterworth filter are shown in Figure 5.13 and the phase responses of the same filters with the same order are shown in Figure 5.14. From these figures, it is seen that the magnitude response

Figure 5.13 Magnitude response of a third-order Bessel filter and Butterworth filter

5.2 Practical Analog Low-Pass Filter Design

249

Figure 5.14 Phase response of a third-order Bessel filter and Butterworth filter

of the Bessel filter is poorer than that of the Butterworth filter, whereas the phase response of the Bessel filter is more linear in the passband than that of the Butterworth filter.

5.2.6

Comparison of Various Types of Analog Filters

The magnitude response and phase response of the normalized Butterworth, Chebyshev Type 1, Chebyshev Type 2, and elliptic filters of the same order are compared with the following specifications: filter order ¼ 8, maximum passband ripple ¼ 1 dB and minimum stopband ripple ¼ 35 dB: The following MATLAB program is used to generate the magnitude and phase responses for these specifications. Program 5.5 Magnitude and Phase Responses of Analog Filters of Order 8 with a Passband Ripple of 1 dB and a Stopband Ripple of 35 dB clear all;clc; [z,p,k]=buttap(8); [num1,den1]=zp2tf(z,p,k);[z,p,k]=cheb1ap(8,1); [num2,den2]=zp2tf(z,p,k);[z,p,k]=cheb2ap(8,35); [num3,den3]=zp2tf(z,p,k); [z,p,k]=ellipap(8,1,35);

250

5 Analog Filters

[num4,den4]=zp2tf(z,p,k); omega=[0:0.01:5]; h1=freqs(num1,den1,omega);h2=freqs(num2,den2,omega); h3=freqs(num3,den3,omega);h4=freqs(num4,den4,omega); ph1=angle(h1);ph1=unwrap(ph1); ph2=angle(h2);ph2=unwrap(ph2); ph3=angle(h3);ph3=unwrap(ph3); ph4=angle(h4);ph4=unwrap(ph4); figure(1),plot(omega,20*log10(abs(h1)),‘-’);hold on plot(omega,20*log10(abs(h2)),‘--’);hold on plot(omega,20*log10(abs(h3)),‘: ’);hold on plot(omega,20*log10(abs(h4)),‘-.’); xlabel(‘Normalized frequency’);ylabel(‘Gain,dB’);axis([0 5 -80 5]); legend(‘Butterworth’,‘Chebyshev Type 1’,‘Chebyshev Type 2’,‘Elliptic’);hold off figure(2),plot(omega,ph1,‘-’);hold on plot(omega,ph2,‘--’);hold on plot(omega,ph3,‘: ’);hold on plot(omega,ph4,‘-.’) xlabel(‘Normalized frequency’);ylabel(‘Phase,radians’);axis([0 5 -8 0]); legend(‘Butterworth’,‘Chebyshev Type 1’,‘Chebyshev Type 2’,‘Elliptic’);

The magnitude and phase responses for the above specifications are shown in Figure 5.15. The magnitude response of Butterworth filter decreases monotonically both in passband and stopband with wider transition band. The magnitude response of the Chebyshev Type 1 exhibits ripples in the passband, whereas the Chebyshev Type 2 has approximately the same magnitude response to that of the Butterworth filter. The transition band of both the Type 1 and Type 2 Chebyshev filters is the same, but less than that of the Butterworth filter. The elliptic filter exhibits an equiripple magnitude response both in the passband and the stopband with a transition width smaller than that of the Chebyshev Type 1 and Type 2 filters. But the phase response of the elliptic filter is more nonlinear in the passband than that of the phase response of the Butterworth and Chebyshev filters. If linear phase in the passband is the stringent requirement, then the Bessel filter is preferred, but with a poor magnitude response. Another way of comparing the various filters is in terms of the order of the filter required to satisfy the same specifications. Consider a low-pass filter that meets the passband edge frequency of 450 Hz, stopband edge frequency of 550 Hz, passband ripple of 1 dB, and stopband ripple of 35 dB. The orders of the Butterworth, Chebyshev Type 1, Chebyshev Type2, and elliptic filters are computed for the above specifications and listed in Table 5.5. From this table, we can see that elliptic filter can meet the specifications with the lowest filter order.

5.2 Practical Analog Low-Pass Filter Design

251

Figure 5.15 A comparison of various types of analog low-pass filters: (a) magnitude response and (b) phase response

252

5 Analog Filters

Table 5.5 Comparison of orders of various types of filters

5.2.7

Filter Butterworth Chebyshev Type 1 Chebyshev Type 2 Elliptic

Order 24 9 9 5

Design of Analog High-Pass, Band-Pass, and Band-Stop Filters

The analog high-pass, band-pass, and band-stop filters can be designed using analog frequency transformations. In this design process, first, the analog prototype lowpass filter specifications are derived from the desired specifications of the analog filter using suitable analog-to-analog transformation. Next, by using the specifications so obtained, a prototype low-pass filter is designed. Finally, the transfer function of the desired analog filter is determined from the transfer function of the prototype analog low-pass transfer function using the appropriate analog-to-analog frequency transformation. The low-pass to low-pass, low-pass to high-pass, low-pass to band-pass, and low-pass to band-stop analog transformations are considered next. Low Pass to Low Pass b p be the passband edge frequencies of the normalized prototype Let Ωp ¼ 1 and Ω low-pass filter and the desired low-pass filter, as shown in Figure 5.16. The transb ¼0 formation from the prototype low pass to the required low pass must convert Ω b to Ω ¼ 0 and Ω ¼ 1 to Ω ¼ 1. The transformation such as s ¼ kbs or Ω b achieves the above transformation for any positive value of k. If k is chosen to ¼ k Ω  b p gets transformed to Ωp ¼ 1, and Ω b s to Ωs ¼ Ω b s =Ω b p . Since b p , then Ω be 1=Ω Ωp ¼ 1 is the passband edge frequency for the normalized Type I Chebyshev and elliptic low-pass filters, we have the design equations for these filters as b s =Ω b p: Ωp ¼ 1, Ωs ¼ Ω ð5:51aÞ

 Also, the transfer function H LP bs for these filters is related to the corresponding normalized low-pass transfer function HN (s) by

 H LP bs ¼ H N ðsÞc

bp s¼b s=Ω

ð5:51bÞ

However, in the case of a Butterworth filter, since Ω  ¼ 1 corresponds to the cutoff frequency of the filter, the transfer function H LP bs for the Butterworth filter is related to the normalized low-pass Butterworth transfer function HN (s) by

5.2 Practical Analog Low-Pass Filter Design

253

Figure 5.16 Low-pass to low-pass frequency transformation. (a) Prototype Low-pass filter frequency response. (b) Low-pass filter frequency response

(a)

(b)

 H LP bs ¼ H N ðsÞc

bc s¼b s=Ω

ð5:51cÞ

b c is the cutoff frequency of the desired Butterworth filter and is given by where Ω

 Eq. (5.19). For similar reasons, the transfer function H LP bs for the Type 2 Chebyshev filter is related to the normalized transfer function HN (s) by

 H LP bs ¼ H N ðsÞc

bs s¼b s=Ω

ð5:51dÞ

Low Pass to High Pass (Figure 5.17) Let the passband edge frequencies of the prototype low-pass and the desired highb p , as shown in Figure 5.17. The transformation from pass filters be Ωp ¼ 1 and Ω b ¼ 0 to Ω ¼ 1 and prototype low pass to the desired high pass must transform Ω b ¼ 1 to Ω ¼ 0. The transformation such as s ¼ k=bs or Ω ¼ k=Ω b achieves the Ω

254

5 Analog Filters

(a)

(b) Figure 5.17 Low-pass to high-pass frequency transformation. (a) Prototype low-pass filter frequency response. (b) High-pass filter frequency response

b p to Ωp ¼ 1, above transformation for any positive value of k. By transforming Ω b the constant k can be determined as k ¼ Ω p .Thus, design equations are b p =Ω b s, ð5:52aÞ Ωp ¼ 1, Ωs ¼ Ω

 and the desired transfer function H HP bs is related to the low-pass transfer function HN (s) by

 H HP bs ¼ H N ðsÞj b s¼Ω p=b s

ð5:52bÞ

5.2 Practical Analog Low-Pass Filter Design

255

(a)

(b)

Figure 5.18 Low-pass to band-pass frequency transformation. (a) Prototype low-pass filter frequency response. (b) Band-pass filter frequency response

The above equations (5.52a) and (5.52b) hold for all filters except for Butterworth and Type 2 Chebyshev filter. For Butterworth

 H LP bs ¼ H N ðsÞc b  s¼bs=Ω c 

 H HP bs ¼ H N ðsÞ b s¼Ω p=b s For Type 2 Chebyshev filter, the design equations are

ð5:53aÞ ð5:53bÞ

256

5 Analog Filters

b s =Ω b p , Ωs ¼ 1 Ωp ¼ Ω

ð5:53cÞ

 H HP bs ¼ H N ðsÞc b s¼Ω s =b s

ð5:53dÞ

and

Example 5.5 Design a Butterworth analog high-pass filter for the following specifications: Passband edge frequency: 30.777 Hz Stopband edge frequency: 10 Hz Passband ripple: 1 dB Stopband ripple: 20 dB Solution For the prototype analog low-pass filter, we have b p =Ω b s ¼ 3:0777, Ωp ¼ 1, Ωs ¼ Ω

αp ¼ 1 dB, αs ¼ 20 dB

Substituting these values in Eq. (5.23), the order of the filter is given by 

N

2



102 1 100:1 1

 log 3:077 1

log

¼ 2:6447

Hence, we choose N ¼ 3. From Table 5.1, the third-order normalized Butterworth low-pass filter transfer function is given by H N ðsÞ ¼

1 ð s þ 1Þ ð s 2 þ s þ 1Þ

Substituting the values of Ωs and N in Eq. (5.20), we obtain   3:0777 6 ¼ 102  1 Ωc Solving for Ωc, we get Ωc ¼ 1.4309. The analog transfer function of the low-pass filter is obtained from the above s transfer function by substituting s ¼ Ωsc ¼ 1:4309 ; hence, H LP ðsÞ ¼

2:93 s3 þ 2:8619s2 þ 4:0952s þ 2:93

From the above transfer function, the analog transfer function of the high-pass _ Ωp 3:0777 ¼ filter can be obtained by substituting s ¼ s s

5.2 Practical Analog Low-Pass Filter Design

H HP ðsÞ ¼

257

s3 s3 þ 4:3017s2 þ 9:2521s þ 9:9499

Example 5.6 Design a Butterworth analog high-pass filter for the specifications of Example 5.5 using MATLAB Solution The following MATLAB code fragments can be used to design HHP (s) [N,Wn]=buttord(1,3.0777,1,20,‘s’); [B,A]=butter(N,Wn,‘s’); [num, den]=1p2hp(B,A,3.0777);

The transfer function HLP(s) of the analog low-pass filter can be obtained by displaying numerator and denominator coefficient vectors B and A and is given by H LP ðsÞ ¼

2:93 s3 þ 2:8619s2 þ 4:0952s þ 2:93

The transfer function HHP (s) of the analog high-pass filter can be obtained by displaying numerator and denominator coefficient vectors num and den and is given by H HP ðsÞ ¼

s3

þ

4:3017s2

s3 þ 9:2521s þ 9:499

Low Pass to Band Pass The prototype low-pass and the desired band-pass filters are shown in Figure 5.18. In b p1 is the lower passband edge frequency, Ω b p2 the upper passband edge this figure, Ω b s2 the upper stopband edge b s1 the lower stopband edge frequency, and Ω frequency, Ω frequency of the desired band-pass filter. Let us denote by Bp the bandwidth of the b mp the geometric mean between the passband edge frequencies of passband and by Ω the band-pass filter, i.e., b p2  Ω b p1 Bp ¼ Ω qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b p1 Ω b p2 b mp ¼ Ω Ω

ð5:54aÞ ð5:54bÞ

Now, consider the transformation

2  b2 bs þ Ω mp S¼ Bpbs

ð5:55Þ

b ¼ 0, Ω b p1 , Ω b mp , Ω b p2 , As a consequence of this transformation, it is seen that Ω and 1 transform to the frequencies Ω ¼ 1, 1, 0, þ1, and 1, respectively, for

258

5 Analog Filters

the normalized low-pass filter. Also, the transformation (5.55) transforms the freb s1 and Ω b s2 to Ω0 and Ω00 , respectively, where quencies Ω s s b2  Ω b p2 b p1 Ω Ω Ω0s ¼ s1 ¼ A1 ðsayÞ  b b s1 b Ω p2  Ω p1 Ω

ð5:56Þ

b2  Ω b p2 b p1 Ω Ω Ω00s ¼ s2 ¼ A2 ðsayÞ  b p2  Ω b s2 b p1 Ω Ω

ð5:57Þ

and

In order to satisfy the stopband requirements and to have symmetry of the stopband edges in the low-pass filter, we choose Ωs to be themin{|A1|, |A2|}. Thus, the spectral transformation (5.55) leads to the following design equations for the normalized low-pass filter (except in the case of the Type 2 Chebyshev filter) Ωp ¼ 1, Ωs ¼ minfjA1 j; jA2 jg

ð5:58aÞ

where A1 and A2 are given by  (5.56) and (5.57), respectively, and the desired highpass transfer function H BP bs can be obtained from the normalized low-pass transfer function HN (s) using (5.55). In the case of the Type 2 Chebyshev filter, the equation corresponding to (5.58a) is Ωp ¼ maxf1=jA1 j; 1=jA2 jg,

Ωs ¼ 1

ð5:58bÞ

Example 5.7 Design a Butterworth IIR digital band-pass filter for the following specifications: Lower passband edge frequency: 41.4 Hz Upper passband edge frequency: 50.95 Hz Lower stopband edge frequency: 7.87 Hz Upper stopband edge frequency: 100 Hz Passband ripple: 2 dB Stopband ripple: 10 dB Solution We have ð0:0787Þ2 þ ð0, 414Þð0:5095Þ ¼ 27:25 0:0787ð0:5095  0, 414Þ 1 þ ð0, 414Þð0:5095Þ A2 ¼ ¼ 8:26 1ð0:5095  0, 414Þ

A1 ¼

For the prototype analog low-pass filter, Ωp ¼ 1, Ωs ¼ min {|A1|, |A2|} ¼ 8.26; αp ¼ 2 dB αs ¼ 10 dB

5.2 Practical Analog Low-Pass Filter Design

259

Substituting these values in Eq. (5.23), the order of the filter is given by h N¼

log10

101 1 100:2 :1

i

2log10 ð8:26Þ

¼ 0:5203

Let us choose N ¼ 1 The transfer function of the first-order normalized Butterworth low-pass filter is given by H N ðsÞ ¼

1 sþ1

Substituting the values of Ωs and N in Eq. (5.20), we obtain   8:26 2 ¼ 101  1 Ωc Solving for Ωc, we get Ωc ¼ 2.7533 The analog transfer function of the low-pass filter can be obtained from the above s transfer function by substituting s ¼ Ωsc ¼ 2:7533 H LP ðsÞ ¼

2:7533 s þ 2:7533

To arrive at the analog transfer function of the band-pass filter, variable s in the above normalized transfer function is to be replaced by

2    b p2 b p1 Ω s þΩ s2 þ 2109:3 ¼ S¼

 b p2  Ω b p1 s 9:55s Ω 26:2943 s H BP ðsÞ ¼ 2 s þ 26:2943s þ 2109:3 Example 5.8 Design a band-pass Butterworth filter for the specifications of Example 5.7 using MATLAB Solution The following MATLAB code fragments can be used to design HBS (s): pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Bandwidth ¼ bw ¼ 50.95–41.4 ¼ 9.55; Ωo ¼ ð50:95Þð41:4Þ ¼ 45:9271. [N,Wn]=buttord(1,8.26,2,10,‘s’); [B,A]=butter(N,Wn,‘s’); [num, den]=1p2bp(B,A, 45.9271,9.55);% [num, den]=1p2bp(B,A, Ωo, bw);

The transfer function HLP(s) of the analog low-pass filter can be obtained by displaying numerator and denominator coefficient vectors B and A. It is given by

260

5 Analog Filters

H LP ðsÞ ¼

2:7533 s þ 2:7533

The transfer function HBP (s) of the analog band-pass filter can be obtained by displaying numerator and denominator coefficient vectors num and den. It is given by H BP ðsÞ ¼

s2

26:2943 s þ 26:2943s þ 2109:3

Low Pass to Band Stop The prototype low-pass and the desired band-stop filters are shown in Figure 5.19. In b p1 is the lower passband edge frequency, Ω b p2 the upper passband edge this figure, Ω b s2 the upper stopband b frequency, Ω s1 the lower stopband edge frequency, and Ω edge frequency of the desired band-stop filter. Let us now consider the transformation kbs S¼

 b2 bs 2 þ Ω ms

ð5:59Þ

b ms is the geometric mean between the stopband edge frequencies of the where Ω band-stop filter, i.e., b ms ¼ Ω

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b s1 Ω b s2 Ω

ð5:60Þ

b ¼ 0 and 1 As a consequence of this transformation, it is seen that Ω b transformed to the frequency Ω ¼ 0 for the normalized low-pass filter. Now, we b s1 to the stopband edge frequency Ωs transform the lower stopband edge frequency Ω of the normalized low-pass filter; hence, Ωs ¼

k b s2  Ω b s1 Ω

¼

k Bs

ð5:61aÞ



b s2  Ω b s1 is the bandwidth of the stopband. Also, the upper stopband where Bs ¼ Ω b s2 is transformed to edge frequency Ω k k  ¼  ¼ Ωs b b Bs Ω s2  Ω s1

ð5:61bÞ

Hence, the constant k is given by 

b s2  Ω b s1 Ωs k ¼ Bs Ωs ¼ Ω As a consequence, the passband edge frequencies transformed to

ð5:61cÞ b p1 and Ω b p2 Ω

are

5.2 Practical Analog Low-Pass Filter Design

261

Ω0p ¼

 b s2  Ω b p1 b s1 Ω Ω 1 Ωs ¼ Ωs b s1 Ω b s2  Ω b2 A 1 Ω p1

ð5:62aÞ

and

Ω00p

 b s2  Ω b p2 b s1 Ω Ω 1 ¼ Ωs ¼ Ωs 2 b s1 Ω b s2  Ω b A2 Ω p2

ð5:62bÞ

In order to satisfy the passband requirement as well as to satisfy the symmetry requirement of the passband   edge  of the normalized low-pass filter, we have to     choose the higher of Ω0p  and Ω00p  as Ωp. Since for the normalized filter (except for the case of Type 2 Chebyshev filter), Ωp ¼ 1, we have to choose Ωs to be the lower of {|A1|, |A2|}. Hence, the design equations for the normalized low-pass filter (except for the Type 2 Chebyshev) (Figure 5.19) are Ωp ¼ 1, Ωs ¼ minfjA1 j; jA2 jg

ð5:63aÞ

where b s1 Ω b s2  Ω b2 Ω p1 A1 ¼

,  b s2  Ω b p1 b s1 Ω Ω

b s1 Ω b s2  Ω b2 Ω p2 A2 ¼

 b s2  Ω b p2 b s1 Ω Ω

ð5:63bÞ

and the transfer function of the required band-stop filter is

 H BS bs ¼ H N ðsÞc ^ s¼

^ Ω s2 Ω s1



Ωs ^s

ð5:63cÞ

^ Ω ^ ^s 2 þΩ s1 s2

For the Type 2 Chebyshev filter, Eq. (5.63a) would be replaced by Ωp ¼ maxf1=jA1 j; 1=jA2 jg, Ωs ¼ 1

ð5:63dÞ

Example 5.9 Design an analog band-stop Butterworth filter with the following specifications: Lower passband edge frequency: 22.35 Hz Upper passband edge frequency: 447.37 Hz Lower stopband edge frequency: 72.65 Hz Upper stopband edge frequency: 137.64 Hz Passband ripple: 3 dB Stopband ripple: 15 dB

262

5 Analog Filters

Figure 5.19 Low-pass to band-stop frequency transformation. (a) Prototype low-pass filter frequency response. (b) Band-stop filter frequency response

(a)

(b)

Solution From Eq. (5.63b), we have b s1 Ω b s2  Ω b2 Ω p1 ¼ 6:5403, A1 ¼

 b s2  Ω b p1 b s1 Ω Ω

b s1 Ω b s2  Ω b2 Ω p2 A2 ¼

¼ 6:5397  b s2  Ω b p2 b s1 Ω Ω

Now using (5.63a), we get the specifications for the normalized analog low-pass filter to be Ωp ¼ 1, Ωs ¼ minfjA1 j; jA2 jg,

αp ¼ 3 dB, αs ¼ 15 dB

5.2 Practical Analog Low-Pass Filter Design

263

Substituting these values in Eq. (5.23), the order of the filter is given by  N

log



101:5 1 100:3 1

2 logð6:5397Þ

¼ 0:9125

We choose N ¼ 1. The transfer function of the first-order normalized Butterworth low-pass filter is H N ðsÞ ¼

1 ð s þ 1Þ

Substituting the values of Ωs and N in Eq. (5.20), we obtain



6:5397 Ωc

2

¼ 101:5  1:

Solving for Ωc, we get Ωc ¼ 1.1818. The analog transfer function of the low-pass s filter is obtained from HN(s) by substituting s ¼ Ωsc ¼ 1:1818 H LP ðsÞ ¼

1:1818 s þ 1:1818

To arrive at the analog transfer function of the band-stop filter, we use, in the above expression, the low-pass to band-stop transformation given by (5.63c), namely,

 b s2  Ω b s1 Ωs s ð64:99Þð6:5397Þs Ω 425s S¼ ¼ 2 ¼ b s2 b s1 Ω s2 þ 10000 s þ 10000 s2 þ Ω to obtain H BS ðsÞ ¼

s2 þ 10000 s2 þ 360s þ 10000

Example 5.10 Design a band-stop Butterworth filter for the specifications of Example 5.9 using MATLAB Solution The following MATLAB code fragments can be used to design HBS(s): pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Bandwidth ¼ bw ¼ 447.37 – 22.35; Ωo ¼ ð137:64Þð72:65Þ ¼ 100. [N,Wn]=buttord(1,6.5397,3,15,‘s’); [B,A]=butter(N,Wn,‘s’); [num, den]=1p2bs(B,A,100,425.02);% [num, den]=1p2bs(B,A, Ωo, bw);

The transfer function HLP (s) of the analog low-pass filter can be obtained by displaying numerator and denominator coefficient vectors B and A and is given by H LP ðsÞ ¼

1:1818 s þ 1:1818

264

5 Analog Filters

The transfer function HBS(s) of the analog band-stop filter can be obtained by displaying numerator and denominator coefficient vectors num and den and is given by H BS ðsÞ ¼

5.3

s2

s2 þ 10000 þ 360s þ 10000

Effect of Poles and Zeros on Frequency Response

Frequency response of a system can be obtained by evaluating H(s) for all values of s ¼ jΩ.

5.3.1

Effect of Two Complex System Poles on the Frequency Response

Consider the following system function with complex poles H ðsÞ ¼

1

s  ðα þ jΩÞ s  ðα  jΩÞ

ð5:64Þ

with placement of poles as shown in Figure 5.20(a). The magnitude response of the system with pole locations shown in Figure 5.20(a) is given by jH ðjΩÞj ¼

1 dd0

ð5:65Þ

and is shown in Figure 5.20(b), and its phase response is shown in Figure 5.20(c)

5.3.2

Effect of Two Complex System Zeros on the Frequency Response

Consider the following system function with complex zeros  HðsÞ ¼ s  ðα þ jΩÞÞðs  ðα  jΩÞÞ

ð5:66Þ

with placement of zeros as shown in Figure 5.21(a). The magnitude response of the system with zeros locations shown in Figure 5.21(a) is given by jH ðjΩÞj ¼ rr 0

ð5:67Þ

and is shown in Figure 5.21(b), and its phase response is shown in Figure 5.21(c)

5.4 Design of Specialized Analog Filters by Pole-Zero Placement

265

(a)

(b)

(c)

Figure 5.20 (a) Pole locations of H(s). (b) Magnitude response. (c) Pole locations of H(s)

5.4

Design of Specialized Analog Filters by Pole-Zero Placement

There are certain specialized filters often used in signal processing applications in addition to the filters designed in the previous sections. These specialized filters can be directly designed based on placement of poles and zeros.

266

5 Analog Filters

(a)

(b)

(c)

Figure 5.21 (a) Zero locations of H(s). (b) Magnitude response. (c) Phase response

5.4.1

Notch Filter

The notch filter removes a single frequency f0, called the notch frequency. The notch filter can be realized with two zeros placed at jΩ0, where

Ω0 ¼ ð2πf 0 Þ

As such a filter does not have unity gain at zero frequency. The notch will not be sharp. By placing two poles close to the two zeros on the semicircle as shown in Figure 5.22(a), the notch can be made sharp with unity gain at zero frequency as shown in Figure 5.22(b).

5.5 Problems

267

Figure 5.22 (a) Placing two poles close the two zeros on the semicircle. (b) Magnitude response of (a)

Example 5.11 Design a second-order notch filter to suppress 50 Hz hum in an audio signal Solution Choose Ω0 ¼ 100π. Place zeros at s ¼ jΩ0 and poles at –Ω0 cos θ jΩ0 sin θ. Then, the transfer function of the second-order notch filter is given by H ðsÞ ¼ ¼

5.5

ðs  jΩ0 Þðs þ jΩ0 Þ ðs þ Ω0 cos θ þ jΩ0 sin θÞðs þ Ω0 cos θ  jΩ0 sin θÞ s 2 þ Ω0 2 s2 þ 98775:5102 ¼ 2 2 2 s þ ð628:57 cos θÞs þ 98775:5102 s þ ð2Ω0 cos θÞs þ Ω0

Problems

1. Test the impulse response of an ideal low-pass filter for the following properties: (i) Real valued (ii) Even (iii) Causal

268

5 Analog Filters

2. Consider the first-order RC circuit shown in the figure below

(i) Determine H(Ω), the transfer function from vs to vc. Sketch the magnitude and phase of H(Ω). (ii) What is the cutoff frequency for H(Ω)? (iii) Consider the following system:

(a) Draw the corresponding RC circuit and determine H(Ω), the transfer function from v to vs. Sketch the magnitude and phase of H(Ω). (b) What is the corresponding cutoff frequency? 3. Design a continuous time low-pass filter with the following transfer function HðΩÞ ¼

α α þ jΩ

with the following specifications

Find the range of values of α that meets the specifications. 4. Consider the following system

Further Reading

269

If H(Ω) is an ideal band-pass filter, determine for what values of α, it will act as an ideal band-stop filter. 5. Design an elliptic analog high-pass filter for the specifications of Example 5.5

Further Reading 1. Raut, R., Swamy, M.N.S.: Modern Analog Filter Analysis and Design: a Practical Approach. Springer, WILEY- VCH Verlag & Co. KGaA, Weinheim, Germany (2010) 2. Antoniou, A.: Digital Filters: Analysis and Design. McGraw Hill Book Co., New York (1979) 3. Parks, T.W., Burrus, C.S.: Digital Filter Design. Wiley, New York (1987) 4. Temes, G.C., Mitra, S.K. (eds.): Modern Filter Theory and Design. Wiley, New York (1973) 5. Vlach, J.: Computerized Approximation and Synthesis of Linear Networks. Wiley, New York (1969) 6. Mitra, S.K.: Digital Signal Processing. McGraw-Hill, New York (2006) 7. Chen, C.T.: Digital Signal Processing, Spectral Computation and Filter Design. Oxford University Press, NewYork/Oxford, UK (2001)

Chapter 6

Discrete-Time Signals and Systems

Discrete-time signals are obtained by the sampling of continuous-time signals. Digital signal processing deals basically with discrete-time signals, which are processed by discrete-time systems. The characterization of discrete-time signals as well as discrete-time systems in time domain is required to understand the theory of digital signal processing. In this chapter, time-domain sampling and the fundamental concepts of discrete-time signals as well as discrete-time systems are considered. First, the sampling process of analog signals is described. Next, the basic sequences of discrete-time systems and their classification are emphasized. The input-output characterization of linear time-invariant (LTI) systems by means of convolution sum is described. Further, sampling of discrete-time signals is introduced. Finally, the state-space representation of discrete-time LTI systems is described.

6.1 6.1.1

The Sampling Process of Analog Signals Impulse-Train Sampling

The acquisition of an analog signal at discrete-time intervals is called sampling. The sampling process mathematically can be treated as a multiplication of a continuoustime signal x(t) by a periodic impulse train p(t) of unit amplitude with period T. For example, consider an analog signal xa(t) as shown in Figure 6.1(a), and a periodic pulse train p(t) of unit amplitude with period T as in Figure 6.1 (b) is referred to as the sampling function, the period T as the sampling period, and the fundamental frequency ωT ¼ (2π/T ) as the sampling frequency in radians. Then, the sampled version xp(t) is shown in Figure 6.1 (c).

© Springer International Publishing AG, part of Springer Nature 2018 K. D. Rao, Signals and Systems, https://doi.org/10.1007/978-3-319-68675-2_6

271

272

6 Discrete-Time Signals and Systems

xa (t )

p(t) 1

.... t

0

0

t

T 2T

(b)

(a) xp(t)

....

t

0 T 2T

(c) Figure 6.1 (a) Continuous-time signal, (b) pulse train, (c) sampled version of (b)

In the time domain, we have xp ðt Þ ¼ xa ðt Þpðt Þ

ð6:1Þ

where pð t Þ ¼

X1 n¼1

δðt  nT Þ

ð6:1aÞ

xp(t) is the impulse train with the amplitudes of the impulses equal to the samples of xa(t) at intervals T, 2T, 3T, . . . . Therefore, the sampled version of signal xp(t) mathematically can be represented as xp ð t Þ ¼

6.1.2

X1

x ðnT Þδðt n¼1 a

 nT Þ

ð6:2Þ

Sampling with a Zero-Order Hold

In Section 6.1.1, the sampling process establishes a fact that the band-limited signal can be uniquely represented by its samples. In a practical setting, it is difficult to generate and transmit narrow large amplitude pulses that approximate impulses.

6.1 The Sampling Process of Analog Signals

Lowpass filter

Sample and hold

Quantizer

2b

x (t)

Analog input

273

FT

• • •

Encoder

Logic circuit

1

x (n)

Digital output code

Figure 6.2 A block diagram representation of an analog-to-digital conversion process.

Hence, it is more convenient to implement the sampling process using a zero-order hold. It samples analog signal at a given sampling instant and holds the sample value until the succeeding sampling instant. A block diagram representation of the analogto-digital conversion (ADC) process is shown in Figure 6.2. The amplitude of each signal sample is quantized into one of the 2b levels, where b is the number of bits used to represent a sample in the ADC. The discrete amplitude levels are encoded into distinct binary word of length b bits. A sequence of samples x(n) is obtained from an analog signal xa(t) according to the relation xðnÞ ¼ xa ðnT Þ

1 < n < 1:

ð6:3Þ

In Eq. (6.2), T is the sampling period, and its reciprocal, FT ¼ 1/T is called the sampling frequency, in samples per second. The sampling frequency FT is also referred to as the Nyquist frequency. Sampling Theorem The sampling theorem states that an analog signal must be sampled at a rate at least twice as large as highest frequency of the analog signal to be sampled. This means that F T  2f max

ð6:4Þ

where fmax is maximum frequency component of the analog signal. The frequency 2fmax is called the Nyquist rate. For example, to sample a speech signal containing up to 3 kHz frequencies, the required minimum sampling rate is 6 kHz, that is, 6000 sample per second. To sample an audio signal having frequencies up to 22 kHz, the required minimum sampling rate is 44 kHz, that is, 44000 samples per second. A signal whose energy is concentrated in a frequency band range fL < |f| < fH is often referred to as a band-pass signal. The sampling process of such signals is generally referred to as band-pass sampling. In the band-pass sampling process, to prevent aliasing effect, the band-pass continuous-time signal can be sampled at sampling rate greater than twice the highest frequency ( fH): F T  2f H

ð6:5Þ

274

6 Discrete-Time Signals and Systems

The bandwidth of the band-pass signal is defined as Δf ¼ f H  f L

ð6:6Þ

Consider that the highest frequency contained in the signal is an integer multiple of the bandwidth that is given as f H ¼ cðΔf Þ

ð6:7Þ

The sampling frequency is to be selected to satisfy the condition as F T ¼ 2ðΔf Þ ¼

6.1.3

fH c

ð6:8Þ

Quantization and Coding

Quantization and coding are two primary steps involve in the process of A/D conversion. Quantization is a nonlinear and non-invertible process that rounds the given amplitude x(n) ¼ x(nT) to an amplitude xk that is taken from the finite set of values at time t ¼ nT. Mathematically, the output of the quantizer is defined as xq ðnÞ ¼ Q½xðnÞ ¼ b xk

ð6:9Þ

The procedure of the quantization process is depicted as

x 1 x1 x 2

x 2 x 3 x3 x 4 x4 x 5 x 5 ………………

The possible outputs of the quantizer (i.e., the quantization levels) are indicated by b x1 b x L where L stands for number of intervals into which the x2 b x3 b x4    b signal amplitude is divided. For uniform quantization, b x kþ1  b xk ¼ Δ k ¼ 1, 2,   , L: xkþ1  xk ¼ Δ for finite xk , xkþ1 : ð6:10Þ where Δ is the quantizer step size. The coding process in an ADC assigns a unique binary number to each quantization level. For L levels, at least L different binary numbers are needed. With word length of n bits, 2n distinct binary numbers can be represented. Then, the step size or the resolution of the A/D converter is given by Δ¼ where A is the range of the quantizer.

A 2n

ð6:11Þ

6.1 The Sampling Process of Analog Signals

275

(a)

(b)

(c) Figure 6.3 (a) Quantizer, (b) mathematical model, (c) power spectral density of quantization noise

Quantization Error Consider an n bit ADC sampling analog signal x(t) at sampling frequency of FTas shown in Figure 6.3(a). The mathematical model of the quantizer is shown in Figure 6.3(b). The power spectral density of the quantization noise with an assumption of uniform probability distribution is shown in Figure 6.3(c). If the quantization error is uniformly distributed in the range (‐Δ/2, Δ/2) as shown in Figure 6.3(b), the mean value of the error is zero, and the variance (the quantization noise power) σ 2e is given by Pqn ¼ σ 2e ¼

ð Δ=2 Δ=2

qe 2 ðnÞPðeÞde ¼

Δ2 12

ð6:12Þ

The quantization noise power can be expressed by σ 2e ¼

quantization step2 A2 1 A2 ¼  2n ¼ 22n 12 12 2 12

ð6:13Þ

276

6 Discrete-Time Signals and Systems

The effect of the additive quantization noise on the desired signal can be quantified by evaluating the signal-to-quantization noise (power) ratio (SQNR) that is defined as Px Pqn

SQNR ¼ 10log10

ð6:14Þ

h i   where Px ¼ σ 2x ¼ E x2 ðnÞ is the signal power and Pqn ¼ σ 2e ¼ E e2q ðnÞ is the quantization noise power.

6.2 6.2.1

Classification of Discrete-Time Signals Symmetric and Anti-symmetric Signals

A real valued signal x(n) is said to be symmetric if it satisfies the condition xðnÞ ¼ xðnÞ

ð6:15aÞ

Example of a symmetric sequence is shown in Figure 6.4 On the other hand, a signal x(n) is called anti-symmetric if it follows the condition xðnÞ ¼ xðnÞ

ð6:15bÞ

An example of anti-symmetric sequence is shown in Figure 6.5.

6.2.2

Finite and Infinite Length Sequences

A signal is said to be of finite length or duration if it is defined only for a finite time interval:

· ···

·

· · ·

·

·

·

·

-1 0

1

· Figure 6.4 An example of symmetric sequence

·

·

· ·

·

···

6.2 Classification of Discrete-Time Signals

· ···

·

277

· ·

· ·

·

-1 0

· 1

·

·

·

·

·

·

···

·

Figure 6.5 An example of anti-symmetric sequence

1 < N 1  n  N 2 < 1

ð6:16Þ

The length of the sequence is N ¼ N2  N1 þ 1. Thus, a finite sequence of length N has N samples. A discrete-time sequence consisting of N samples is called a Npoint sequence. Any finite sequence can be viewed as an infinite length sequence by adding zero-valued samples outside the range (N1, N2). Also, an infinite length sequence can be truncated to produce a finite length sequence.

6.2.3

Right-Sided and Left-Sided Sequences

A right-sided sequence is an infinite sequence x(n) for which x(n) ¼ 0 for n < N1, where N1 is a positive or negative integer. If N1  0, the right-sided sequence is said to be causal. Similarly, if x(n) ¼ 0 for n > N2, where N2 is a positive or negative integer, then the sequence is called a left-sided sequence. Also, if N2  0, then the sequence is said to be anti-causal.

6.2.4

Periodic and Aperiodic Signals

A sequence x(n) ¼ x(n + N ) for all n is periodic with a period N, where N is a positive integer. The smallest value of N for which x(n) ¼ x(n + N ) is referred as the fundamental period. A sequence is called aperiodic, if it is not periodic. An example of a periodic sequence is shown in Figure 6.6. Proposition 6.1 A discrete-time sinusoidal sequence x(n) ¼ A sin (ω0n + θ) is periodic if and only if ω2π0 is a rational number. The rational number is defined as the ratio of two integers. For the given periodic signal x(n) ¼ A sin (ω0n þ θ), its fundamental period N is obtained from the following relationship

278

6 Discrete-Time Signals and Systems

·

·

· ·

·

·

···

··· ·

· -5

-4

-3

·

·

·

-2

-1

0

1

2

-3

· ·

·

·

4

5

6

· 7

8

-9

·

10 11

· 12

Figure 6.6 An example of a periodic sequence

ω0 m ¼ 2π N 2π N¼ m ω0 The fundamental period of a discrete-time sinusoidal sequence satisfying the proposition 6.1 is calculated by setting m equal to a small integer that results in an integer value for N. The fundamental period of a discrete-time complex exponential sequence can also be calculated satisfying the proposition 6.1. Example 6.1 Determine if the discrete-time sequences are periodic:   (i) xðnÞ ¼ cos πn 4 (ii) x(n) ¼ sin2n   (iii) xðnÞ ¼ sin πn 4 þ cos 2n: jð5πn þθ (iv) xðnÞ ¼ e 8 Þ Solution (i) The value of ω0 in x(n) is π4. Since ω2π0 ¼ 18 is a rational number, it is periodic discrete-time sequence. The fundamental period of x(n) is given by N ¼ ω2π0 m:   For m ¼ 1, N ¼ 2π π4 ¼ 8. Hence, xðnÞ ¼ cos πn 4 is periodic with fundamental period N ¼ 8, (ii) x(n) ¼ sin 2n is aperiodic because ω0N ¼ 2N ¼ 2πm is not satisfied for any integer πnvalue of m in making N to be an integer. (iii) sin 4 is periodic and cos 2n is aperiodic. Since the sum of periodic and   aperiodic signals is aperiodic, the signal xðnÞ ¼ sin πn 4 þ cos 2n is aperiodic. ω0 5π 5 (iv) The value of ω0 in x(n) is 8 : Since 2π ¼ 16 is a rational number, it is periodic discrete-time sequence. The fundamental period of x(n) is given by N ¼ ω2π0 m: 5πn For m ¼ 5, N ¼ 8 2π 5 ¼ 16: Hence, xðnÞ ¼ ejð 8 þθÞ is periodic with funda5π

mental period N ¼ 16.

6.2 Classification of Discrete-Time Signals

6.2.5

279

Energy and Power Signals

The total energy of a signal x(n), real or complex, is defined as E¼

1 X

jxðnÞj2

ð6:17Þ

n¼1

By definition, the average power of an aperiodic signal x(n) is given by N X 1 jxðnÞj2 N!1 2N þ 1 n¼N

P ¼ Lt

ð6:18aÞ

The signal is referred to as an energy signal if the total energy of the signal satisfies the condition 0 < E < 1. It is clear that for a finite energy signal, the average power P is zero. Hence, an energy signal has zero average power. On the other hand, if E is infinite, then P may be finite or infinite. If P is finite and nonzero, then the signal is called a power signal. Thus, a power signal is an infinite energy signal with finite average power. The average power of a periodic sequence x(n) with a period I is given by P¼

I1 1X jxðnÞj2 I n¼0

ð6:18bÞ

Hence, periodic signals are power signals. Example 6.2 Determine whether the sequence x(n) ¼ anu(n) is an energy signal or a power signal or neither for the following cases: ðaÞjaj < 1, ðbÞjaj ¼ 1, ðcÞjaj > 1: Solution For x(n) ¼ anu(n), E is given by E¼ P ¼ limN!1

X1  X1  2  xðnÞ2 ¼  an  1 0

X1  XN    1 xðnÞ2 ¼ limN!1 1 a2n  1 0 2N þ 1 2N þ 1

(a) For |a| < 1, E¼ P ¼ limN!1

X1  X1  xðnÞ2 ¼ jan j2 ¼ 1 0

1 1  j aj 2

is finite

XN   1 1  jaj2ðNþ1Þ a2n  ¼ limN!1 1 ¼0 n¼0 2N þ 1 2N þ 1 1  jaj2

280

6 Discrete-Time Signals and Systems

The energy E is finite and the average power P is zero. Hence, the signal x(n) ¼ an u(n) is an energy signal for |a| < 1. (b) For |a| ¼ 1, P n 2 E¼ 1 0 ja j ! 1 1 X N  2n  N þ1 1 ¼ a ¼ limn!1 P ¼ limN!1 n¼0 2N þ 1 2N þ 1 2 The energy E is infinite, and the average power P is finite. Hence, the signal x (n) ¼ anu(n) is a power signal for |a| ¼ 1. (c) For |a| > 1, E¼ P ¼ limN!1

X1 0

jan j2 ! 1

XN   1 jaj2ðNþ1Þ  1 a2n  ¼ limN!1 1 !1 n¼0 2N þ 1 2N þ 1 jaj2  1

The energy E is infinite and also the average power P is infinite. Hence, the signal x(n) ¼ anu(n) is neither an energy signal nor a power signal for |a| > 1. Example 6.3 Determine whether the following sequences (i) x(n) ¼ e–nu(n), (ii) x(n) ¼ enu(n), (iii) x(n) ¼ nu(n), and (iv) x(n) ¼ cosπn u(n) are energy or power signals or neither energy nor power signals. Solution (i) x(n) ¼ enu(n). Hence, E and P are given by X1 1 2 x ð n Þ ¼ e2n ¼ is finite j j 1 0 1  e2 1 XN P ¼ limN!1 j x ð nÞ j 2 n¼0 2N þ 1 1 X N 2n ¼ limN!1 e n¼0 2N þ 1

E¼

X1

¼ limN!1

1 1  e2ðNþ1Þ ¼0 2N þ 1 1  e2

The energy E is finite and the average power P is zero. Hence, the signal x(n) ¼ enu(n) is an energy signal.

6.3 Discrete-Time Systems

281

(ii) x(n) ¼ e+nu(n). Therefore, E and P are given by E¼

P1

1

j x ð nÞ j 2 ¼

P1 0

e2n ! 1

N M 1 X 1 X 1 e2ðNþ1Þ  1 !1 e2n ¼ lim jxðnÞj2 ¼ lim n!1 2N þ 1 N!1 2N þ 1 N!1 2N þ 1 e2  1 n¼0 n¼0

P ¼ lim

The energy E is infinite and also the average power P is infinite. Hence, the signal x(n) ¼ enu(n) is neither an energy signal nor a power signal. (iii) x(n) ¼ nu(n). Hence, E and P are given by E¼

P1

1

jxðnÞj2 ¼

P1 0

n2 ! 1

1 X1 jxðnÞj2 1 2N þ 1 1 XN 2 NðN þ 1Þð2N þ 1Þ !1 ¼ limN!1 n ¼ limN!1 n¼0 2N þ 1 6ð2N þ 1Þ

P ¼ limN!1

The energy E is infinite and also the average power P is infinite. Hence, the signal x(n) ¼ nu(n) is neither an energy signal nor a power signal. (iv) x(n) ¼ cos πn u(n). Sincecosπn ¼ (1)n, E and P are given by E¼

P1

1

jxðnÞj2 ¼

P1 0

j cos πnj2 ¼

P1 0

ð1Þ2n ! 1

X1 1 jxðnÞj2 1 2N þ 1 XN 1 Nþ1 1 ¼ ¼ limN!1 ð1Þ2n ¼ limN!1 n¼0 2N þ 1 2N þ 1 2

P ¼ limN!1

The energy E is not finite and the average power P is finite. Hence, the signal x(n) ¼ cos πnu(n) is a power signal.

6.3

Discrete-Time Systems

A discrete-time system is defined mathematically as a transformation that maps an input sequence x(n) into an output sequence y(n). This can be denoted as yðnÞ ¼ ℜ½xðnÞ where ℜ is an operator.

ð6:19Þ

282

6.3.1

6 Discrete-Time Signals and Systems

Classification of Discrete-Time Systems

Linear Systems A system is said to be linear if and only if it satisfies the following conditions: ℜ½axðnÞ ¼ aℜ½xðnÞ

ð6:20Þ

ℜ½x1 ðnÞ þ x2 ðnÞ ¼ ℜ½x1 ðnÞ þ ℜ½x2 ðnÞ ¼ y1 ðnÞ þ y2 ðnÞ

ð6:21Þ

where a is an arbitrary constant and y1(n) and y2(n) are the responses of the system when x1(n) and x2(n) are the respective inputs. Equations (6.20) and (6.21) represent the homogeneity and additivity properties, respectively. The above two conditions can be combined into one representing the principle of superposition as ℜ½ax1 ðnÞ þ bx2 ðnÞ ¼ aℜ½x1 ðnÞ þ bℜ½x2 ðnÞ

ð6:22Þ

where a and b are arbitrary constants. Example 6.4 Check for linearity of the following systems described by the following input-output relationships: Xn (i) yðnÞ ¼ xðk Þ k¼1 2 (ii) y(n) ¼ x (n) (iii) y(n) ¼ x(n  n0), where n0 is an integer constant Solution (i) The outputs y1(n) and y2(n) for inputs x1(n) and x2(n) are, respectively, given by y1 ðnÞ ¼ y2 ðnÞ ¼

n X

x1 ðkÞ

k¼1 n X

x2 ðkÞ

k¼1

The output y(n) due to an input x(n) ¼ ax1(n) þ bx2(n) is then given by y ð nÞ ¼

n X k¼1

ax1 ðk Þ þ bx2 ðkÞ ¼ a

n X k¼1

x1 ð k Þ þ b

n X

x2 ð k Þ

k¼1

¼ ay1 ðnÞ þ by2 ðnÞ Xn xðkÞ is a linear system. Hence the system described by yðnÞ ¼ k¼1

6.3 Discrete-Time Systems

283

(ii) The outputs y1(n) and y2(n) for inputs x1(n) and x2(n) are given by y1 ðnÞ ¼ x21 ðnÞ y2 ðnÞ ¼ x22 ðnÞ The output y(n) due to an input x(n) ¼ ax1(n) þ bx2(n) is then given by yðnÞ ¼ ðax1 ðnÞ þ bx2 ðnÞÞ2 ¼ a2 x21 ðnÞ þ 2abx1 ðnÞx2 ðnÞ þ b2 x22 ðnÞ ay1 ðnÞ þ by2 ðnÞ ¼ ax21 ðnÞ þ bx22 ðnÞ 6¼ yðnÞ Therefore, the system y(n) ¼ x2(n) is not linear. (iii) The outputs y1(n) and y2(n) for inputs x1(n) and x2(n), respectively, are given by y 1 ð nÞ ¼ x 1 ð n  n0 Þ y 2 ð nÞ ¼ x 2 ð n  n0 Þ The output y(n) due to an input x(n) ¼ ax1(n) þ bx2(n) is then given by yðnÞ ¼ ax1 ðn  n0 Þ þ bx2 ðn  n0 Þ ¼ ay1 ðnÞ þ by2 ðnÞ Hence, the system y(n) ¼ x(n  n0) is linear. Time-Invariant Systems A time-invariant system (shift invariant system) is one in which the internal parameters do not vary with time. If y1(n) is output to an input x1(n), then the system is said to be time invariant if, for all n0, the input sequence x1(n) ¼ x(n  n0) produces the output sequence y1(n) ¼ y(n  n0), i.e., ℜ½xðn  n0 Þ ¼ yðn  n0 Þ where n0 is a positive or negative integer. Example 6.5 Check for time-invariance of the system defined by P1 k¼1 xðk Þ

y ð nÞ ¼

Solution From given Eq., the output y(n) of the system delayed by n0 can be written as nn X0

y ð n  n0 Þ ¼

xð k Þ

k¼1

For example, for an input x1(n) ¼ x(n  n0), the output y1(n) can be written as y 1 ð nÞ ¼

n X k¼1

x ð k  n0 Þ

284

6 Discrete-Time Signals and Systems

Substitution of the change of variables k1 ¼ k  n0 in the above summation yields y 1 ð nÞ ¼

nn X0

x ð k 1 Þ ¼ y ð n  n0 Þ

k1 ¼1

Hence, it is a time-invariant system. Example 6.6 Check for time-invariance of the down-sampling system with a factor of 2, defined by the relation yðnÞ ¼ xð2nÞ

1 < n < 1

Solution For an inputx1(n) ¼ x(n  n0), the output y1(n) of the compressor system can be written as y1 ðnÞ ¼ xð2n  n0 Þ From given equation, y ð n  n0 Þ ¼ x ð 2ð n  n0 Þ Þ Comparing the above two equations, it can be observed that y1(n) 6¼ y(n  n0). Thus, the down-sampling system is not time invariant. Causal System A system is said to be causal if its output at time instant n depends only on the present and past input values, but not on the future input values. For example, a system defined by yðnÞ ¼ xðn þ 2Þ  xðn þ 1Þ is not causal, as the output at time instant n depends on future values of the input. But, the system defined by y ð nÞ ¼ x ð nÞ  x ð n  1Þ is causal, since its output at time instant n depends only on the present and past values of the input. Stable System A system is said to be stable if and only if every bounded-input sequence produces a bounded-output sequence. The input x(n) is bounded if there exists a fixed positive finite value βx such that jxðnÞj  βx < 1

for all n

ð6:23Þ

Similarly, the output y(n) is bounded if there exists a fixed positive finite value βy such that for all n ð6:24Þ jyðnÞj  βy < 1 and this type of stability is called bounded-input bounded-output (BIBO) stability.

6.3 Discrete-Time Systems

285

Example 6.7 Check for stability of the system described by the following inputoutput relation y ð nÞ ¼ x 2 ð nÞ Solution Assume that the input x(n) is bounded such that |x(n)|  βx < 1 for all n jyðnÞj ¼ jxðnÞj2  β2x < 1

Then,

Hence, y(n) is bounded and the system is stable. Example 6.8 Check for stability, causality, linearity, and time-invariance of the system described by ℜ[x(n)] ¼ (1)nx(n) This transformation outputs the current value of x(n) multiplied by either 1. It is stable, since it does not change the magnitude of x(n) and hence satisfies the conditions for bounded-input bounded-output stability. It is causal, because each output depends only on the current value of x(n). Let

y1 ðnÞ ¼ ℜ½x1 ðnÞ ¼ ð1Þn x1 ðnÞ

y2 ðnÞ ¼ ℜ½x2 ðnÞ ¼ ð1Þn x2 ðnÞ Then,

ℜ½ax1 ðnÞ þ bx2 ðnÞ ¼ ð1Þn ax1 ðnÞ þ ð1Þn bx2 ðnÞ ¼ ay1 ðnÞ þ by2 ðnÞ

Hence, it is linear. yðnÞ ¼ ℜ½xðnÞ ¼ ð1Þn xðnÞ

ℜ½xðn  1Þ ¼ ð1Þn xðn  1Þ

ℜ½xðn  1Þ 6¼ yðn  1Þ Therefore, it is not time invariant. Example 6.9 Check for stability, causality, linearity, and time-invariance of the system described by ℜ[x(n)] ¼ x(n2) Solution Stable, since if x(n) is bounded, x(n2) is also bounded. It is not causal, since, for example, if n ¼ 4, then the output y(n) depends upon the future input because y(4) ¼ ℜ[x(4)] ¼ x(16) y1 ¼ ℜ½x1 ðnÞ ¼ x1 ðn2 Þ; y2 ðnÞ ¼ ℜ½x2 ðnÞ ¼ x2 ðn2 Þ; ℜ½ax1 ðnÞ þ bx2 ðnÞ ¼ ax1 ðn2 Þ þ bx2 ðn2 Þ ¼ ay1 ðnÞ þ by2 ðnÞ Therefore, it is linear. yðnÞ ¼ ℜ½xðnÞ ¼ xðn2 Þ ℜ½xðn  1Þ 6¼ yðn  1Þ Hence, it is not time invariant.

286

6.3.2

6 Discrete-Time Signals and Systems

Impulse and Step Responses

Let the input signal x(n) be transformed by the system to generate the output signal y(n). This transformation operation is given by y ð nÞ ¼ ℜ½ x ð nÞ 

ð6:25Þ

If the input to the system is a unit sample sequence (i.e., impulse input δ(n)), then the system output is called as impulse response and denoted by h(n). If the input to the system is a unit step sequence u(n), then the system output is called its step response. In the next section, we show that a linear time-invariant discrete-time system is characterized by its impulse response or step response.

6.4

Linear Time-Invariant Discrete-Time Systems

Linear time-invariant systems have significant signal processing applications, and hence it is of interest to study the properties of such systems.

6.4.1

Input-Output Relationship

An arbitrary sequence x(n) can be expressed as a weighted linear combination of unit sample sequences given by x ð nÞ ¼

X1 k¼1

xðkÞδðn  k Þ

ð6:26Þ

Now, the discrete-time system response y(n) is given by y ð nÞ ¼ ℜ½ x ð nÞ  ¼ ℜ

hX1

xðkÞδðn  k Þ k¼1

i

ð6:27Þ

From the principle of superposition, the above equation can be written as yðnÞ ¼

X1 k¼1

xðkÞℜ½δðn  kÞ

Let the response of the system due to input δ(n  k) be hk(n), that is, hk ðnÞ ¼ ℜ½δðn  kÞ Then, the system response y(n) for an arbitrary input x(n) is given by yðnÞ ¼

X1 k¼1

xðkÞhk ðnÞ

ð6:28Þ

6.4 Linear Time-Invariant Discrete-Time Systems

287

Since δ(n  k) is a time-shifted version of δ(n), the response hk(n) is the timeshifted version of the impulse response h(n), since the operator is time invariant. Hence,hk(n) ¼ h(n ‐ k). Thus, y ð nÞ ¼

X1 k¼1

xðk Þhðn  kÞ

ð6:29Þ

The above equation for y(n) is commonly called the convolution sum and represented by yðnÞ ¼ xðnÞ∗hðnÞ

ð6:29aÞ

where the symbol * stands for convolution. The discrete-time convolution operates on the two sequences x(n) and h(n) to produce the third sequence y(n). Example 6.10 Determine discrete convolution of the following sequences for large value of n: hðnÞ ¼

1n 5

uð nÞ

xðnÞ ¼ ð1Þn uðnÞ yðnÞ ¼ xðnÞ∗hðnÞ P ¼ 1 k¼1 xðk Þhðn  k Þ 1 n  k X 1k X 1 ¼ uðk Þð1Þnk uðn  kÞ ¼ ð1Þn ð1Þk 5 5 k¼1 k¼0  1nþ1

  n 1  5 X 1 k   ¼ ð1Þn  ¼ ð1Þn 1 5 k¼0 1  5  1nþ1

1  5 ¼ ð1Þn 1 1þ 5  1nþ1 tends to zero and hence, For large n, 5

Solution

yðnÞ ¼ ð1Þn

1 1:2

Example 6.11 Determine discrete convolution of the following two finite duration sequences:  n 1 uð nÞ 3  n 1 uð nÞ x ð nÞ ¼ 5

hð nÞ ¼

288

6 Discrete-Time Signals and Systems

Solution The impulse response h(n) ¼ 0 for n < 0; hence the given system is causal; and x(n) ¼ 0 for n < 0, therefore the sequence x(n) is causal sequence: yðnÞ ¼ xðnÞ∗hðnÞ ¼

P n 1k 1nk k¼0 5

3

¼

1n P n 3k 3

k¼0 5

 n 1  ð3=5Þnþ1 ¼ 13 1  ð3=5Þ

6.4.2

Computation of Linear Convolution

Matrix Method If the input x(n) is of length N1 and the impulse sequence h(n) is of length N2, then the convolution sequence is of length N1 + N2  1. Thus, the linear convolution given by Eq. (6.29) can be written in matrix form as 3 3 2 2 xð0Þ 0 0  0 yð0Þ 7 7 6 6 7 6 xð1Þ 6 xð0Þ 0  0 7 yð1Þ 7 7 6 6 7 7 6 6 7 7 6 xð2Þ 6 xð1Þ xð0Þ    0 yð2Þ 7 7 6 6 7 7 6 6 7 6 6 ⋮ xð2Þ xð1Þ  ⋮ 7 yð3Þ 7 7 6 6 7 7 6 6 7 7 ¼ 6 xðN 1  1Þ 6 ⋮ ⋮ xð2Þ    ⋮ 7 7 6 6 7 7 6 6 6 yðN 1  1Þ 7 6 0 xðN 1  1Þ ⋮  ⋮ 7 7 7 6 6 7 7 6 6 7 7 6 6 yðN 0 0 xðN Þ  1Þ    ⋮ 1 1 7 7 6 6 7 7 6 6 7 6 6 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 7 5 5 4 4 yðN 1 þ N 2  2Þ 0 0 0    xð0Þ 3 2 hð0Þ 7 6 6 hð1Þ 7 7 6 7 6 6 hð2Þ 7 7 6 7 6 6 hð3Þ 7 7 6 7 6 7 6 ⋮ 7 6 7 6 6 hðN 2  1Þ 7 7 6 7 6 7 6 0 7 6 7 6 7 6 ⋮ 5 4 0 ð6:30Þ The following example illustrates the above procedure for computation of linear convolution.

6.4 Linear Time-Invariant Discrete-Time Systems

289

Example 6.12 Find the convolution of the sequences x(n) ¼ {6, 3} and h(n) ¼ {3, 6, 3}. Solution Using Eq. (6.19), the linear convolution of x(n) and h(n) is given by 2

yð0Þ

2

3

6

7 6 6 6 yð1Þ 7 6 3 7 6 6 7¼6 6 6 yð2Þ 7 6 0 5 4 4 0

yð3Þ

0

0

6

0

3

6

0

3

0

32

3

3

2

18

3

7 6 76 7 7 6 6 7 0 7 76 6 7 6 45 7 7¼6 76 7 7 6 6 7 0 7 54 3 5 4 0 5 9 0 6

Thus, yðnÞ ¼ xðnÞ∗hðnÞ ¼ f18; 45; 0; 9g Graphical Method for Computation of Linear Convolution Evaluation of sum at any sample n consists of the following four important operations: (i) Time reversing or reflecting of the sequence h(k) about k ¼ 0 sample to give h(–k). (ii) Shifting the sequence h(–k) to the right by n samples to obtain h(n – k). (iii) Forming the product x(k)h(n – k) sample by sample for the desired value of n. (iv) Summing the product over the index k in y(n) for the desired value of n. The length of the convolution sum sequence y(n) is given by n ¼ N1 + N2  1, where N1 is length of the sequence x(n) and N2 is length of the sequence h(n). Example 6.13 Compute the convolution of the sequences of Example 6.12 using the graphical method. Solution The sequences x(n) and h(n) are shown in Figures 6.7.

6

6

h (n )

x ( n)

3 0 n

0

1

1 -3

-3 Figure 6.7 Sequences x(n) and h(n)

2

n

290 Figure 6.8 Convolution of sequences x(n) and h(n)

6 Discrete-Time Signals and Systems

6.4 Linear Time-Invariant Discrete-Time Systems

291

Figure 6.9 Sequence generated by the convolution

6.4.3

Computation of Convolution Sum Using MATLAB

The MATLAB function conv(a,b) can be used to compute convolution sum of two sequences a and b as illustrated in the following example. Example 6.14 Compute convolution sum of the sequences x(n) ¼ {2,1,0,0} and h(n) ¼ {1,2,1}, using MATLAB. Program 6.1. Illustration of convolution a=[ 2 -1 0 0 ];% first sequence b=[-1 2 1];% second sequence c=conv(a,b);% convolution of first sequence and second sequence len=length(c)-1; n=0:1:len; stem(n,c) xlabel(‘Time index n’); ylabel(‘Amplitude’); axis([0 5 -3 5])

6.4.4

Some Properties of the Convolution Sum

Starting with the convolution sum given by (6.30), namely, y(n) ¼ x(n) * h(n), we can establish the following properties:

292

6 Discrete-Time Signals and Systems

1) The convolution sum obeys the commutative law xðnÞ∗hðnÞ ¼ hðnÞ∗xðnÞ

ð6:31aÞ

2) The convolution sum obeys the associative law ðxðnÞ∗h1 ðnÞÞ∗h2 ðnÞ ¼ xðnÞ∗ðh1 ðnÞ∗h2 ðnÞÞ

ð6:31bÞ

3) The convolution sum obeys the distributive law xðnÞ∗ðh1 ðnÞ þ h2 ðnÞÞ ¼ xðnÞ∗h1 ðnÞ þ xðnÞ∗h2 ðnÞ

ð6:31cÞ

Let us now interpret the above relations physically. 1) The commutative law shows that the output is the same if we interchange the roles of the input and the impulse response. This is illustrated in Figure 6.10. 2) To interpret the associative law, we consider a cascade of two systems whose impulse responses are h1(n) and h2(n). Then y1(n) ¼ x(n) * h(n) if x(n) is the input to the system with the impulse response h1(n). If y1(n) is now fed as the input to the system with impulse response h2(n), then the overall system output is given by yðnÞ ¼ y1 ðnÞ∗h2 ðnÞ ¼ ½xðnÞ∗h1 ðnÞ∗h2 ðnÞ ¼ xðnÞ∗½h1 ðnÞ∗h2 ðnÞ, by associative law

¼ xðnÞ∗hðnÞ

ð6:32Þ

This equivalence is shown in Figure 6.11. Hence, if two systems with impulse responses h1(n) and h2(n) are cascaded, then the overall system response is given by Figure 6.10 Interpretation of the commutative law

y1 ( n)

x( n)

h1 (n)

y (n )

h2( n)

≡

Figure 6.11 Interpretation of the associative law

x( n)

h(n) = h1(n)* h2 (n)

y(n)

6.4 Linear Time-Invariant Discrete-Time Systems

h1( n)

y1 ( n)

y(n)

x( n) h2 ( n)

293

≡

x( n)

h1(n) + h2 (n)

y(n)

y2 ( n)

Figure 6.12 Interpretation of distributive law

x ( n)

Figure 6.13 Input-output relations for Example 6.15

y(- n)

x (n)

h (n)

h (n) h1 (n)

hðnÞ ¼ h1 ðnÞ∗h2 ðnÞ

y (n )

y1 ( n)

y2 ( n)

ð6:33Þ

This can be generalized to a number of LTI systems in cascade. 3) We now consider the distributive law given by (6.31c). This can be easily interpreted as two LTI systems in parallel and that the overall system impulse response h(n) of the two systems in parallel is given by h ð n Þ ¼ h1 ð nÞ þ h 2 ð nÞ

ð6:34Þ

This is illustrated in Figure 6.12. Example 6.15 Consider the system shown in Figure 6.13 with h(n) being real. If y2(n) ¼ y1(–n), find the overall impulse response h1(n) that relates y2(n) to x(n). Solution

yðnÞ ¼ xðnÞ∗hðnÞ

From Figure 6.13, we have the following relations: y1(n) ¼ y(n) ∗ h(n) y2 ðnÞ ¼ y1 ðnÞ ¼ yðnÞ∗hðnÞ ¼ ðxðnÞ∗hðnÞÞ∗hðnÞ ¼ xðnÞ∗ðhðnÞ∗hðnÞÞ ¼ xðnÞ∗h1 ðnÞ Hence, the overall impulse response ¼ h1(n) ¼ h(n) ∗ h(–n)

294

6 Discrete-Time Signals and Systems

x ( n)

h1 (n )

h2 (n )

h2 (n )

y ( n)

Figure 6.14 Interconnection of three causal LTI systems

Example 6.16 Consider the cascade interconnection of three causal LTI systems as shown in Figure 6.14. The impulse response h2(n) is given by h2 ð nÞ ¼ uð nÞ  uð n  2Þ and the overall impulse response h(n) ¼ {1,5,10,11,8,4,1}. Determine the impulse response h1(n). Solution Let the overall impulse response of the cascaded system be h(n). Hence, hðnÞ ¼ h1 ðnÞ∗h2 ðnÞ∗h2 ðnÞ Since the convolution is associative in nature, hðnÞ ¼ h1 ðnÞ∗ðh2 ðnÞ∗h2 ðnÞÞ Let h3(n) ¼ h2(n) * h2(n). Since h2(n) is nonzero for n ¼ 0 and 1 only, h3(n) can be written as h3 ð nÞ ¼

1 X

h2 ðkÞh2 ðn  kÞ

k¼0

Therefore, h3 ð0Þ ¼

X1 k¼0

h3 ð 1Þ ¼

h2 ðkÞh2 ðkÞ ¼ 1:1 þ 1:0 ¼ 1

1 X

h2 ðkÞh2 ð1  k Þ ¼ 1:1 þ 1:1 ¼ 2

k¼0

h3 ð 2Þ ¼

1 X

h2 ðkÞh2 ð2  k Þ ¼ 1:0 þ 1:1 ¼ 1

k¼0

Thus, we obtain h3 ðnÞ ¼ f1; 2; 1g Now, h(n) is nonzero in the interval 0 to 6 and h3(n) is nonzero in the interval 0 to 2: hðnÞ ¼ h1 ðnÞ∗h3 ðnÞ Hence, h1(n) will be nonzero in the interval 0 to 4. Then, we have

6.4 Linear Time-Invariant Discrete-Time Systems

hðnÞ ¼ h1 ðnÞ∗h3 ðnÞ ¼

295 4 X

h1 ðkÞh3 ðn  k Þ

k¼0

Let h1(n) ¼ {a1, a2, a3, a4, a5}. Therefore, we have hð 0Þ ¼

4 X

h1 ðkÞh3 ðk Þ ¼ a1  1 ¼ 1

k¼0

hð 1Þ ¼

4 X

) a1 ¼ 1: h1 ðkÞh3 ð1  k Þ ¼ a1  1 þ a2  2 ¼ 5

k¼0

hð 2Þ ¼

4 X

) a2 ¼ 3 h1 ðkÞh3 ð2  k Þ ¼ a1  1 þ a2  2 þ a3  1 ¼ 101

k¼0

hð 3Þ ¼

4 X

) a3 ¼ 3: h1 ðkÞh3 ð3  k Þ ¼ a2  1 þ a3  2 þ a4  1 ¼ 11

k¼0

hð 4Þ ¼

4 X

) a4 ¼ 2: h1 ðkÞh3 ð4  k Þ ¼ a3  1 þ a4  2 þ a5  1 ¼ 8

k¼0

) a5 ¼ 1:

Thus, h1 ðnÞ ¼ f1; 3; 3; 2; 1g

6.4.5

Stability and Causality of LTI Systems in Terms of the Impulse Response

The output of a LTI system can be expressed as    X 1 1 X   hðkÞxðn  k Þ  j y ð nÞ j ¼  jhðkÞjjxðn  kÞj  k¼1 k¼1 For bounded input x(n) j x ð nÞ j  β x < 1 we have

296

6 Discrete-Time Signals and Systems 1 X

jyðnÞj  βx

jhðk Þj

ð6:35Þ

k¼1

X1 It is seen from (6.35) that y(n) is bounded if and only if jhðk Þj is k¼1 bounded. Hence, the necessary and sufficient condition for stability is that S¼

X1 k¼1

jhðkÞj < 1:

ð6:36Þ

The output y(n0) of a LTI causal system can be expressed as y ð n0 Þ ¼

1 X

hðkÞxðn0  kÞ

k¼1

¼ hð1Þxðn0 þ 1Þ þ . . . . . . : þ hð2Þxðn0 þ 2Þ þ hð1Þxðn0 þ 1Þ þhð0Þxðn0 Þ þ hð1Þxðn0  1Þ þ hð2Þxðn0  2Þ þ . . . :: For a causal system, the output at n ¼ n0 should not depend on the future inputs. Hence, in the above equation, h(k) ¼ 0 for k < 0. Thus, it is clear that for causality of a LTI system, its impulse response sequence hðnÞ ¼ 0

for n < 0:

ð6:37Þ

Example 6.17 Check for the stability of the systems with the following impulse responses: (i) Ideal delay, h(n) ¼ δ(n  nd); (ii) forward difference, h(n) ¼ δ(n þ 1)  δ(n). (iii) Backward difference, h(n) ¼ δ(n)  δ(n  1); (iv) h(n) ¼ u(n). (v) h(n) ¼ anu(n), where |a| < 1, and (vi) h(n) ¼ anu(n), where |a|  1. Solution Given impulse responses of the systems, stability of each system can be tested by computing the sum S¼

X1 k¼1

jhðk Þj

In case of (i), (ii), and (iii), it is clear that S < 1. As such, the systems corresponding to (i), (ii), and (iii) are stable. For the impulse response given in (iv), the system is unstable since S¼

1 X

uðnÞ ¼ 1:

n¼0

This is an exampleX of an infinite-duration impulse response (IIR) system. 1 In case of (v), S ¼ jajn . For |a| < 1, S < 1, and hence the system is stable. n¼0 This is an example of a stable IIR system. Finally, in case of (vi), |a|  1, and the sum is infinite, making the system unstable.

6.5 Characterization of Discrete-Time Systems

297

Example 6.18 Check the followingsystems for causality: n n  n (i) hðnÞ ¼ 34 uðnÞ, (ii) hðnÞ ¼ 12 uðn þ 2Þ þ 34 uðnÞ, 1n 3jnj (iii) hðnÞ ¼ 2 uðn  1Þ, (iv) hðnÞ ¼ 4 , and (v) h(n) ¼ u(n þ 1)  u(n) Solution (i) h(n) ¼ 0 for n < 0; hence the system is causal. (ii) h(n) 6¼ 0 for n < 0; hence the system is not causal. (iii) h(n) 6¼ 0 for n < 0; thus, the system is not causal.  jnj (iv) hðnÞ ¼ 34 ; hence h(n) 6¼ 0 for n < 0; so, the system is not causal. (v) h(n) ¼ u(n þ 1) – u(n), and h(n) 6¼ 0 for n < 0; so, the system is not causal. Example 6.19  n Check the following systems for stability: (i) hðnÞ ¼ 13 uðn  1Þ, (ii) h(n) ¼ u(n þ 2)  u(n  5), (iii) h(n) ¼ 5nu(n  3),   1jnj   (iv) hðnÞ ¼ sin nπ cos πn 4 uðnÞ, and (v) hðnÞ ¼ 2 4 Solution

X1 (i) The system is stable, since S ¼ jhðk Þj < 1: k¼1 (ii) h(n) ¼ u(n þ 2) – u(n – 5). The system is stable, since S is finite. 3 1  n X X X 1 (iii) h(n) ¼ 5nu(–n – 3). Hence, 5n ¼ < 1: Therefore, jhðnÞj ¼ 5 n n¼1 n¼3 the system is nπstable.  (iv) hðnÞ ¼ sin 4 uðnÞ

Summing |h(n)| over all positive n, we see that S tends to infinity. Hence, the system is not stable.  jnj   (v) hðnÞ ¼ 12 cos πn 4 X1  jnj |h(n)| is upper bounded by 12 . Thus, S ¼ jhðk Þj < 1: Hence the k¼1 system is stable.

6.5

Characterization of Discrete-Time Systems

Discrete-time systems are characterized in terms of difference equations. An important class of LTI discrete-time systems is one that is characterized by a linear difference equation with constant coefficients. Such a difference equation may be of two types, namely, non-recursive and recursive.

298

6.5.1

6 Discrete-Time Signals and Systems

Non-Recursive Difference Equation

A non-recursive LTI discrete-time system is one that can be characterized by a linear constant coefficient difference equation of the form 1 X

y ð nÞ ¼

bm x ð n  m Þ

ð6:38Þ

m¼1

where bm’s represent constants. By assuming causality, the above equation can be written as y ð nÞ ¼

1 X

bm xðn  mÞ

ð6:39Þ

m¼0

In addition, if x(n) ¼ 0 for n < 0 and bm ¼ 0 for m > N, then Eq. (6.39) becomes y ð nÞ ¼

N X

bm xðn  mÞ

ð6:40Þ

m¼0

Thus an LTI, causal, non-recursive system can be characterized by an Nth-order linear non-recursive difference equation. The Nth-order non-recursive difference equation has a finite impulse response (FIR). Therefore, an FIR filter is characterized by a non-recursive difference equation.

6.5.2

Recursive Difference Equation

The response of a discrete-time system depends on the present and previous values of the input as well as the previous values of the output. Hence a linear time-invariant causal, recursive discrete-time system can be represented by the following Nth-order linear recursive difference equation: y ð nÞ ¼

N X m¼0

bm xðn  mÞ 

N X

am yðn  mÞ

ð6:41Þ

m¼1

where am and bm are constants. An Nth-order recursive difference equation has an infinite impulse response. Hence, an infinite impulse response (IIR) filter is characterized by a recursive difference equation. Example 6.20 An initially relaxed LTI system was tested with an input signal x(n) ¼ u(n) and found to have a response as shown in Table 6.1. (i) Obtain the impulse response of the system. (ii) Deduce the difference equation of the system.

6.5 Characterization of Discrete-Time Systems

299

Table 6.1 Response of an LTI system for an input x(n) ¼ u(n) n y(n)

1 1

2 2

3 4

4 6

. . .. . . . . . .. . . .

5 10

. . .. . . . . . .. . . .

100 10

Solution (i) From Table 6.1, it can be observed that the response y(n) for an input x(n) ¼ u(n) is given by yðnÞ ¼ f1; 2; 4; 6; 10; 10; 10; . . . . . . ::g Similarly, for an input x(n) ¼ u(n1), the response y(n-1) is given by yðn  1Þ ¼ f0; 1; 2; 4; 6; 10; 10; 10; . . . . . . ::g For an input x(n) ¼ u(n)u(n1), the response of an LTI system is the impulse response h(n) given by hðnÞ ¼ yðnÞ  yðn  1Þ ¼ f1; 1; 2; 2; 4g (ii) The difference equation is given by y ð nÞ ¼

4 X

hðmÞxðn  mÞ

m¼0

Hence, the difference equation of the system can be written as yðnÞ ¼ xðnÞ þ 1xðn  1Þ þ 2xðn  2Þ þ 2xðn  3Þ þ 4xðn  4Þ

6.5.3

Solution of Difference Equations

A general linear constant coefficient difference equation can be expressed as y ð nÞ ¼ 

XN

a yð n  k Þ þ k¼1 k

XM k¼0

bk x ð n  k Þ

ð6:42Þ

The solution of the difference equation is the output response y(n). It is the sum of two components which can be computed independently as yðnÞ ¼ yc ðnÞ þ yp ðnÞ

ð6:43aÞ

where yc(n) is called the complementary solution and yp(n) is called the particular solution.

300

6 Discrete-Time Signals and Systems

The complementary solution yc(n) is obtained by setting x(n) ¼ 0 in Eq. (6.42). Thus yc(n) is the solution of the following homogeneous difference equation: N X

ak y ð n  k Þ ¼ 0

ð6:43bÞ

k¼0

where a0 ¼ 1. To solve the above homogeneous difference equation, let us assume that y c ð nÞ ¼ λ n

ð6:43cÞ

where the subscript c indicates the solution to the homogeneous difference equation. Substituting yc(n) in Eq. (6.43b), the following equation can be obtained: PN k¼0

ak λnk ¼ 0

  ¼ λnN λN þ a1 λN1 þ . . . :: þ aN1 λ þ aN ¼ 0

ð6:44Þ

which takes the form: λN þ a1 λN1 þ . . . :: þ aN1 λ þ aN ¼ 0

ð6:45Þ

The above equation is called the characteristic equation, which consists of N roots represented by λ1, λ2, . . . . . . , λN. If the N roots are distinct, then the complementary solution can be expressed as yc ðnÞ ¼ α1 λ1n þ α2 λ2n þ . . . :: þ αN λNn

ð6:46Þ

where α1, α2, . . . . . . , αN are constants which can be obtained from the specified initial conditions of the discrete-time system. For multiple roots, the complementary solution yc(n) assumes a different form. In the case when the root λ1 of the characteristic equation is repeated m times, but λ2, . . . . . . , λN are distinct, then the complementary solution yc(n) assumes the form   n λ1n α1 þ α2 n þ . . . :: þ αm nm1 þ β2 λ2n þ . . . þ βNM λNM

ð6:47Þ

In case the characteristic equation consists of complex roots λ1, λ2 ¼ a  jb, then the complementary solution results in yc(n) ¼ (a2 + b2)n/2 (C1 cos nq + C2 sin nq), where q ¼ tan–1b/a and C1 and C2 are constants. We now look at the particular solution yp (n) of Eq. (6.42) The particular solution yp(n) is any solution that satisfies the difference equation for the specific input signal x(n), for n  0, i.e., y ð nÞ þ

XN

a yð n  k Þ ¼ k¼1 k

XM k¼0

bk x ð n  k Þ

ð6:48Þ

6.5 Characterization of Discrete-Time Systems

301

The procedure to find the particular solution yp(n) assumes that yp(n) depends on the form of x(n). Thus, if x(n) is a constant, then yp(n) is implicitly a constant. Similarly, if x(n) is a sinusoidal sequence, then yp(n) is implicitly a sinusoidal sequence and so on. In order to find out the overall solution, the complementary and particular solutions must be added. Hence, yðnÞ ¼ yc ðnÞ þ yp ðnÞ

ð6:49Þ

Example 6.21 Determine impulse response for the case of x(n) ¼ δ(n) of a discretetime system characterized by the following difference equation: yðnÞ þ 2yðn  1Þ  3yðn  2Þ ¼ xðnÞ

ð6:50Þ

Solution First, we determine the complementary solution by setting x(n) ¼ 0 and y(n) ¼ λn in Eq. (6.50), which gives us   λn þ 2λn1  3λn2 ¼ λn2 λ2 þ 2λ  3 ¼ λn2 ðλ  1Þðλ þ 3Þ ¼ 0 Hence, the zeros of the characteristic polynomial λ2 þ 2λ  3 are λ1 ¼ 3 and λ2 ¼ 1. Therefore, the complementary solution is of the form yc ðnÞ ¼ α1 ð3Þn þ α2 ð1Þn

ð6:51Þ

For impulse x(n) ¼ δ(n), x(n) ¼ 0 for n > 0 and x(0) ¼ 1. Substituting these relations in Eq. (6.50) and assuming that y(1) ¼ 0 and y(2) ¼ 0, we get yð0Þ þ 2yð1Þ  3yð2Þ ¼ xð0Þ ¼ 1, i.e., y(0) ¼ 1. Similarly y(1) þ 2y(0) – 3y(–1) ¼ x(1) ¼ 0 yields y(1) ¼ –2. Thus, from Eq. (6.51), we get α1 þ α2 ¼ 1 and 3α1 þ α2 ¼ 2 Solving these two equations, we obtain α1 ¼ 3/4 and α2 ¼ 1/4. Since x(n) ¼ 0 for n > 0, there is no particular solution. Hence, the impulse response is given by hðnÞ ¼ yc ðnÞ ¼ 0:75ð3Þn þ 0:25ð1Þn

ð6:52Þ

302

6 Discrete-Time Signals and Systems

Example 6.22 A discrete-time system is characterized by the following difference equation: yðnÞ þ 5yðn  1Þ þ 6yðn  2Þ ¼ xðnÞ

ð6:53Þ

Determine the step response of the system, i.e., x(n) ¼ u(n). Solution For the given difference equation, total solution is given by yðnÞ ¼ yc ðnÞ þ yp ðnÞ First, we determine the complementary solution by setting x(n) ¼ 0 and y(n) ¼ λn in Eq. (6.53), which give us   λn þ 5λn1 þ 6λn2 ¼ λn2 λ2 þ 5λ þ 6 ¼ 0 Hence, the zeros of the characteristic polynomial λ2 + 5λ þ 6 are λ1 ¼ –3 and λ2 ¼ –2. Therefore, the complementary solution is of the form yc(n) ¼ α1(3)n + α2(2)n The particular solution for the step input is of the form yp(n) ¼ K For n > 2, substituting yp(n) ¼ K and x(n) ¼ 1 in Eq. (6.53), we get 1 1 K þ 5K þ 6K ¼ 1; K ¼ 12 , and yp ðnÞ ¼ 12 . Therefore, the solution for given difference equation is yðnÞ ¼ α1 ð3Þn þ α2 ð2Þn þ

1 12

ð6:54Þ

For n ¼ 0, Eq. (6.53) becomes y(0) þ 5y(–1) þ 6y(–2) ¼ x(0) Assuming y(–1) ¼ y(–2) ¼ 0, from the above equation, we get y(0) ¼ x(0) ¼ 1 and for n ¼ 1, y(1) þ 5y(0) þ 6y(–1) ¼ x(1) ¼ 1, i.e., y(1) ¼ –4. Then, we get from Eq. (6.54) 1 α1 þ α2 þ 12 ¼1 3α1  2α2 þ

1 ¼ 4 12

16 Solving these equations, we arrive at α1 ¼ 27 12 and α2 ¼ 12 . Then, the step response is given by

y ð nÞ ¼

27 16 1 ð3Þn  ð2Þn þ 12 12 12

ð6:55Þ

6.5 Characterization of Discrete-Time Systems

303

Example 6.23 A discrete-time system is characterized by the following difference equation: yðnÞ  2yðn  1Þ þ yðn  2Þ ¼ xðnÞ  xðn  1Þ

ð6:56Þ

Determine the response y(n), n  0 when the system input is x(n) ¼ (–1)nu(n) and the initial conditions are y(1) ¼ 1 and y(2) ¼ 1. Solution For the given difference equation, the total solution is given by yðnÞ ¼ yc ðnÞ þ yp ðnÞ First, determine the complementary solution by setting x(n) ¼ 0 and y(n) ¼ λn in Eq. (6.56); this gives   λn  2λn1 þ λn2 ¼ λn2 λ2  2λ þ 1 ¼ 0 Hence, the zeros of the characteristic polynomial λ2  2λ þ 1 are λ1 ¼ λ2 ¼ 1. It has repeated roots; thus, the complementary solution is of the form yc(n) ¼ 1n(α1 + nα2). The particular solution for the step input is of the form yp(n) ¼ K(1)nu(n). Substituting x(n) ¼ (–1)nu(n) and yp(n) ¼ K(-1)nu(n) in Eq. (6.56), we get K(–1)nu(n) – 2K(–1)n–1u(n – 1) þ K(–1)n–2u(n – 2) ¼ (–1)nu(n) – (–1)n–1u(n – 1) For n ¼ 2, the above equation becomes K + 2K + K ¼ 2; K ¼ 12. Therefore, the particular solution is given by 1 yp ðnÞ ¼ ð1Þn uðnÞ 2 Then, the total solution for given difference equation is 1 yðnÞ ¼ 1n ðα1 þ nα2 Þ þ ð1Þn uðnÞ: 2

ð6:57Þ

For n ¼ 0, Eq. (6.56) becomes yð0Þ  2yð1Þ þ yð2Þ ¼ 1 Using the initial conditions y(–1) ¼ 1 and y(–2) ¼ –1, we get y(0) ¼ 4. Then, for n ¼ 1, from Eq. (6.56), we get y(1) ¼ 5. Thus, we get from Eq. (6.57) α1 þ ð1=2Þ ¼ 4 α1 þ α2  ð1=2Þ ¼ 5 Solving these two equations, we arrive at α1 ¼ (7/2) and α2 ¼ 2. Thus, the response of the system for the given input is

304

6 Discrete-Time Signals and Systems

  7 1 þ 2n þ ð1Þn uðnÞ y ð nÞ ¼ 1 2 2 n

6.5.4

ð6:58Þ

Computation of Impulse and Step Responses Using MATLAB

The impulse and step responses of LTI discrete-time systems can be computed using MATLAB function: y ¼ filter ðb; a; xÞ where b and a are the coefficient vectors of difference equation describing the system, x is the input data vector, and y is the vector generated assuming zero initial conditions. The following example illustrates the computation of the impulse and step responses of an LTI system. Example 6.24 Determine the impulse and step responses of a discrete-time system described by the following difference equation: yðnÞ  2yðn  1Þ ¼ xðnÞ þ 0:1xðn  1Þ  0:06xðn  2Þ Solution Program 6.2 is used to compute and plot the impulse and step responses, which are shown in Figure 6.15a and b, respectively. Program 6.2: Illustration of Impulse and Step Response Computation clear;clc; flag=input(‘enter 1 for impulse response, and 2 for step response’); len=input(‘enter desired response length=‘); b=[l -2];%b coefficients of the difference equation a=[l 0.l -0.06]; %a coefficients of the difference equation if flag==l; x=[l,zeros(l,len-l)]; end if flag==2; x=[ones(1,len)]; end y=filter(b,a,x); n=0:1:len-1; stem(n,y) xlabel(‘Time index n’); ylabel(‘Amplitude’);

6.6 Sampling of Discrete-Time Signals

305

Figure 6.15 (a) Impulse response and (b) step response for Example 6.24

6.6

Sampling of Discrete-Time Signals

It is often necessary to change the sampling rate of a discrete-time signal, i.e., to obtain a new discrete-time representation of the underlying continuous-time signal of the form x’(n) ¼ xa(nT’). One approach to obtain the sequence x’(n) from x(n) is to reconstruct xa(t) from x(n) and then resample xa(t) with period T’ to obtain x’(n).

306

6 Discrete-Time Signals and Systems

x ( n)

xd (n) = x(nM )

M

Sampling period T ' = MT

Sampling period T

Figure 6.16 Block diagram representation of a down sampler

x ( n)

L

xe (n) = x(n / L) Sampling period T ' = TL

Sampling period T

Figure 6.17 Block diagram representation of an up-sampler

Often, however, this is not a desirable approach, because of the non-ideal analog reconstruction filter, DAC, and ADC that would be used in a practical implementation. Thus, it is of interest to consider methods that involve only discrete-time operation.

6.6.1

Discrete-Time Down Sampler

The block diagram representation of a down sampler, also known as a sampling rate compressor, is depicted in Figure 6.16. The down-sampling operation is implemented by defining a new sequence xd(n) in which every Mth sample of the input sequence is kept and (M1) in-between samples are removed to obtain the output sequence, i.e., xd(n) is identical to the sequence obtained from xa(t) with a sampling period T’ ¼ MT xd ðnÞ ¼ xðnM Þ

ð6:59Þ

For example, if x(n) ¼ {2,6,3,0,1,2,–5,2,4,7,–1,1,–2,. . .}, then xd(n) ¼ {2, 1, 4, –2, . . .} for M ¼ 4, i.e., M1 ¼ 3 samples are left in between the samples of x(n) to get xd(n).

6.6.2

Discrete-Time Up-Sampler

The block diagram representation of an up-sampler, also called a sampling rate expander or simply an interpolator, is shown in Figure 6.17. The output of an up-sampler is given by x e ð nÞ ¼

1 X

xðk Þδðn  kLÞ ¼ x

n

k¼1

¼0

L

n ¼ 0,  L,  2L, . . . . . . :: otherwise

ð6:60Þ

6.7 State-Space Representation of Discrete-Time LTI Systems

307

Eq. (6.60) implies that the output of an up-sampler can be obtained by inserting (L – 1) equidistant zero-valued samples between two consecutive samples of the input sequence x(n), i.e., xe(n) is identical to the sequence obtained from xa(t) with a sampling period T’ ¼ T/L. For example, if xe(n) ¼ {2,1,4,–2, . . . .}, then xe(n) ¼ {2,0,0,0,1,0,0,0,4,0,0,0,–2,0,0,0, . . .} for L ¼ 4, i.e., L1 ¼ 3 zerovalued samples are inserted in between the samples of x(n) to get xe(n).

6.7 6.7.1

State-Space Representation of Discrete-Time LTI Systems State-Space Representation of Single-Input Single-Output Discrete-Time LTI Systems

Consider a single-input single-output discrete-time LTI system described by the following Nth-order difference equation: yðnÞ þ a1 yðn  1Þ þ a2 yðn  2Þ þ . . . þ aN yðn  N Þ ¼ ℧ðnÞ

ð6:61Þ

where y(n) is the system output and ℧(n) is the system input. Define the following useful set of state variables: x1 ðnÞ ¼ yðn  N Þ, x2 ðnÞ ¼ yðn  N þ 1Þ, x3 ðnÞ ¼ yðn  N þ 2Þ, . . . , x N ð nÞ ¼ y ð n  1Þ

ð6:62Þ

Then from Eqs. (6.61) and (6.62), we get x 1 ð n þ 1Þ ¼ x 2 ð nÞ x 2 ð n þ 1Þ ¼ x 3 ð nÞ ⋮ xN1 ðn þ 1Þ ¼ xN ðnÞ xN ðn þ 1Þ ¼ aN x1 ðnÞ  aN1 x2 ðnÞ      a1 xN ðnÞ þ ℧ðnÞ

ð6:63aÞ

and y ð nÞ ¼ x N ð n þ 1Þ Eqs. (6.63a) and (6.63b) can be written in matrix form as

ð6:63bÞ

308

2

6 Discrete-Time Signals and Systems

x1 ð n þ 1 Þ

3

2

1

0



0

1



⋮

⋮

⋱

0

0



aN1

aN2



0

7 6 6 6 x2 ð n þ 1 Þ 7 6 0 7 6 6 7 6 6 7¼6 ⋮ 6 ⋮ 7 6 6 7 6 6 6 xN1 ðn þ 1Þ 7 6 0 5 4 4 aN xN ð n þ 1 Þ

2

yðnÞ ¼ ½aN aN1 aN2

32

0

x1 ðnÞ

3

2

0

3

7 6 7 76 7 6 7 6 0 7 76 x2 ðnÞ 7 6 0 7 7 6 7 76 7 6 7 6 ⋮ 7 76 ⋮ 7 þ 6 0 7℧ðnÞ 7 6 7 76 7 6 7 6 1 7 54 xN1 ðnÞ 5 4 ⋮ 5 a1

x 1 ð nÞ

xN ðnÞ

3

2

0

3

7 6 7 6 6 x 2 ð nÞ 7 6 0 7 7 6 7 6 7 6 7 6    a1 6 ⋮ 7 þ 6 0 7℧ðnÞ 7 6 7 6 6 x ðnÞ 7 6 ⋮ 7 4 N1 5 4 5 x N ð nÞ

1 ð6:64aÞ

ð6:64bÞ

1

Define a Nx1 dimensional vector called state vector as 2

x 1 ð nÞ

3

7 6 6 x 2 ð nÞ 7 7 6 7 6 X ð nÞ ¼ 6 ⋮ 7 7 6 6 x ð nÞ 7 4 N1 5

ð6:65Þ

x N ð nÞ More compactly Eqs. (6.64a) and (6.64b) can be written as

where 2

0

6 6 0 6 6 A¼6 ⋮ 6 6 0 4 aN

X ðn þ 1Þ ¼ AX ðnÞ þ b℧ðnÞ

ð6:66aÞ

yðnÞ ¼ cX ðnÞ þ d℧ðnÞ

ð6:66bÞ

1

0



0

1



⋮

⋮

⋱

0

0



aN1

aN2

0

3

2

0

3

7 6 7 6 0 7 0 7 7 6 7 7 6 7 ⋮ 7; b ¼ 6 0 7; 7 6 7 6⋮7 1 7 5 4 5

   a1

1

c ¼ ½aN  aN1  aN2     a1 ; d ¼ 1: Equation (6.66a) and (6.66b) is called N-dimensional state-space representation or state equations of the discrete-time system.

6.7 State-Space Representation of Discrete-Time LTI Systems

309

Example 6.25 Obtain the state-space representation of a discrete-time system described by the following differential equation: yðn  3Þ þ 2yðn  2Þ þ 3yðn  1Þ þ 4yðnÞ ¼ ℧ðnÞ Solution The order of the differential equation is three. We have to choose three state variables: Let x1(n) ¼ y(n  3), x2(n) ¼ y(n  2), x3(n) ¼ y(n  1) Then x 1 ð n þ 1Þ ¼ x 2 ð nÞ x 2 ð n þ 1Þ ¼ x 3 ð nÞ 1 1 3 1 x3 ðn þ 1Þ ¼  x1 ðnÞ  x2 ðnÞ  x3 ðnÞ þ ℧ðnÞ 4 2 4 4 y ð nÞ ¼ x 3 ð n þ 1Þ The state-space representation in matrix form is given by 2 3 0 76 6 7 6 7 6 7 6 7 6 0 7 607 0 1 7 76 7 ℧ð n Þ 6 x 2 ð n þ 1Þ 7 ¼ 6 6 x 2 ð nÞ 7 þ 6 6 7 7 4 5 4 4 5 6 5 4 1 1 3 15    x 3 ð n þ 1Þ x 3 ð nÞ 4 2 4 4 2 3 x 1 ð nÞ  7 1 1 1 3 6 6 7 y ð nÞ ¼    6 x 2 ð nÞ 7 þ ℧ð n Þ 5 4 2 4 4 4 2

x 1 ð n þ 1Þ

3

2

0

1

0

32

x 1 ð nÞ

3

x 3 ð nÞ

6.7.2

State-Space Representation of Multi-input Multi-output Discrete-Time LTI Systems

The state-space representation of discrete-time system with m inputs and 1 output and N state variables can be expressed as

where

X ðn þ 1Þ ¼ AX ðnÞ þ B℧ðnÞ

ð6:67aÞ

yðnÞ ¼ CX ðnÞ þ D℧ðnÞ

ð6:67bÞ

310

6 Discrete-Time Signals and Systems

2

a11

6 6 a21 A¼6 6⋮ 4



a22



⋮

⋱

a1N

3

c11 6 6 c21 C¼6 6⋮ 4

c12 ⋮

cl1

cl2



aN1

aN2 c22

2

7 a2N 7 7 ⋮ 7 5

   aNN 3    c1N 7    c2N 7 7 ⋱ ⋮7 5

2

6.8

a12

clN

b11

6 6 b21 B¼6 6⋮ 4 NN

lN

2

bN1

b12



b22



⋮

⋱

bN2



b1m

3

7 b2m 7 7 ⋮ 7 5 bNm 3

d11 6 6 d21 D¼6 6⋮ 4

d 12



d 22



⋮

⋱

7 d2m 7 7 ⋮7 5

d l1

d l2



dlm

Nm

d1m

lm

Problems

1. Determine if the following discrete-time signals are periodic:   π (i) xðnÞ ¼ sin πn 6 þ3  3πn (ii) xðnÞ ¼ cos 10 þ =0   (iii) xðnÞ ¼ cos n2 þ θ   (iv) xðnÞ ¼ ej πn 0 4 þ= n (v) x(n) ¼ 6(1)   63πn (vi) xðnÞ ¼ sin 3πn 8  cos 64 3πn  (vii) xðnÞ ¼ sin 8 þ cos 63πn 64     j 3πn (viii) xðnÞ ¼ ej 7πn 4 þe 4 2. Determine if the following discrete-time signals are energy or power signals or neither. Calculate the energy and power of the signals in each case:   3πn (ix) xðnÞ ¼ cos πn 2 þ sin 4 (x) x(n) ¼ (1)n 8 n 0  n  10 ð 3Þ > < (xi) xðnÞ ¼ 2 11  n  15 > : 8 0 otherwise

< cos πn  10  n  0 15 (xii) xðnÞ ¼ : 0 otherwise   π (xiii) xðnÞ ¼ ej πn þ 2 8 3. Determine if the following discrete-time signals are even, odd, or neither even nor odd:   (i) xðnÞ ¼ sin ð4nÞ þ cos 2πn 3

6.8 Problems

311

  2πn (ii) xðnÞ ¼ sin πn 30 þ cos 3  3πn (iii) xðnÞ ¼ sin 8 þ cos 3πn 4

ð1Þn n  0 (iv) xðnÞ ¼ 0 n<0 4. Check the following for linearity, time-invariance, and causality: (i) y(n) ¼ 5nx2(n). (ii) y(n) ¼ x(n)sin2n. (iii) y(n) ¼ e–nx(n + 3)  n1 5. Given the input x(n) ¼ u(n) and the output yðnÞ ¼ 12 uðn  1Þ of a system, (i) Determine the impulse response h(n) (ii) Is the system stable? (iii) Is the system causal? 6. Check for stability and causality of a system for the following impulse responses:   πn (i) hðnÞ ¼ e2n sin πn 2 uðn  1Þ (ii) hðnÞ ¼ sin 2 uðnÞ 7. Determine if the following signals are periodic, and if periodic, find its period:   3πn (ii) sin n (b) e jπn/3 (c) sin πn 4 þ sin 4 8. Determine the convolution of the sum of the two sequences: x1(n) ¼ (3,2,1,2) and x2(n) ¼ (1,2,1,2). 9. Determine the convolution of the sum of the two sequences x1(n) and x2(n), if x1(n) ¼ x2(n) ¼ cnu(n) for all n, where c is a constant. 10. Determine the impulse response (i.e., when x(n) ¼ δ(n) of a discrete-time system characterized by the following difference equation: yðnÞ þ yðn  1Þ  6yðn  2Þ ¼ xðnÞ 11. A discrete-time system is characterized by the following difference equation: 6yðnÞ  yðn  1Þ  yðn  2Þ ¼ 6xðnÞ Determine the step response of the system, i.e., x(n) ¼ u(n), given the initial conditions y(-1) ¼ 1 and y(-2) ¼ -1. 12. A discrete-time system is characterized by the following difference equation: yðnÞ  5yðn  1Þ þ 6yðn  2Þ ¼ xðnÞ Determine the response of the system for x(n) ¼ nu(n) and initial conditions y(-1) ¼ 1 and y(-2) ¼ 0.

312

6 Discrete-Time Signals and Systems

13. Determine the response of the system described by the following difference equation: yðnÞ þ yðn  1Þ ¼ sin 3n uðnÞ 14. Obtain the state-space representation of a discrete-time system described by the following differential equation: 2yðnÞ þ 3yðn  1Þ þ yðn  2Þ ¼ ℧ðnÞ

6.9

MATLAB Exercises

1. Using the function impz, write a MATLAB program to determine the impulse response of a discrete-time system represented by yðnÞ  5yðn  1Þ þ 6yðn  2Þ ¼ xðnÞ  2xðn  1Þ 2. Write a MATLAB program to illustrate down-sampling by an integer factor of 4 of a sum of two sinusoidal sequences, each of length 50, with normalized frequencies of 0.2 Hz and 0.35 Hz. 3. Write a MATLAB program to illustrate up-sampling by an integer factor of 4 of a sum of two sinusoidal sequences, each of length 50, with normalized frequencies of 0.2 Hz and 0.35 Hz.

Further Reading 1. Linden, D.A.A.: Discussion of sampling theorem. Proceedings of the IRE. 47, 1219–1226 (1959) 2. Proakis, J.G., Manolakis, D.G.: Digital Signal Processing Principles, Algorithms and Applications, 3rd edn. Prentice-Hall, India (2004) 3. Crochiere, R.E., Rabiner, L.R.: Multirate Digital Signal Processing. Prentice-Hall, Englewood Cliffs (1983) 4. Hsu, H.: Signals and Systems, Schaum’s Outlines, 2nd edn, Mc Graw Hill, New York (2011) 5. Mandal, M., Asif, A.: Continuous and Discrete Time Signals and Systems. Cambridge, UK; New York: Cambridge University Press, (2007)

Chapter 7

Frequency Domain Analysis of DiscreteTime Signals and Systems

This chapter discusses the transform domain representation of discrete-time sequences by discrete-time Fourier series (DTFS) and discrete-time Fourier transform (DTFT) in which a discrete-time sequence is mapped into a continuous function of frequency. We first obtain the discrete-time Fourier series (DTFS) expansion of a periodic sequence. The periodic convolution and the properties of DTFS are discussed. The Fourier transform domain representation of discrete-time sequences are described along with the conditions for the existence of DTFT and its properties. Later, the frequency response of discrete-time systems, frequency domain representation of sampling process, and reconstruction of band-limited signals from its samples are discussed.

7.1

The Discrete-Time Fourier Series

If a sequence x(n) is periodic with period N, then x(n) ¼ x(n + N ) for all n. In analogy with the Fourier series representation of a continuous periodic signal, we can look for a representation of x(n) in terms of the harmonics corresponding to the fundamental frequency of (2π/N ). Hence, we may write x(n) in the form X x ð nÞ ¼ b ej2πkn=N ð7:1aÞ k k It can easily be verified from Eq. (7.1a) that x(n) ¼ x(n + N ). Also, we know that there are only N distinct values for e j2πkn/N, corresponding to k ¼ 0, 1, . . . . N  1, these being 1, e j2πn/N, . . ., e j2πn(N  1)/N. Hence, we may rewrite (7.1a) as x ð nÞ ¼

XN1 k¼0

ak ej2πkn=N

ð7:1bÞ

It should be noted that the summation can be taken over any N consecutive values of k. Eq. (7.1b) is called the discrete-time Fourier series (DTFS) of the periodic © Springer International Publishing AG, part of Springer Nature 2018 K. D. Rao, Signals and Systems, https://doi.org/10.1007/978-3-319-68675-2_7

313

314

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

sequence x(n) and ak as the Fourier coefficients. We will now obtain the expression for the Fourier coefficients ak. It can easily be shown that {e j2πkn/N} is an orthogonal sequence satisfying the relation XN1 n¼0

e

j2πkn=N j2πl n=N

e

 ¼

0 N

k¼ 6 l k¼l

ð0  k; l  ðN  1Þ

ð7:2Þ

Now, multiplying both sides of (7.1b) by ej2πln/N and summing over n between 0 and (N  1), we get PN1 n¼0

xðnÞ ej2πln=N ¼ ¼

PN1 PN1 n¼0

PN1 k¼0

k¼0

ak

ak ej2πkn=N ej2πln=N

PN1 n¼0

ej2πkn=N ej2πln=N

¼ al N, using Eq:ð7:2Þ: Hence, ak ¼

1 XN1 xðnÞej2πkn=N , n¼0 N

k ¼ 0, 1, 2, . . . , N  1

ð7:3Þ

It is common to associate the factor (1/N ) with x(n) rather than ak. This can be done by denoting Nak by X(k); in such a case, we have x ð nÞ ¼

1 XN1 X ðkÞ ej2πkn=N k¼0 N

ð7:4Þ

where the Fourier coefficients X(k) are given by X ðk Þ ¼

XN1 n¼0

x ð nÞ e 

j2πkn N

,

k ¼ 0, 1, 2, . . . , N  1

ð7:5Þ

It is easily seen that X(k + N ) ¼ X(k), that is, the Fourier coefficient sequence X(k) is also periodic of period N. Hence, the spectrum of a signal x(n) that is periodic with period N is also a periodic sequence with the same period. It is also noted that since the Fourier series of a discrete periodic signal is a finite sequence, the series always converges, and the Fourier series gives an exact alternate representation of the discrete sequence x(n).

7.1.1

Periodic Convolution

In the case of two periodic sequences x1(n) and x2(n) having the same period N, linear convolution as defined by Eq. (6.29) does not converge. Hence, we define a different form of convolution for periodic signals by the relation

7.1 The Discrete-Time Fourier Series

315

Table 7.1 Some important properties of DTFS Property Linearity Time shifting

Periodic sequence ax1(n) þ bx2(n) a and b are constants x(n  m)

DTFS coefficients aX1(k) þ bX2(k) 2π ejð N Þkm X ðk Þ

Frequency shifting

X(k  l )

Periodic convolution

ejð N Þln xðnÞ N1 X x1 ðmÞx2 ðn  mÞ

Multiplication

x1(n)x2(n)

N1 1 X X 1 ðlÞX 2 ðk  lÞ N l¼0

x*(n) x*(n) RefxðnÞg j lmfxðnÞg

X*(k) X*(k)

2π

X1(k)X2(k)

m¼0

Symmetry properties

1 X e ðk Þ ¼ ðX ðk Þ þ X ∗ ðk ÞÞ 2 1 X o ðk Þ ¼ ðX ðk Þ  X ∗ ðk ÞÞ 2 RefX ðk Þg j Im fX ðk Þg

x e ð nÞ 1 ¼ ½xðnÞ þ x∗ ðnÞ 2 xo ðnÞ 1 ¼ ½xðnÞ  x∗ ðnÞ 2 If xðnÞis real 1 xe ðnÞ ¼ ½xðnÞ þ xðnÞ 2 1 xo ðnÞ ¼ ½xðnÞ  xðnÞ 2

y ð nÞ ¼

XN1

x ðmÞx2 ðn  mÞ ¼ m¼0 1

RefX ðk Þg j Im fX ðk Þg

XN1

x ðn m¼0 1

 mÞx2 ðmÞ

ð7:6Þ

The above convolution is called periodic convolution. It may be observed that y (n) ¼ y(n + N ), that is, the periodic convolution is itself periodic of period N. Some important properties of the DTFS are given in Table 7.1. In this table, it is assumed that x1(n) and x2(n) are periodic sequences having the same period N. The proofs are omitted here, since they are similar to the ones that will be given in Section 7.2 for the corresponding properties of the DTFT. Example 7.1 Determine the Fourier series representation for the following discretetime signal: xðnÞ ¼ 3 sin

  πn 2πn sin 4 5

316

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

  2πn 4 5      3 3n π 13nπ cos ¼  cos 2 20 20   3nπ 13nπ 13nπ 3 j 3nπ j 20 j 20 j 20 20 ¼ e þe e e 4   3nπ 17nπ 13nπ 7nπ 3 ¼ ej 20 þ ej 20  ej 20  ej 20 4

Solution

xðnÞ ¼ 3 sin

πn

sin

X(0) ¼ X(1) ¼ X(2) ¼ X(4) ¼ X(5) ¼ X(6) ¼ X(8) ¼ X(9) ¼ X(10) ¼ X (11) ¼ X(12) ¼ X(14) ¼ X(15) ¼ X(16) ¼ X(18) ¼ X(19) ¼ 0, X(3) ¼ Xð17Þ ¼ 32, X(7) ¼ Xð13Þ ¼ 32 Example 7.2 Discrete-time signal x(n) is periodic of period 8, and x(n) ¼ n for 0  n  7. Solution The sequence is periodic with period N ¼ 8. X ðk Þ ¼

XN1 n¼0

xðnÞej2πkn=N , k ¼ 0, 1, 2, . . . , N  1

Using above equation, the DTFS coefficients are computed as X(0) ¼ 28., X(1) ¼ 4þ 9.6569 j, X(2) ¼ 4 þ 4 j X(3) ¼ 4 þ 1.6569 j,

X(4) ¼ 4 X(5) ¼ 4  1.6569 j X(6) ¼ 4  4 j X(7) ¼ 4  9.6569 j

X(k) ¼ {28, 4þ 9.6569j, 4 þ 4 j, 4 þ 1.6569 j, 4, 4  1.6569 j, 44i,4  49.6569 j}

7.2 7.2.1

Representation of Discrete-Time Signals and Systems in Frequency Domain Fourier Transform of Discrete-Time Signals

The discrete-time Fourier transform (DTFT) of a finite energy sequence x(n) is defined as   X1 xðnÞeðjωnÞ ð7:7Þ F½xðnÞ ¼ X ejω ¼ n¼1

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

317

From X(e jω), x(n) can be computed as 1 x ð nÞ ¼ 2π

ðπ

  X ejω eðjωnÞ dω

π

ð7:8Þ

Eq. (7.8) is called the inverse Fourier transform. Convergence of the DTFT The existence of DTFT of x(n) depends on the convergence of the series in Eq. (7.7). Now, we look at theX condition for convergence.  jω  1 Let X k e ¼ xðnÞeðjωnÞ denote the partial sum of the weighted k¼1 complex exponentials in Eq. (7.7). Then for uniform convergence of X(e jω),     lim X k ejω ¼ X ejω

k!1

ð7:9Þ

Hence, for uniform convergence of X(e jω), x(n) must be absolutely summable, i.e., 1 X

jxðnÞj < 1,

ð7:10Þ

n¼1

Then 1 1 1 X X  jω  X jωn X e ¼ xðnÞe jxðnÞj ejωn  jxðnÞj < 1  n¼1 n¼1 n¼1

ð7:11Þ

guaranteeing the existence of X(e jω), for all values of ω. Consequently, Eq. (7.10) is only a sufficient condition for the existence of the DTFT, but is not a necessary condition.

7.2.2

Theorems on DTFT

We will now consider some important theorems concerning DTFT that can be used in digital signal processing. All these properties can be proved using the definition of DTFT. The following notation is adopted for convenience:   X ejω ¼ F½xðnÞ

  xðnÞ ¼ F1 X ejω

ð7:12aÞ ð7:12bÞ

Linearity If x1(n) and x2(n) are two sequences with Fourier transforms X1(e jω) and X2(e jω), then the Fourier transform of a linear combination of x1(n) and x2(n) is given by

318

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

    F½a1 x1 ðnÞ þ a2 x2 ðnÞ ¼ a1 X 1 ejω þ a2 X 2 ejω

ð7:13Þ

where a1and a2 are arbitrary constants. Time Reversal If x(n) is a sequence with Fourier transform X(e jω), then the Fourier transform of time reversed sequence x(n) is given by   F½xðnÞ ¼ X ejω

ð7:14Þ

Time Shifting If x(n) is a sequence with Fourier transform X(e jω), then the Fourier transform of the delayed sequence x(n  k), where k an integer, is given by   F½xðn  k Þ ¼ ejωk X ejω

ð7:15Þ

Therefore, time shifting results in a phase shift in the frequency domain. Frequency Shifting If x(n) is a sequence with Fourier transform X(e jω), then the Fourier transform of the sequence ejω0 n x(n) is given by  i

 F ejω0 n xðnÞ ¼ X ejðωω0 Þ

ð7:16Þ

Thus, multiplying a sequence x(n) by a complex exponential ejω0 n in the time domain corresponds to a shift in the frequency domain. Differentiation in Frequency If x(n) is a sequence with Fourier transform X(e jω), then the Fourier transform of the sequence nx(n) is given by F½nxðnÞ ¼ j

d  jω  X e dω

ð7:17Þ

Convolution Theorem If x1(n) and x2(n) are two sequences with Fourier transforms X1(e jω)and X2(e jω), then the Fourier transform of the convolution of x1(n) and x2(n) is given by     F½x1 ðnÞ∗x2 ðnÞ ¼ X 1 ejω X 2 ejω

ð7:18Þ

Hence, convolution of two sequences x1(n) and x2(n) in the time domain is equal to the product of their frequency spectra. In the above equation, since X1(e jω) and X2(e jω) are periodic in ω with period 2π, the convolution is a periodic convolution. Windowing Theorem If x(n) and w(n) are two sequences with Fourier transforms X(e jω) and W(e jω), then the Fourier transform of the product of x(n) and w(n) is given by     1 F½xðnÞwðnÞ ¼ X ejω ∗W ejω ¼ 2π

ðπ

    X ejθ W ejðωθÞ dθ

π

The above result is called the windowing theorem.

ð7:19Þ

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

319

Correlation Theorem If x1(n) and x2(n) are two sequences with Fourier transforms X1(e jω) and X2(e jω), then the Fourier transform of the correlation r x1 x2 ðlÞ of x1(n) and x2(n) defined by r x1 x2 ðlÞ ¼

X1

x ðnÞx2 ðn n¼1 1

 lÞ

ð7:20aÞ

is given by hX1 i     x ð n Þx ð n  l Þ ¼ X 1 ejω X 2 ejω F ½ r x1 x2 ð l Þ  ¼ F 1 2 n¼1

ð7:20bÞ

which is called the cross energy density spectrum of the signals x1(n) and x2(n). Parseval’s Theorem If x(n) is a sequence with Fourier transform X(e jω), then the energy E of x(n) is given by E¼

X1

1 jxðnÞj ¼ 1 2π 2

ðπ

 jω  2 X e dω

π

ð7:21Þ

where |X(e jω)|2 is called the energy density spectrum. Proof The energy E of x(n) is defined as P ∗ jxðnÞj2 ¼ 1 1 xðnÞx ðnÞ ð P 1 π ∗  jω  jωn ¼ 1 x ð n Þ X e e dω 1 2π π

E¼

P1

1

ð7:22Þ

using Eq. (7.8). Interchanging the integration and summation signs, the above equation can be rewritten as E¼

1 2π

¼

1 2π

¼

1 2π

ðπ π

1  X X ∗ ejω xðnÞ ejωn dω 1

ðπ

X π ðπ

 ∗

   ejω X ejω dω

 jω  2 X e dω

π

Thus, E¼

X1

1 jxðnÞj ¼ 1 2π 2

ðπ π

 jω  2 X e dω

ð7:23Þ

320

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Table 7.2 Some properties of discrete-time Fourier transforms Property Linearity Time shifting Time reversal Frequency shifting

Sequence a1x1(n) þ a2x2(n) x(n  k) x(n) ejω0 n xðnÞ

DTFT a1X1(e jω) þ a2X2(e jω) ejωkX(e jω) X(ejω)   X ejðωω0 Þ

Differentiation in the frequency domain

nx(n)

Convolution theorem Windowing theorem Correlation theorem

x1(n) * x2(n) x1(n)x2(n) X1 x ðnÞx2 ðn  lÞ 1 1

d jdω X ðejω Þ X1(e jω) X2(e jω) X1(e jω) ∗ X2(e jω) X1(e jω)X2(ejω)

Table 7.3 Some useful DTFT pairs

x(n) δ(n) 1 (1 < n < 1)

DTFT 1 X1 k¼1 1 1aejω

anu(n), |a| < 1



sin ðωc nÞ πn



2πδðω þ 2πk Þ

1 jωj < ωc 0 ωc < jωj < π

sin ωðLþ1Þ=2 jωL=2 e sin ω=2

1 0nL 0 otherwise

X1

ejω0 n

k¼1

2πδðω  ω0 þ 2πk Þ

The above theorems concerning DTFT are summarized in Table 7.2 Parseval’ s theorem

X1

jxðnÞj2 ¼ 1

1 2π

ðπ

 jω  2 X e dω

π

Using the definitions of DTFT pair given by (7.7) and (7.8), we may establish the DTFT pairs for some useful functions. These are given in Table 7.3

7.2.3

Some Properties of the DTFT of a Complex Sequence x (n)

From Eq. (7.7), the DTFT of a time reversed sequence x(n) can be written as F½xðnÞ ¼

X1 n¼1

xðnÞejωn ¼

X1 l¼1

  xðlÞejωl ¼ X ejω

ð7:24aÞ

Similarly, the DTFT of the complex conjugate sequence x*(n) can be expressed as

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

F ½ x ∗ ð nÞ  ¼

X1

x∗ ðnÞejωn ¼ n¼1

X1

xðnÞejωn n¼1

∗

321

  ¼ X ∗ ejω ð7:24bÞ

From the above two equations, it can be easily shown that   F½x∗ ðnÞ ¼ X ∗ ejω

ð7:25Þ

The sequence x(n) can be represented as a sum of conjugate symmetric sequence xe (n) and a conjugate antisymmetric sequence xo(n) as xðnÞ ¼ xe ðnÞ þ xo ðnÞ

ð7:26Þ

1 xe ðnÞ ¼ ½xðnÞ þ x∗ ðnÞ 2

ð7:27Þ

1 xo ðnÞ ¼ ½xðnÞ  x∗ ðnÞ 2

ð7:28Þ

where

and

The DTFT X(e jω) can be split into       X ejω ¼ X e ejω þ X o ejω

ð7:29Þ

where Xe(e jω) and Xo(e jω) are the DTFTs of xe(n) and xo(n), respectively. Using Eqs. (7.7), (7.25), and (7.27), Xe(e jω) can be expressed as   X e ejω ¼ F½xe ðnÞ  

  1 1   ¼ ðF½xðnÞþF½x∗ ðnÞÞ ¼ X ejω þ X ∗ ejω ¼ Re X ejω ð7:30Þ 2 2 In a similar way, using Eqs. (7.7), (7.25), and (7.28), Xo(e jω) can be written as X o ðejω Þ ¼ F½xo ðnÞ

 1 1 ¼ ðF½xðnÞ  F½x∗ ðnÞ ¼ ½Xðejω Þ  X ∗ ðejω Þ ¼ jIm½Xðejω Þ 2 2 ð7:31Þ

A complex sequence x(n)can be decomposed into a sum of its real and imaginary parts as xðnÞ ¼ xR ðnÞ þ jxI ðnÞ

ð7:32Þ

322

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

where 1 xR ðnÞ ¼ ½xðnÞ þ x∗ ðnÞ 2

ð7:33Þ

1 jxI ðnÞ ¼ ½xðnÞ  x∗ ðnÞ 2

ð7:34Þ

and

The DTFT of xR (n) can be written as

1 F½ReðxðnÞ ¼ F ðxðnÞ þ x∗ ðnÞÞ 2   1  jω  ¼ X e þ X ∗ ejω 2

ð7:35Þ

Similarly, the DTFT of jxI (n) can be expressed as

1 F½jImðxðnÞÞ ¼ F ðxðnÞ  x∗ ðnÞÞ 2 1 jω ¼ ½Xðe Þ  X ∗ ðejω Þ 2

ð7:36Þ

The above properties of the DTFT of a complex sequence are summarized in Table 7.4.

7.2.4

Some Properties of the DTFT of a Real Sequence x(n)

Since e–jωn ¼ cosωn – jsinωn, the DTFT X(e jω) given by Eq. (7.7) can be expressed as X1   X1 ð Þ X ejω ¼ x n cos ωn  j xðnÞ sin ωn n¼1 n¼1

ð7:37Þ

The Fourier transform X(e jω) is a complex function of ω and can be written as the sum of the real and imaginary parts as Table 7.4 Some properties of DTFT of a complex sequence

Sequence x∗(n) x∗(n) xR(n) ¼ Re [x(n)]

DTFT X∗(ejω) X∗(e jω) ∗ jω 1 jω Þ 2 ½X ðe Þ þ X ðe

jxI(n) ¼ j Im [x(n)]

1 jω 2 ½X ðe Þ  jω

xe ðnÞ ¼ x0 ðnÞ ¼

1 ∗ 2 ½xðnÞ þ x ðnÞ 1 ∗ 2 ½xðnÞ þ x ðnÞ

X ∗ ðejω Þ

Re[X(e )]

j Im [X(e jω)]

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

      X ejω ¼ X R ejω þ j X I ejω

323

ð7:38Þ

From Eq. (7.37), the real and imaginary parts of X(e jω) are given by   X1 X R ejω ¼ xðnÞ cos ωn n¼1

ð7:39Þ

X1   X I ejω ¼  xðnÞ sin ωn n¼1

ð7:40Þ

and

Since cos(ωn) ¼ cosωn and sin(ωn) ¼ sinωn, we can obtain the following relations from Eqs. (7.39) and (7.40):   X1  jω  X R ejω ¼ x ð n Þ cos ωn ¼ X e R n¼1   X1   X I ejω ¼ xðnÞ sin ωn ¼ X I ejω n¼1

ð7:41aÞ ð7:41bÞ

indicating that the real part of DTFT is an even function of ω, while the imaginary part is an odd function of ω. Thus,     X ejω ¼ X ∗ ejω

ð7:42Þ

In polar form, X(e jω) can be written as     X ejω ¼ X ejω ejθω

ð7:43Þ

ffi  jω  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi X e ¼ ½X R ðejω Þ2 þ ½X I ðejω Þ2

ð7:44Þ

where

and     X I ðejω Þ θðωÞ ¼ ∠X ejω ¼ phase of X ejω ¼ tan 1 X R ðejω Þ

ð7:45Þ

Using the above relations, it can easily be seen that |X (e jω)| is an even function of ω, whereas the function θ(ω)is an odd function of ω. Now, the DTFT of xe (n), the even part of the real sequence x(n) is given by     1 1   F½xe ðnÞ ¼ ðF½xðnÞ þ F½xðnÞÞ ¼ X ejω þ X ejω ¼ X R ejω 2 2

ð7:46Þ

Thus, the DTFT of even part of a real sequence is the real part of X (e jω). Similarly, the DTFT of xo (n), the odd part of the real sequence x(n), is given by

324

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Table 7.5 Some properties of DTFT of a real sequence

F½xðnÞ ¼ X ðejω Þ ¼ X R ðejω Þ þ jX I ðejω Þ F½xe ðnÞ ¼ X R ðejω Þ F½xo ðnÞ ¼ jX I ðejω Þ XR(e jω) ¼ XR(ejω) XI(e jω) ¼  XI(ejω) X(e jω) ¼ X∗(ejω) |X(e jω)| ¼ |X(ejω)| ∠X(e jω) ¼  ∠ X(ejω)

F½xo ðnÞ ¼

    1  jω  X e  X ejω ¼ jX I ejω 2

ð7:47Þ

Hence, the DTFT of the odd part of a real sequence is jXI (ejω). The above properties of the DTFT of a real sequence are summarized in Table 7.5. Example 7.3 A causal LTI system is represented by the following difference equation: yðnÞ  ayðn  1Þ ¼ xðn  1Þ (i) Find the impulse response of the system h(n), as a function of parameter a. (ii) For what range of values would the system be stable? Solutions (i) Given yðnÞ  ayðn  1Þ ¼ xðn  1Þ Taking Fourier transform on both sides of above equation, we get       Y ejω  aejω Y ejω ¼ ejω X ejω From the above relation, we arrive at   Y ðejω Þ ejω ¼ H ejω ¼ X ðejω Þ 1  aejω P P n jωn jω n F½an uðnÞ ¼ 1 ¼ 1 Þ n¼1 a e n¼1 ðae 1 ¼ 1  aejω From the above equation and time shifting property, the impulse response is given by    hðnÞ ¼ F1 H ejω ¼ F1



ejω 1  aejω

 ¼ an1 uðn  1Þ

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

325

(ii) Now, X1

j hð nÞ j ¼ n¼1

X1 n¼1

jajn1 < 1

for jaj < 1:

Thus, the system is stable for |a| < 1. Example 7.4 Find the impulse response of a system described by the following difference equation: 5 1 1 y ð nÞ  y ð n  1Þ þ y ð n  2Þ ¼ x ð n  1Þ 6 6 3 Solution Taking Fourier transform on both sides of given difference equation, we get 5 1 1 Yðejω Þ  ejω Yðejω Þ þ e2jω Yðejω Þ ¼ ejω Xðejω Þ 6 6 3 From the above relation, we arrive at Y ðejω Þ ð1=3Þejω ¼ jω X ðe Þ 1  ð5=6Þejω þ ð1=6Þe2jω 2 2 ¼  1  ð1=2Þejω 1  ð1=3Þejω

H ðejω Þ ¼

The impulse response h(n) is given by 

2 1  ð1=2Þejω

 n  n  ¼ 2 12  13 uðnÞ

hðnÞ ¼ F1



 F1



2 1  ð1=3Þejω



Þ! n Example 7.5 Find the DTFT of xðnÞ ¼ ðn!nþm1 ðm1Þ! a uðnÞ, jaj < 1

Solution Let x1(n) ¼ anu(n) The Fourier transform of x1(n) is given by 1 1  X   X n ðaÞn ejωn ¼ aejω ¼ X 1 ejω ¼ n¼0

n¼0

1 1  aejω

For m ¼ 2, xðnÞ ¼ ðn þ 1Þan uðnÞ Using the differentiation property of DTFT, the Fourier transform of nanu(n) is given by

326

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

  dX 1 ðejω Þ d 1 aejω ¼j j ¼ dω dω 1  aejω ð1  aejω Þ2 Using linearity property of the DTFT, the Fourier transform of x(n) is denoted by   X ejω ¼

aejω ð1  aejω Þ2

þ

1 1 ¼ ð1  aejω Þ ð1  aejω Þ2

For m ¼ 3, x ð nÞ ¼ ¼

  ðn þ 2Þðn þ 1Þ n n2 þ 3n þ 2 n a uð nÞ a uð nÞ ¼ 2 2  1 2 n n a uðnÞ þ 3nan uðnÞ þ 2an uðnÞ 2

Using the differentiation and linearity properties of DTFT, the Fourier transform of x(n) is given by ! " # 1 d aejω 3aejω 2 X ðe Þ ¼ j þ þ 2 dω ð1  aejω Þ2 ð1  aejω Þ2 ð1  aejω Þ " # 1 aejω ð1 þ aejω Þ 3aejω 2 ¼ þ þ 2 ð1  aejω Þ3 ð1  aejω Þ2 ð1  aejω Þ " # 1 2 1 ¼ ¼ 2 ð1  aejω Þ3 ð1  aejω Þ3 jω

In general, for m ¼ k, the Fourier transform of x(n) is given by   X ejω ¼

1 ð1  aejω Þk

,

where k is any integer value:

Example 7.6 Let G1(e jω) denote the DTFT of the sequence g1(n) shown in Figure 7.1 (a). Express the DTFT of the sequence g2(n) in Figure 7.1b in terms of G1(e jω). Do not evaluate G1(e jω). Solution From Figure 7.1(b), g2(n) can be expressed in terms of g1(n) as g2 ð nÞ ¼ g1 ð nÞ þ g1 ð n  4Þ Applying DTFT on both sides, we obtain           G2 ejω ¼ G1 ejω þ ej4ω G1 ejω ¼ 1 þ ej4ω G1 ejω

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

327

g1 (n)

4 3

g 2 ( n)

2 1 0

1

2

n

3

0

1

2

(a)

3

4

5

6

7

(b)

Figure 7.1 (a) Sequence g1(n). (b) Sequence g2(n)

Example 7.7 Evaluate the inverse DTFT of each of the following DTFTs: (a) X 1 ðejω Þ ¼

1 P

δðω þ 2πkÞ

k¼1

jω

αe (b) X 2 ðejω Þ ¼ ð1αe , jω Þ2

jα j < 1

1   X Solution (a) X 1 ejω ¼ δðω þ 2πk Þ k¼1

From Table 7.3, Fð1Þ ð1 < n < 1Þ ¼

X1 k¼1

2πδðω þ 2πkÞ

Hence, F1 ½δðω þ 2πkÞ ¼

1 , ð1 < n < 1Þ 2π

jω

αe (b) X 2 ðejω Þ ¼ ð1αe , jαj < 1 jω Þ2

From Example 7.5, 1 ðn þ m  1Þ! n α uð nÞ $ n!ðm  1Þ! ð1  αejω Þm For m ¼ 2, 1 ð1  αejω Þ2 1 ð1  αejω Þ2

$

ðn þ 1Þ! n α uð nÞ n!ð1Þ!

$ ðn þ 1Þαn uðnÞ

n

328

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

· ·

4

3

1

· -2

-3

·

1

0

1

·

-1

2

3

4

5

·

-1

·

-2

·

-3

Figure 7.2 A length-9 sequence x(n)

Then α ð1  αejω Þ2

$ ðn þ 1Þαnþ1 uðnÞ

αejω ð1  αejω Þ2

$ nαn uðn  1Þ

Example 7.8 A length-9 sequence x(n) is shown in Figure 7.2 If the DTFT of x(n) is X(e jω), calculate the following functions without computing X(e jω). ðπ j0

(a) X(e )

jπ

(b) X(e )



ðπ

 X e dω (d)

(c)

jω

π

 jω  2 X e dω (e)

π

ðπ dX ðejω Þ 2 d dω

π

Solution From the given data, x(3) ¼3, x(2) ¼0, x(1) ¼ 1, x(0) ¼ 2, x(1) ¼ 3, x (2) ¼ 4, x (3) ¼ 1, x (4) ¼ 0, x(5) ¼ 1 (a) X(e j0) From the definition of Fourier transform, X ðejω Þ ¼

1 X

xðnÞejωn

n¼1

X ðej0 Þ ¼

1 X

x ð nÞ

n¼1

¼ ½ 3 þ 0 þ 1  2  3 þ 4 þ 1 þ 0  1 ¼ 3

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain

(b) X(e jπ) From the definition of Fourier transform, 1 X

X ðejπ Þ ¼

xðnÞejπ

n¼1

X ðejπ Þ ¼ 

1 X

xðnÞ ¼ 3

n¼1

ðπ (c)

  X ejω dω

π

From the definition of inverse Fourier transform, 1 x ð nÞ ¼ 2π

ðπ

  X ejω ejωn dω

π

Hence, ðπ π

ðπ (d)

  X ejω ejωn dω ¼ 2πxð0Þ ¼ 4π

 jω  2 X e dω

π

From the definition of Parseval’s theorem, 1 X

1 j x ð nÞ j ¼ 2π n¼1

ðπ

2

 jω  2 X e dω

π

Hence, Ðπ

π

2

jX ðejω Þj dω ¼ 2π

P1 n¼1

jxðnÞj2

¼ 2π ð9 þ 0 þ 1 þ 4 þ 9 þ 16 þ 1 þ 0 þ 1Þ ¼ 82π ðπ dX ðejω Þ 2 (e) dω dω π

From differentiation property and Parseval’s theorem of DTFT,

329

330

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

( )

H1 e jw

( )

H 2 e jw

1

1 p/2

w

p

p /3

(a)

(b)

Figure 7.3 (a) Fourier transform of h1(n) (b) Fourier transform of h2 (n)

( )

H1 e jw

( )

H 2 e jw

1

1

p /3

w

p /2

(a)

p

(b)

Figure 7.4 (a) Fourier transform of h1(n) (b) Fourier transform of h2(n)

ðπ 1 X dX ðejω Þ 2 jnxðnÞj2 dω dω ¼ 2π n¼1

π

¼ 2π ½81 þ 0 þ 1 þ 0 þ 9 þ 64 þ 9 þ 0 þ 25 ¼ 189π

Example 7.9 (a) The Fourier transforms of the impulse responses, h1(n) and h2 (n), of two LTI systems are as shown in Figure 7.3. Find the Fourier transform of the impulse response of the overall system, when they are connected in cascade. (b) The Fourier transforms of the impulse responses h1(n) and h2(n) of two LTI systems are as shown in Figure 7.4. Find the Fourier transform of the overall system, when they are connected in parallel.

Solution (a) The impulse response h(n) of the overall system is given by hðnÞ ¼ h1 ðnÞ∗h2 ðnÞ

7.2 Representation of Discrete-Time Signals and Systems in Frequency Domain Figure 7.5 (a) Fourier transform of the impulse response of the cascade system (b) Fourier transform of the impulse response of the parallel system

331

( )

H e Jw

1

p

p

3

2

w

(a)

( )

H e Jw

1.0

p

p

3

2

w

(b) Then, by the convolution property of the Fourier transform, the Fourier transform of the impulse response of the cascade system is given by     H 1 ejω H 2 ejω The Fourier transform of impulse response of the cascade system is shown in Figure 7.5(a). (b) The impulse response h(n) of the overall system is given by hð nÞ ¼ h1 ð nÞ þ h 2 ð nÞ Hence, the Fourier transform of impulse response of the cascade system is given by     H 1 ejω þ H 2 ejω The Fourier transform of the impulse response of the parallel system is shown in Figure 7.5(b).

332

7.3

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Frequency Response of Discrete-Time Systems

For an LTI discrete-time system with impulse response h(n) and input sequence x(n), the output y(n) is the convolution sum of x(n) and h(n) given by y ð nÞ ¼

1 X

hðkÞxðn  k Þ

ð7:48Þ

k¼1

To demonstrate the eigenfunction property of complex exponential for discretetime systems, consider the input x(n) of the form xðnÞ ¼ ejωn ,

1 < n < 1

ð7:49Þ

Then from Eq. (7.48), the output is given by y ð nÞ ¼

1 X

hðkÞe

jωðnk Þ

¼

k¼1

1 X

! hðkÞe

jωk

ejωn

ð7:50Þ

k¼1

The above equation can be rewritten as   yðnÞ ¼ H ejω ejωn ,

ð7:51aÞ

1 X   H ejω ¼ hðnÞejωn :

ð7:51bÞ

where

n¼1

H(e jω) is called the frequency response of the LTI system whose impulse response is h(n), e jωn is an eigenfunction of the system, and the associated eigenvalue is H(e jω). In general H(e jω) is complex and is expressed in terms of real and imaginary parts as       H ejω ¼ H R ejω þ jH I ejω

ð7:52Þ

where HR(e jω) and HI(e jω) are the real and imaginary parts of H(e jω), respectively. Furthermore, due to convolution, the Fourier transforms of the system input and output are related by       Y ejω ¼ H ejω X ejω

ð7:53Þ

where X(e jω) and Y(e jω) are the Fourier transforms of the system input and output, respectively. Thus,   Y ðejω Þ H ejω ¼ X ðejω Þ

ð7:54Þ

7.3 Frequency Response of Discrete-Time Systems

333

The frequency response function H(e jω) is also known as the transfer function of the system. The frequency response function provides valuable information on the behavior of LTI systems in the frequency domain. However, it is very difficult to realize a digital system since it is a complex function of the frequency variable ω. In polar form, the frequency response can be written as     H ejω ¼ H ejω ejθðωÞ

ð7:55aÞ

where |H(e jω)|, the amplitude response term, and θ(ω), the phase-response term, are given by jHðejω Þj2 ¼ jH R ðejω Þj2 þ jH I ðejω Þj2   jω 1 H I ðe Þ θðωÞ ¼ tan H R ðejω Þ

ð7:55bÞ ð7:55cÞ

Phase and Group Delays If the input is a sinusoidal signal given by xðnÞ ¼ cos ðωnÞ,

for  1 < n < 1,

then from Eq. (7.55a), the output is   y½n ¼ H ejω0 cos ðωn þ θðωÞÞ

ð7:56aÞ

ð7:56bÞ

The above equation can be rewritten as     jω  θ ð ωÞ 0 , y ½ n ¼ H e cos ω n þ ω  jω     ¼ H e 0 cos ω n  τp ðωÞ

ð7:57aÞ

It can be clearly seen that the above equation expresses the phase response as a time delay in seconds which is called as phase delay and is defined by τ P ð ωÞ ¼ 

θ ð ωÞ ω

ð7:57bÞ

An input signal consisting of a group of sinusoidal components with frequencies within a narrow interval about ω experiences different phase delays when processed by an LTI discrete-time system. As such, the signal delay is represented by another parameter called group delay defined as τ g ð ωÞ ¼ 

dθðωÞ dω

ð7:57cÞ

334

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Example 7.10 Determine the magnitude  n and phase response of a system whose impulse response is given by hðnÞ ¼ 12 uðnÞ  n Solution For hðnÞ ¼ 12 uðnÞ, the frequency response is given by  1  n 1  X X 1 1 jω n jωn e H ðe Þ ¼ e ¼ 2 2 n¼1 n¼1 1 1 ¼ ¼ jω 1  0:5e 1  0:5 cos ω þ j0:5 sin ω jω

The magnitude response is given by  jω  1 1 H e ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ¼ r ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1  0:5 cos ωÞ2 þ ð0:5Þ2 sin 2 ω 1 þ ð0:5Þ2  2ð0:5Þ cos ω The phase response is θðωÞ ¼  tan 1

0:5 sin ω 1  0:5 cos ω

The magnitude and phase values are tabulated in Table 7.6 for various values of ω and plotted in Figure 7.6(a) and (b), respectively.

Table 7.6 Magnitude and phase ω |H(e jω)| θ(ω)

0 2 00

π 4

π 2

3π 4

1.3572 28.675

0.8944 26.565

0.7148 14.640

2

30

1.8

20 Phase, degrees

1.6 Magnitude

π 0.67 00

1.4 1.2

3π 2

7π 4

0.715 14.640

0.894 26.565

1.3572 28.6750

2π 2 00

10 0

-10

1

-20

0.8 0

5π 4

0.2

0.4

0.6

0.8

1 1.2 ω/π

(a)

1.4

1.6

1.8

2

-30 0

0.2 0.4 0.6 0.8

1 1.2 1.4 1.6 1.8 ω/π

(b)

Figure 7.6 (a) Magnitude and (b) phase responses of h(n) of Example 7.10

2

7.3 Frequency Response of Discrete-Time Systems Figure 7.7 (a) Impulse response of h1(n) (b) impulse response of h2(n)

335

h2 (n)

1

h1 (n)

3

2

0

2

2

1

1

3

2

2

1

n

4

-2

(a)

0

2

n

(b)

Example 7.11 Compute the magnitude and phase responses of the impulse responses given in Figure 7.7, and comment on the results. Solution Since h1(n) is an even function of time, it has a real DTFT indicating that the phase is zero, that is, the phase is a horizontal line; h2(n) is the right-shifted version of h1(n). Hence, from time shifting property of DTFT, the transform of h2(n) is obtained by multiplying the transform of h1(n) by e–j2ω. This changes the slope of the phase linearly and can be verified as follows: The frequency response of h1(n) is H 1 ðejω Þ ¼ e2jω þ 2ejω þ 3 þ 2ejω þ e2jω ¼ ðe2jω þ e2jω Þ þ 2ðejω þ ejω Þ þ 3 ¼ 2 cos 2ω þ 4 cos ω þ 3 The magnitude response of H1(e jω) is  jω  H 1 e ¼ 2 cos 2ω þ 4 cos ω þ 3 The phase response of H1(e jω) is zero. The frequency response of h2(n) is H 2 ðejω Þ ¼ e2jω H 1 ðejω Þ ¼ e2jω ð2 cos 2ω þ 4 cos ω þ 3Þ The magnitude response of H2(e jω) is  jω  H 2 e ¼ 2 cos 2ω þ 4 cos ω þ 3 The phase response of H2(e jω) is given by   ∠H 2 ejω ¼ ∠e2jω ¼ 2ω: The magnitude and phase responses of h1(n) and h2(n) are shown in Figure 7.8(a), (b), (c), and (d). From the magnitude and phase responses of h1(n) and h2(n), it is observed that h1(n) has zero phase and h2(n) has a linear phase response, whereas both h1(n) and h2(n) have the same magnitude responses.

336

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Figure 7.8 (a) Magnitude response of h1(n). (b) Phase response of h1(n). (c) Magnitude response of h2(n). (d) Phase response of h2(n)

Example 7.12 The trapezoidal integration formula is represented by a recursive difference equation as y(n) – y(n – 1) ¼ 0.5x(n) þ 0.5x(n – 1). Determine H(e jω) of the trapezoidal integration formula.

7.3 Frequency Response of Discrete-Time Systems

Figure 7.8 (continued)

Solution Given yðnÞ  yðn  1Þ ¼ 0:5xðnÞ þ 0:5xðn  1Þ

337

338

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Taking Fourier transform on both sides of the above equation, we get Y ðejω Þ  ejω Y ðejω Þ ¼ 0:5X ðejω Þ þ 0:5ejω X ðejω Þ Y ðejω Þð1  ejω Þ ¼ 0:5X ðejω Þð1 þ ejω Þ   Y ðejω Þ ð1 þ ejω Þ ¼ 0:5 H ejω ¼ X ðejω Þ ð1  ejω Þ " #  

ejω=2 ejω=2 þ ejω=2 cos ðω=2Þ ¼ j0:5 ¼ 0:5 jω=2 jω=2 sin ðω=2Þ ðe  ejω=2 Þ e The magnitude response is given by  jω  H e ¼ 0:5 cos ðω=2Þ sin ðω=2Þ The phase response is given as follows:  If  0 < ω < π, then both cos ω/2 and sin ω/2 are positive, and hence the phase is π 2 .  If π < ω < 2π, then cos ω/2 is negative, but sin ω/2 is positive; hence the phase is π2 .

7.3.1

Frequency Response Computation Using MATLAB

The M-file function freqz(h, w) in MATLAB can be used to determine the values of the frequency response of an impulse response vector h at a set of given frequency points ω. Similarly, the M-file function freqz(b, a, ω) can also be used to find the frequency response of a system described by the recursive difference equation with the coefficients in vectors b and a. From frequency response values, the real and imaginary parts can be computed using MATLAB functions real and imag, respectively. The magnitude and phase of the frequency response can be determined using the functions abs and angle as illustrated in the following examples: Example 7.13 Determine the magnitude and phase response of a system described by the difference equation, y(n) ¼ 0.5x(n) þ 0.5x(n – 2). Solution If x(n) ¼ δ(n), then the impulse response h(n) is given by hðnÞ ¼ 0:5δðnÞ þ 0:5δðn  2Þ Hence, h(n) sequence is [0.5 0 0.5]. When this sequence is used in Program 7.1 given below, the resulting magnitude and phase responses are as shown in Figure 7.9 (a) and (b), respectively.

7.3 Frequency Response of Discrete-Time Systems

Figure 7.9 (a) Magnitude response of h(n) sequence. (b) Phase response of h(n) sequence

339

340

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Program 7.1 clear;clc; w=0:0.05:pi; h=exp(j*w); %set h=exp(jw) num=0.5+0*h.^-1+0.5*h.^-2; den=1; %Compute the frequency responses H=num/den; %Compute and plot the magnitude response mag=abs(H); figure(1),plot(w/pi,mag); ylabel(‘Magnitude’);xlabel(‘\omega/\pi’); %Compute and plot the phase responses ph=angle(H)*180/pi; figure(2),plot(w/pi,ph); ylabel(‘Phase, degrees’); xlabel(‘\omega/\pi’)

Example 7.14 Determine the magnitude and phase responses of a system described by the following difference equation: yðnÞ  2:1291yðn  1Þ þ 1:7834yðn  2Þ  0:5435yðn  3Þ ¼ 0:0534xðnÞ  0:0009xðn  1Þ  0:0009xðn  2Þ þ 0:0534xðn  3Þ Comment on the frequency response of the system. Solution The following MATLAB program 7.2 is used and the resultant magnitude response and phase response are shown in Figure 10(b) and (b), respectively. Program 7.2 clear;close all; num=[0.0534 -0.0009 -0.0009 0.0534];% numerator coefficients den=[1 -2.1291 1.7834 -0.5435];% denominator coefficients w=0:pi/255:pi; %Compute the frequency responses H=freqz(num,den,w); %Compute and plot the magnitude response mag=abs(H); figure(1),plot(w/pi,mag); ylabel(‘Magnitude’);xlabel(‘\omega/\pi’); %Compute and plot the phase responses ph=angle(H)*180/pi; figure(2),plot(w/pi,ph); ylabel(‘Phase, degrees’);xlabel(‘\omega/\pi’);

7.3 Frequency Response of Discrete-Time Systems

341

Figure 7.10 (a) Magnitude response (b) phase response

The frequency response shown in Figure 7.10 characterizes a low-pass filter with nonlinear phase. Example 7.15 Determine the magnitude and phase responses of a system described by the following difference equation:

342

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

yðnÞ  3:0538yðn  1Þ þ 3:8281yðn  2Þ  2:2921yðn  3Þ þ 0:5507yðn  4Þ ¼ xðnÞ  4xðn  1Þ þ 6xðn  2Þ  4xðn  3Þ þ xðn  4Þ: Comment on the frequency response of the system. Solution Program 7.2 with variables num ¼ [ 1 4 6 4 1] and den ¼ [1 3.0538 3.8281 2.2921 0.5507] is used, and the resultant magnitude and phase responses are shown in Figure 7.11(a) and (b), respectively. It is observed from this figure that the frequency response characterizes a narrowband band-pass filter. Example 7.16 An LTI system is described by the following difference equation: yðnÞ ¼ xðnÞ þ 2xðn  1Þ þ xðn  2Þ (a) Find the frequency response H(e jω) and group delay grd [H(e jω)] of the system. (b) Determine the difference equation of a new system such that the frequency response H1(e jω) of the new system is related to H(e jω) as H1(e jω) ¼ H(e j(ω þ π)). Solution (a) yðnÞ ¼ xðnÞ þ 2xðn  1Þ þ xðn  2Þ hðnÞ ¼ δðnÞ þ 2δðn  1Þ þ δðn  2Þ H ðejω Þ ¼ 1 þ 2ejω þ e2jω  

  1 1 ¼ 2ejω ðejω Þ þ 1 þ ðejω Þ 2 2 ¼ 2ejω ð cos ω þ 1Þ Hence, jH ðejω Þj ¼ 2ð cos ω þ 1Þ ∠H ðejω Þ ¼ ω Therefore,

  d∠H ðejω Þ ¼1 group delay ¼ grad H ejω ¼  dω (b) By frequency shifting property, ejπnh(n) $ H(e j(ω+π)). Therefore, h1 ðnÞ ¼ ejπn hðnÞ ¼ ð1Þn hðnÞ ¼ δðnÞ  2δðn  1Þ þ δðn  2Þ Hence, the difference equation of the new system is yðnÞ ¼ xðnÞ  2xðn  1Þ þ xðn  2Þ:

7.3 Frequency Response of Discrete-Time Systems

Figure 7.11 (a) Magnitude response (b) phase response

343

344

7.4

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

Representation of Sampling in Frequency Domain

As mentioned in Section 6.1, mathematically, the sampling process involves multiplying a continuous-time signal xa(t) by a periodic impulse train p(t) pð t Þ ¼

X1 n¼1

δðt  nT Þ

ð7:58Þ

As a consequence, the multiplication process gives an impulse train xp(t), which can be expressed as xp ðt Þ ¼ xa ðt Þpðt Þ P1 ¼ 1 xa ðt Þδðt  nT Þ

ð7:59Þ

Since xa(t) δ(t – nT) ¼ xa(nT) δ(t – nT), the above reduces to xp ð t Þ ¼

X1

x ðnT Þδðt 1 a

 nT Þ

ð7:60Þ

If we now take the Fourier transform of (7.59), and use the multiplication property of the Fourier transform, we get X p ðjΩÞ ¼

1 ½X a ðjΩÞ∗PðjΩÞ 2π

ð7:61Þ

where * denotes the convolution in the continuous-time domain and Xp(jΩ), Xa(jΩ), and P(jΩ) are the Fourier transforms of xp(t), xa(t), and p(t), respectively. Since p(t) is periodic with a period T, it can be expressed as a Fourier series pð t Þ ¼

1 2π 1X ejð T Þkt T 1

Since the Fourier transform of f ðt Þ ¼ ejΩT t is given by F(jΩ) ¼ 2πδ(Ω – ΩT), we see that the Fourier transform of p(t) is given by PðjΩÞ ¼

2π X1 δðΩ  kΩT Þ k¼1 T

ð7:62Þ

where ΩT ¼ 2π T : Substitution of (7.62) in (7.61) yields X p ðjΩÞ ¼

i X1 1h X a ðjΩÞ∗ δ ð Ω  kΩ Þ T k¼1 T

ð7:63Þ

Since the convolution of Xa(jΩ) with a shifted impulse δ(Ω – kΩT) is the shifted function Xa(j(Ω – kΩT)), the above reduces to X p ðjΩÞ ¼

1 X1 X ðjΩ  jkΩT Þ k¼1 a T

ð7:64Þ

7.4 Representation of Sampling in Frequency Domain

345

Eq. (7.64) shows that the spectrum of xp(t) consists of an infinite number of shifted copies of the spectrum of xa(t), and the shifts in frequency are multiples of ΩT; that is, Xp(jΩ) is a periodic function with a period of ΩT ¼ 2π/T. Since the continuous Fourier transform of δ(t – nT) is given by F½δðt  nT Þ ¼ ejΩTn ,

ð7:65Þ

we have from Eq.(7.60) that X p ðjΩÞ ¼

X1

x ðnT Þe n¼1 a

jΩTn

ð7:66Þ

Since xðnÞ ¼ xa ðnT Þ,

1 < n < 1

and the fact that the DTFT of the sequence x(n) is given by   X1 X ejω ¼ xðnÞejωn , n¼1

ð7:67Þ

  X ejω ¼ X p ðjΩÞ Ω¼ω=T

ð7:68aÞ

  X p ðjΩÞ ¼ X ejω ω¼ΩT

ð7:68bÞ

we obtain

or equivalently

Hence, we have from (7.68a) and (7.64) that     1 X1 1 X1 ω 2πk  j X ð jΩ  jkΩ Þ ¼ X j ð7:69Þ X ejω ¼ T k1 a k1 a T T T T Ω¼ω=T On the other hand, the above equation can also be expressed as   1 X1 X ejΩT ¼ X ðjΩ  jkΩT Þ k1 a T

ð7:70Þ

From Eq.(7.69) or (7.70), it can be observed that X(e jω) is obtained by frequency scaling Xp ( jΩ) using Ω ¼ ω/T. As mentioned earlier, the continuous-time Fourier transform Xp(jΩ) is periodic with respect to Ω having a period of ΩT ¼ (2π/T). In view of the frequency scaling, the DTFT X(e jω) is also periodic with respect to ω with a period of 2π.

346

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

(a)

(b)

Figure 7.12 (a) Spectrum of an analog signal (b) spectrum of the pulse train

Figure 7.13 Spectrum of an undersampled signal, showing aliasing (fold-over region). Signals in the fold-over region are not recoverable

7.4.1

Sampling of Low-Pass Signals

Sampling Theorem If the highest component of frequency in analog signal xa(t) is Ωm, then xa(t) is uniquely determined by its samples xa(nT), provided that ΩT  2Ωm

ð7:71Þ

where ΩT is called the sampling frequency in radians. Eq. (7.71) is often referred as the Nyquist condition. The spectra of the analog signal xa(t) and the impulse train p(t) with a sampling period T ¼ 2π/ΩT are shown in Figure 7.12(a) and (b), respectively. Undersampling If ΩT < 2Ωm, then the signal is undersampled, and the corresponding spectrum Xp(jΩ) is as shown in Figure 7.13. In this figure, the image frequencies centered at ΩT will alias into the baseband frequencies, and the information of the desired signal is indistinguishable from its image in the fold-over region.

7.5 Reconstruction of a Band-Limited Signal from Its Samples

347

Figure 7.14 Spectrum of an oversampled signal

Oversampling If ΩT > 2Ωm, then the signal is oversampled, and its spectrum is shown in Figure 7.14. Its spectrum is the same as that of the original analog signal, but repeats itself at every multiple of ΩT. The higher-order components centered at multiples of ΩT are called image frequencies.

7.5

Reconstruction of a Band-Limited Signal from Its Samples

According to the sampling theorem, samples of a continuous-time band-limited signal (i.e., its Fourier transform Xa(jΩ) ¼ 0 for |Ω| > |Ωm|) taken frequently enough are sufficient to represent the signal exactly. The original continuous-time signal xa(t) can be fully recovered by passing the modulated impulse train xp(t) through an ideal low-pass filter, HLP(jΩ), whose cutoff frequency satisfies Ωm  Ωc  ΩT/2. Consider a low-pass filter with a frequency response:  H LP ðjΩÞ ¼

jΩj  Ωc jΩj > Ωc

T 0

ð7:72Þ

Applying the inverse continuous-time Fourier transform to HLP(jΩ), we obtain the impulse response hLP(t) of the ideal low-pass filter given by 1 hLP ðtÞ ¼ 2π

ð1

T H LP ðjΩÞe dΩ ¼ 2π 1 jΩt

ð Ωc Ωc

ejΩt dΩ ¼

sin ðΩc tÞ , 1
For a given sequence of samples x(n), we can form an impulse train xp(t) in which successive impulses are assigned an area equal to the successive sequence values, i.e., xp ð t Þ ¼

X1 n¼1

xðnÞδðt  nT Þ

ð7:74Þ

The nth sample is associated with the impulse at t ¼ nT, where T is the sampling period associated with the sequence x(n). Therefore, the output xa(t) of the ideal low-pass filter is given by the convolution of xp(t) with the impulse response hLP(t) of the analog low-pass filter:

348

7 Frequency Domain Analysis of Discrete-Time Signals and Systems

xa ( t (

x (n(

ADC

H ( e Jw (

y ( n(

T1 = 0.0001sec

DAC

yr (t (

T2 = 0.0001sec

(a)

1

X a ( jW)

-10000p

10000p

W

(b) Figure 7.15 (a) Discrete time system (b) spectrum of input xa(t)

xa ð t Þ ¼

X1 n¼1

xðnÞhLP ðt  nT Þ

ð7:75Þ

Substituting hLP(t) from Eq.(7.73) in Eq. (7.75) and assuming for simplicity that Ωc ¼ ΩT/2 ¼ π/T, we get xa ðt Þ ¼

X1 n¼1

x ð nÞ

sin ½π ðt  nT Þ=T  π ðt  nT Þ=T

ð7:76Þ

The above expression indicates that the reconstructed continuous-time signal xa(t) is obtained by shifting in time the impulse response hLP(t) of the low-pass filter by an amount nT and scaling it in amplitude by the factor x(n) for all integer values of n in the range –1 < n < 1 and then summing up all the shifted versions. Example 7.17 Consider the system shown in Figure 7.15(a), where H(e jω) is an ideal LTI low-pass filter with cutoff of π /8 rad/sec, and the spectrum of xa(t) is shown in Figure 7.15(b). (i) What is the maximum value of T to avoid aliasing in the ADC? (ii) If 1/T ¼ 10 kHz, then what will be the spectrum of yr(t). Solution (i) From Figure 7.15(b), Ωm ¼ 10 k π. The given T1 ¼ 0.0001 sec. Then ΩT ¼ T2π1 ¼ 20 kπ. The condition to avoid aliasing in the ADC is ΩT ¼ 2Ωm (Figure 7.16) 1 ¼ 0:0001 sec (ii) T ¼ 10K

7.6 Problems

349

Figure 7.16 1

-W

-10000

W

10000

(a) jWT 1 X (e ) T

-W

-20000

-10000

20000

10000

W

(b) X (e jw ) 1 T

w

H (e jw )

1

·

-2p

-

-p -p 8

p p 8

2p

w = WT

(c) 1 T

-2p

-

Y (e jw )

p

p

8

8

2p

w

(d) Yr (e jWT )

H r ( jW)

T 1 T

-

2p 8T

10000p

(e) 1

-

p 8

2p 8T

T

r

p 8

T

(f )

7.6

Problems

1. Obtain the DTFS representation of the periodic sequence shown in Figure P7.1

350

7 Frequency Domain Analysis of Discrete-Time Signals and Systems 4

4

3

3 2

2

……. 1

0

1

1

2

3

4

5

6 7

8

n

9

Figure P7.1 Periodic sequence with period N ¼ 5

2. Find the Fourier coefficients in DTFS representation of the sequence  n xðnÞ ¼ sin 5π 4 3. Find the DTFT for the following sequences: (a) x1(n) ¼ u(n) – u(n – 5) (b) x2(n) ¼ αn(u(n) – u(n – 8)), |α| < 1  jnj (c) x3 ðnÞ ¼ n 12 (d) x4(n) ¼ |a|nsin ωn, |α| < 1 4. Let G1(e jω) denote the DTFT of the sequence g1(n) shown in Figure P7.2(a). Express the DTFTs of the remaining sequences in Figure P7.2 in terms of G1(e jω). Do not evaluate G1(e jω).

g1 (n)

4 3 2 1 0

1

2

n

3

(a)

g 2 ( n)

0

1

2

3

4

5

6

7

n

Figure P7.2 Sequences g1(n), g2(n), and g3(n)

g 3 ( n)

0

1

2

3

4

5

6

7

n

Further Reading

351

5. Determine the inverse DTFT of each of the following DTFTs: (a) H1(e jω) ¼ 1 þ 4 cos ω þ 3 cos 2ω (b) H2(e jω) ¼ (3 þ 2 cos ω þ 4 cos (2ω)) cos (ω/2)ejω/2 (c) H3(e jω) ¼ ejω/4 (d) H4(e jω) ¼ ejω[1 þ 4 cos ω] 6. A continuous-time signal xa(t) has its spectrum Xa(jΩ) as shown in Figure P7.3(a). The signal xa(t) is input to the system shown in Figure P7.3(b). H(e jω) in Figure P7.3(b) is an ideal LTI low-pass filter with a cutoff frequency of (π/2). Sketch the spectrums of x(n), y(n), and yr(t).

1

- 5000p

5000p

W

(a) xa (t)

ADC

x(n )

( )

H e Jw

T1 = 0.0001sec

y (n )

DAC

yr (t)

T2 = 0.0001sec

(b) Figure P7.3 (a) Spectrum of signal. (b) Signal reconstruction

Further Reading 1. Morrison, N.: Introduction to Fourier Analysis. Wiley, New York (1994) 2. Mitra, S.K.: Digital Signal Processing. McGraw-Hill, New York (2006) 3. Oppenheim, A.V., Schafer, W.: Discrete-Time Signal Processing, 2nd edn. Prentice-Hall, Upper Saddle River (1999)

Chapter 8

The z-Transform and Analysis of Discrete Time LTI Systems

The DTFT may not exist for all sequences due to the convergence condition, whereas the z-transform exists for many sequences for which the DTFT does not exist. Also, the z-transform allows simple algebraic manipulations. As such, the z-transform has become a powerful tool in the analysis and design of digital systems. This chapter introduces the z-transform, its properties, the inverse z-transform, and methods for finding it. Also, in this chapter, the importance of the z-transform in the analysis of LTI systems is established. Further, one-sided z-transform and the solution of statespace equations of discrete-time LTI systems are presented. Finally, transformations between continuous-time systems and discrete-time systems are discussed.

8.1

Definition of the z-Transform

The z-transform of an arbitrary discrete-time signal x(n) is defined as X ðzÞ ¼ Z ½xðnÞ ¼

X1 n¼1

xðnÞzn

ð8:1Þ

where z is a complex variable. For the existence of theX z-transform, Eq. (8.1) should 1 xðnÞzn is absolutely converge. It is known from complex variables that if n¼1 convergent, then Eq. (8.1) is convergent. Eq. (8.1) can be rewritten as X ðzÞ ¼

X1 n¼0

xðnÞzn þ

X1 n¼1

xðnÞzn

ð8:2Þ

By ratio test, the first series is absolutely convergent if     xðn þ 1Þ 1  xðn þ 1Þ zðnþ1Þ   z  < 1   limn !1  ¼ limn!1  x ð nÞ x ð nÞ  zn 

© Springer International Publishing AG, part of Springer Nature 2018 K. D. Rao, Signals and Systems, https://doi.org/10.1007/978-3-319-68675-2_8

353

354

8 The z-Transform and Analysis of Discrete Time LTI Systems

or    x ð n þ 1Þ    ¼ r 1 ðsayÞ jzj > limn!1  x ð nÞ 

ð8:3aÞ

Similarly, the second series in Eq. (8.2) is absolutely convergent if   xðn þ 1Þ 1  z  < 1 limn!1  x ð nÞ  or    x ð n þ 1Þ   ¼ r 2 ðsayÞ jzj < limn!1  x ð nÞ 

ð8:3bÞ

Thus, in general, Eq. (8.1) is convergent in some annulus r 1 < j zj < r 2

ð8:4Þ

The set of values of z satisfying the above condition is called the region of convergence (ROC). It is noted that for some sequences r1 ¼ 0 or r2 ¼ 1. In such cases, the ROC may not include z ¼ 0 or z ¼ 1, respectively. Also, it is seen that no z-transform exists if r1 > r2. The complex variable z in polar form may be written as z ¼ rejω

ð8:5Þ

where r and ω are the magnitude and the angle of z, respectively. Then, Eq. (8.1) can be rewritten as 1 1 X X   X rejω ¼ xðnÞðreÞjωn ¼ xðnÞejωn r n n¼1

ð8:6Þ

n¼1

When r ¼ 1, that is, when the contour |z| ¼ 1, a unit circle in the z-plane, then Eq. (8.5) becomes the DTFT of x(n). Rational z-Transform In LTI discrete-time systems, we often encounter with a z-transform which is a ratio of two polynomials in z: X ðzÞ ¼

N ðzÞ b0 þ b1 z1 þ b2 z2 þ    þ bM zM ¼ DðzÞ 1 þ a1 z1 þ a2 z2 þ    þ aN zN

ð8:7Þ

The zeros of the numerator polynomial N(z) are called the zeros of X(z) and those of the denominator polynomial D(z) as the poles of X(z). The numbers of finite zeros and poles in Eq. (8.7) are M and N, respectively. For example, the function X ðzÞ ¼ ðz1Þzðz2Þ has a zero at z ¼ 0 and two poles at z ¼ 1 and z ¼ 2.

8.2 Properties of the Region of Convergence for the z-Transform

8.2

355

Properties of the Region of Convergence for the z-Transform

The properties of the ROC are related to the characteristics of the sequence x(n). In this section, some of the basic properties of ROC are considered. Property 1:

ROC should not contain poles.

In the ROC, X(z) should be finite for all z. If there is a pole p in the ROC, then X(z) is not finite at this point, and the z-transform does not converge at z ¼ p. Hence, ROC cannot contain any poles. Property 2:

The ROC for a finite duration causal sequence is the entire z-plane except for z ¼ 0.

A causal finite duration sequence of length N is such that x(n) ¼ 0 for n < 0 and for n > N  1. Hence X(z) is of the form P n XðzÞ ¼ N1 n¼0 xðnÞz ¼ xð0Þ þ xð1Þz1 þ    þ xðN  1ÞzNþ1

ð8:8Þ

It is clear from the above expression that X(z) is convergent for all values of z except for z ¼ 0, assuming that x(n) is finite. Hence, the ROC is the entire z-plane except for z ¼ 0 and is shown as shaded region in Figure 8.1. Property 3:

The ROC for a noncausal finite duration sequence is the entire z-plane except for z ¼ 1.

A noncausal finite duration sequence of length N is such that x(n) ¼ 0 for n  0 and for n  N. Hence, X(z) is of the form X ðzÞ ¼

P1 n¼N

xðnÞzn

¼ xðN ÞzN þ    þ xð2Þz2 þ xð1Þz

Figure 8.1 ROC of a finite duration causal sequence

ð8:9Þ

Im(z)

Re(z)

356

8 The z-Transform and Analysis of Discrete Time LTI Systems

Figure 8.2 ROC of a finite duration noncausal sequence

Im(z)

Re(z)

It is clear from the above expression that X(z) is convergent for all values of except for z ¼ 1, assuming that x(n) is finite. Hence, the ROC is the entire z-plane except for z ¼ 1 and is shown as shaded region in Figure 8.2. Property 4:

The ROC for a finite duration two-sided sequence is the entire z-plane except for z ¼ 0 and z ¼ 1.

A finite duration of length (N2 + N1 þ 1) is such that x(n) ¼ 0 for n < N1 and for n > N2, where N1 and N2 are positive. Hence, x(z) is of the form X ðzÞ ¼

PN 2 n¼N 1

xðnÞzn

¼ xðN 1 ÞzN 1 þ    þ xð1Þz þ xð0Þ þ xð1Þz1 þ    þ xðN 2 ÞzN 2

ð8:10Þ

It is seen that the above series is convergent for all values of z except for z ¼ 0 and z ¼ 1. Property 5:

The ROC for an infinite duration right-sided sequence is the exterior of a circle which may or may not include z ¼ 1.

For such a sequence, x(n) ¼ 0 for n < N. Hence, X(z) is of the form X ðzÞ ¼

X1 n¼N

xðnÞzn

ð8:11Þ

If N  0, then the right-sided sequence corresponds to a causal sequence and the above series converges if Eq (8.3a) is satisfied, that is,   xðn þ 1Þ   ¼ r1 ð8:12Þ jzj > limn!1  x ð nÞ  Hence, in this case the ROC is the region exterior to the circle |z| ¼ r1 or the region |z| > r1 including the point at z ¼ 1. However, if N is a negative integer, say, N ¼ N1, then the series (8.12) will contain a finite number of terms involving positive powers of z. In this case, the series is not convergent for z ¼ 1, and hence the ROC is the exterior of the circle |z| ¼ r1 but will not include the point at z ¼ 1.

8.2 Properties of the Region of Convergence for the z-Transform Figure 8.3 ROC of an infinite duration causal sequence

357

Region of Convergence

Im

r1

Re

As an example of an infinite duration causal sequence, consider ( x ð nÞ ¼ Then

X ðzÞ ¼

1 X

r 1n zn

r 1n

n  0,

n < 0: 1  X n ¼ r 1 z1 ¼ 0

n¼0

n¼0

1 1  r 1 z1

ð8:13Þ

Eq. (8.13) holds only if |r1z1| < 1. Hence, the ROC is |z| > r1. The ROC is indicated by the shaded region shown in Fig. 8.3 and includes the region |z| > r1. It can be seen that X(z) has a zero at z ¼ 0 and pole at z ¼ r1. The zero is denoted by O and the pole by X. Property 6:

The ROC for an infinite duration left-sided sequence is the interior of a circle which may or may not include z ¼ 0.

For such a sequence, x(n) ¼ 0 for n > N. Hence, X(z) is of the form X ðzÞ ¼

XN n¼1

xðnÞzn

ð8:14Þ

If N < 0, then the left-sided sequence corresponds to a noncausal sequence and the above series converges if Eq. (8.3b) is satisfied, that is,    x ð n þ 1Þ    ¼ r2 ð8:15Þ jzj < limn!1  x ð nÞ  Hence, in this case, the ROC is the region interior to the circle |z| ¼ r2 or the region |z| < r2 including the point at z ¼ 0. However, if N is a positive integer, then the series (8.14) will contain a finite number of terms involving negative powers of z. In this case, the series is not convergent for z ¼ 0, and hence the ROC is the interior of the circle |z| ¼ r2 but will not include the point at z ¼ 0.

358

8 The z-Transform and Analysis of Discrete Time LTI Systems

Im

Figure 8.4 ROC of an infinite duration noncausal sequence

Region of Convergence

r2 Re

As an example of an infinite duration noncausal sequence, consider ( x ð nÞ ¼ Then, X ðzÞ ¼ X ðzÞ ¼

P1 n¼1

0

n  0,

r 2n

n  1:

n r n ¼ r 1 2 z 2 z

1 z ¼ 1  r 2 z1 z  r 2

ð8:16Þ

P1

m m m¼0 r 2 z

for jzj < r 2

ð8:17Þ

Hence, the ROC is |z| < r2, that is, the interior of the circle |z| ¼ r2. The ROC and the pole and zero of X(z) are shown in Fig. 8.4. Property 7:

The ROC of an infinite duration two-sided sequence is a ring in the z-plane.

In this case, the z-transform X(z) is of the form X ðzÞ ¼

X1 n¼1

xðnÞzn

ð8:18Þ

and converges in the region r1 < |z| < r2, where r1 and r2 are given by (8.3a) and (8.3b), respectively. As mentioned before, the z-transform does not exist if r1 > r2. As an example, consider the sequence ( x ð nÞ ¼

r 1n

n  0,

r 2n

n < 1:

ð8:19Þ

Then, X ðzÞ ¼

z z zð2z  r 1  r 2 Þ þ ¼ z  r 1 z  r 2 ðz  r 1 Þ ðz  r 2 Þ

ð8:20Þ

8.2 Properties of the Region of Convergence for the z-Transform

359

Im

Region of Convergence

Figure 8.5 ROC of an infinite duration two-sided sequence

r1

r2 Re

where the region of convergence is r1< |z| < r2. Thus, the ROC is a ring with a pole on the interior boundary and a pole on the exterior boundary of the ring, without any pole in the ROC. There are two zeros, one being located at the origin and the other in the ROC. The poles and zeros as well as the ROC are shown in Figure 8.5. Example 8.1 Determine the z-transform and the ROC for the following sequence: xðnÞ ¼ 2n

for n  0

Solution From the definition of the z-transform, X ðzÞ ¼

1 X

xðnÞzn ¼

n¼1

¼

1 , 1  2z1

1 1  X X n 2n zn ¼ 2 z1 n¼0

n¼0

 1  2z  < 1

Thus, the ROC is |z| > 2. Example 8.2 Determine the z-transform and the ROC for the following sequence: 8  n 1 > > > < 5 x ð nÞ ¼  n > 1 > > : 3

Solution

for n  0 for n < 0

  n 1  1 X X 1 n n 1 X ðzÞ ¼ xðnÞz ¼  z þ  zn 5 3 n¼1 n¼1 n¼0      1  1 1 1     < z and z < ¼ þ , for j j j j    3 1 þ ð1=5Þz1 1  ð1=3Þz1 5 1 X

respectively.   Thus, the ROC is 15 < jzj < 13

n

360

8 The z-Transform and Analysis of Discrete Time LTI Systems

Properties of the z-Transform

8.3

Properties of the z-transform are very useful in digital signal processing. Some important properties of the z-transform are stated and proved in this section. We will denote in the following the ROC of X(z) by R(r1 < |z| < r2) and those of X1(z) and X2(z) by R1 and R2, respectively. Also, the region (1/(r2) < |z| < 1/(r1) is denoted by (1/R). Linearity If x1(n) and x2(n) are two sequences with z-transforms X1(z) and X2(z) and ROCs R1 and R2, respectively, then the z-transform of a linear combination of x1(n) and x2(n) is given by Zfa1 x1 ðnÞ þ a2 x2 ðnÞg ¼ a1 X 1 ðzÞ þ a2 X 2 ðzÞ

ð8:21Þ

whose ROC is at least (R1 \ R1) and a1 and a2 being arbitrary constants. Proof Zfa1 x1 ðnÞ þ a2 x2 ðnÞg ¼ ¼

P1

n¼1 fa1 x1 ðnÞ þ P n a1 1 n¼1 x1 ðnÞz

a2 x2 ðnÞgzn P n þ a2 1 n¼1 x2 ðnÞz

¼ a1 X 1 ð z Þ þ a2 X 2 ð z Þ

ð8:22Þ ð8:23Þ

The result concerning the ROC follows directly from the theory of complex variables concerning the convergence of a sum of two convergent series. Time Reversal If x(n) is a sequence with z-transform X(z) and ROC R, then the z-transform of the time reversed sequence x(n) is given by   Z fxðnÞg ¼ X z1

ð8:24Þ

whose ROC is 1/R. Proof From the definition of the z-transform, we have Z ½xðnÞ ¼

P1 n¼1

xðnÞzn ¼ ¼

P1 m¼1

xðmÞ zm

m¼1

xðmÞðz1 Þ

P1

m

ð8:25Þ

Hence,   Z ½xðnÞ ¼ X z1

ð8:26Þ

Since (r1 < |z| < r2), we have (1/(r2) < |z1| < 1/(r1)). Thus, the ROC of Z [x(n)] is 1/R.

8.3 Properties of the z-Transform

361

Time Shifting If x(n) is a sequence with z-transform X(z) and ROC R, then the z-transform of the delayed sequence x(n  k), k being an integer, is given by Z ½xðn  kÞ ¼ zk X ðzÞ

ð8:27Þ

whose ROC is the same as that of X(z) except for z ¼ 0 if k > 0 and z ¼ 1 if k < 0 Proof

Z fx ð n  k Þ g ¼

X1 n¼1

xðn  kÞzn

ð8:28Þ

Substituting m ¼ n  k, Z ½ xð n  k Þ  ¼ ¼

P1

ð m þ k Þ m¼1 xðmÞz P m zk 1 m¼1 xðmÞz

¼ zk

P1 m¼1

¼ zk X ðzÞ

xðmÞzm

ð8:29Þ ð8:30Þ

It is seen from Eq. (8.30) that in view of the factor zk, the ROC of Z [x(n  k)] is the same as that of X(z) except for z ¼ 0 if k > 0 and z ¼ 1 if k < 0. It is also observed that in particular, a unit delay in time translates into the multiplication of the z-transform by z1. Scaling in the z-Domain If x(n) is a sequence with z-transform X(z), then Z{anx(n)} ¼ X(a1z) for any constant a, real or complex. Also, the ROC of Z{anx(n)} is |a|R, i.e., |a|r1 < |z| < |a|r2. Proof Z fan x ð nÞ g ¼

1 X

an xðnÞzn

ð8:31Þ

n¼1

¼

1 X

x ð nÞ

n¼1

 z n

¼X

a

z a

ð8:32Þ

Since the ROC of X(z) is r1 < |z| < r2, the ROC of X(a1z) is given by r1 < |a–1z| < r2, that is, jajr 1 < jzj < jajr 2 Differentiation in the z-Domain If x(n) is a sequence with z-transform X(z), then Z fnxðnÞg ¼ z

dX ðzÞ dz

whose ROC is the same as that of X(z). Proof From the definition, Z ½ x ð nÞ  ¼

X1 n¼1

xðnÞzn

ð8:33Þ

362

8 The z-Transform and Analysis of Discrete Time LTI Systems

Differentiating the above equation with respect to z, we get 1 X dX ðzÞ ¼ ðnÞxðnÞ zn1 dz n¼1

ð8:34Þ

Multiplying the above equation both sides by z, we obtain z

1 X dX ðzÞ ¼ z ðnÞxðnÞzn1 dz n¼1

ð8:35Þ

which can be rewritten as z

1 X dX ðzÞ ¼ nxðnÞzn ¼ Z fnxðnÞg dz n¼1

ð8:36aÞ

Now, the region of convergence ra < |z| < rb of the sequence nx(n) can be found using Eqs. (8.3a) and (8.3b).     ðn þ 1Þxðn þ 1Þ  x ð n þ 1Þ      r a ¼ limn!1   ¼ limn!1  xðnÞ  ¼ r 1 nxðnÞ and   ðn þ 1Þxðn þ 1Þ  ¼n r b ¼ limn!1   nxðnÞ

   x ð n þ 1Þ    ¼ r2 lim n!1  xðnÞ 

Hence, the ROC of Z[nx(n)] is the same as that of X(z). By repeated differentiation of Eq. (8.36a), we get the result 

dfX ðzÞg k Z nk xðnÞ ¼ z dz

ð8:36bÞ

It is to be noted that the ROC of Z[nkx(n)] is also the same as that of X(z). Convolution of Two Sequences If x1(n) and x2(n) are two sequences with z-transforms X1(z) and X2(z), and ROCs R1 and R2, respectively, then Z ½x1 ðnÞ∗x2 ðnÞ ¼ X 1 ðzÞX 2 ðzÞ

ð8:37Þ

whose ROC is at least R1 \ R2. Proof X ðzÞ ¼

1 X n¼1

xðnÞzn

ð8:38Þ

8.3 Properties of the z-Transform

363

The discrete convolution of x1(n) and x2(n) is given by x1 ðnÞ∗x2 ðnÞ ¼

X1

x ðkÞx2 ðn k¼1 1

 kÞ ¼

X1

x ðk Þx1 ðn k¼1 2

 kÞ

ð8:39Þ

Hence, the z-transform of the convolution is Z ½x1 ðnÞ∗x2 ðnÞ ¼

X1

hX1

i x ð k Þx ð n  k Þ zn 2 1 k¼1

n¼1

ð8:40Þ

Interchanging the order of summation, the above equation can be rewritten Z ½x1 ðnÞ∗x2 ðnÞ ¼ ¼ ¼

P1

P1 n k¼1 x1 ðk Þ n¼1 x2 ðn  k Þz P1 P1 ðmþk Þ k¼1 x1 ðk Þ m¼1 x2 ðmÞz P1 P1 k m k¼1 x1 ðk Þz m¼1 x2 ðmÞz

ð8:41Þ

Hence, Z ½x1 ðnÞ∗x2 ðnÞ ¼ X 1 ðzÞX 2 ðzÞ

ð8:42Þ

Since the right side of Eq. (8.42) is a product of the two convergent sequences X1(z) and X2(z) with ROCs R1 and R1, it follows from the theory of complex variables that the product sequence is convergent at least in the region R1 \ R2. Hence, the ROC of Z[x1(n) ∗ x2(n)] is at least R1 \ R2. Correlation of Two Sequences If x1(n) and x2(n) are two sequences with z-transforms X1(z) and X1(z), and ROCs R1 and R2, respectively, then   Z ½r x1 x2 ðlÞ ¼ X 1 ðzÞX 2 z1

ð8:43Þ

whose ROC is at least R1 \ (1/R2) Proof Since r x1 x2 ðlÞ ¼ x1 ðlÞ∗x2 ðlÞ,

ð8:44Þ

Z ½r x1 x2 ðlÞ ¼ Z ½x1 ðlÞ∗x2 ðlÞ, ¼ Z ½x1 ðlÞZ ½x2 ðlÞ, 1

¼ X 1 ðzÞX 2 ðz Þ,

using Equation ð8:37Þ

using Equation ð8:24Þ

ð8:45Þ

Since the ROC of X2(z) is R2, the ROC of X2(z1) is 1/R2 from the property concerning time reversal. Also, since the ROC of X1(z) is R1, it follows from Eq. (8.45) that the ROC of Z ½r x1 x2 ðlÞ is at least R1 \ (1/R2).

364

8 The z-Transform and Analysis of Discrete Time LTI Systems

Conjugate of a Complex Sequence If x(n) is a complex sequence with the z-transform X(z), then Z ½x∗ðnÞ ¼ ½X ðz∗ Þ

∗

ð8:46Þ

with the ROCs of both X(z) and Z[x*(n)] being the same Proof The z-transform of x*(n) is given by Z ½x∗ðnÞ ¼ ¼

X1 n¼1

x∗ ðnÞzn

hX1

ð8:47Þ i n ∗

xðnÞðz∗ Þ n¼1

ð8:48Þ

In the R.H.S. of the above equation, the term in the brackets is equal to x (z*). Therefore, Eq. (8.48) can be written as Z½x∗ ðnÞ ¼ ½Xðz∗ Þ∗ ¼ X ∗ ðz∗ Þ

ð8:49Þ

It is seen from Eq. (8.49) that the ROC of the z-transform of conjugate sequence is identical to that of X(z). Real Part of a Sequence If x(n) is a complex sequence with the z-transform X(z), then 1 Z ½RefxðnÞg ¼ ½X ðzÞ þ X ∗ ðz∗ Þ 2

ð8:50Þ

whose ROC is the same as that of X(z). Proof



1 Z ½RefxðnÞg ¼ Z fxðnÞ þ x∗ ðnÞg 2

ð8:51Þ

Since the z-transform satisfies the linearity property, we can write Eq. (8.51) as 1 1 Z ½RefxðnÞg ¼ Z ½xðnÞ þ Z ½x∗ ðnÞ 2 2 1 ¼ ½X ðzÞ þ X ∗ ðz∗ Þ, using ð8:49Þ 2 It is clear that the ROC of Z[Re{x(n)}] is the same as that of X(z).

ð8:52Þ ð8:53Þ

8.4 z-Transforms of Some Commonly Used Sequences

365

Imaginary Part of a Sequence If x(n) is a complex sequence with the z-transform X(z), then Z ½ImfxðnÞg ¼

1 ½X ðzÞ  X ∗ ðz∗ Þ 2j

ð8:54Þ

whose ROC is the same as that of X(z). Proof Now xðnÞ  x∗ ðnÞ ¼ 2jImfxðnÞg

ð8:55Þ

Thus, ImfxðnÞg ¼

1 fx ð nÞ  x ∗ ð nÞ g 2j

ð8:56Þ

Hence,  Z½ImfxðnÞg ¼ Z

1 fxðnÞ  x∗ ðnÞg 2j

ð8:57Þ

Again, since the z-transform satisfies the linearity property, we can write Eq. (8.57) as 1 1 Z ½ImfxðnÞg ¼ Z ½xðnÞ  Z ½x∗ ðnÞ 2j 2j 1 ¼ ½X ðzÞ  X ∗ ðz∗ Þ, using ð8:49Þ 2j

ð8:58Þ

Again, it is evident that the ROC of the above is the same as that of X(z). The above properties of the z-transform are all summarized in Table 8.1.

8.4

z-Transforms of Some Commonly Used Sequences

Unit Sample Sequence The unit sample sequence is defined by

δðnÞ ¼

1 0

for n ¼ 0 elsewhere

ð8:59Þ

By definition, the z-transform of δ(n) can be written as X ðzÞ ¼

X1 n¼1

xðnÞzn ¼ 1z0 ¼ 1

It is obvious from (8.60) that the ROC is the entire z-plane.

ð8:60Þ

366

8 The z-Transform and Analysis of Discrete Time LTI Systems

Table 8.1 Some properties of the z-transform Property Linearity Time shifting

Sequence a1x1(n) þ a2x2(n) x(n  k)

z-Transform a1X1(z) þ a2X2(z) zkX(z).

ROC At least R1 \ R2 Same as R except for z ¼ 0 if k > 0 and for z ¼ 1 if k < 0

Time reversal

x(n)

X(z1)

1 R

1

n

Scaling in the z-domain Differentiation in the z-domain Convolution theorem Correlation theorem

a x(n)

X(a z)

|a|R

nx(n)

z dXdzðzÞ

R

x1(n) ∗ x2(n)

X1(z)X2(z)

At least R1 \ R2

X1(z)X2(z1)

At least R1 \ 1/R2

Conjugate complex sequence Real part of a complex sequence Imaginary part of a complex sequence Time reversal of a complex conjugate sequence

x∗(n)

[X(z∗)]∗

R

Re[x(n)]

1 2 ½X ðzÞ

þ X ∗ ðz∗ Þ

At least R

Im[x(n)]

1 2j ½X ðzÞ

 X ∗ ðz∗ Þ

At least R

x∗(n)

X∗(1/z∗)

r x1 x2 ðlÞ ¼

1 P

x1 ðnÞx2 ðn  lÞ

n¼1

1 R

Unit Step Sequence The unit step sequence is defined by

uð nÞ ¼

1 0

for n  0 elsewhere

ð8:61Þ

The z-transform of x(n) by definition can be written as X ðzÞ ¼ ¼

P1 n¼1

1 1  z1

xðnÞzn ¼ 1 þ z1 þ z2 þ      z for z1  < 1 ¼ z1

ð8:62Þ

Hence, the ROC for X(z) is |z| > 1 Example 8.3 Find the z-transform of x(n) ¼ δ(n  k) Solution By using the time shifting property, we get Z ½δðn  kÞ ¼ zk Z ½δðnÞ ¼ zk

ð8:63Þ

The ROC is the entire z-plane except for z ¼ 0 if k is positive and for z ¼ 1 if k is negative

8.4 z-Transforms of Some Commonly Used Sequences

367

Example 8.4 Find the z-transform of x(n) ¼ u(n  1) z Solution We know that Z ½uðnÞ ¼ z1 for jzj > 1 from Eq. (8.62) Hence, using the time shifting property

Z ½uðn  1Þ ¼ z1

z 1 ¼ z1 z1

for jzj > 1

ð8:64Þ

Now, using the time reversal property (Table 8.1), we get Z½uðn  1Þ ¼

1 z ¼ 1 1z

for jzj < 1

z1

Hence, Z ½uðn  1Þ ¼

z z1

for jzj < 1

ð8:65Þ

Example 8.5 Find the z-transform of the sequence x(n) ¼ {bnu(n)} z Solution Let x1(n) ¼ u(n). From Eq. (8.62), Z ½uðnÞ ¼ X 1 ðzÞ ¼ z1 for jzj > 1 Using the scaling property, we get

  Z ½bn uðnÞ ¼ X 1 b1 z ¼

z zb

for jzj > jbj

Example 8.6 Find the z-transform of x(n) ¼ nu(n) Solution Let x1(n) ¼ u(n). Again, z Z ½uðnÞ ¼ X 1 ðzÞ ¼ z1 for jzj > 1 Using the differentiation property, Z ½nxðnÞ ¼ z

using

Eq.

(8.62),

we

dX ðzÞ dz

we get Z ½nuðnÞ ¼ z

  dX 1 ðzÞ d z z ¼ z ¼ dz dz z  1 ðz  1Þ2

for jzj > 1

Example 8.7 Obtain the z-transform of the following sequence: ( xðnÞ ¼

n2 u ð n Þ 0

elsewhere

have

368

8 The z-Transform and Analysis of Discrete Time LTI Systems

Solution

X ðzÞ ¼

X1 n¼1

xðnÞzn ¼

X1 n¼0

n2 uðnÞzn

Let x(n) ¼ n2x1(n), where x1(n) ¼ u(n). Then X 1 ðzÞ ¼

z z1

for jzj > 1

Using the differentiation property that   d 2 ZT ZT if xðnÞ $ X ðzÞ, then n2 xðnÞ $ X z X ðzÞ dz we get " #   d d d z z ð z þ 1Þ z ½X 1 ðzÞ ¼ z X ðzÞ ¼ z ¼ dz dz dz ðz  1Þ2 ðz  1Þ3 The ROC of X(z) is the same as that of u(n), namely, |z| > 1 Example 8.8 Find the z-transform of x(n) ¼ sin ωn u(n) Solution

jωn  e  ejωn 1 Zfsin ωn uðnÞg ¼ Z uðnÞ ¼ ½Zfejωn uðnÞg  Zfejωn uðnÞg 2j 2j Using the scaling property, we get    jωn  1 1  jωn z z Z e uð nÞ  Z e uð nÞ ¼  2j 2j z  ejω z  ejω z sin ω for jzj > 1 ¼ 2 z  2z cos ω þ 1 Therefore, Z f sin ωn uðnÞ ¼

z2

z sin ω  2z cos ω þ 1

for jzj > 1

Example 8.9 Find the z-transform of x(n) ¼ cos ωn u(n). n jωn jωn i uðnÞ ¼ 12 ½Z fejωn uðnÞg þ Z fejωn uðnÞg Solution Z f cos ωn uðnÞ ¼ Z e þe 2 Using the scaling property, we get  1 1 z z jωn jωn ½Zfe uðnÞg þ Zfe uðnÞg ¼ þ 2 2 z  ejω z  ejω ¼

zðz  cos ωÞ z2  2zcos ω þ 1

for jzj > 1

8.4 z-Transforms of Some Commonly Used Sequences

369

Therefore, Z f cos ωn uðnÞ ¼

z2

zðz  cos ωÞ  2z cos ω þ 1

for jzj > 1

Example 8.10 Find the z-transform of the sequence x(n) ¼ [u (n)  u (n  5)] Solution XðzÞ ¼

4 X

zn ¼ 1 þ z1 þ z2 þ z3 þ z4 ¼

n¼ 0

z 1 z5  1 ð1  z5 Þ ¼ 4 ðz  1Þ z z1

The ROC is the entire z-plane except for z ¼ 0 Example 8.11 Determine X(z) for the function xðnÞ ¼ 

1 n 2

uðn  1Þ

Solution From Eq. (8.65), we have Z ½uðn  1Þ ¼

z z1

for jzj < 1

Now using the scaling property (Table 8.1),

 n  1 2z Z  uðn  1Þ ¼ 2 2z  1

for jzj <

1 2

Thus the ROC is jzj < 12 Example 8.12 Consider a system with input x(n) and output y(n). If its impulse response h(n) ¼ Ax(L  n), where L is an integer constant, and A is a known constant, find Y(z) in terms of X(z). Solution

hðnÞ ¼ AxðL  nÞ yðnÞ ¼ xðnÞ∗hðnÞ

By the convolution property of the z-transform, we have Y ðzÞ ¼ H ðzÞX ðzÞ where H ðzÞ ¼ Z fAxðL  nÞg ¼ A

X1 n¼1

xððn  LÞÞzn

Letting n  L ¼ m in the above, we have H ðzÞ ¼ A

P1

m ¼ 1

P m xðmÞzðmþLÞ ¼ AzL 1 m¼1 xðmÞz P m ¼ AzL 1 m¼1 xðmÞz ¼ AzL X ðz1 Þ ¼ AzL X ð1=zÞ

370

8 The z-Transform and Analysis of Discrete Time LTI Systems

Hence, Y ðzÞ ¼ AzL X ð1=zÞX ðzÞ A list of some commonly used z-transform pairs are given in Table 8.2 Initial Value Theorem If a sequence x(n) is causal, i.e., x(n) ¼ 0 for n < 0, then xð0Þ ¼ Lt X ½z

ð8:66Þ

z!1

Proof Since x(n) is causal, its z-transform X[z] can be written as X ½z ¼

1 X

xðnÞ:zn ¼ xð0Þ þ xð1Þz1 þ xð2Þz2 þ   

ð8:67Þ

n¼0

Now, taking the limits on both sides Lt X ðzÞ ¼ Lt

z!1

z!1



 xð0Þ þ xð1Þz1 þ xð2Þz2 þ    ¼ xð0Þ

ð8:68Þ

Hence, the theorem is proved. Example 8.13 Find the initial value of a causal sequence x(n) if its z-transform X(z) is given by X ðzÞ ¼

0:5z2 ðz  1Þðz2  0:85z þ 0:35Þ

Table 8.2 Some commonly used z-transform pairs x(n) δ(n) u(n) nu(n) anu(n  1) nan{u(n  1)}

X(z) 1 1 1  z1 z1 ð1  z1 Þ2 1 1  az1 az1

ROC Entire z-plane |z| > 1 |z| > 1 |z| < |a| |z| < |a|

az1 Þ2

{cosωn} u(n) {sinωn} u(n)

ð1  1  z1 cos ω 1  2z1 cos ω þ z2 z1 sin ω 1  2z1 cos ω þ z2

|z| > 1 |z| > 1

8.5 The Inverse z-Transform

371

Solution The initial value x(0) is given by xð0Þ ¼ lim X ðzÞ ¼ lim z!1

8.5

n!1

ðz 

0:5z2 0:5z2 ¼ lim ¼0  0:85z þ 0:35Þ z!1 zðz2 Þ

1Þ ð z 2

The Inverse z-Transform

The z-transform of a sequence x(n), Z[x(n)], defined by Eq. (8.1), is X ðzÞ ¼

X1 m¼1

xðmÞzm

ð8:69Þ

Multiplying the above equation both sides by zn  1 and integrating both sides on a closed contour C in the ROC of the z-transform X(z) enclosing the origin, we get Þ

C X ðzÞz

n1

dz ¼ ¼

Þ P1 C

m¼1

xðmÞzm zn1 dz

C

m¼1

xðmÞzmþn1 dz

Þ P1

ð8:70Þ

1 Multiplying both sides of Eq. (8.70) by 2πj , we arrive at

1 2πj

þ X ðzÞzn1 dz ¼ C

1 2πj

þ X 1 C

m¼1

xðnÞzmþn1 dz

ð8:71Þ

By Cauchy integral theorem, we have 1 2πj

1 for m ¼ n mþn1 z dz ¼ m¼1 0 for m ¼ 6 n þ 1 X ðzÞzn1 dz ¼ xðnÞ: 2πj C

þ X 1 C

ð8:72Þ

Thus, the inverse z-transform of X(z), denoted by Z1[X(z)], is given by þ 1 1 Z ½ X ð z Þ  ¼ x ð nÞ ¼ X ðzÞzn1 dz ð8:73Þ 2πj C It should be noted that given the ROC and the z-transform X(z), the sequence x(n) is unique. Table 8.2 can be used in most of the cases for obtaining the inverse transform. We will consider in Section 8.6 different methods of finding the inverse transform.

372

8 The z-Transform and Analysis of Discrete Time LTI Systems

8.5.1

Modulation Theorem in the z-Domain

The z-transform of the product of two sequences (real or complex) x1(n) and x2(n) is given by þ z 1 Z ½x1 ðnÞx2 ðnÞ ¼ X 1 ðvÞX 2 ð8:74Þ v1 dv 2πj C v where C is a closed contour which   encloses the origin and lies in the ROC that is common to both X1(v) and X 2 vz . Proof Let x(n) ¼ x1(n)x2(n) The inverse z-transform of x1(n) is given by þ 1 x 1 ð nÞ ¼ X 1 ðvÞ vn1 dv 2πj C

ð8:75Þ

Using Eq. (8.75), we get xðnÞ ¼ x1 ðnÞx2 ðnÞ ¼

1 2πj

þ X 1 ðvÞ vn1 x2 ðnÞdv

ð8:76Þ

C

Taking the z-transform of Eq. (8.76), we obtain XðzÞ ¼

 þ P1 1 n1 n xðnÞz ¼ X ðvÞ v x ðnÞdv zn 1 2 n¼1 n¼1 2πj C þ X1 1 ¼ X 1 ðvÞ½ n¼1 vn x2 ðnÞzn v1 dv 2πj C

P1

Using the scaling property, we have that X1

vn x2 ðnÞ zn ¼ X 2 n¼1

z v

Hence, Eq. (8.77) becomes 1 X ðzÞ ¼ 2πj

þ X 1 ðvÞX 2 C

z v

v1 dv

which is the required result.

8.5.2

Parseval’s Relation in the z-Domain

If x1(n) and x2(n) are complex valued sequences, then

ð8:77Þ

8.5 The Inverse z-Transform

X1 n¼1

373

½x1 ðnÞx2

∗

þ

1 ð nÞ  ¼ 2πj

X 1 ð vÞ X 2

∗



 1 1 v dv v∗

C

ð8:78Þ

where C is a contour contained in the ROC common to the ROCs of X1(v) and X ∗ 2   1 . v∗ Proof From Eq. (8.77), we have Z ½x1 ðnÞx2 ðnÞ ¼

1 2πj

þ X 1 ðvÞX 2 C

z v1 dv v

Hence, Z ½x1 ðnÞx2 ∗ ðnÞ ¼

1 2πj

þ

X 1 ðvÞX 2 ∗ C



 z∗ 1 v dv v∗

ð8:79Þ

where we have used the result concerning the z-transform of a complex conjugate (see Table 8.1). That is, X1

½x ðnÞx2 ∗ ðnÞ zn ¼ n¼1 1

1 2πj

þ

X 1 ðvÞX 2 ∗



C

 z∗ 1 v dv v∗

ð8:80Þ

Letting z ¼ 1 in Eq. (8.80), we get X1

½x ðnÞx2 ∗ ðnÞ ¼ n¼1 1

1 2πj

þ

X 1 ðvÞX 2 ∗

C



 1 1 v dv v∗

Hence, the theorem. If x1(n) ¼ x2(n) ¼ x(n) and the unit circle is included by the ROC of X(z), then by letting v ¼ e jω in (8.78), we get the energy of sequence in the z-domain to be X1

j x ð nÞ j 2 ¼ n¼1

1 2πj

þ

X ðzÞX ∗ C

  1 1 z dz z∗

ð8:81Þ

For the energy of real sequences in the z-domain, the above expression becomes þ X1  1  1 1 2 z dz x ð n Þ ¼ X ð z ÞX z ð8:82Þ j j n¼1 2πj C The Parseval’s relation in the frequency domain is given by X1

jxðnÞj2 ¼ n¼1

1 2π

ðπ π

jXðejω Þj2 dω

374

8 The z-Transform and Analysis of Discrete Time LTI Systems

Thus, X1

1 jxðnÞj ¼ n¼1 2πj

8.6 8.6.1

þ

2

1 XðzÞXðz Þz dz ¼ 2π C 1

1

ðπ π

jXðejω Þj2 dω

ð8:83Þ

Methods for Computation of the Inverse z-Transform Cauchy’s Residue Theorem for Computation of the Inverse z-Transform

By Cauchy’s residue theorem, the integral in Eq. (8.73) for rational z-transforms yields Z1[X(z)] ¼ x(n) ¼ sum of the residues of the function [X(z)zn1] at all the poles pi enclosed by a contour C that lies in the ROC of X(z) and encloses the origin. The residue at a simple pole pi is given by

res X ðzÞzn1 ¼ lim ðz  pi Þ z¼p

z!pi

X ðzÞzn1



ð8:84Þ

while for a pole pi of multiplicity m, the residue is given by

res X ðzÞzn1 ¼ z¼p

1 d m1 lim m1 ðz  pi Þm X ðzÞzn1 ðm  1Þ! z!pi dz

ð8:85Þ

We will now consider a few examples of finding the inverse z-transform using the residue method. Example 8.14 Assuming the sequence x(n) to be causal, find the inverse z-transform of X ðzÞ ¼

z ð z þ 1Þ ð z  1Þ 3

Solution Since the sequence is causal, we have to consider the poles of X(z)zn1 for only n  0. For n  0, the function X(z)zn1 has only one pole at z ¼ 1 of multiplicity 3. Thus, the inverse z-transform is given by " # 1 d2 z ð z þ 1 Þ x ð nÞ ¼ lim ð z  1Þ 3 zn1 ð3  1Þ! z!1 dz2 ðz  1Þ3 " # 1 d2 3 zðz þ 1Þ n1 lim x ð nÞ ¼ ð z  1Þ z ð3  1Þ! z!1 dz2 ðz  1Þ3 ¼



1 d2 1 lim 2 ½ðz þ 1Þzn  ¼ lim nðn þ 1Þzn1 þ nðn  1Þzn2 2! z!1 dz 2 z!1

¼ n2

8.6 Methods for Computation of the Inverse z-Transform

375

It should be mentioned that if x(n) were not causal, then X(z)zn1 would have had a multiple pole of order n at the origin, and we would have to find the residue of X(z)zn1 at the origin to evaluate x(n) for n < 0. Example 8.15 If x(n) is causal, find the inverse z-transform of X ðzÞ ¼

1 2ðz  0:8Þðz þ 0:4Þ

Solution Since the sequence is causal, we have to consider the poles of X(z)zn1 for only n  0. Hence X ðzÞzn1 ¼ 2ðz0:81Þðzþ0:4Þ zn1 , we see that for n  1, X(z)zn1 has two simple poles at 0.8 and 0.4. However for n ¼ 0, we have an additional pole at the origin. Hence, we evaluate x(0) separately by evaluating the residues of X ðzÞz1 ¼ 2ðz0:81Þðzþ0:4Þ. Thus, xð 0Þ ¼ ¼

1 1 1  jz¼0 þ jz¼0:4 þ jz¼0:8 2zðz  0:8Þ 2zðz þ 0:4Þ 2ðz  0:8Þ z þ 0:4 1 1 1 þ þ ¼0 2ð0:8Þð0:4Þ 2ð0:4Þð1:2Þ 2ð0:8Þð1:2Þ

For n > 0, zn1 zn1 jz¼0 :4 þ jz¼0 2ðz  0:8Þ 2ðz þ 0:4Þ :8  ð0:4Þn1 0:8n1 1  n1 : 0:8 þ ¼  ð0:4Þn1 ¼ 2ð1:2Þ 2ð1:2Þ 2:4

x ð nÞ ¼

Hence for any n  0, x ð nÞ ¼

8.6.2

 1  n1 : 0:8  ð0:4Þn1 uðn  1Þ 2:4

Computation of the Inverse z-Transform Using the Partial Fraction Expansion

Partial fraction expansion is another technique that is useful for evaluating the inverse z-transform of a rational function and is a widely used method. To apply the partial fraction expansion method to obtain the inverse z-transform, we may consider the z-transform to be a ratio of two polynomials in either z or in z1. We now consider a rational function X(z) as given in Eq. (8.7). It is called a proper rational function if M > N; otherwise, it is called an improper rational function. An improper rational function can be expressed as a proper rational function by dividing

376

8 The z-Transform and Analysis of Discrete Time LTI Systems

the numerator polynomial N(z) by its denominator polynomial D(z) and expressing X (z) in the form X ðzÞ ¼

MN X

f k zk þ

k¼0

N 1 ðzÞ DðzÞ

ð8:86Þ

where the order of the polynomial N1(z) is less than that of the denominator polynomial. The partial fraction expansion can be now made on N1(z)/D(z). The z inverse z-transform of the terms in the sum is obtained from the pair δ½n $ 1 (see Table 8.1) and the time-shift property (see Table 8.2). Let X(z) be a proper rational function expressed as X ðzÞ ¼

N ðzÞ b0 þ b1 z1 þ b2 z2 þ    þ bM zM ¼ DðzÞ 1 þ a1 z1 þ a2 z2 þ    þ aN zN

ð8:87Þ

For simplification, eliminating negative powers, Eq. (8.87) can be rewritten as XðzÞ ¼

NðzÞ b0 zN þ b1 zN1 þ b2 zN2 þ    þ bM zNM ¼ DðzÞ zN þ a1 zN1 þ a2 z2 þ    þ aN

ð8:88Þ

Since X(z) is a proper fraction, so will be [X(z)/z]. If all the poles pi are simple, then, [X(z)/z] can be expanded in terms of partial fractions as N X ðzÞ X ci ¼ z z  pi i¼1

ð8:89Þ

where c i ¼ ð z  pi Þ

 X ðzÞ z z¼v

ð8:90Þ

If [X(z)/z] has a multiple pole, say at pj, with a multiplicity of k, in addition to (Nk) simple poles at pi, then the partial fraction expansion given in Eq. (8.89) has to be modified as follows. XNk ci X ðzÞ cj1 cj2 cjk ¼ þ 2 þ    þ  k þ i¼1 z  p z z  pj i z  pj z  pj

ð8:91Þ

where ci is still given by (8.90) and cjk by cjk ¼ Hence,

  k X ðzÞ  1 dðkjÞ   z  p j ðk  jÞ! dzkj z z¼pj

ð8:92Þ

8.6 Methods for Computation of the Inverse z-Transform

X ðzÞ ¼

XNk ci z cj1 z cj2 z cjk z þ 2 þ    þ  k þ i¼1 z  p z  pj i z  pj z  pj

377

ð8:93Þ

Then inverse z-transform is obtained for each of the terms on the right-hand side of (8.91) by the use of Tables 8.1 and 8.2. We will now illustrate the method by a few examples. Example 8.16 Assuming the sequence x(n) to be right-sided, find the inverse z-transform of the following: X ðzÞ ¼

z ð z  aÞ ð z  bÞ

Solution The given function has poles at z ¼ a and z ¼ b. Since X(z) is a right-sided sequence, the ROC of X(z) is the exterior of a circle around the origin that includes both the poles. Now X(z)/z can be expressed in partial fraction expansion as X ðzÞ a 1 b 1 ¼  z a  bz  a a  bz  b Hence, X ðzÞ ¼

a 1 b 1  a  b 1  az1 a  b 1  bz1

We can now find the inverse transform of each term using Table 8.2 as x ð nÞ ¼

a b an uð nÞ  bn uðnÞ ab ab

Example 8.17 Assuming the sequence x(n) to be causal, find the inverse z-transform of the following: X ðzÞ ¼

10z2  3z 10z2  9z þ 2

Solution Dividing the numerator and denominator by z2, we can rewrite X(z) as 10  3z1 10  9z1 þ 2z2 4 5 ¼  1 2z 5  2z1 2 1 ¼  1  0:5z1 1  0:4z1 ¼

378

8 The z-Transform and Analysis of Discrete Time LTI Systems

Each term in the above expansion is a first-order z-transform and can be recognized easily to evaluate the inverse transform as Z 1 fX ðzÞg ¼ xðnÞ ¼ 2ð0:5Þn uðnÞ  ð0:4Þn uðnÞ: Example 8.18 Assuming the sequence x(n) to be causal, determine the inverse z-transform of the following: X ðzÞ ¼

z ð z þ 1Þ ð z  1Þ 3

Solution Since X(z)/z can be written in partial fraction expansion as X ðzÞ zþ1 A B C ¼ þ ¼ þ z ðz  1Þ3 z  1 ðz  1Þ2 ðz  1Þ3 Solving for A, B, and C, we get A ¼ 0, B ¼ 1, C ¼ 2. Hence, X(z) can be expanded as X ðzÞ ¼

z ð z  1Þ

2

þ

2z ðz  1Þ3

Making use of Table 8.2, the inverse z-transform of X(z) can be written as Z 1 fX ðzÞg ¼ xðnÞ ¼ nuðnÞ þ nðn  1ÞuðnÞ ¼ n2 uðnÞ Example 8.19 If x(n) is a right-handed sequence, determine the inverse z-transform for the function: X ðzÞ ¼ Solution X ðzÞ ¼

1 þ 2z1 þ z3 ð1  z1 Þð1  0:5z1 Þ

1 þ 2z1 þ z3 z3 þ 2z2 þ 1 ¼ ð1  z1 Þð1  0:5z1 Þ zðz  1Þðz  0:5Þ

Now, X(z)/z can be written in partial fraction expansion form as X ðzÞ z3 þ 2z2 þ 1 A B C D ¼ 2 þ ¼ þ 2þ z z ðz  1Þðz  0:5Þ z z ðz  1Þ ðz  0:5Þ Solving for A, B, C, and D, we get A ¼ 6, B ¼ 2, C ¼ 8, D ¼ 13. Hence, X ðzÞ ¼

z3 þ 2z2 þ 1 2 8z 13z ¼6þ þ  zðz  1Þðz  0:5Þ z ðz  1Þ ðz  0:5Þ

8.6 Methods for Computation of the Inverse z-Transform

379

Since the sequence is right-handed and the poles of X(z) are located z ¼ 0, 0.5, and 1, the ROC of X(z) is |z| > 1. Thus, from Table 8.2, we have Z 1 fX ðzÞg ¼ xðnÞ ¼ 6δðnÞ þ 2δðn  1Þ þ 8uðnÞ  13ð0:5Þn uðnÞ Example 8.20 Assuming h(n) to be causal, find the inverse z-transform of H ðzÞ ¼

ð z  1Þ 2 ðz2  0:1z  0:56Þ

Solution Expanding H(z)/z as H ðzÞ ð z  1Þ 2 A B C ¼ þ ¼ þ z zðz  0:8Þðz þ 0:7Þ z ðz  0:8Þ ðz þ 0:7Þ Solving for A, B, and C, we get A ¼ 1.78, B ¼ 0.033, and C ¼ 2.75 Therefore, H(z) can be expanded as H ðzÞ ¼ 1:7857 þ

0:0333z 2:7524z þ ðz  0:8Þ ðz þ 0:7Þ

Hence, Z 1 fH ðzÞg ¼ hðnÞ ¼ 1:7857δðnÞ þ 0:0333ð0:8Þn uðnÞ þ 2:7524ð0:7Þn uðnÞ

8.6.3

Inverse z-Transform by Partial Fraction Expansion Using MATLAB

The M-file residue z can be used to find the inverse z-transform using the power Series expansion. The coefficients of the numerator and denominator polynomial written in descending powers of z for Example 8.20 can be num= [1 -2 1]; den= [1 -0.1 -0.56];

The following MATLAB statement determines the residue (r), poles (p), and direct terms (k) of the partial fraction expansion of H(z). [r,p,k]= residuez(num,den);

380

8 The z-Transform and Analysis of Discrete Time LTI Systems

After execution of the above statements, the residues, poles, and constants obtained are Residues: 0.0333 2.7524 Poles: 0.8000 –0.7000 Constants: 1.7857 The desired expansion is H ðzÞ ¼ 1:7857 þ

8.6.4

0:0333z 2:7524z þ ðz  0:8Þ ðz þ 0:7Þ

ð8:94Þ

Computation of the Inverse z-Transform Using the Power Series Expansion

The z-transform of an arbitrary sequence defined by Eq. (8.1) implies that X(z) can be expressed as power series in z1 or z. In this expansion, the coefficient of the term indicates zn the value of the sequence x(n). Long division is one way to express X(z) in power series. Example 8.21 Assuming h(n) to be causal, find the inverse z-transform of the following: H ðzÞ ¼

z2 þ 2z þ 1 z2 þ 0:4z  0:12

Solution We obtain the inverse z-transform by long division of the numerator by the denominator as follows: 1 þ 1:6z1 þ 0:48z2 þ 0z3 þ 0:0576z4 þ    z þ 0:4z  0:12 2

j

z2 þ 2z þ 1 z2 þ 0:4z  0:12 1:6z þ 1:12 1:6z þ 0:64  0:192z1 0:48 þ 0:192z1 0:48 þ 0:19z1  0:0576z2 0:0576z2 0:0576z2 þ 0:02304z3  0:006912z4 ............

 0:02304z3 þ 0:006912z4

8.6 Methods for Computation of the Inverse z-Transform

381

Hence, H(z) can be written as H ðzÞ ¼ 1:0 þ 1:6z1 þ 0:48z2 þ 0z3 þ 0:0576z4 þ    implying that fh½ng ¼ f1:0;

1:6;

0:48;

0:0576; . . .g

0

for n  0

Example 8.22 Find the inverse z-transform of the following:   X ðzÞ ¼ log 1 þ bz1 ,

jbj < jzj

Solution We know that power series expansion for log(1 þ u) is u2 u3 u4 u5 logð1 þ uÞ ¼ u  þ  þ     2 3 4 5 1 X ð1Þnþ1 un , j uj < 1 ¼ n n¼1 Letting u ¼ bz1, X(z) can be written as 1   X ð1Þnþ1 bn zn X ðzÞ ¼ log 1 þ bz1 ¼ , n n¼1

jbj < jzj

From the definition of z-transform of x(n), we have X ðzÞ ¼

1 X

xðnÞzn

n¼1

Comparing the above two expressions, we get x(n), i.e., the inverse z-transform of X(z) ¼ log (1 þ bz1) to be ( x ð nÞ ¼

bn

ð1Þnþ1 n 0

n>0 n0

ð8:95Þ

Example 8.23 Find the inverse z-transform of X ðzÞ ¼

z , zb

for jzj > jbj

Solution The sequence is a right-sided causal sequence as the region of convergence is |z| > |b|. We can use the long division as we did in Example 8.21 to express z/ (zb) as a series in powers of z1. Instead, we will use binomial expansion.

382

8 The z-Transform and Analysis of Discrete Time LTI Systems

X ðzÞ ¼

z 1 ¼ z  b 1  bz1

  ¼ 1 þ bz1 þ b2 z2 þ    for bz1  < 1 P n 1 ¼ 1 for jzj > jbj n¼0 b z Hence, Z 1 fX ðzÞg ¼ xðnÞ ¼ Z 1

z zb

 ¼ bn uðnÞ:

Example 8.24 Find the inverse z-transform of X ðzÞ ¼

z , for jzj < jbj zb

Solution Since the region of convergence is |z| < |b|, the sequence is a left-sided sequence. We can use the long division to obtain z/(zb) as a power series in z. However, we will use the binomial expansion. z z 1 ¼ zb b1  ðz=bÞ  z z  z 2 ¼ 1þ þ þ    for b b b P n n ¼ 1 for jzj < jbj n¼1 b z

X ðzÞ ¼

z    <1 b

Hence, Z 1 fX ðzÞg ¼ xðnÞ ¼ Z 1

z zb

 ¼ bn uðn  1Þ

Example 8.25 Using the z-transform, find the convolution of the sequences: x1 ðnÞ ¼ f1; 3; 2g and x2 ðnÞ ¼ f1; 2; 1g Solution Step 1: Determine z-transform of individual signal sequences X 1 ð z Þ ¼ Z ½ x 1 ð nÞ  ¼

P2

¼ 1  3z1 þ 2z

n¼0 x1 ðnÞz 2

1

¼ x1 ð0Þ þ x1 ð1Þz1 þ x1 ð2Þz2

and X 2 ð z Þ ¼ Z ½ x 2 ð nÞ  ¼

P2

¼ 1 þ 2z1 þ z

n¼ 0 x2 ðnÞz 2

1

¼ x2 ð0Þ þ x2 ð1Þz1 þ x2 ð2Þz2

8.6 Methods for Computation of the Inverse z-Transform

383

Step 2: Obtain X(z) ¼ X1(z)X2(z) X ðzÞ ¼ ð1  3z1 þ 2z2 Þð1 þ 2z1 þ z2 Þ ¼ 1  z1  3z2 þ z3 þ 2z4 Step 3: Obtain the inverse z-transform of X(z)

xðnÞ ¼ Z 1 1  z1  3z2 þ z3 þ 2z4 ¼ f1; 1; 3; 1; 2g

8.6.5

Inverse z-Transform via Power Series Expansion Using MATLAB

The M-file impz can be used to find the inverse z-transform using the power series expansion. The coefficients of the numerator and denominator polynomial for Example 8.21 can be written as num = [1 2 1]; den = [1 0.4 -0.12];

The following statement can be run to obtain the coefficients of the inverse ztransform: h = impz(num,den);

where h is the vector containing the coefficients of the inverse z-transform. The first 11 coefficients of the inverse z-transform of Example 8.21 obtained after execution of the above MATLAB statements are Columns 1 through 9 1.0000 1.6000 0.4800 0 0.0576 -0.0230 0.0161 -0.0092 0.0056 Columns 10 through 11 -0.0034 0.0020

8.6.6

Solution of Difference Equations Using the z-Transform

Example 8.26 Determine the impulse response of the system described by the difference equation: yðnÞ  3yðn  1Þ  4yðn  2Þ ¼ xðnÞ þ 2xðn  1Þ: Assume that the system is relaxed initially.

384

8 The z-Transform and Analysis of Discrete Time LTI Systems

Solution Let X(z) ¼ Z[x(n)] and Y(z) ¼ Z[y(n)]. Taking z-transform on both sides and using the time shifting property, we get 

   1  3z1  4z2 Y ðzÞ ¼ 1 þ 2z1 X ðzÞ

Since X(z) ¼ 1, we have 1 þ 2z1 1  3z1 þ 4z2 YðzÞ zþ2 ð6=5Þ ð1=5Þ ¼ ¼  z ðz  4Þðz þ 1Þ z  4 z þ 1 ð6=5Þ ð1=5Þ YðzÞ ¼  1  4z1 1 þ z1 YðzÞ ¼

We now take inverse transform of the above and use Table 8.2 to obtain y(n), which is the impulse response of the system as hðnÞ ¼ yðnÞ ¼ ð6=5Þ4n uðnÞ  ð1=5Þð1Þn uðnÞ Example 8.27 Determine the response y(n), n  0 of the system described by the second-order difference equation yðnÞ  3yðn  1Þ  4yðn  2Þ ¼ xðnÞ þ 2xðn  1Þ for the input x(n) ¼ 4n u(n) Solution Applying z-transform to both sides of the equation, we have



Y ðzÞ 1  3z1  4z2 ¼ X ðzÞ 1 þ 2z1 Given that x(n) ¼ 4n u(n), we have X ðzÞ ¼

1 1  4z1

Substituting for X(z) in the expression for Y(z) and simplifying, we get Y ðzÞ ð z 2 þ 2Þ ¼ z ðz  4Þ2 ðz þ 1Þ or Y ðzÞ 1 26 24 ¼ þ þ z 25ðz þ 1Þ 25ðz  4Þ 5ðz  4Þ2

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

385

Hence, Y ðzÞ ¼

z 26z 24z þ þ 25ðz þ 1Þ 25ðz  4Þ 5ðz  4Þ2

By applying inverse z-transforms, we get y ð nÞ ¼

1 6 26 ð1Þn uðnÞ þ nð4Þn uðnÞ þ ð4Þn uðnÞ 25 5 25

Example 8.28 Find the impulse response of the system yðnÞ ¼ 3yðn  1Þ þ 2yðn  2Þ þ xðnÞ Solution Taking z-transforms on both sides of the above equation, and using the factZ[δ(n)] ¼ 1, we get z2 1   3z  2 0:86 0:135 þ Y ðzÞ ¼ 1  3:56z1 1 þ 0:56z1 Y ðzÞ ¼

1

3z1

2z2

¼

z2

Hence, the impulse response is given by hðnÞ ¼ yðnÞ ¼ 0:86ð3:56Þn uðnÞ þ 0:135ð0:561Þn uðnÞ

8.7 8.7.1

Analysis of Discrete-Time LTI Systems in the z-Transform Domain Transfer Function

It was stated in Chapter 6 that an LTI system can be completely characterized by its impulse response h(n). The output signal y(n) of a LTI system and the input signal x (n) are related by convolution as yðnÞ ¼ hðnÞ∗xðnÞ

ð8:96Þ

Taking z-transform on both sides of the above equation and using the convolution property, we get Y ðzÞ ¼ H ðzÞX ðzÞ

ð8:97Þ

indicating the z-transform of the output sequence y(n) is the product of the z-transforms of the impulse response h(n) and the input sequence x(n). The quantities h(n) and H(z) are two equivalent descriptions of a system in the time domain and

386

8 The z-Transform and Analysis of Discrete Time LTI Systems

z-domain, respectively. The transform H(z) is called the transfer function or the system function and expressed as H ðzÞ ¼

Y ðzÞ X ðzÞ

ð8:98aÞ

Or equivalently, PM H ðzÞ ¼

1þ

k¼0 P N

bk zk

k¼1

ak zk



ð8:98bÞ

where the constants ak and bk are real. The above transfer function is a ratio of polynomials in z1 and, hence, is a rational transfer function or system function. Example 8.29 The following are known about a LTI discrete-time system: (i) y(n) ¼ δ(n) þ a(0.25)nu(n) for x(n) ¼ (0.5)nu(n) (ii) y(n) ¼ 0 for all n if x(n) ¼ (2)n for all n Find the value of the constant a. Solution It is given that for the input x(n) ¼ (0.5)nu(n), the output of the LTI system is y(n) ¼ δ(n) þ a(0.25)nu(n). From this fact, the transfer function H(z) is given by H ðzÞ ¼ ¼

Y ðzÞ X ðzÞ

1 þ a  0:25z1 ð1  0:25z1 Þð1  0:5z1 Þ

It is also given that the output y(n) ¼ 0 for the input x(n) ¼ (2)n for all n. Since the function z0n is an eigenfunction for a discrete-time LTI system, the output to this input is H ðz0 Þz0n . From this, it can be inferred that H(2) ¼ 0. Using this in the above transfer function, the value of a is calculated to be 1.125

8.7.2

Poles and Zeros of a Transfer Function

As mentioned earlier, the zeros of a system function H(z) are the values of z for which H(z) ¼ 0, while the poles are the values of z for which H(z) ¼ 1. Since H(z) is a rational transfer function, the number of finite zeros and the number of finite poles are equal to the degrees of the numerator and denominator polynomials, respectively. In MATLAB, tf2zp command can be used to find the zeros, poles, and gains of a rational transfer function. z plane command can be used for plotting pole-zero plot of a rational transfer function.

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

387

Example 8.30 Determine the pole-zero plot using MATLAB for the system described by the system function H ðzÞ ¼

Y ðzÞ z1 ¼ 2 X ðzÞ 8z  6z þ 1

Solution The coefficients of the numerator and denominator polynomial can be written as numerator = [0 1 -1]; denominator = [8 -6 1];

The following MATLAB statement yields the poles and zeros and gain of the system: [z,p,gain] = tf2zp (numerator, denominator) zeros, z = 1 poles, p = [0.500 0.250] and gain = 0.1250

The MATLAB command z-plane (z, p) plots the poles and zeros as shown in Figure 8.6.

Figure 8.6 Pole-zero plot of Example 8.30

388

8.7.3

8 The z-Transform and Analysis of Discrete Time LTI Systems

Frequency Response from Poles and Zeros

By factorizing the numerator and denominator polynomials of Eq. (8.98b), the transfer function can be written in pole-zero form as M Q

HðzÞ ¼

i¼1 b0 zðNMÞ N Q

ðz  zi Þ ð8:99Þ ðz  pi Þ

i¼1

where zi and pi are the zeros and poles of H(z). It should be noted that the zeros are either real or occur in conjugate pairs. The frequency response of the system can be obtained by letting z ¼ e jω in the transfer function H(z), that is,    H ejω ¼ H ðzÞz¼ejω Hence, M Q

Hðejω Þ ¼

b0 ejωðNMÞ i¼1 N Q

ðejω  zi Þ ð8:100Þ ðejω  pi Þ

i¼1

The contribution of the zeros and poles to the system frequency response can be visualized from the above expression. The magnitude of the frequency response can be expressed by M Q

jHðe Þj ¼ jω

jb0 jjejω jðNMÞ i¼1 N Q

jðejω  zi Þj ð8:101Þ jðejω

 pi Þj

i¼1

The zeros contribute to pulling down the magnitude of the frequency response, whereas the poles contribute to pushing up the magnitude of the frequency response. The size of decrease or increase in the magnitude response depends on how far the zero or the pole is from the unit circle. A peak in |H(e jω)| appears at the frequency of a pole very close to the unit circle. To illustrate this, consider the following example. Example 8.31 Consider a system with the transfer function H ðzÞ ¼

0:1ðz2 þ 2z þ 1Þ 1:2z2 þ 1

ð8:102Þ

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

389

The numerator and denominator polynomials coefficients in descending powers of z can be written as num=[1 2 1]; den=[1.2 0 1];

Then, as used in Example 8.30, using the MATLAB commands tf2zp and z-plane, the pole zero plot can be obtained as shown in Figure 8.7(a). The magnitude and phase responses of the above system transfer function are obtained using the above num and den vectors using the MATLAB command freqz. The magnitude and phase responses are shown in Figure 8.7(b) and (c), respectively. Figure 8.7(a) indicates that the system has zeros of order 2 at z ¼ 1 and two poles on the imaginary axis close to the unit circle. In the magnitude response of Figure 8.7(b), a peak occurs at ω ¼ π/2. This can be attributed to the fact that the frequency of the poles is π/2. The magnitude response is small at high frequencies due to the zeros.

8.7.4

Stability and Causality

The stability of a LTI system can be expressed in terms of the transfer function or the impulse response of the system. It is known from Section 6.4.5 that a necessary and sufficient condition for a LTI system to be BIBO (bounded-input bounded-output) stable is that its impulse response be absolutely summable, i.e., 1 X

jhðnÞj < 1

ð8:103Þ

n¼1

H ðzÞ ¼ j H ðzÞj 

1 X

X1 n¼1

hðnÞ zn

jhðnÞzn j ¼

n¼1

1 X

jhðnÞjjzn j

ð8:104Þ ð8:105Þ

n¼1

On the unit circle (i.e., |z| ¼ 1), the above expression becomes jH ðzÞj 

1 X

j hð nÞ j

ð8:106Þ

n¼1

Therefore, for a stable system, the ROC of its transfer function H(z) must include the unit circle. Thus we have the following theorem. BIBO Stability Theorem A discrete LTI system is BIBO stable if and only if the ROC of its system function includes the unit circle, |z| ¼ 1.

390 Figure 8.7 (a) Pole-zero plot, (b) magnitude response, (c) phase response

8 The z-Transform and Analysis of Discrete Time LTI Systems

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

391

We know from Section 6.4.5 that for a discrete LTI system to be causal h(n) ¼ 0 for n < 0. Thus, the sequence should be right-sided. We also know from Section 8.2 that the ROC of a right-sided sequence is the exterior of a circle whose radius is equal to the magnitude of the pole that is farthest from the origin. At the same time, we also know that for a right-sided sequence, the ROC may or may not include the point z ¼ 1. But we know from Section 8.2 that a causal system cannot have a pole at infinity. Thus, in a causal system, the ROC should include the point z ¼ 1. Thus, we may summarize the result for causality by the following theorem: Causality Theorem A discrete LTI system is causal if and only if the ROC of its system function is the exterior of a circle including z ¼ 1. An alternate way of stating this result is that a system is causal if and only if its ROC contains no poles, finite or infinite. Thus the conditions for stability and causality are quite different. A causal system could be stable or unstable, just as a noncausal system could be stable or unstable. Also, a stable system could be causal or noncausal just as an unstable system could be causal or noncausal. However, we can conclude from the above two theorems that a causal stable system must have a system function whose ROC is |z| ¼ r, where r < 1. Hence, we can summarize this result as follows. Condition for a System to Be Both Causal and Stable A causal LTI system is BIBO stable if and only if all its poles are within the unit circle. As a consequence, for a LTI system with a system function H(z) to be stable and causal, it is necessary that the degree of the numerator polynomial in z not exceed that of the denominator polynomial. As such, an FIR system is always stable, whereas if an IIR system is not designed properly, it may be unstable. Example 8.32 Given the system function H ðzÞ ¼ 

zð4z  3Þ  z  13 ðz  4Þ

Find the various regions of convergence for H(z), and state whether the system is stable and/or causal in each of these regions. Also, find the impulse response h(n) in each case. Solution The system function can be expressed in partial fraction in the form z 3z 1 1 þ þ3 ¼ H ðzÞ ¼  1 1 1 ð z  4Þ 1  4z1 z3 1  3z The system function has two zeros, viz., z ¼ 0, 34, and two poles at z ¼ 13 , 4: Hence, there are three regions of convergence: (i) jzj < 13, (ii) 13 < jzj < 4, and (iii) |z| > 4. Let us consider each of these regions separately.

392

8 The z-Transform and Analysis of Discrete Time LTI Systems

(i) jzj < 13 In this region, there are no poles including the origin, but has poles exterior to it. Hence, the system is noncausal. Also, it is an unstable system, since the ROC does not include the unit circle. By using Table 8.2, we get  n 1 n hðnÞ ¼  þ 3ð4Þ uðn  1Þ 3 (ii)

1 3

< j zj < 4

This region includes the unit circle and hence the system is stable. However, since the pole |z| ¼ 4 is exterior to this region, it is noncausal, and the corresponding sequence is two-sided. Again by using Table 8.2, we have hð nÞ ¼

 n 1 uðnÞ  3ð4Þn uðn  1Þ 3

(iii) |z| > 4 This region does not include the unit circle, and hence the system is unstable. However, in this region, there are no poles, finite or infinite, and hence, the system is causal. The impulse response of the system is obtained from H(z) using Table 8.2 as  n 1 hð nÞ ¼ uð nÞ þ 3ð 4Þ n uð nÞ 3 Example 8.33 The rotational motion of a satellite was described by the difference equation yðnÞ ¼ yðn  1Þ  0:5 yðn  2Þ þ 0:5 xðnÞ þ 0:5 xðn  1Þ Is the system stable? Is the system causal? Justify your answer. Solution Taking the z-transform on both sides of the given difference equation, we get Y ðzÞ ¼ z1 Y ðzÞ  0:5z2 Y ðzÞ þ 0:5X ðzÞ þ 0:5z1 X ðzÞ H ðzÞ ¼

Y ðzÞ 0:5ð1 þ z1 Þ 0:5ðz þ 1Þz ¼ ¼ X ðzÞ 1  z1 þ 0:5z2 ðz2  z þ 0:5Þ

The poles of the system are at z ¼ 0.5  0.5j as shown in Figure 8.8 All poles of the system are inside the unit circle. Hence, the system is stable. It is causal since the output only depends on the present and past inputs.

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

393

Figure 8.8 Poles of Example 8.33

0.5 -0.5

´

0.5 ´

Example 8.34 Consider the difference equation 7 2 yðnÞ  yðn  1Þ þ yðn  2Þ ¼ xðnÞ 3 3 (a) Determine the possible choices for the impulse response of the system. Each choice should satisfy the difference equation. Specifically indicate which choice corresponds to a stable system and which choice corresponds to a causal system. (b) Can you find a choice which implies that the system is both stable and causal? If not, justify your answer. Solution (a) Taking the z-transform on both sides and using the shifting theorem, we get   7 2 1  z1 þ z2 Y ðzÞ ¼ X ðzÞ 3 3 Y ðzÞ ¼ X ðzÞ

1 7 1 2 2 1 z þ z 3 3 2 z   H ðzÞ ¼ 1 ð z  2Þ z  3

The system function H(z) has a zero of order 2 at z ¼ 0 and two poles at z ¼ 1/3, 2. Hence, there are three regions of convergence, and thus, there are three possible choices for the impulse response of the system. The regions are (i) R1: jzj < 13,

(ii) R2: 13 < jzj < 2, and

(iii) R3: |z| > 2.

The region R1 is devoid of any poles including the origin and hence corresponds to an anti-causal system, which is not stable since it does not include the unit circle. Region R2 does include the unit circle and hence corresponds to a stable system; however, it is not causal in view of the presence of the pole z ¼ 2. Finally, the region R3 does not have any poles including at infinity and hence corresponds to a causal system; however, since R3 does not include the unit circle, the system is not stable.

394

8 The z-Transform and Analysis of Discrete Time LTI Systems

(b) There is no ROC that would imply that the system is both stable and causal. Therefore, there is no choice for h(n) which make the system both stable and causal. Example 8.35 A system is described by the difference equation yðnÞ þ yðn  1Þ ¼ xðnÞ, yðnÞ ¼ 0,

for n < 0:

(i) Determine the transfer function and discuss the stability of the system. (ii) Determine the impulse response h(n) and show that it behaves according to the conclusion drawn from (i). (iii) Determine the response when x(n) ¼ 10 for n  0. Assume that the system is initially relaxed. Solution (i) Taking the z-transforms on both sides of the given equation, we get Y ðzÞ þ Y ðzÞz1 ¼ X ðzÞ Hence, H ðzÞ ¼

Y ðzÞ z ¼ X ðzÞ z þ 1

The pole is at z ¼ 1, that is, on the unit circle. So the system is marginally stable or oscillatory. (ii) Since h(n) ¼ 0 for n < 0, hðnÞ ¼ Z 1



z ¼ ð1Þn uðnÞ zþ1

This impulse response confirms that the impulse response is oscillatory. (iii) Since xðnÞ ¼ 10 X ðzÞ ¼

for n  0,

10z z1

Thus, Y ðzÞ ¼ H ðzÞX ðzÞ ¼ or

z 10 zþ 1z 1

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

395

Y ðzÞ 5 5 ¼ þ z zþ1 z1 Therefore, yðnÞ ¼ Z 1 ½Y ðzÞ ¼ ½5ð1Þn þ 5uðnÞ

8.7.5

Minimum-Phase, Maximum-Phase, and Mixed-Phase Systems

A causal stable transfer function with all its poles and zeros inside the unit circle is called a minimum-phase transfer function. A causal stable transfer function with all its poles inside the unit circle and all the zeros outside the unit circle is called a maximum-phase transfer function. A causal stable transfer function with all its poles inside the unit circle and with zeros inside and outside the unit circle is called a mixed-phase transfer function. For example, consider the systems with the following transfer functions: H 1 ðzÞ ¼

Y ðzÞ z þ 0:4 ¼ X ðzÞ z þ 0:3

ð8:107Þ

H 2 ðzÞ ¼

Y ðzÞ 0:4z þ 1 ¼ X ðzÞ z þ 0:5

ð8:108Þ

Y ðzÞ ð0:4z þ 1Þ ðz þ 0:4Þ ¼ X ðzÞ ðz þ 0:5Þ ðz þ 0:3Þ

ð8:109Þ

H 3 ðzÞ ¼

The pole-zero plot of the above transfer functions are shown in Figure 8.9 (a), (b), and (c), respectively. The transfer function H1(z) has a zero at z ¼ 0.4 and a pole at z ¼ 0.3, and they are both inside the unit circle. Hence, H1(z) is a minimum-phase function. The transfer function H2(z) has a pole inside the unit circle, at z ¼ 0.5, and a zero at z ¼ 2.5, outside the unit circle. Thus, H2(z) is a maximum-phase function. The transfer function H3(z) has two poles, one at z ¼ 0.3 and the other at z ¼ 0.5, and two zeros one at z ¼ 0.4, inside the unit circle, and the other at z ¼ 2.5, outside the unit circle. Hence, H3(z) is a mixed-phase function.

8.7.6

Inverse System

Let H(z) be the system function of a linear time-invariant system. Then its inverse system function HI(z) is defined, if and only if the overall system function is unity when H(z) and HI(z) are connected in cascade, that is, H(z) HI(z) ¼ 1, implying

396

8 The z-Transform and Analysis of Discrete Time LTI Systems

Figure 8.9 Pole-zero plot of (a) a minimum-phase function, (b) a maximum-phase function, and (c) a mixed-phase function

H I ðzÞ ¼

1 H ðzÞ

ð8:110Þ

In the time domain, this is equivalently expressed as hI ðnÞ∗hðnÞ ¼ δðnÞ

ð8:111Þ

Example 8.36 A system is described by the following difference equation: yðnÞ ¼ xðnÞ  e8α xðn  8Þ where the constant α > 0. Find the corresponding inverse system function to recover x(n) from y(n). Check for the stability and causality of the resulting recovery system, justifying your answer.

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

Solution Y ðzÞ ¼ X ðzÞ  e8α z8 X ðzÞ;

397

 Y ðzÞ  ¼ 1  e8α z8 X ðzÞ

The corresponding inverse system H 1 ðzÞ ¼

1 X ðzÞ ¼ ð1  e8α z8 Þ Y ðzÞ

The recovery system is both stable and causal, since all the poles of the system HI(z) are inside the unit circle.

8.7.7

All-Pass System

Consider a causal stable Nth-order transfer function of the form aN þ aN1 z1 þ    þ zN M ðzÞ ¼ D ðzÞ 1 þ a1 z1 þ    þ aN zN

ð8:112Þ

Dðz1 Þ ¼ 1 þ a1 z þ a2 z2 þ    þ aN zN ¼ zN ½aN þ aN1 z1 þ    þ zN  ¼ zN MðzÞ

ð8:113Þ

H ðzÞ ¼  Now

or   M ðzÞ ¼ zN D z1

ð8:114Þ

Hence, Dðz1 Þ D ðzÞ

ð8:115Þ

  D ðzÞ H z1 ¼ zN Dðz1 Þ

ð8:116Þ

  H ðzÞ H z1 ¼ 1

ð8:117Þ

    jH ðωÞj2 ¼ H ejω H ejω ¼ 1

ð8:118Þ

H ðzÞ ¼ zN and

Therefore,

Thus,

for all values of ω.

398

8 The z-Transform and Analysis of Discrete Time LTI Systems

In other words, H(z) given by (8.112) passes all the frequencies contained in the input signal to the system, and hence such a transfer function is an all-pass transfer function, and the corresponding system is an all-pass system. It is also seen from (8.112) that if z ¼ pi is a zero of D(z), then z ¼ (1/pi) is a zero of M(z). That is, the poles and zeros of an all-pass function are reciprocal of one another. Since all the poles of H(z) are located within the unit circle, all the zeros are located outside the unit circle. If x(n) is the input sequence and y(n) the output sequence for an all-pass system, then Y ðzÞ ¼ H ðzÞX ðzÞ:

ð8:119Þ

      Y ejω ¼ H ejω X ejω :

ð8:120Þ

  jω    jω  Y e  ¼ X e 

ð8:121Þ

Thus,

Since |H(e jω)| ¼ 1, we get

We know from Parseval’s relation that the output energy of a LTI system is given by ð 1 π   jω 2 Y e dω jyðnÞj ¼ n¼1 2π π ð 1 π   jω 2 ¼ X e dω 2π π

X1

2

ð8:122Þ ð8:123Þ

Hence, X1

jyðnÞj2 ¼ n¼1

X1 n¼1

jxðnÞj2

ð8:124Þ

Thus, the output energy is equal to the input energy for an all-pass system. Hence, an all-pass system is that it is a lossless system. Example 8.37 A discrete-time system with poles at z ¼ 0.6 and z ¼ 0.7 and zeros at z ¼ 1/0.6 and z ¼ 1/0.7 is shown in Figure 8.10. Demonstrate algebraically that magnitude response is constant. Solution For given pole-zero pattern, the system function is given by H ap ðzÞ ¼

0:42 þ 1:3z1 þ z2 1 þ 1:3z1 þ 0:42z2

Substituting z ¼ e jω in the above transfer function, we get

8.7 Analysis of Discrete-Time LTI Systems in the z-Transform Domain

399

Figure 8.10 Pole-zero plot of a second-order all-pass system

  0:42 þ 1:3ejω þ e2jω H ap ejω ¼ 1 þ 1:3ejω þ 0:42e2jω 0:42 þ 1:3ejω þ e2jω H ap ðejω Þ ¼ 1 þ 1:3ejω þ 0:42e2jω  2 H ap ðωÞ ¼ H ðejω ÞH ðejω Þ ¼ 1

8.7.8

All-Pass and Minimum-Phase Decomposition

Consider an Nth-order mixed-phase system function H(z) with m zeros outside the unit circle and (nm) zeros inside the unit circle. Then H(z) can be expressed as   1   1  H ðzÞ ¼ H 1 ðzÞ z1  a∗  a∗ z  a∗ 1 2  z m

ð8:125Þ

where H1(z) is a minimum-phase function as its N poles and (nm) zeros are inside the unit circle. Eq. (8.125) can be equivalently expressed as H ðzÞ ¼ H 1 ðzÞð1  z1 a1 Þð1  z1 a2 Þ  ð1  z1 am Þ  1  1   1  z  a∗ z  a∗  a∗ 1 1  z m  ð1  z1 a1 Þð1  z1 a2 Þ  ð1  z1 am Þ

ð8:126Þ

In the above equation, the factor H1(z)(1z1a1)(1z1a2)  (1z1am) is also a minimum-phase function, since |a1|, |a2|, . . ., |am| are less than 1, the zeros are inside

400

8 The z-Transform and Analysis of Discrete Time LTI Systems



 1   1  z1  a∗  a∗ z  a∗ 1 2  z m is all-pass. ð1  z1 a1 Þð1  z1 a2 Þ  ð1  z1 am Þ Thus, any transfer function H(z) can be written as

the unit circle, and the factor

H ðzÞ ¼ H min ð zÞH ap ðzÞ

ð8:127Þ

Hmin(z) has all the poles and zeros of H(z) that are inside the unit circle in addition to the zeros that are conjugate reciprocals of the zeros of H(z) that are outside the unit circle, while Hap(z) is an all-pass function that has all the zeros of H(z) that lie outside the unit circle along with poles to cancel the conjugate reciprocals of the zeros of H(z) that lie outside the unit circle, which are now contained as zeros in Hmin(z). Example 8.38 A signal x(n) is transmitted across a distorting digital channel characterized by the following system function: H d ðzÞ ¼

ð1  0:5z1 Þð1  1:25ej0:8π z1 Þð1  1:25ej0:8π z1 Þ ð1  0:81z2 Þ

Consider the compensating system shown in Figure 8.11. Find H1C(z) such that the overall system function G1(z) is an all-pass system. Solution H d ðzÞ ¼ H dmin1 ðzÞH ap ðzÞ    ð1  0:5z1 Þ H dmin1 ðzÞ ¼ ð1:25Þ2 1  0:8ej0:8π z1 1  0:8ej0:8π z1 ð1  0:8z2 Þ ðz1  0:8ej0:8π Þðz1  0:8ej0:8π Þ H ap ðzÞ ¼   1  0:8ej 0:8π z1 ð1  0:8ej0:8π z1 Þ ð1  0:81z2 Þ 1 ¼ H 1C ðzÞ ¼ H dmin1 ðzÞ ð1:25Þ2 ð1  0:5z1 Þð1  0:8eJ0:8π z1 Þð1  0:8eJ0:8π z1 Þ Then, G1 ðzÞ ¼ H d ðzÞH 1C ðzÞ ¼ H ap ðzÞ is an all-pass system.

G1 ( z )

Figure 8.11 Compensating system x (n )

H d (z )

xd (n)

H1C (z)

xca (n)

8.8 One-Sided z-Transform

8.8

401

One-Sided z-Transform

The unilateral or one-sided z-transform, which is appropriate for problems involving causal signals and systems, is evaluated using the portion of a signal associated with nonnegative values of time index (n  0). It gives considerable meaning to assume causality in many applications of the z-transforms. Definition The one-sided z-transform of a signal x[n] is defined as Z þ ½xðnÞ ¼ X þ ðzÞ ¼

X1 n¼0

xðnÞzn

ð8:128Þ

which depends only on x(n) for (n  0). It should be mentioned that the two-sided ztransform is not useful in the evaluation of the output of a non-relaxed system. The one-sided transform can be used to solve for systems with nonzero initial conditions or for solving difference equations with nonzero initial conditions. The following special properties of X+ (z) should be noted. 1. The one-sided transform X+ (z) of x(n) is identical to the two-sided transform X(z) of the sequence x(n)u(n). Also, since x(n)u(n) is always causal, its ROC and hence that of X+ (z) are always the exterior of a circle. Hence, it is not necessary to indicate the ROC of a one-sided z-transform. 2. X+ (z) is unique for a causal signal, since such a signal is zero for n < 0. 3. Almost all the properties of the two-sided transform are applicable to the one-sided transform, one major exception being the shifting property. Shifting Theorem for X+ (z) When the Sequence is Delayed by k If Z þ ½xðnÞ ¼ X þ ðzÞ, then Xk Z þ ½xðn  kÞ ¼ zk X þ ðzÞ þ xðnÞzn , k > 0 n¼1

ð8:129Þ

However, if x(n) is a causal sequence, then the result is the same as in the case of the two-sided transform and Z þ ½xðn  kÞ ¼ zk X þ ðzÞ Proof By definition, Z þ ½xðn  kÞ ¼

X1 n¼0

xðn  kÞ zn

Letting (nk) ¼ m, the above equation may be written as

ð8:130Þ

402

8 The z-Transform and Analysis of Discrete Time LTI Systems

i X1 m m x ð m Þ z þ x ð m Þ z m¼0 m¼k h i Xk n ¼ zk X þ ðzÞ þ x ð n Þz n¼1

Z þ ½xðn  kÞ ¼ zk

hX1

which proves (8.129). If the sequence x(n) is causal, then the second term on the right side of the above equation is zero, and hence we get the result (8.130). Shifting Theorem for X+ (z) When the Sequence is Advanced by k If Z þ ½xðnÞ ¼ X þ ðzÞ, then h i Xk1 n Z þ ½xðn þ kÞ ¼ zk X þ ðzÞ  x ð n Þz , n¼0

k>0

ð8:131Þ

Proof By definition Z þ ½xðn þ kÞ ¼

X1 n¼0

xðn þ kÞzn

Letting (n + k) ¼ m, the above equation may be written as Z þ ½xðn þ kÞ ¼ zk ¼ zk

P1 m¼k

hP 1

xðmÞzm



m  m¼0 xðmÞz

Pk1

m m¼0 xðmÞz i

i

h P n ¼ zk X þ ðzÞ  k1 n¼0 xðnÞz thus establishing the result (8.131). Final Value Theorem If a sequence x(n) is causal, i.e., x(n) ¼ 0 for n < 0, then

limn!1 xðnÞ ¼ limz!1 ðz  1ÞX ðzÞ

ð8:132Þ

The above limit exists only if the ROC of (z1) X(z) exists. Proof Since the sequence x(n) is causal, we can write its z-transform as follow Z ½xðnÞ ¼

X1 n¼0

xðnÞ zn ¼ xð0Þ þ xð1Þz1 þ xð2Þz2 þ   :

ð8:133Þ

Also Z ½xðn þ 1Þ ¼

X1 n¼0

xðn þ 1Þ zn ¼ xð1Þ þ xð2Þz1 þ xð3Þz2 þ   : ð8:134Þ

8.8 One-Sided z-Transform

403

Hence, we see that Z ½xðn þ 1Þ ¼ z½Z ½xðnÞ  xð0Þ

ð8:135Þ

Thus, Z ½xðn þ 1Þ  Z ½xðnÞ ¼ ðz  1ÞZ ½xðnÞ  zxð0Þ Substituting (8.133) and (8.135) for the L.H.S., we have ½xð1Þ  xð0Þ þ ½xð2Þ  xð1Þz þ ½xð3Þ  xð2Þz2 þ    ¼ ðz  1ÞX ðzÞ  zxð0Þ Taking the limit as z ! 1, we get ½xð1Þ  xð0Þ þ ½xð2Þ  xð1Þ þ ½xð3Þ  xð2Þ þ    ¼ limz!1 ðz  1ÞX ðzÞ  xð0Þ Thus, xð0Þ þ xð1Þ ¼ limz!1 ðz  1ÞX ðzÞ  xð0Þ or xð1Þ ¼ limz!1 ðz  1ÞX ðzÞ Hence, limn!1 xðnÞ ¼ limz!1 ðz  1ÞX ðzÞ It should be noted that the limit exists only if the function (z1) X(z) has an ROC that includes the unit circle; otherwise, system would not be stable and the limn!1x (n) would not be finite. Example 8.39 Find the final value of x(n) if its z-transform X(z) is given by X ðzÞ ¼

ðz 

1Þ ð z 2

0:5z2  0:85z þ 0:35Þ

Solution The final value or steady value of x(n) is given by xðnÞ ¼ limz!1 ðz  1ÞX ðzÞ ¼

0:5 ¼1 ð1  0:85 þ 0:35Þ

The result can be directly verified by taking the inverse transform of the given X(z) Example 8.40 The following facts are given about a discrete-time signal x(n) with X (z) as its z-transform: (i) x(n) is real and right-sided. (ii) X(z) has exactly two poles.

404

8 The z-Transform and Analysis of Discrete Time LTI Systems

(iii) X(z) has two zeros at the origin. π (iv) X(z) has a pole at z ¼ 0:5ej3 . (v) X ð1Þ ¼ 83. Determine X(z) and specify its region of convergence. Solution It is given that x(n) is real and that X(z) has exactly two poles with one of jπ the pole at z ¼ 12 e 3 . Since, x [n] is real, the two poles must be complex conjugates of jπ

each other. Thus, the other pole of the system is at Z ¼ 12 e 3 . Also, it is given that X (z) has two zeros at the origin. Therefore, X(z) will have the following form: Kz2   X ðzÞ ¼  1 jπ 1 jπ 3 3 z e z e 2 2 ¼

Kz2 1 1 z2  z þ 2 4

for some constant K to be determined. Finally, it is given that Xð1Þ ¼ 83. Substituting this in the above equation, we get K ¼ 2. Thus, X ðzÞ ¼

2z2 z2  12 z þ 14

Since x[n] is right-sided, its ROC is jzj > 12 (note that both poles have magnitude 12).

8.8.1

Solution of Difference Equations with Initial Conditions

The one-sided z-transform is very useful in obtaining solutions for difference equations which have initial conditions. The procedure is illustrated with an example. Example 8.41 Find the step response of the system   1 y ð nÞ  y ð n  1Þ ¼ x ð nÞ 2 with the initial condition y(1) ¼ 1 Solution Taking one-sided z-transforms on both sides of the given equation and using (8.129), we have

8.9 Solution of State-Space Equations Using z-Transform

405

  1 1 þ Y ðzÞ  ½z Y ðzÞ þ yð1Þ ¼ X þ ðzÞ 2 þ

Substituting for X+ (z) and y(1), we have    1 1 þ 1 1 1 z Y ðzÞ ¼ þ 2 2 1  z1 Hence, 1 1 1 þ  Y þ ðzÞ ¼  1 1 1 1 2 1 z 1 z ð1  z1 Þ 2 2 2 1 1  ¼   1 1 1z 2 1  z1 2 Taking the inverse transform, we get " Z 1 þ

8.9

#

  1 YðzÞ ¼ yðnÞ ¼ 2  Þnþ1 uðnÞ 2

Solution of State-Space Equations Using z-Transform

For convenience, the state-space equations of a discrete-time LTI system from Chapter 6 are repeated here: X ðn þ 1Þ ¼ A X ðnÞ þ b℧ðnÞ

ð8:136aÞ

yðnÞ ¼ cX ðnÞ þ d℧ðnÞ

ð8:136bÞ

Taking unilateral z-transform Eqs. (8.136a) and (8.136b), we obtain zX ðzÞ  zX ð0Þ ¼ AX ðzÞ þ b℧ðzÞ

ð8:137aÞ

Y ðzÞ ¼ cX ðzÞ þ d℧ðzÞ

ð8:137bÞ

where 2

3 X 1 ðzÞ 6 X 2 ðzÞ 7 6 7 7 X ðzÞ ¼ 6 6 ⋮ 7 4 X N1 ðzÞ 5 X N ðzÞ Eq. (8.137a) can be rewritten as

406

8 The z-Transform and Analysis of Discrete Time LTI Systems

½zI  AX ðzÞ ¼ zX ð0Þ þ b℧ðzÞ

ð8:138Þ

where I is the identity matrix. From Eq. (8.138), we get X ðzÞ ¼ ½zI  A1 zX ð0Þ þ ½zI  A1 b℧ðzÞ

ð8:139Þ

The inverse one-sided z-transform of Eq. (8.139) yields h i h i z1 ½X ðzÞ ¼ z1 ½zI  A1 zX ð0Þ þ z1 ½zI  A1 b℧ðzÞ h i z1 ½zI  A1 zX ð0Þ ¼ An X ð0Þ

ð8:140Þ ð8:141Þ

By using convolution theorem, we obtain h i Xn1 z1 ½zI  A1 b℧ðzÞ ¼ An1k b℧ðk Þ k¼0

ð8:142Þ

Thus, z1 ½X ðzÞ ¼ X ðnÞ ¼ An X ð0Þ þ

Xn1 k¼0

An1k b℧ðkÞ

n > 0:

ð8:143Þ

Substituting Eq. (8.143) into Eq. (8.136b), we get yðnÞ ¼ cAn X ð0Þ þ

Xn1 k¼0

An1k b℧ðkÞ þ d℧ðnÞ

n > 0:

ð8:144Þ

Example 8.42 Consider an initially relaxed discrete time system with the following state-space representation. Find y(n). "

x 1 ð n þ 1Þ x 2 ð n þ 1Þ

#

2

0 ¼4 1  3

3" # " # 1 x 1 ð nÞ 0 5 þ ℧ðnÞ 4 x 2 ð nÞ 1 3

  1 4 x1 ðnÞ þ ℧ðnÞ yðnÞ ¼  3 3 x2 ðnÞ

8.9 Solution of State-Space Equations Using z-Transform

Solution

"

#   0 1 0 1 4 1 4 ; b¼ A¼ ; d ¼ 1: ; c¼   1 3 3 3 3 2 3 " " 0 1 ## z 1 z 0 1 4 ½zI  A ¼  ¼ 41 45  z 0 z 3 3 3 3 2 3 4 " #1 1 z  z 1 1 7 3  6 ½zI  A1 ¼ 1 ¼ 4 5 4 1 1 z  3 3  z ð z  1Þ z  3 3 3 2 4 z 7 6 1 6 3   7 7 6 1 1 7 6 ð z  1Þ z  ðz  1Þ z  6 3 3 7 ¼6 7 7 6 1=3 z 7 6    7 6 4 1 1 5 ð z  1Þ z  ðz  1Þ z  3 3 3 2 1=2 3=2 3=2 3=2 þ  6 z1 1 z1 17 6 z z 7 6 3 37 7 ¼6 6 1=2 1=2 3=2 1=2 7 7 6 þ  4 1 z1 15 z1 z z 3i 3 h 1 n 1 A ¼ z ½zI  A z 3 2 1 3 3 3 z z z z 6 2 þ 2 2  2 7 6 1 z1 17 7 6z  1 z z 7 6 7 6 3 3 ¼ z1 6 7 1 1 3 1 7 6 6 z z z z 7 6 2 þ 2 2  2 7 4 1 z1 15 z1 z z 3 3 2     3 1 3 1 n 3 3 1 n  6 2 þ 2 3 2 2 3 7 6 7 ¼6     7 4 1 1 1 n 3 1 1 n5   þ 2 2 3 2 2 3 2 1 33 2 3 3 3  6 7  n 6 7 ¼ 4 2 2 5 þ 13 4 2 2 5 1 3 1 1  2 2 2 2

407

408

8 The z-Transform and Analysis of Discrete Time LTI Systems

Since the system is initially relaxed, cAnX(0) ¼ 0 82 3 2 39 1 3 3 3 > > >  > <6  1n1k 6 2 2 7= 1 4 2 27 n1k 6 7 6 7 cA b¼  4 1 35 þ 3 41 1 5> 3 3 > > ; :   > 2 2 2 2 2 3 2  1 3 " #  3 7 0  n1k 1 4 6 1 4 6 6 2 27 62 ¼  þ 13  4 5 1 3 3 3 3 3 41 1  2 2 2  n1k 3 1 1 ¼  2 6 3

3 3 " #  7 0 27 15 1  2

Hence,   n1 X 3 1 1 n1k  ð1Þk þ 1 2 6 3 k¼0 "   # n 1 X 3 1 1 nk  ¼ ð1Þk þ 1 2 2 3 k¼0 n1 n1  nk X 3 1X 1  þ1 ¼ 2 2 3 k¼0 k¼0   P 3 1 1 n Xn1 k  ¼ n1 3 þ1 k¼0 k¼0 2 2 3     3 1 1 n 1  3n ¼ n þ 1 n > 0: 2 2 3 13  n 3 1 1 ð1  3n Þ þ 1 n > 0: ¼ nþ 2 4 3

y ð nÞ ¼

8.10

Transformations Between Continuous-Time Systems and Discrete-Time Systems

The transformation of continuous-time system to discrete-time system arises in various situations. In this section, two techniques, namely, (i) Impulse invariance technique (ii) Bilinear transformation technique are discussed for transforming continuous-time system to discrete-time system.

8.10

Transformations Between Continuous-Time Systems and Discrete-Time Systems

409

8.10.1 Impulse Invariance Method In this method, the impulse response of an analog filter is uniformly sampled to obtain the impulse response of the digital filter, and hence this method is called the impulse invariance method. The process of designing an IIR filter using this method is as follows: Step 1: Design an analog filter to meet the given frequency specifications. Let Ha(s) be the transfer function of the designed analog filter. We assume for simplicity that Ha(s) has only simple poles. In such a case, the transfer function of the analog filter can be expressed in partial fraction form as H a ðsÞ ¼

XN k¼1

Ak s  pk

ð8:145Þ

where Ak is the residue of H(s) at the pole pk. Step 2: Calculate the impulse response h(t) of this analog filter by applying the inverse Laplace transformation on H(s). Hence, ha ðt Þ ¼

XN k¼1

Ak epk t ua ðt Þ

ð8:146Þ

Step 3: Sample the impulse response of the analog filter with a sampling period T. Then, the sampled impulse response h(n) can be expressed as hðnÞ ¼ ha ðt Þjt¼nT PN n ¼ k¼1 ðAk epk T Þ uðnÞ

ð8:147Þ

Step 4: Apply the z-transform on the sampled impulse response obtained in Step 3, to form the transfer function of the digital filter, i.e., H(z) ¼ Z[h(n)]. Thus, the transfer function H(z) for the impulse invariance method is given by H ðzÞ ¼

XN k¼1

Ak 1  epk T z1

ð8:148Þ

This impulse invariant method can be extended for the case when the poles are not simple. Example 8.43 Consider a continuous system with the transfer function: H ðsÞ ¼

sþb ðs þ bÞ2 þ c2

410

8 The z-Transform and Analysis of Discrete Time LTI Systems

Solution The inverse Laplace transform of H(s) yields

hð t Þ ¼

ebt cos ðct Þ 0

for t  0 otherwise

Sampling h(t) with sampling period T, we get ( hðnT Þ ¼ H ðzÞ ¼

ebnT cos ðcnT Þ

for n  0

0

otherwise

1 X

ebnT cos ðcnT Þzn

n¼0

1  X  1 ¼ ebnT zn ejcnT þ ejcnT 2 n¼0 1 h  n  n i 1X eðbjcÞT Z 1 þ eðbþjcÞT Z 1 ¼ 2 n¼0  1 1 1  ¼ 2 1  eðbjcÞTz1 1  eðbþjcÞTz1 ¼

1  ebT cos ðcT Þz

1 2

1  2ebT cos ðcT Þz1 þe2bTz

Disadvantage of Impulse Invariance Method The frequency responses of the digital and analog filters are related by     1 X1 ω þ 2πk H j H ejω ¼ a k¼1 T T

ð8:149Þ

From Eq. (8.149), it is evident that the frequency response of the digital filter is not identical to that of the analog filter due to aliasing in the sampling process. If the analog filter is band-limited with ω   ω   Ha j ð8:150Þ ¼ 0   ¼ jΩj  π=T T T then the digital filter frequency response is of the form   1  ω H ejω ¼ H a j T T

jωj  π

ð8:151Þ

In the above expression, if T is small, the gain of the filter becomes very large. This can be avoided by introducing a multiplication factor T in the impulse invariant transformation. In such a case, the transformation would be hðnÞ ¼ Tha ðnT Þ and H(z) would be

ð8:152Þ

8.10

Transformations Between Continuous-Time Systems and Discrete-Time Systems

H ðzÞ ¼ T

XN k¼1

Ak 1  epk T z1

411

ð8:153Þ

Also, the frequency response is  ω   1 H ejω ¼ H a j T T

j ωj  π

ð8:154Þ

Hence, the impulse invariance method is appropriate only for band-limited filters, i.e., low-pass and band-pass filters, but not suitable for high-pass or band-stop filters where additional band limiting is required to avoid aliasing. Thus, there is a need for another mapping method such as bilinear transformation technique which avoids aliasing.

8.10.2 Bilinear Transformation In order to avoid the aliasing problem mentioned in the case of the impulse invariant method, we use the bilinear transformation, which is a one-to-one mapping from the s-plane to the z-plane; that is, it maps a point in the s-plane to a unique point in the z-plane and vice versa. This is the method that is mostly used in designing an IIR digital filter from an analog filter. This approach is based on the trapezoidal rule. Consider the bilinear transformation given by S¼

2 ð z  1Þ T ð z þ 1Þ

ð8:155Þ

Then a transfer function Ha (s) in the analog domain is transformed in the digital domain as  H ðzÞ ¼ H a ðsÞ 2 ðz1Þ ð8:156Þ S¼ T ðzþ1Þ

Also, from Eq. (8.155), we have Z¼

2 ð1 þ sÞ T ð1  sÞ

ð8:157Þ

We now study the mapping properties of the bilinear transformation. Consider a point s ¼ σ + jΩ in the left half of the s-plane. Then, from Eq. (8.157),   ð1  σ þ jΩÞ  >1 ð8:158Þ jzj ¼  ð1 þ σ  jΩÞ Hence, the left half of the s-plane maps into the interior of the unit circle in the z-plane (see Figure 8.12). Similarly, it can be shown that the right-half of the s-plane

412

8 The z-Transform and Analysis of Discrete Time LTI Systems Left half s-plane

Im z

jΩ

z-plane -1

1

0

ߪ

Figure 8.12 Mapping of the s-plane into the z-plane by the bilinear transformation

maps into the exterior of the unit circle in the z-plane. For a point z on the unit circle, z ¼ e jω, we have from Eq. (8.155) S¼

2 ðejω  1Þ 2 ω ¼ j tan jω T ð e þ 1Þ T 2

ð8:159Þ

2 ω tan T 2

ð8:160Þ

Thus, Ω¼ or ω ¼ 2 tan 1



ΩT 2

 ð8:161Þ

showing that the positive and negative imaginary axes of the s-plane are mapped, respectively, into the upper and lower halves of the unit circle in the z-plane. We thus see that the bilinear transformation avoids the problem of aliasing encountered in the impulse invariant method, since it maps the entire imaginary axis in the s-plane onto the unit circle in the in the z-plane. Further, in view of the mapping, this transformation converts a stable analog filter into a stable digital filter. Example 8.44 Design a low-pass digital filter with 3 dB cutoff frequency at 50 Hz and attenuation of at least 10 dB for frequency larger than 100 Hz. Assume a suitable sampling frequency. Solution Assume the sampling frequency as 500 Hz. Then, ωc ¼

2π f c 2π  50 ¼ 0:2π ¼ 500 FT

ωs ¼

2π f s 2π  100 ¼ 0:4π ¼ 500 FT

T ¼ 1=500 ¼ 0:002 Prewarping of the above normalized frequencies yields

8.11

Problems

413

  0:2π 2 tan ð0:1π Þ ¼ 325 ΩC ¼ tan ¼ 2 T   0:4π 2 tan ð0:2π Þ Ωs ¼ tan ¼ 727 ¼ 2 T Substituting these values in ðΩs =Ωc Þ2N ¼ 100:1αs  1 and solving for N, we get   log 101  1 0:9542 ¼ ¼ 1:3643: N¼ 2logð0:727=0:325Þ 0:6993 Hence, the order of the Butterworth filter is 2. The normalized low-pass Butterworth filter for N ¼ 2 is given by H N ðsÞ ¼

s2

1 pffiffiffi þ 2s þ 1

The transfer function Hc(s) corresponding to Ωc ¼ 0.325 is obtained by substituting s ¼ (s/Ωc) ¼ (s/0.325) in the expression for HN(s); hence, H a ðsÞ ¼

s2

0:1056 þ 0:4595s þ 0:1056

The digital transfer function H(z) of the desired filter is now obtained by using the bilinear transformation (8.157) in the above expression: H ðzÞ ¼ H a ðsÞj H ðzÞ ¼

8.11

2 ðz1Þ s¼ T ðzþ1Þ

0:1056z2 þ 0:2112z þ 0:1056 1000459:6056z2  1999999:7888z þ 999540:6056

Problems

1. Find the z-transform of the sequence x(n) ¼ {1,2,3,4,5,6,7}. 2. Find the z-transform and ROC of the sequence x(n) tabulated below. n x(n)

2 1

1 2

0 3

1 4

2 5

3 6

4 7

3. Find the z-transform of the signal x(n) ¼ [3(3)n4(2)n]. 4. Find the z-transform of the sequence x(n) ¼ (1/3)n1u(n1).

414

8 The z-Transform and Analysis of Discrete Time LTI Systems

5. Find the z-transform of the sequence

x ð nÞ ¼

1, 0,

0nN1 otherwise

6. Find the z-transform of the following discrete-time signals, and find the ROC for each.  n  n (i) xðnÞ ¼ 12 uðnÞ þ 3 14 uðn  1Þ   (ii) xðnÞ ¼ 14 δðnÞ þ δðn  2Þ  13 δðn  3Þ  n  (iii) xðnÞ ¼ ðn þ 0:5Þ 12 uðn  1Þ  13 δðn  3Þ 7. 8. 9. 10.

Find the z-transform of the sequence x(n) ¼ nan1u(n1). Find the z-transform of the sequence x(n) ¼ (1/4)n+1u(n). Find the z-transform of the signal x(n) ¼ [(4)n+13(2)n1]. Determine the z-transform and the ROC for the following time signals. Sketch the ROC, poles, and zeros in the z-plane.   π (i) xðnÞ ¼ sin 3π 4 n  8 u½ n  1   π (ii) xðnÞ ¼ ðn þ 1Þ sin 3π 4 n þ 8 u½n þ 2:

11. Find the inverse z-transform of the following, using partial fraction expansions: zþ0:5 (i) X ðzÞ ¼ ðzþ0:2 Þðz2Þ ,

jzj > 2

1

jzj > 2

(ii) X ðzÞ ¼ 1þ3z1þz 1 þ2z2 , (iii) X ðzÞ ¼ (iv) X ðzÞ ¼

z2 þz ðz3Þðz2Þ , zðzþ1Þ

ðz12Þðz13Þ

jzj > 3 ,

jzj > 12

12. Find the inverse z-transform of the following using the partial fraction expansion. (i) X ðzÞ ¼ ðz1Þzðz4Þ ,

jzj < 1

(ii) X ðzÞ ¼ ðz1zÞðþ2z3 z3Þðz4Þ , 2

z , (iii) X ðzÞ ¼ 3z2 4zþ1

for ðaÞ jzj > 4 and ðbÞ jzj < 1

jzj < 13

13. Determine all the possible signals that can have the following z-transform: X ðzÞ ¼

z2 z2  0:8z þ 0:15

8.11

Problems

415

14. Find the stability of the system with the following transfer function: H ðzÞ ¼

z3



1:4z2

z þ 0:65z  0:1

15. The transfer function of a system is given as H ðzÞ ¼

z þ 0:5 ðz þ 0:4Þðz  2Þ

Specify the ROC of H(z) and determine h(n) for the following conditions: (i) The system is causal. (ii) The system is stable. (iii) Can the given system be both causal and stable? 16. A causal LTI system is described by the following difference equation: 1 1 y ð nÞ  y ð n  2Þ ¼ x ð n  2Þ  x ð nÞ 4 4 Determine whether the system is an all-pass system. 17. In the system shown in Figure P8.1, if S1 is a causal LTI system with system function,     1 1 3 1  H ðzÞ ¼ 1  z 1 z 1  3z1 2 4 Determine the system function for a system S2 so that the overall system is an all-pass system. x (n )

S1

S2

y (n )

Figure P8.1 Cascade connection of two systems S1 and S2

18. The transfer function of a system is given by H ðzÞ ¼

1 z2 þ 5z þ 6

Determine the response when x(n) ¼ u(n). Assume that the system is initially relaxed.

416

8 The z-Transform and Analysis of Discrete Time LTI Systems

19. A causal LTI system is described by the following difference equation: y ð nÞ  y ð n  1Þ  y ð n  2Þ ¼ x ð n  1Þ Is it a stable system? If not, find a noncausal stable impulse response that satisfies the difference equation. 20. Using the one-sided z-transform, solve the following difference equation:   1 y ð nÞ  yðn  2Þ ¼ uðnÞ, 9

yð1Þ ¼ 0, yð2Þ ¼ 2

21. Consider a continuous system with the transfer function H ðsÞ ¼

1 ð s þ 1Þ ð s 2 þ s þ 1Þ

Determine the transfer function and pole-zero pattern for the discrete-time system by using the impulse invariance technique. 22. Design a low-pass Butterworth filter using Bilinear transformation for the following specifications: Passband edge frequency: 1000 Hz Stopband edge frequency: 3000 Hz Passband ripple: 2 dB Stopband ripple: 20 dB Sampling frequency: 8000 Hz 23. Consider an initially relaxed discrete time with the following state-space representation. Find y(n). "

8.12

3 0 1 " x ð nÞ # " 0 # 1 ¼ 4 1 35 þ ℧ðnÞ  x 2 ð n þ 1Þ x 2 ð nÞ 1 8 4 #  " 1 3 x 1 ð nÞ y ð nÞ  þ ℧ð n Þ 8 4 x 2 ð nÞ x 1 ð n þ 1Þ

#

2

MATLAB Exercises

1. Write a MATLAB program using the command residuez to find the inverse of the following by partial fraction expansion:

Further Reading

417

X ðzÞ ¼

16  4z1 þ z2 8 þ 2z1  2z2

2. Write a MATLAB program using the command impz to find the inverse of the following by power series expansion: X ðzÞ ¼

15z3

15z3 þ 5z2  3z  1

3. Write a MATLAB program using the command z-plane to obtain a pole-zero plot for the following system: H ðzÞ ¼

1 þ 13 z1 þ 57 z2  32 z3 1 þ 52 z1  13 z2  35 z3

4. Write a MATLAB program using the command freqz to obtain magnitude and phase responses of the following system: H ðzÞ ¼

1  3:0538z1 þ 3:8281z2  2:2921z3 þ 0:5507z4 1  4z1 þ 6z2  4z3 þ z4

Further Reading 1. Lyons, R.G.: Understanding Digital Signal Processing. Addison-Wesley, Reading (1997) 2. Oppenheim, A.V., Schafer, R.W.: Discrete-Time Signal Processing, 2nd edn. Prentice-Hall, Upper Saddle River (1999) 3. Mitra, S.K.: Digital Signal Processing. McGraw-Hill, New York (2006) 4. Hsu, H.: Signals and Systems, Schaum’s Outlines, 2nd edn. McGraw-Hill, New York (2011) 5. Kailath, T.: Linear Systems. Prentice-Hall, Englewood Cliffs (1980) 6. Zadeh, L., Desoer, C.: Linear System Theory. McGraw-Hill, New York (1963)

Index

A Aliasing, 273, 347, 410–412 All-pass decomposition, 399–400 All-pass system, 397–400, 415 Amplitude, 5 Amplitude demodulation, 163–165 Amplitude modulation (AM), 155, 162–164 Analog filter design, 237, 238, 240, 242, 244, 250, 252–255, 258, 260–263 band-pass, 227, 230, 252–264, 269, 411 band-stop, 31, 227, 231, 252–264, 411 Butterworth low-pass filter, 62, 163, 228, 232–237, 249, 263, 413 Chebyshev analog low-pass filter type 1 Chebyshev low-pass filter, 237, 238, 240, 255, 258, 261 type 2 Chebyshev filter, 242, 244, 250, 253, 261 elliptic analog low-pass filter, 227, 245–247, 251 high-pass, 31, 227–229, 231, 252–264, 269, 411 low-pass, 227, 229, 231–264 notch, 31, 266–267 specifications of low-pass filter, 232, 259, 261, 263 transformations low-pass to band-pass, 252 low-pass to band-stop, 252, 260, 262, 263 low-pass to high-pass, 252, 254 low-pass to low-pass, 252, 253 Analog filter types comparison, 249–252 Application examples, vii, 10, 30, 162–164 Associative property, 51–52, 292

B Band-pass filter, 230, 257–260, 269, 411 Band-stop filter, 231, 260, 261, 263, 264 Basic continuous-time signals complex exponential function, 24 ramp function, 22 real exponential function, 23–24 rectangular pulse function, 22–23 signum function, 23 sinc function, 24–27, 137 unit impulse function, 21–22, 136, 187 unit step function, 20, 22, 29, 98, 188 Basic sequences arbitrary, 21, 50, 77, 113, 139, 176, 178, 282, 286, 318, 353, 360, 380 exponential and sinusoidal, 277, 278, 301 unit sample, 286, 365 unit step, 20, 49, 68, 286 Bessel filter, 248–250 BIBO stability theorem, 284, 285, 389, 391 Bilateral Laplace transform, 171, 172 Bilinear transformation, 411, 413, 416 Block diagram representation described by differential equations, 82–93 Butterworth analog low-pass filter, 233–237

C Cauchy’s residue theorem, 374–375 Causal and stable conditions, 391 Causality, 41, 77, 82, 86–87, 107, 204–206, 208, 285, 297, 298, 311, 391, 401 Causality for LTI systems, 77, 294–297 Causality theorem, 391 Characteristic equation, 300

© Springer International Publishing AG, part of Springer Nature 2018 K. D. Rao, Signals and Systems, https://doi.org/10.1007/978-3-319-68675-2

419

420 Chebyshev analog low-pass filter, 237–245 Classification of signals analog and digital signals, 5, 271–276, 346 causal, non-causal and anti-causal signals, 12 continuous time and discrete time signals, 5 deterministic and random signals, 20 energy and power signals, 13–20 even and odd signals, 9–12, 141 periodic and aperiodic signals, 6–9 Commutative property, 46 Complex Exponential Fourier Series, 111–128 Computation of convolution integral using MATLAB, 70–74 Computation of convolution sum using MATLAB, 291 Computation of linear convolution graphical method, 289 matrix method, 288 Conjugate of complex sequence, 364 Continuous Fourier Transform convergence of Fourier transform, 135–136 Continuous Fourier transform properties convolution property, 151 differentiation in frequency, 146, 147 differentiation in time, 143 duality, 154 frequency shifting, 143 integration, 148 linearity, 139, 151, 158, 160 modulation, 155, 158 Parseval’s theorem, 149, 150, 157 symmetry properties, 119, 158 time and frequency scaling, 143, 158 time reversal, 158 time shifting, 142, 158, 168 Continuous time signal, 113–133 complex exponential Fourier series, 111–128 convergence of Fourier series, 113 properties of Fourier series, 113–128 trigonometric Fourier series, 128–133, 166 symmetry conditions, 129–133 Continuous-time systems causal system, 48, 77 invertible system, 49, 79 linear systems, 42–48 memory and memoryless system, 49 stable system, 49, 78 time–invariant system, 43–48, 105 Convergence of the DTFT, 317 Convolution integral associative property, 51–52 commutative property, 50 distributive property, 50–51, 75 graphical convolution, 58–70

Index Convolution of two sequences, 151, 318, 362 Convolution sum, 271, 287, 289, 332 Convolution theorem, 220, 318, 320, 366, 406 Correlation, 30, 31, 33, 319, 363, 366 Correlation of discrete-time signals, vii Correlation of two sequences, 363 Correlation theorem, 319

D Direct form I, 96 Direct form II, 95–97 Discrete-time Fourier series (DTFS) Fourier coefficients, 350 multiplication, 315 periodic convolution, 313–316, 318 symmetry properties, 315 Discrete-time Fourier transform (DTFT) linearity, 315 Discrete time LTI systems in z-domain, 385–400 Discrete-time signal, 271, 315–331, 414 Discrete-time signals classification energy and power signals, 13, 279–281 finite and infinite length, 276 periodic and aperiodic, 6–9, 278 right-sided and left-sided, 277 symmetric and anti-symmetric, 276 Discrete-time system characterization non-recursive difference equation, 298 recursive difference equation, 298, 336 Discrete-time systems classification causal, 284, 298 linear, 282, 286–289, 291–297 stable, 284 time-invariant, 283–284 Discrete transformation, 281 Distributive property, 50–51, 75 Down-sampler, 306

E Elementary operations on signals, 1–5 Elliptic analog low-pass filter, 246 Energy and power signals, 13–20, 279 Examples of real world signals and systems audio recording system, 32 global positioning system, 33 heart monitoring system, 34–36 human visual system, 36 location-based mobile emergency services system, 33–34 magnetic resonance imaging, 36–37

Index F Filtering, 31, 227, 233, 235, 236 Final value theorem, 187, 200, 222 Fourier transform, 318 Fourier transform of discrete-time signals, 158, 315, 317–331, 335, 342 convergence of the DTFT, 317 properties of DTFT for a complex sequence, 320–322 for a real sequence, 322–331 theorems on DTFT convolution theorem, 318, 320 correlation theorem, 319, 320 differentiation in frequency, 158, 318 frequency shifting, 315, 318, 320, 342 linearity, 317, 320, 326 Parseval’s theorem, 319, 320, 328, 329 time reversal, 318, 320 time shifting, 320, 324, 335 windowing theorem, 318 Frequency division multiplexing (FDM), 32, 164, 166 Frequency response computation using MATLAB, 338, 341, 342, 346 Frequency response from poles and zeros, 264–265, 388–389 Frequency response of continuous time systems, 111, 159–162 Frequency response of discrete-time systems computation using MATLAB, 338, 341, 342, 346 Frequency shifting, 115, 143, 145, 158, 315, 318, 342

G Generation of continuous-time signals using MATLAB, 28–30 Graphical convolution, 58–70

H Half-wave symmetry, 119–124, 126, 127, 132 High-pass filter, 231, 257

I Imaginary part of a sequence, 365 Impulse and step responses, 286 Impulse and step responses computation using MATLAB, vii, 92, 225, 304, 305, 312

421 Impulse response, 49, 53–57, 62, 66–68, 73–75, 77–79, 88, 92–95, 106, 109, 153, 161, 193, 202, 204, 217, 227, 229, 267, 286–288, 294–297, 324, 330, 332 Impulse step responses, 304, 305 Initial Value Theorem, 186–187, 370–371 Input-output relationship, 271, 282, 286–288, 293 Interconnected systems, 74–76 Inverse discrete Fourier transform, 111, 135, 136, 138 Inverse Fourier transform, 317 Inverse Laplace transform partial fraction expansion, 194–202, 209, 211, 375–379, 416 partial fraction expansion using MATLAB, 201–202 partial fraction expansion with multiple poles, 195–201 partial fraction expansion with simple poles, 195, 376 Inverse system, 79–81, 205, 395, 396 Inverse z-transform Cauchy’s residue theorem, 374–375 modulation theorem, 372 Parseval’s relation, 126, 372–374, 398 partial fraction expansion, 375–379, 414 partial fraction expansion using MATLAB, 379–380 power series expansion, 379–383 power series expansion using MATLAB, 383

L Laplace transform, 117, 144, 151, 155, 172, 174–191, 200, 208, 215, 222, 402, 403 block diagram representation, 218–219 definition of, 171–225 existence of, 172 inter connection of systems, 218 properties, 171, 178–187 convolution in the frequency domain, 117, 155, 181, 182 convolution in the time domain, 151, 181 differentiation in the s-domain, 180, 184, 190 differentiation in the time domain, 144, 179, 184 division by t, 180 final value theorem, 187, 200, 222, 402, 403

422 Laplace transform (cont.) initial value theorem, 186–187 integration, 184, 187 linearity, 178, 184, 208 properties of even and odd functions, 182–183 shifting in the s-domain, 178, 184, 189–191 time scaling, 179, 184 time shifting, 178, 184, 188 region of convergence, 174–176, 203, 208, 223 finite duration signal, 174 left sided signal, 175–177 right sided signal, 172, 175–177 strips parallel to the jΩ axis, 174 two sided signal, 176, 177 region of convergence (ROC), 173 relationship to Fourier transform, 172–173 representation of Laplace transform in the s-plane, 173, 188 system function, 202, 204 block diagram representation, 218–219 interconnection of systems, 218 table of properties, 184 transfer Function, 202–204, 223, 235 unilateral Laplace transform, 171, 172, 183–186, 215 differentiation property, 183–186, 215 Linear constant coefficient difference equations, 298, 299 Linear constant-coefficient differential equations, 82–84, 159, 171, 207–210 Linearity, 41, 43, 44, 50, 86, 105, 113, 119, 139, 140, 151, 158, 160, 178, 184, 193, 208, 217, 282, 285, 311, 315, 317, 320, 326, 360, 364–366 Linearity property of the Laplace transform, 217 Low-pass to band-pass, 257 Low-pass filter, 164, 203, 225, 227–229, 231–264, 267, 268, 341, 351 LTI discrete-time systems, vii, 271, 286–289, 291–297, 304, 332, 333, 354, 386 LTI systems with and without memory, 77

M Matrix method, 288 Maximum-phase systems, 395 Minimum-phase decomposition, 399–400 Minimum-phase systems, 395, 399 Mixed-phase systems, 395 Modulation, 158 Modulation and demodulation, 31, 170

Index Modulation property, 155–158 Modulation theorem, 372 Multiplexing and demultiplexing, 32

N Nonperiodic signals, 111, 133–158 Non-recursive difference equation, 298 Notch filter, 266–267

O One-sided z-transform, 384, 393, 401, 404–405 properties shifting theorem, 384, 393, 401 final value theorem, 402–404 solution of linear difference equations with initial conditions, 401, 404–405

P Parseval’s relation, 126, 372–374, 398 Parseval’s theorem, 118, 149, 150, 157, 319, 320, 328 Partial fraction expansion, 378 Partial fraction expansion using MATLAB, 201–202, 379–380 Particular solution, 85, 89, 90, 300–303 Periodic convolution, 76, 107, 117, 119, 313–316, 318 Phase and group delays, 333 Poles and zeros, 173, 192, 227, 235, 240, 242, 243, 246, 265, 386–388, 395, 398, 400, 414 Pole-zero pairing, 388 Pole-zero placement, 227, 264–267 Power series expansion using MATLAB, 417 Properties of the convolution integral, 50–57, 151 Properties of the convolution sum, 291–295 Properties of the impulse function sampling property, 21, 52, 67, 136 scaling property, 22 shifting property, 21, 324, 335 Proposition, 7, 18, 278

Q Quantization and coding, 274–276 Quantization error, 275

R Rational transfer function, 386 Rational z-transform, 354, 374 Real part of a sequence, 364

Index Reconstruction of a band-limited signal from its samples, 350 Recursive difference equation, 298, 336, 338 Region of convergence (ROC), 173–181, 184, 185, 188, 192, 195, 197, 199, 203–208, 210, 222, 354–364, 366, 369–371, 373, 377, 379, 389, 391, 392, 401–404, 413–415 Representation of signals in terms of impulses, 41–42

S Sampling continuous time signals, vii, 5, 271, 273, 305, 344 discrete time signals, 271–276, 305–307 frequency domain, vii, 306, 344–347 Nyquist rate, 273 Sampling frequency, 271, 273–275, 346, 412, 416 Sampling in frequency-domain aliasing, 273, 347, 410 over-sampling, 347 sampling theorem, 273, 346 under sampling, 346 Sampling period, 271, 273, 306, 346, 409 Sampling theorem, 273–274, 346, 347 Scalar multiplication, 93 S-domain, 178, 180, 184, 189–191 Single-side-band (SSB) AM, 164 Singularity functions, 41, 95 Solution of difference equations characteristic equation, 300 complementary solution, 85, 90, 300 using MATLAB, 91–92 particular solution, 300–303 Solution of linear differential equations using MATLAB, 216 Stability and causality, 208, 295–297, 311, 389, 391, 396 Stability and causality of LTI systems in terms of the impulse response, 295–297 Stability for LTI systems, 77–79 Step and impulse responses, 49 Step response, 49, 68, 89, 91–94, 106, 107, 110, 286, 302, 304, 305, 311, 404 Systems all-pass, 397–400, 415 causal, 48, 77, 78, 86, 107, 172, 204–210, 219, 223, 284, 288, 294–298, 324, 391–394, 397, 415, 416

423 inverse, 79–82, 205, 395, 396 maximum-phase, 395 minimum-phase, 395 mixed-phase, 395, 399 stable, 49, 78, 204–210, 218, 219, 223, 285, 296, 297, 311, 324, 325, 389, 391–394, 397, 415, 416

T Theorems on DTFT, 317–320, 329 Time-invariant, 41, 43–48, 50, 77–82, 88, 283, 286–289, 291–297, 395 Time reversal, 3–5, 115, 119, 158, 318, 320, 360 Time scaling, 2–3, 115, 119, 179, 184 Time shifting, 2, 17, 18, 113, 119, 120, 142, 158, 168, 178, 184, 188, 315, 318, 324, 335 Transfer function, 202–204, 223, 227, 234–241, 243–246, 248, 252, 254, 256–259, 261, 263, 267, 385–386, 388, 389, 394, 395, 397–399, 409, 411, 413, 415 Transformation, 31, 42, 111, 227, 252–255, 257, 260, 263, 281, 285, 286, 325, 408–413, 416

U Under sampling, 346 Unilateral Laplace transform differentiation property, 183–186, 208 Unit doublet, 95, 97 Unit impulse response, 95 Unit ramp, 22 Unit sample sequence, 286, 365 Unit step response, 68 Unit step sequence, 286 Unstable system, 78, 391, 392 Up-sampling, 312

W Windowing theorem, 318, 320

Z Zero locations, 206, 264, 266 Zero-order hold, 272–274 Z-transform definition of, 353, 359, 360, 381 properties of, 360–366

424 Z-transform (cont.) region of convergence (ROC), 354–359, 361–366, 368–371, 374, 377, 379, 413, 414 Z-transform properties conjugate of a complex sequence, 364 convolution of two sequences, 362 correlation of two sequences, 363 differentiation in the z-domain, 361

Index imaginary part of a sequence, 365 linearity, 360, 364–366 real part of a sequence, 364 scaling in the z-domain, 361 time reversal, 360, 363, 366 time shifting, 361, 366 Z-transforms of commonly-used sequences unit step sequence, 366