Unit - 1 Amplitude Modulation

Question 1 Define modulation. Explain different types of analog modulation. Ans.

Modulation is the process of varying some characteristics (amplitude, frequency or phase) of a high frequency carrier wave in accordance with the instantaneous value of the low frequency modulating signal. In the modulation process, the baseband or modulating signal (such as voice, video etc.) modifies another higher frequency signal called the carrier signal. The carrier is usually a sinusoidal wave that is higher in frequency than the highest baseband signal frequency. The baseband or modulating signal modifies the amplitude or frequency or phase of the carrier.

There are three types of analog modulation : 1.

Amplitude Modulation : Amplitude modulation is defined as the modulation in which the amplitude of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal, keeping its (carrier) frequency and phase constant.

2.

Frequency Modulation : Frequency modulation is defined as the modulation in which the frequency of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal keeping its (carrier) amplitude and phase constant.

3.

Phase Modulation : Phase modulation is defined as the modulation in which the phase of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal, keeping its (carrier) amplitude and frequency constant.

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Question 2 Explain the need of modulation briefly. Ans. Modulation is the process of varying some characteristics (amplitude, frequency or phase) of a high frequency carrier wave in accordance with the instantaneous value of the low frequency modulating signal. The baseband or modulating signals are incompatible for direct transmission over the medium and therefore modulation technique is used for the communication of modulating signal. The need of modulation are : 1. Modulation reduces the height of antenna : The height of the antenna required for transmission and reception of radio waves in radio transmission is a function of wavelength  of the frequency used. The minimum height of the antenna is given by  / 4 .  c c  Height of antenna   4 4f f  Where, c  3 10 m/sec is the velocity of light and

f

is the transmitting

frequency. Consider the modulating signal with f  15 kHz Height of antenna 

 c 3  108    5000 meters 4 4 f 4  15  103

This 5000 m height of a vertical antenna is unthinkable and impracticable. Consider the modulated signal with f  1 MHz Height of antenna 

2.

 c 3  108    75 meters 4 4 f 4  1 106

This height of antenna is practicable and such antenna can be installed. Modulation avoids mixing of signals : All sound signals range from 20 Hz to 20 kHz. The transmission of baseband signal from various sources causes the mixing of signal and then it is difficult to separate these signals at the receiver end. In order to separate the various signals, it is necessary to translate them all to different portions of the electromagnetic spectrum or channel; each must be given its own bandwidth commonly known as channel bandwidth. This can be achieved by taking different carrier frequency for different signal source. Once the signals have been transmitted, a tuned circuit at the receiving end selects the portion of the channel it is tuned for. Therefore, modulating different signal sources by different carrier frequencies avoid mixing of signals.

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Amplitude Modulation

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3.

Modulation increases the range of communication : Low frequency signals have poor radiation and they get highly attenuated. Therefore baseband signals cannot be transmitted directly over long distances. Modulation effectively increases the frequency of the signals and thus they can be transmitted over long distances. 4. Modulation allows multiplexing of signals : Modulation allows the multiplexing to be used. Multiplexing means transmission of two or more signals simultaneously over the same communication channel. For example number of TV channel operating simultaneously. The different signals from different stations can be separated in the receiver since the carrier frequencies for these signals are different. 5. Modulation allows adjustments in the bandwidth : Bandwidth of a modulated signal may be made smaller or larger than the original signal. Signal to noise ratio in the receiver which is the function of the signal bandwidth can thus be improved by proper control of the bandwidth at the modulating stage. 6. Modulation improves quality of reception : Modulation techniques like frequency modulation, pulse code modulation reduces the effect of noise to great extent. Reduction of noise improves the quality of reception. Question 3 What is demodulation? [CSVTU May 2011] Ans. Demodulation is the process in which the original signal is recovered from the transmitted signal or the modulating signal is recovered from the modulated signal. Demodulation is the process in which high frequency is converted to a low frequency signal. Demodulation performs inverse function of modulation.

Demodulator is the main block of receiver which performs demodulation.

Double Sideband Full Carrier Question 4 Define amplitude modulation.

[CSVTU May 2011, Dec 2010, May 2009]

Or A signal Ac cos(c t ) is amplitude modulated by another signal Am cos(m t ) . Amplitude sensitivity is K a . What will be the amplitude of modulated signal? Ans.

Amplitude modulation is defined as the modulation in which the amplitude of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal, keeping its (carrier) frequency and phase constant. Instantaneous value of modulating signal, m(t )  Am cos(2f mt ) Instantaneous value of carrier signal, c(t )  Ac cos(2f c t )

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Time domain equation of amplitude modulated (AM) wave is given by, s(t )  Ac cos(2fct )  Ac Ka m(t )cos(2fct )  Ac 1  Ka m(t ) cos(2fct )

s(t )  Ac 1  ma cos(2f mt ) cos(2fct )  Ac cos(2fct )  ma Ac cos(2f mt )cos(2fct )

Where K a is a constant called the amplitude sensitivity of the modulator and ma is modulation index. Question 5 Give time domain and frequency domain representation of DSB-FC amplitude modulation. Ans. In DSB-FC AM, the amplitude of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating signal, keeping its (carrier) frequency and phase constant. Time domain representation of DSB-FC : Instantaneous value of modulating signal, m(t )  Am cos(2f mt ) Instantaneous value of carrier signal, c(t )  Ac cos(2f c t ) Where Am and Ac are maximum amplitudes of modulating and carrier signal respectively. f m and f c are frequency of modulating and carrier signal respectively. Time domain equation of amplitude modulated (AM) wave is given by, s(t )  Ac cos(2fct )  Ac Ka m(t )cos(2fct )  Ac 1  Ka m(t ) cos(2fct )

s(t )  Ac 1  ma cos(2f mt ) cos(2fct )  Ac cos(2fct )  ma Ac cos(2f mt )cos(2fct ) Where K a is a constant called the amplitude sensitivity of the modulator and ma is modulation index.

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Amplitude Modulation

When | K a m(t ) |  1 , the modulating signal is clearly identified as the envelope of a conventional AM signal and the modulated signal s (t ) has the same shape as the modulating signal m(t ) . When | K a m(t ) |  1 , the resulting AM wave is said to have envelope distortion. Frequency domain representation of DSB-FC : The amplitude modulated carrier has new signals at different frequencies, called side frequencies or sidebands, occur in the frequency spectrum directly above and below the carrier frequency.

s(t )  Ac 1  ma cos(2f mt ) cos(2fct )  Ac cos(2fct )  ma Ac cos(2f mt )cos(2fct )

ma Ac m A cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 Upper side band frequency, fUSB  f c  f m s(t )  Ac cos(2f c t ) 

Lower sideband frequency, f LSB  fc  f m Bandwidth, BW  fUSB  f LSB  ( f c  f m )  ( f c  f m )  2 f m The bandwidth required for the amplitude modulation is twice the frequency of the modulating signal. Bandwidth is defined as the portion of the electromagnetic spectrum occupied by a signal. It is the frequency range over which an information signal is transmitted. It is the difference between the upper and lower frequency limits of the signal. Question 6 Draw the single sided spectra, double sided spectra and phasor diagram for DSBFC. Or What are the frequency components in an AM wave? Ans.

The amplitude modulated carrier has new signals at different frequencies, called side frequencies or sidebands, occur in the frequency spectrum directly above and below the carrier frequency.

s(t )  Ac 1  ma cos(2f mt ) cos(2fct )  Ac cos(2fct )  ma Ac cos(2f mt )cos(2fct )

ma Ac m A cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 Upper side band frequency, fUSB  f c  f m s(t )  Ac cos(2f c t ) 

Lower sideband frequency, f LSB  fc  f m Bandwidth, BW  fUSB  f LSB  ( f c  f m )  ( f c  f m )  2 f m The frequency spectrum of AM DSB-FC waveform has (1) A component of carrier frequency, f c . (2) An upper side band (USB) whose highest frequency component is at f c  f m . (3) A lower side band (LSB) whose highest frequency component is at fc  f m . (4) The bandwidth of the modulated waveform is twice the modulating signal bandwidth.

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Single sided spectrum of DSB-FC :

Double sided spectrum of DSB-FC :

In double sided spectrum, amplitude gets halved wrt single sided spectrum. Phasor diagram of DSB-FC :

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Amplitude Modulation

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The waveform of AM signal can be interpreted as superposition of three waveforms (1) Carrier wave Ac cos(2fc t ) represented by OA ma Ac cos 2( f c  f m )t represented by AB 2 m A (3) Lower sideband a c cos 2( f c  f m )t represented by AC 2

(2) Upper sideband

Resultant waveform of AM signal is represented by OR Question 7 Define envelope of the AM wave. Ans.

Time domain equation of amplitude modulated (AM) wave is given by,

s(t )  Ac 1  Ka m(t ) cos(2fct )  E(t )cos(2fct ) The AM wave has time varying amplitude called as the envelope of the AM wave. E (t ) is the envelope of the AM wave. This envelope consists of the baseband or modulating signal m(t ) . The baseband signal can be recovered from an AM wave by detecting the envelope. Question 8 Explain points mentioned below briefly :

[CSVTU Dec 2009]

(1) Under modulation

(2) Critical modulation

(3) Over modulation

(4) Percentage of modulation

(5) Recovery of the baseband signal Or Define modulation index.

[CSVTU May 2012] Or

Show the effect of different modulation indices on DSB-FC waveform. Ans.

(1) Modulation index The ratio of amplitude of modulating signal to the amplitude of carrier wave is known as modulation index. It is also known as modulation function or modulation coefficient or depth of modulation or degree of modulation. It is denoted by ma or  . ma is a dimensionless constant. Its value lies between 0 and 1. If ma  0 , there is no modulation while ma  1 corresponds to maximum modulation that can occur without distortion. Modulation index,   ma 

Am Ac

Modulation index in terms of K a , ma  K a Am Time domain equation of DSB-FC (AM) wave is given by,

s(t )  Ac 1  K a m(t )  cos(2f c t )  Ac 1  K a Am cos(2f m t )  cos(2f c t )  Ac 1  ma cos(2f m t ) cos(2f c t )

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(2) Percentage of modulation The ratio of amplitude of modulating signal to the amplitude of carrier wave expressed in percentage is known as percentage of modulation. A  Percentage modulation index is given by, % ma   m   100%  Ac  (3) Under modulation In under modulation, the modulation index is less than 1 ( ma  1 ) and the amplitude of modulating signal is less than the amplitude of carrier signal ( Am  Ac ).

As is evident from the waveform, the envelope is not reaching the zero amplitude axis of the AM wave, and the baseband signal is fully preserved in the envelope of the AM wave. An envelope detector can recover the baseband signal without distortion. (4) Ideal modulation or critical modulation In ideal modulation, the modulation index is equal to 1 ( ma  1 ) and the amplitude of modulating signal is equal to the amplitude of carrier signal ( Am  Ac ).

Here, the waveform envelope just touches the zero amplitude axis. The baseband signal remains preserved in the envelope and can be recovered by using an envelope detector. However, this is a limiting factor as even a small variation in the magnitude of baseband signal will cause envelope distortion. (5) Over modulation In over modulation, the modulation index is more than 1 ( ma  1 ) and the amplitude of modulating signal is more than the amplitude of carrier signal ( Am  Ac ).

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Amplitude Modulation

Here, the portion of the envelope crosses zero amplitude axis. The envelope and the baseband are not the same. This is called envelope distortion. An envelope detector provides distorted baseband signal. An over modulated signal can be recovered using a costly and complex technique synchronous detection. (6) Recovery of the baseband signal The modulated signal is used at the transmitter for shifting the baseband signal (with maximum frequency f m ) to a much higher frequency f c . The modulated signal is transmitted from the transmitter and it reaches the receiver via a propagating media. At the receiver, the original baseband signal is desired to be recovered from modulated signal. This is achieved by retranslating the baseband signal from a higher spectrum to the original spectrum. This process of retranslation is known as demodulation or detection. The original baseband is recovered from the modulated signal by the detection process. Question 9 Draw the frequency spectrum diagram of DSB-FC amplitude modulated signal for periodic and non-periodic modulating signal (baseband signal). [CSVTU Dec 2011] Sol. Time domain equation of AM wave, s(t )  Ac 1  Ka m(t ) cos(2fc t ) Frequency spectrum diagram of DSB-FC amplitude modulated signal for periodic modulating signal (baseband signal)

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Frequency spectrum diagram of DSB-FC amplitude modulated signal for non-periodic modulating signal (baseband signal)

Question 10 Prove that ma 

Emax  Emin Emax  Emin

where Emax and Emin are the maximum and minimum value of the envelope of the Ans.

AM signal. Emax : Maximum value of the envelope of the modulated wave

Emin : Minimum value of the envelope of the modulated wave Em : Amplitude of modulating signal Ec : Amplitude of carrier signal

Emax  Ec (1  ma )

…..(i)

Emin  Ec (1  ma )

…..(ii)

From (i) and (ii), Emax 1  ma   Emin 1  ma

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On cross multiplication, Emax  ma Emax  Emin  ma Emin



ma 



Emax  Emin  ma Emin  ma Emax

Emax  Emin Emax  Emin

Proved.

On adding (i) and (ii), Ec  Now, ma  So, Em 

Amplitude Modulation

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Em Ec

Emax  Emin 2

 E  Emin Em  ma Ec   max  Emax  Emin



  Emax  Emin    2  

Emax  Emin 2

Question 11 Derive the equation for power content in AM wave and prove that  m2  PT  PC 1  a  2  

Ans.

Time domain equation of amplitude modulated (AM) wave is given by,

s(t )  Ac cos(2fct )  Ac Ka m(t )cos(2 fct )  Ac 1  Ka m(t ) cos(2f ct )

ma Ac m A cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 The AM wave has three components : unmodulated carrier, lower sideband and upper sideband. Therefore, the total power of AM wave is the sum of the carrier power PC and power in the two sidebands i.e. PUSB and PLSB . s(t )  Ac cos(2f c t ) 

PT  PC  PUSB  PLSB  PC  PSB The average carrier power is given by, PC  P  Ac cos(2 f c t )  

Ac2 2

The average upper sideband power is given by, 2

m2 A2 m A  1m A  PUSB  P  a c cos 2( f c  f m )t    a c   a c 8  2  2 2 

The average lower sideband power is given by, m A  PLSB  P  a c cos 2( f c  f m )t   2  

2

m2 A2 1  ma Ac   a c   2 2  8

ma2 Ac2 ma2 PC  8 4 2 2 2 2 2 2 m A m A m A m2 P Total sideband power, PSB  PUSB  PLSB  a c  a c  a c  a C 8 8 4 2 2 2 2 2 2 2 2 A m A m A A  m m2  Total power, PT  PC  PUSB  PLSB  c  a c  a c  c 1  a  a  2  4 4  2 8 8 The average sideband power is given by, PUSB  PLSB 

 m2   Total transmitted power, PT  PC 1  a  2  

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 12  For 100% modulation ma  1 , PT  PC 1    1.5PC  2

Carrier power, PC 

1 PT  0.666 PT 1.5

Total sideband power, PSB  PUSB  PLSB 

ma2 PC 12 1  PC  PT  0.333PT 2 2 3

In amplitude modulated wave, the 66.66% of the transmitted power is used by the carrier signal and remaining 33.33% of the power is used by the sidebands  PUSB and PLSB  . Question 12 For signal tone modulation show that ratio of total sideband power to the total 2 power is equal to where μ is modulation factor. [CSVTU May 2008] 2  2 Or Define the transmission efficiency for amplitude modulation wave. Ans.

Transmission efficiency is defined as the ratio of the power carried by the sidebands to the total transmitted power. Transmission efficiency,  

PUSB  PLSB PT

The average carrier power is given by, PC  P  Ac cos(2 f c t )   The average sideband power is given by, PUSB  PLSB 

Ac2 2

ma2 Ac2 ma2 PC  8 4

 m2  Total transmitted power, PT  PC  PUSB  PLSB  PC 1  a  2  

Total sideband power, PUSB  PLSB 



ma2 Ac2 ma2 Ac2 ma2 Ac2 ma2 PC    8 8 4 2

PUSB  PLSB ma2 PC / 2 ma2 / 2   2 PT PC (1  ma / 2) 1  ma2 / 2

 Transmission efficiency,  

PUSB  PLSB m2  2 a  100% PT ma  2

For signal tone modulation show that ratio of total sideband power to the total power 2 is equal to where   ma is modulation factor. 2  2 For ma  0.5 ,   11.11% For ma  1 ,   33.33%

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Amplitude Modulation

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Question 13 In a multitone modulated signal, a carrier Ac cos c t is modulated by a modulating signal m(t )  Am1 cos(2f m1 t )  Am 2 cos(2f m 2 t ) . Derive expressions for

Sol.

(a) total modulated power (b) net modulation index of the AM wave Time domain equation of amplitude modulated (AM) wave is given by, s(t )  Ac cos(2fct )  Ac Ka m(t )cos(2 fct )  Ac 1  Ka m(t ) cos(2f ct )



s (t )  Ac 1  K a m1 (t )  m2 (t ) cos(2f c t ) s(t )  Ac 1  K a  Am1 cos(2f m1t )  Am 2 cos(2f m 2t ) cos(2f c t )

s(t )  Ac 1  ma1 cos(2fm1t )  ma 2 cos(2fm2t ) cos(2fc t )

s(t )  Ac cos(2fc t )  ma1 Ac cos(2fc t ).cos(2f m1t )  ma 2 Ac cos(2f c t ).cos(2f m 2t ) 2cos A cos B  cos( A  B)  cos( A  B) ma1 Ac m A cos 2( f c  f m1 )t  a1 c cos 2( f c  f m1 )t 2 2 ma 2 Ac ma 2 Ac  cos 2( f c  f m 2 )t  cos 2( f c  f m 2 )t 2 2

s (t )  Ac cos(2f c t ) 

When there are two modulating frequencies, we get, two upper sidebands (USB) fc  f m1 and f c  f m 2 and two lower sidebands (LSB) fc  f m1 and f c  f m 2 . Am Amplitude of first sideband  c a1 2 Am Amplitude of second sideband  c a 2 2 Single sided spectrum :

Total transmitted power : The total power in the amplitude modulated wave is calculated as follows :

PT  PC  PUSB1  PLSB1  PUSB 2  PLSB 2 2

2

2

A2 1  m A  1  m A  1  m A  1  m A  PT  c   a1 c    a1 c    a 2 c    a 2 c  2 2 2  2 2  2 2  2 2  Ac2 ma21 Ac2 ma22 Ac Ac2  ma21 ma22      1   2 4 4 2  2 2   m2  A2 2 2 2 PT  PC 1  at  where PC  c and mat  ma1  ma 2 2 2  

PT 



2

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Net modulation index : mat2  ma21  ma22 2 In general, net modulation index is given by mat  ma21  ma22  ma23  ........  man

Question 14 Why AM is called a linear modulation? Ans.

In a linear modulation scheme the frequency content of the modulating signal is translated to occupy a different portion of the frequency spectrum. In AM, low frequency of the modulating signal is translated in accordance with high frequency of carrier signal but no new components are generated. Hence, AM is called a linear modulation.

Question 15 Why AM is called wasteful of power & bandwidth? Or Explain the disadvantages or limitation of AM wave (DSB-FC). Ans.

Disadvantages of AM (DSB-FC) wave : 1. Power is wasted in the transmitted signal : Most of the transmitted power is in the carrier, which does not carry any information. The information is contained in the two sidebands only. But the sidebands are images of each other and hence both of them contain the same information. For 100% modulation i.e. ma  1, only 33.33% (1 / 3rd) of the total power will be in sideband and 66.67% (2 /3rd) of the total power will be in the carrier, which does not contain any information. The total power transmitted by an AM wave is given by

PT  PC  PUSB  PLSB  PC 

ma2 m2 PC  a PC 4 4

In this equation, carrier component does not contain any information and one sideband is redundant because both sidebands carry the same informaton. So out of the total power, PT , the wastage power is given by ma2 PC 4 The DSB-FC system is bandwidth inefficient system : The transmitted signal requires twice the bandwidth of the message signal ( f m ) i.e. BW  2 f m . This is PC 

2.

3.

due to the transmission of both the sideband, out of which only one sideband is sufficient to convey all the information. Thus, the bandwidth of DSB-FC is double than actually required. AM wave gets effected due to noise : When the AM wave travels from the transmitter to receiver over a communication channel, noise gets added to it. The noise will change the amplitude of the envelope of AM in a random manner. As the information is contained in the amplitude variations of the AM wave, the noise will contaminate the information contents in the AM. Hence the performance of AM is very poor in presence of noise.

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Amplitude Modulation

Question 16 Write the advantages, disadvantages and applications of AM (DSB-FC). Sol.

Advantages of AM : 1. AM transmitters are less complex. 2. AM receivers are simple, detection is easy. 3. AM receivers are cost efficient. 4. AM waves can travel a longer bandwidth. Disadvantages of AM : 1. Power is wasted in the transmitted signal. 2. AM needs large bandwidth. 3. AM wave gets affected due to noise. Applications of AM : 1. Radio broadcasting. 2. Picture transmission in a TV system.

Double Sideband Suppressed Carrier Question 17 What is DSB-SC modulation ? What are the advantages and limitations of DSB-SC as compared to standard AM ? Ans.

From the AM wave or DSB-FC wave, if only the carrier is suppressed then the remaining signal is called as DSB-SC (Double sideband suppressed carrier) signal. Thus, a DSB-SC signal only has upper and lower sidebands. Advantages of DSB-SC : 1. Carrier wave is suppressed. 2. Power saving due to suppression of carrier. 3. The modulation system is simple. 4. Efficiency is more than AM. 5. Linear modulation type is required. 6. It can be used for point to point communication. Disadvantages of DSB-SC : 1. Design of receiver is complex. 2. Bandwidth required is same as that of AM. Application of DSB-SC : 1. Analogue TV systems to transmit color information

Question 18 Give time domain representation and frequency domain representation of DSB-SC amplitude modulation. Ans.

From the AM wave or DSB-FC wave, if only the carrier is suppressed then the remaining signal is called as DSB-SC (Double sideband suppressed carrier) signal. Thus, a DSB-SC signal only has upper and lower sidebands.

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Time domain representation of DSB-SC : Instantaneous value of modulating signal, m(t )  Am cos(2f mt ) Instantaneous value of carrier signal, c(t )  Ac cos(2f c t ) Where Am and Ac are maximum amplitudes of modulating and carrier signal respectively. f m and f c are frequency of modulating and carrier signal respectively. Time domain equation of DSB-SC wave is given by, s(t )  Ac Ka m(t ) cos(2f c t )

s(t )  Ac Ka Am cos(2f mt )cos(2f c t )  ma Ac cos(2f mt )cos(2f ct ) Where K a is a constant called the amplitude sensitivity of the modulator and ma is the modulation index.

DSB-SC is an overmodulated wave. Carrier phase reversals in the DSB-SC signal occur at the time instants where the modulating signal crosses zero. Frequency domain representation of DSB-SC : The modulated carrier has new signals at different frequencies, called side frequencies or sidebands, occur in the frequency spectrum directly above and below the carrier frequency. s(t )  Ac Ka Am cos(2f mt )cos(2f c t )  ma Ac cos(2f mt )cos(2f ct ) ma Ac m A cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 Upper side band frequency, fUSB  f c  f m s(t ) 

Lower sideband frequency, f LSB  fc  f m Bandwidth, BW  fUSB  f LSB  ( f c  f m )  ( f c  f m )  2 f m

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Amplitude Modulation

The bandwidth required for the DSB-SC is twice the frequency of the modulating signal which is same as that of DSB-FC or AM wave. Question 19 Draw the single sided spectra, double sided spectra and phasor diagram for DSBSC. Ans. Frequency domain representation of DSB-SC : The modulated carrier has new signals at different frequencies, called side frequencies or sidebands, occur in the frequency spectrum directly above and below the carrier frequency. s(t )  Ac Ka Am cos(2f mt )cos(2f c t )  ma Ac cos(2f mt )cos(2f ct ) ma Ac m A cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 Upper side band frequency, fUSB  f c  f m s(t ) 

Lower sideband frequency, f LSB  fc  f m Bandwidth, BW  fUSB  f LSB  ( f c  f m )  ( f c  f m )  2 f m The frequency spectrum of DSB-SC waveform has (1) No component of carrier frequency, f c i.e. carrier component is suppressed. (2) An upper side band (USB) whose highest frequency component is at f c  f m . (3) A lower side band (LSB) whose highest frequency component is at fc  f m . (4) The bandwidth of the modulated waveform is twice the modulating signal bandwidth. The bandwidth of DSB-SC signal is same as that of DSB-FC signal i.e. BW  2 f m . Hence, DSB-SC signal is equally bad when it comes to the bandwidth requirement concern.

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Single sided spectrum of DSB-SC :

Double sided spectrum of DSB-SC :

Phasor diagram of DSB-SC :

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Amplitude Modulation

The waveform of DSB-SC can be interpreted as superposition of two waveforms m A (1) Upper sideband a c cos 2( f c  f m )t represented by AB 2 ma Ac (2) Lower sideband cos 2( f c  f m )t represented by AC 2 Resultant waveform of AM signal is represented by AR Question 20 Derive the equation for percentage power saving in DSB-SC. Ans. From the AM wave or DSB-FC wave if only the carrier is suppressed then the remaining signal is called as DSB-SC (Double sideband suppressed carrier) signal. The spectrum of DSB-SC signal contains only two sidebands but the carrier is not present. Hence some power saving does take place.  m2  Total transmitted power in an AM (DSB-FC) wave is, PT  PC 1  a  2   m A m A s AM (t )  Ac cos(2f c t )  a c cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 2 A Suppressed carrier power  PC  c 2 P 2  100% Percentage power saving in DSB-SC  C  100%  PT 2  ma2

For 100% modulation ma  1 , Percentage power saving in DSB-SC 

2 100%  66.67 % 2 1

Single Sideband Suppressed Carrier Question 21 What is SSB-SC modulation? What are the advantages and limitations of SSB-SC as compared to standard AM? Ans.

From the AM wave or DSB-FC wave, if the carrier and one of the sidebands are suppressed then the remaining signal is called as SSB-SC (Single sideband suppressed carrier) signal. Thus, a SSB-SC signal only has either an upper sideband or a lower sideband. Advantages of SSB-SC :

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Carrier wave and one sideband are suppressed. Power saving due to suppression of carrier and one sideband. SSB-SC signal is a bandwidth efficient system. Less bandwidth requirement, bandwidth  f m , which is half of that required by DSB-FC and DSB-SC signals.

Carrier component does not contain any information. The information is contained in the sidebands only. But the sidebands are images of each other and hence both of them contain the same information. Thus only one sideband is necessary for transmission of information and if the carrier and one sideband are suppressed at the transmitter, no information is lost. Therefore, channel requires the same bandwidth as the message signal. 4. Reduced interference of noise due to reduced bandwidth. 5. Fading does not occur in SSB transmission. (Fading means that a signal alternately increases and decreases in strength as it is picked up by the receiver. It occur because the carrier and sideband may reach the receiver shifting in time and phase wrt each other). Disadvantages of SSB - SC : 1. 2.

The generation and reception of SSB signal is a complex process. Since carrier is absent, the SSB transmitter and receiver need to have an excellent frequency stability. 3. The SSB modulation is expensive and complex to implement. Applications of SSB - SC : 1.

SSB transmission is used in the application where the power saving is required.

2. SSB is also used in application in which bandwidth requirements are low. Ex. Point to point communication, mobile communications, TV, telemetry, military communications, and radio navigation. Question 22 Give time domain representation of SSB-SC amplitude modulation. Ans.

From the AM wave or DSB-FC wave if the carrier and one of the sidebands are suppressed then the remaining signal is called as SSB-SC (Single sideband suppressed carrier) signal. Thus, a SSB-SC signal only has either an upper sideband or a lower sideband. In DSB-FC and DSB-SC, half the transmission bandwidth is occupied by the upper sideband of the modulated wave, whereas the other half of the transmission bandwidth is occupied by the lower sideband of the modulated wave. The upper and lower sidebands are uniquely related to each other by virtue of their symmetry, the carrier frequency ‘ f c ’. Carrier component does not contain any information. The information is contained in the sidebands only. But the sidebands are images of each other and hence both of them contain the same information. Thus only one sideband is necessary for transmission of information and if the carrier and one sideband are suppressed at the transmitter, no information is lost. Therefore, channel requires the same bandwidth as the message signal.

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Amplitude Modulation

Time domain representation of SSB-SC : m A m A DSB-FC signal, s(t )  Ac cos(2f c t )  a c cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 ma Ac Upper sideband SSB signal, SSBUSB  cos 2( f c  f m )t 2 m A Lower sideband SSB signal, SSBLSB  a c cos 2( f c  f m )t 2 The modulated carrier has new signal either at upper sideband frequency or at lower sideband frequency. Bandwidth of SSB, BW  f m The bandwidth required for the SSB-SC is equal to the frequency of the modulating signal which is half of that required by DSB-FC and DSB-SC signals. Thus, SSB-SC signal is a bandwidth efficient system. Question 23 Draw the single sided spectra, double sided spectra and phasor diagram for SSBSC. m A Ans. Upper sideband SSB signal, SSBUSB  a c cos 2( f c  f m )t 2 ma Ac Lower sideband SSB signal, SSBLSB  cos 2( f c  f m )t 2 Bandwidth of SSB, BW  f m The frequency spectrum of SSB-SC waveform has (1) No component of carrier frequency, f c i.e. carrier component is suppressed. (2) Only an upper side band (USB) whose highest frequency component is at f c  f m when lower sideband is suppressed. (3) Only a lower side band (LSB) whose highest frequency component is at fc  f m when upper sideband is suppressed. (4) The bandwidth of the modulated waveform is equal to the modulating signal bandwidth. SSB-SC signal is a bandwidth efficient system.

Analog Communication

Single sided spectrum of SSB-SC :

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Amplitude Modulation

Double sided spectrum of SSB-SC :

Phasor diagram of SSB-SC :

Question 24 Derive the equation for percentage power saving in SSB-SC. Ans.

From the AM wave or DSB-FC wave if the carrier and one of the sidebands are suppressed then the remaining signal is called as SSB-SC (Single sideband suppressed carrier) signal. Thus, a SSB-SC signal only has either an upper sideband or a lower sideband. The spectrum of SSB-SC signal contains only one sideband but the carrier and one sideband is not present. Hence some power saving does take place.  m2  Total transmitted power in an AM (DSB-FC) wave is, PT  PC 1  a  2   m A m A s AM (t )  Ac cos(2f c t )  a c cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2

Analog Communication

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Suppressed carrier power  PC 

Ac2 2

ma2 Ac2 ma2 PC  8 4 P  PSSB 4  ma2 Percentage power saving in SSB-SC  C  100%   100% PT 2  2  ma2 

Suppressed single sideband power 

For 100% modulation, ma  1 Percentage power saving in SSB-SC 

4 1 100%  83.33% 2(2  1)

Numericals Question 25 A 400 V carrier is modulated to a depth of 75%. Calculate the total power in the modulated wave in the following forms of AM. (1) DSB-SC (2) AM Sol. Carrier voltage, Ac  400 V Modulation index, ma  0.75 (1) DSB-SC Time domain equation of DSB-SC wave is given by, mA mA s(t )  a c cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 2

2

1m A  1m A  m A  Total power in DSB-SC, PT   a c    a c    a c  2 2  2 2   2 

2

 0.75  400   PT     22.5 kW 2   (2) AM or DSB-FC Time domain equation of AM wave is given by, m A m A s(t )  Ac cos(2f c t )  a c cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 2

2

Total power in DSB-SC, PT 



PT 

Ans.

2

Ac2 1  ma Ac  1  ma Ac  Ac2 ma2 Ac2      2 2  2  2  2  2 4

Ac2  ma2  4002  0.752  1    1    102.5 kW 2  2  2  2 

Ans.

Question 26 Determine the saving in signal power in the case of a 50% modulation AM signal, if the carrier is suppressed before transmission. Sol. From the AM wave or DSB-FC wave if only the carrier is suppressed then the remaining signal is called as DSB-SC (Double sideband suppressed carrier) signal. Modulation index, ma  0.5

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 m2  With carrier, total transmitted power is Ptotal  PC 1  a   PC 2  

Amplitude Modulation  0.52  1    1.125PC 2  

PC ma2 0.52 PC   0.125 PC 2 2  1.125PC  0.125PC  PC Watts

Without carrier, transmitted power is Ptransmitted  Saving in power  Ptotal  Ptransmitted Percentage of power saving  % Power saving 

P  Ptransmitted Saving in power  total 100% Total transmitted power Ptotal

PC 100%  88.9% 1.125PC

Ans.

or  m2  Total transmitted power in an AM (DSB-FC) wave is, PT  PC 1  a  2   m A m A s AM (t )  Ac cos(2f c t )  a c cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 2 A Suppressed carrier power  PC  c 2 P 2 100% Percentage power saving in DSB-SC  C 100%  PT 2  ma2

% Power saving 

2  100 %  88.9 % 2  0.52

Ans.

Question 27 Calculate the percentage saving when the carrier and one of the sidebands are suppressed in an AM modulated wave to depth of 80%. Sol. From the AM wave or DSB-FC wave if the carrier and one of the sidebands are suppressed then the remaining signal is called as SSB-SC (Single sideband suppressed carrier) signal.  m2  Total transmitted power in an AM (DSB-FC) wave is, PT  PC 1  a  2   m A m A s AM (t )  Ac cos(2f c t )  a c cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 Ac2 Suppressed carrier power  PC  2 m2 A2 m2 P Suppressed single sideband power  a c  a C 8 4 P  PSSB 4  ma2 Percentage power saving in SSB-SC  C  100 %   100 % PT 2  2  ma2 

For 80% modulation, ma  0.8 % Power saving 

4  0.82

2  2  0.82 

 100 %  87.87 %

Ans.

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Question 28 The output current of 60 percent modulated AM generator is 1.5 A. If another sine wave corresponding to 70% modulation is transmitted simultaneously. Calculate the percentage power saving if only carrier is suppressed. Ans. Modulation index, ma1  0.6 and ma 2  0.7 Modulation index for multitone modulation, mat  ma21  ma22  0.62  0.72  0.92 From the AM wave or DSB-FC wave if only the carrier is suppressed then the remaining signal is called as DSB-SC (Double sideband suppressed carrier) signal.  m2  Total transmitted power in an AM (DSB-FC) wave is, PT  PC 1  a  2   m A m A s AM (t )  Ac cos(2f c t )  a c cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 Ac2 Suppressed carrier power  PC  2 P 2 100% Percentage power saving in DSB-SC  C 100%  PT 2  ma2

% Power saving 

2  100 %  70.26 % 2  0.922

Ans.

Question 29 Determine  and the percentage of the total power carried by the sidebands of the AM wave for single tone modulation when (a) ma  0.5 and (b) ma  0.3 Sol.

Transmission efficiency is defined as the ratio of the power carried by the sidebands to the total transmitted power. ma2  100 % Transmission efficiency of an AM wave,   2  ma2 For ma  0.5 ,  

(0.5) 2  100 %  11.11% 2  (0.5) 2

Ans.

Hence, total power carried by the sidebands is 11.11% of the total transmitted power. Ans. For ma  0.3 ,  

(0.3) 2  100 %  4.3% 2  (0.3) 2

Ans.

Hence, total power carried by the sidebands is 4.3% of the total transmitted power. Question 30 An AM broadcast station operates at its maximum allowed total output of 50 kW and with 95% modulation? Estimate the power in the intelligence part of the transmitted signal? Sol. Transmitted power, PT  50 kW Modulation index, ma  0.95  m2  PT  PC 1  a  2  



 0.952  50 103  PC 1   2  

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Amplitude Modulation

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3  Carrier power, PC  34.5 10 W

Power in sidebands  PT  PC  (50  34.5) kW Power in the intelligence part = Power in sidebands  15.5 kW

Ans.

Question 31 The total power content of an AM signal is 1000 W. Determine the power being transmitted at the carrier frequency and at each of the sidebands when the percent modulation is 100%. Sol.

Transmitted power, PT  1000 W

Modulation index, ma  1

 m   PT  PC 1   2    Carrier power, PC  666.67 W

 12  1000  PC 1    2

2 a

Ans.

Power in sidebands  PT  PC  1000  666.67  333.33 W

PUSB  PLSB  2PUSB  2PLSB  333.33 W Power in each sideband, PUSB  PLSB  166.66 W

Ans.

Or P m2  C a 4

Each sideband power, PUSB  PLSB



PUSB  PLSB  666.67 

12  166.66 W 4

Question 32 Determine the power content of the carrier and each of the sidebands for an AM signal having a percent modulation of 80% and a total power of 2500 W. Sol.

Transmitted power, PT  2500 W

Modulation index, ma  0.8

 m   PT  PC 1   2    Carrier power, PC  1893.94 W

 0.82  2500  PC 1    1.32 PC 2  

2 a

Each sideband power, PUSB  PLSB 



PUSB  PLSB  1893.94 

Ans. PC ma2 4

0.82  303.03 W 4

Ans.

Question 33 The total power of transmitter is 10 kW. Determine the power content of the carrier and each of the sidebands if the percentage modulation is 80%. Sol.

Modulation index, ma  0.8

Transmitted power, PT  10 kW  m  PT  PC 1   2   2 a



 Carrier power, PC  7.575 kW

 0.82  10  PC 1    1.32 PC 2  

Ans.

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Each sideband power, PUSB  PLSB 



PUSB  PLSB  7.575 

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PC ma2 4

0.82  1.212 kW 4

Ans.

Question 34 The power content of the carrier of an AM wave is 5 kW. Determine the power content of each of the sidebands and the total power transmitted when the carrier is modulated upto 75%. Sol. Carrier power, PC  5 kW Modulation index, ma  0.75  m2  PT  PC 1  a  2  

 0.752  PT  5 1   2  



 Transmitted power, PT  6.4 kW Each sideband power, PUSB  PLSB 



PUSB  PLSB  5 

Ans. 2 a

PC m 4

0.752  703.125 W 4

Ans.

Question 35 An amplitude modulated wave has a power content of 800 W at its carrier frequency. Determine the power content of each of the sidebands for a 90% modulation. Sol. Carrier power, PC  800 W Modulation index, ma  0.9 Each sideband power, PUSB  PLSB 



PUSB  PLSB  800 

PC ma2 4

0.92  162 W 4

Ans.

Question 36 Determine the percent modulation of an amplitude modulated wave which has a power content at the carrier of 8 kW and 2 kW in each of its sidebands when the carrier is modulated by a simple audio tone. Sol. Carrier power, PC  8 kW Each sideband power, PUSB  PLSB  2 kW PC ma2  4  Modulation index, ma  100% PUSB  PLSB 

2

8  ma2 4

Ans.

Question 37 The total power content of an AM wave is 600 W. Determine the percent modulation of the signal if each of the sideband contains 75 W. Sol. Transmitted power, PT  600 W Each sideband power, PUSB  PLSB  75 W

PT  PC  PUSB  PLSB



600  PC  75  75

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Amplitude Modulation

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Carrier power, PC  450 W PUSB  PLSB 

PC ma2 4



75 

450  ma2 4

 Modulation index, ma  81.65%

Ans.

Question 38 An AM transmitter radiates 9 K watts of power when the carrier is unmodulated and 10.125 K watts when the carrier is sinusoidally modulated. Find the modulation index as percentage of modulation. Now, if another sine wave, corresponding to 40% modulation is transmitted simultaneously, then calculate the total radiated power. Sol.

Unmodulated carrier power, PC  9 kW Modulated total power, PT  10.125 kW  m2  PT  PC 1  a  2  



 m2  10.125  9 1  a  2  

ma2  0.25



ma  0.5

Modulation index, ma  50%

Ans.

With simultaneous 40% modulation, mat  ma21  ma22  0.52  0.42  0.64 The power content of the carrier of an AM signal remains the same regardless of multitone modulation. Total radiated power at 64% modulation,  m2  PT  PC 1  at  2  



 0.642  PT  9 1   2  



PT  10.84 kW

Ans.

Question 39 The percent modulation of an AM wave changes from 40% to 60%. Originally, the power content at the carrier frequency was 900 W. Determine the power content at the carrier frequency and within each of the sidebands after the percent modulation has risen to 60%. Sol.

Modulation index, ma1  0.4 and ma 2  0.6 Carrier power at 40% modulation, PC  900 W In standard AM transmission, carrier power remains the same, regardless of percent modulation. Carrier power at 60% modulation, PC  900 W Each sideband power, PUSB  PLSB 



PUSB  PLSB  900 

0.62  81 W 4

Ans.

PC ma2 4

Ans.

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Question 40 An AM signal in which carrier is modulated 70% contains 1500 W at the carrier frequency. Determine the power contents of upper and lower sidebands for this percent modulation. Calculate the power at the carrier and the power contents of each of the sidebands when the percent modulation drops to 50%. Sol.

Carrier power, PC  1500 W

Modulation index, ma  0.7 2 a

PC m 4

Each sideband power, PUSB  PLSB 



PUSB  PLSB  1500 

0.7 2  183.75 W 4

Ans.

New modulation index, ma  0.5 In standard AM transmission, carrier power remains the same, regardless of percent modulation. Ans.  Carrier power, PC  1500 W PC ma2 4

Each sideband power, PUSB  PLSB 



PUSB  PLSB  1500 

0.52  93.75 W 4

Ans.

Question 41 An amplitude modulated amplifier has a radio frequency output of 50 W at 100% modulation. The internal loss in the modulator is 10 W. (A) What is the unmodulated carrier power? (B) What power output is required from the modulator (baseband signal)? Sol.

Output power, Pout  50 W

Power loss, Ploss  10 W

Input power, Pin  Pout  Ploss  50  10  60 W Modulation index, ma  1 (A) Unmodulated carrier power PC  m2  Pin  PC 1  a  2  



 1 3 60  PC 1    PC  2 2



PC  40 W

Ans.

(B) Power required from the modulator (baseband signal) is obtained by subtracting unmodulated power from the total input power  Pm  Pin  PC  60  40  20 W Ans. Question 42 An audio signal 20sin 500t 80sin2  105 t Calculate : (1) Modulation index

is used to amplitude modulate a carrier of

(3) Amplitude of each sideband frequency

(2) Sideband frequencies (4) Bandwidth required

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Sol.

Amplitude Modulation

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Modulating signal, m(t )  20sin 500t  Am sin(2f mt )

Am  20 V

f m  250 Hz

Carrier signal, c(t )  80sin 2 105 t  Ac sin(2f c t )

Ac  80 V (1) Modulation index, ma 

f c  105 Hz  100 kHz Am 20   0.25 Ac 80

Ans.

(2) Sideband frequencies Upper sideband frequency, fUSB  fc  f m  100 kHz  250 Hz  100.25 kHz Lower sideband frequency, f LSB  fc  f m  100 kHz  250 Hz  99.75 kHz (3) Amplitude of each sideband frequency 

Ans. Ans.

ma Ac 0.25  80   10 V 2 2

Or AM signal, sAM (t )  Ac 1  ma sin(2f mt ) sin(2fc t )

sAM (t )  801  0.25sin500t  sin 2 105 t s AM (t )  80sin 2 105 t  0.25  80sin 500t.sin 2 105 t

2sin A sin B  cos( A  B)  cos( A  B) s AM (t )  80sin 2 105 t  10 cos 2(105  250)t  10 cos 2(105  250)t Amplitude of each sideband frequency is 10 V. (4) Bandwidth required, BW  fUSB  f LSB  2 f m  2  250  500 Hz

Ans. Ans.

Question 43 An audio frequency signal 5sin2(1000)t is used to amplitude modulate a carrier of

100sin 2(106 )t . Assume modulation index of 0.4. Find :

Sol.

(1) Sideband frequencies (2) Bandwidth required (3) Amplitude of each sideband (4) Total power delivered to a load of 100 Ω Modulating signal, m(t )  5sin 2(1000)t  Am sin(2f mt )

Am  5 V

f m  1 kHz

Carrier signal, c(t )  100sin 2(106 )t  Ac sin(2f c t )

Ac  100 V

f c  1 MHz

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Modulation index, ma  0.4 (1) Sideband frequencies Upper sideband frequency, fUSB  fc  f m  1 MHz  1 kHz  1.001 MHz

Ans.

Lower sideband frequency, f LSB  fc  f m  1 MHz  1 kHz  0.999 MHz

Ans.

(2) Bandwidth, BW  fUSB  f LSB  2 f m  2 1 kHz  2 kHz (3) Amplitude of each sideband frequencies 

Ans.

ma Ac 0.4 100   20 V 2 2

 Amplitude of upper and lower sideband is 20 V.

Ans.

(4) Total power delivered to a load of 100   m2  A2  m2  (100)2  (0.4)2  PT  PC 1  a   c 1  a   1 +  2  2R  2  2 × 100  2  



PT  54 W

Ans.

Question 44 The output voltage of a transmitter is given by v  500 1  0.4sin(3140t ) sin  6.28  107 t  . This voltage is fed to a load of 600  resistance. Determine (a) f c (b) f m (c ) PC (d) PT (e) peak output power. Sol.

Load resistance, RL  600  Time domain equation of amplitude modulated (AM) wave is given by, s(t )  Ac 1  ma sin(2f mt ) sin(2fc t ) Modulation index, ma  0.4

6.28 107  10 MHz 2 3140  500 Hz (b) Modulating frequency, f m  2 (c) Carrier amplitude, Ac  500 V (a) Carrier frequency, fc 

Carrier power for resistive load, Pc 



Pc 

5002  208.33 W 2  600

Ans. Ans.

Ac2 2R

Ans.

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 m2  (d) Total transmitted power for resistive load, PT  PC 1  a  2  



 0.42  PT  208.33 1    225 W 2  

Ans.

(e) Peak voltage, VP  Vc (1  ma )  500(1  0.4)  700 V Peak power for resistive load, Ppeak 

VP2 7002   408.3 W 2R 2  600

Ans.

Question 45 A given AM broadcast station transmits a total power of 50 kW when the carrier is modulated by a sinusoidal signal with a modulation index of 0.707. Calculate : (i) The carrier power (ii) The transmission efficiency (iii) The peak amplitude of the carrier assuming the antenna to be represented by a (50  j0)  load. Sol.

Modulation index, ma  0.707

Transmitted power, PT  50 kW

(i) The total power transmitted by the AM broadcast station is given by

 m2  PT  Pc 1  a  2  



 0.707 2  50  PC 1    1.25PC 2  

 Carrier power, PC  40 kW

Ans.

(ii) The transmission efficiency is given by

% 

ma2  100% 2  ma2



% 

0.7072 100% 2  0.7072

 Transmission efficiency, %   20%

Ans.

(iii) The carrier power with resistive load RL  50  is given by

PC 

Ac2 A2  c 2 R 2  50



Ac2  2  50  40000

 Peak carrier amplitude, Ac  2 kV

Ans.

Question 46 The AM signal is given by :

sAM (t )  51  0.5cos(1000t )  0.5cos(2000t ) cos(10000t )

Find the modulation index, total power, the sideband power and power efficiency assuming unit ohm resistive load. Sol.

Time domain equation of multitone AM signal is given by,

sAM (t )  Ac 1  ma1 cos(2f m1t )  ma 2 cos(2f m2t ) cos(2fct )

sAM (t )  51  0.5cos(1000t )  0.5cos(2000t ) cos(10000t ) Modulation index, ma1  0.5 and ma 2  0.5 Net modulation index, mat  ma21  ma22  0.52  0.52  0.707

Ans.

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Carrier amplitude, Ac  5 V  m2  A2  m2  Total modulated power, PT  PC 1  at   c 1  at  2  2  2  



PT 

52  0.707 2  1    15.625 W 2  2 

Ans.

Ac2 52   12.5 W 2 2 Sideband power  PT  PC  15.625  12.5  3.125 W

Carrier power, PC 

Power efficiency, %  



% 

Ans.

2 at

m  100 % 2  mat2

0.7072 100%  20% 2  0.7072

Ans.

Question 47 A carrier signal Vc ( t )  10cos 4  106 t is modulated by a message signal having three

Sol.

frequencies 5 kHz, 15 kHz and 125 kHz. The corresponding modulation indices are 0.3, 0.5 and 0.7. Determine the bandwidth, carrier and sideband powers. Carrier signal, Vc (t )  10cos 4106 t  Ac cos(2f c t ) Time domain equation of multitone AM wave is given by, S AM (t )  Ac 1  ma1 cos(2f m1t )  ma 2 cos(2f m2t )  ma3 cos(2f m3t ) cos(2fct ) Modulating signal frequencies are f m1  5 kHz , f m 2  15 kHz and f m3  125 kHz Modulation indices are ma1  0.3 , ma 2  0.5 and ma 3  0.7 Bandwidth is the twice of maximum modulating signal frequency.  Bandwidth  2 125  250 kHz 2 c

A 10   50 W 2 2 S AM (t )  Ac 1  ma1 cos(2f m1t )  ma 2 cos(2f m2t )  ma3 cos(2f m3t ) cos(2fct )

Carrier power, PC 

Ans.

2

S AM (t )  Ac cos(2f c t )  ma1 Ac cos(2f m1t ) cos(2f c t )  ma 2 Ac cos(2f m 2t ) cos(2f c t )  ma 3 Ac cos(2f m3t ) cos(2f c t )

2cos A cos B  cos( A  B)  cos( A  B) ma1 Ac m A cos 2( f c  f m1 )t  a1 c cos 2( f c  f m1 )t  2 2 ma 2 Ac ma 2 Ac cos 2( f c  f m 2 )t  cos 2( f c  f m 2 )t  2 2 ma 3 Ac m A cos 2( f c  f m3 )t  a 3 c cos 2( f c  f m3 )t 2 2 Total power, PT  PC  PUSB1  PLSB1  PUSB 2  PLSB 2  PUSB3  PLSB3 S AM (t )  Ac cos(2f c t ) 

2

m 2 A2 1m A  First sideband power, PUSB1  PLSB1   a1 c   a1 c 2 2  8

Ans.

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

PUSB1  PLSB1 

Amplitude Modulation

1 - 35

0.32 102  1.125 W 8

Ans. 2

Second sideband power, PUSB 2  PLSB 2



PUSB1  PLSB1 

m2 A2 1m A    a2 c   a2 c 2 2  8

0.52 102  3.125 W 8

Ans. 2

m 2 A2 1m A  Third sideband power, PUSB 3  PLSB 3   a 3 c   a 3 c 2 2  8



PUSB1  PLSB1 

0.72 102  6.125 W 8

Ans.

Question 48 An AM waveform has the form X c (t )  101  0.5cos(2000t )  0.5cos(4000t ) cos(20000t )

Sol.

(a) Sketch the amplitude spectrum of AM signal (b) Find the fraction power content of sidebands. For a multitone modulated signal, s(t )  AC 1  ma1 cos(2f m1t )  ma 2 cos(2f m2t ) cos(2fct )

s(t )  Ac cos(2fc t )  ma1 Ac cos(2fc t ).cos(2f m1t )  ma 2 Ac cos(2f c t ).cos(2f m 2t ) 2cos A cos B  cos( A  B)  cos( A  B) ma1 Ac m A cos 2( f c  f m1 )t  a1 c cos 2( f c  f m1 )t 2 2 ma 2 Ac ma 2 Ac  cos 2( f c  f m 2 )t  cos 2( f c  f m 2 )t 2 2

s (t )  Ac cos(2f c t ) 

X c (t )  101  0.5cos(2000t )  0.5cos(4000t ) cos(20000t ) 20000  10 kHz 2 2000  1 kHz Modulating Frequency, f m1  2 4000  2 kHz Modulating Frequency, f m 2  2 Modulation index, ma1  0.5 and ma 2  0.5

Carrier Frequency, f c 

Upper sideband frequencies, USB1  fc  f m1  11 kHz

USB2  fc  f m2  12 kHz

Lower sideband frequencies, LSB1  fc  f m1  9 kHz

LSB2  fc  f m 2  8 kHz

In single sided spectrum, carrier amplitude, Ac  10 V Ac ma1 10  0.5   2.5 V 2 2 Am 10  0.5 Amplitude of second sideband  c a 2   2.5 V 2 2

Amplitude of first sideband 

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(a) Single sided spectrum

2 2 2 2 (b) Net modulation index, mat  ma1  ma 2  0.5  0.5  0.707

 m2   0.7072  Total transmitted power, PT  PC 1  at   PC 1    1.25PC 2  2   

PC mat2 0.707 2 PC   0.25PC 2 2 P 0.25PC 100  20% Fraction sideband power, SB 100  PT 1.25PC

Total sideband power, PSB  PT  PC 

Ans.  Sideband power is 20% of total transmitted power. Question 49 An AM signal is represented by s(t )  0.11  0.1cos 2512t  0.5cos6280t  sin(107 t  450 ) Sketch the amplitude spectrum Sol.

of the signal. For a multitone modulated signal,

s(t )  AC 1  ma1 cos(2f m1t )  ma 2 cos(2f m2t ) sin(2fct  )

Carrier signal may be sine or cosine and it may have a phase shift but phase shift does not affect amplitude modulated wave.

107  1590 kHz 2 2512  400 Hz Modulating Frequency, f m1  2 6280  1000 Hz Modulating Frequency, f m 2  2 Modulation index, ma1  0.1 and ma 2  0.5 Carrier Frequency, f c 

Upper sideband frequencies, USB1  fc  f m1  1590.4 kHz

USB2  fc  f m2  1591 kHz

Lower sideband frequencies, LSB1  fc  f m1  1589.6 kHz

LSB2  fc  f m2  1589 kHz

In single sided spectrum, carrier amplitude, Ac  0.1 V Amplitude of first sideband 

Ac ma1 0.1 0.1   0.005 V 2 2

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Amplitude of second sideband 

Ac ma 2 0.1 0.5   0.025 V 2 2

Question 50 A conventional single tone AM signal is given by the expression sAM (t )  5cos 450t  20cos 550t  5cos 650t

Obtain the carrier amplitude, modulating signal frequency, carrier frequency, modulation index and peak envelope power. Sol.

sAM (t )  5cos 450t  20cos 550t  5cos 650t

sAM (t )  20cos550t  5cos(550 100)t  cos(550  100)t 

cos( A  B)  cos( A  B)  2cos A cos B sAM (t )  20cos550t  10cos550t.cos100t

sAM (t )  20 1  0.5cos100t  cos550t Time domain equation of AM signal is given by,

sAM (t )  Ac 1  ma cos(2f mt ) cos(2fct )

Carrier amplitude, Ac  20 V

Ans.

Modulating signal frequency, f m  Carrier frequency, f c 

100  50 Hz 2

550  275 Hz 2

Ans. Ans.

Modulation index, ma  0.5

Ans.

Peak envelope amplitude  Ac (1  ma )  20(1  0.5)  30 V Peak envelope power, Ppeak 

302  450 W 2

Ans.

Question 51 Consider the AM signal sAM (t )   Ac  m(t )cos5000t where the modulating signal is given by m(t )  3cos50t  5cos150t . Let the modulation index be 0.8. Find (a) the amplitude of the carrier, Ac Sol.

(c) the transmission efficiency | m(t ) |max (a) Modulation index, ma  Ac

(b) the carrier power

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Since the tone signals are harmonically related, | m(t ) |max  3  5  8

8 Ac

0.8 



Ac  10 V

 Carrier amplitude, Ac  10 V 2 c

Ans.

2

A 10   50 W Ans. 2 2 (c) Transmission efficiency is the ratio of sideband power to the total power expressed as a percentage. P PSB %   SB 100%  100% PT PC  PSB

(b) Carrier power, PC 

sAM (t )  10  3cos50t  5cos150t  cos5000t

sAM (t )  10cos5000t  3cos50t.cos5000t  5cos150t.cos5000t

2cos A cos B  cos( A  B)  cos( A  B) s AM (t )  10cos 5000t  1.5cos 5050t  1.5cos 4950t  2.5cos 5150t  2.5cos 4850t Carrier

Sideband terms

Sideband power, PSB 



% 

2

2

2

1.5 1.5 2.5 2.52     8.5 W 2 2 2 2

8.5  100 %  14.53% 50  8.5

Ans.

Question 52 A scheme for generating a conventional AM signal is shown in below figure. Let m( t )  cos  2.103 t  and c( t )  cos  2.106 t  .

Sol.

(a) Obtain the expression for the modulation index of the AM wave. (b) For a modulation index of 90% and peak envelope power (normalized across 1 ohm) of 100 W, find the values of the amplifier gains A and B. The AM output from the figure is, s(t )  Am(t ).c(t )  Bc(t )  A cos  2.103 t  .cos  2.106 t   B cos  2.106 t 

s(t )   B  A cos  2.103 t  cos  2.106 t 





(a) Envelope of AM signal is E (t )   B  A cos 2.103 t  Emax   B  A Emin   B  A Modulation index, ma 



ma 

A B

Emax  Emin  B  A   B  A  Emax  Emin  B  A   B  A

Ans.

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(b) Modulation index, ma 

A  0.9 B

Peak envelope amplitude,  Emax   B  A

E2  B  A  100 W  max  2 2 2

Peak envelope power, Ppeak

 B  0.9B 

2

A  0.9B



2

 100

A  0.9B  0.9  7.44  6.7 Amplifier gain, A  6.7 and B  7.44

Ans.

Question 53 A single tone sinusoidal wave operating at frequency f m Hz is applied to the input of the modulation in AM broadcasting having Emax  150 mV and Emin  30 mV . However carrier is operating at frequency f c Hz. (i) Work out expression of modulated signal and modulation index. (ii) Total average power of modulated signal. (iii) What is the peak envelope power across the 60  load ? (iv) Modulation efficiency (v) Amplitude and phase of the additional carrier which must be added to the waveform to attain a modulation index of 80%. Sol.

Maximum value of the envelope of the modulated wave, Emax  150 mV Minimum value of the envelope of the modulated wave, Emin  30 mV (i) Modulation index

ma 

Emax  Emin 150  30 120 2    Emax  Emin 150  30 180 3

Modulating signal voltage, Em  Carrier signal voltage, EC 

Ans.

Emax  Emin 150  30   60 mV 2 2

Emax  Emin 150  30   90 mV 2 2

Expression for modulated signal is given by : s(t )  EC 1  ma cos(2f mt ) cos(2fct )



 2  s(t )  90 1  cos(2f mt )  cos(2f c t )  3 

Ans.

(ii) Total average power of modulated signal 3  m2  E 2  m2   90  10  PT  PC 1  a   C 1  a   2 2  2  2  



PT  4.95 mW

2

 (2 / 3) 2  1   2  

Ans.

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(iii) Peak value of AM signal  EC (1  ma ) Peak envelope power, Ppeak 

 90 10  

3 2

Ppeak



(1  2 / 3) 2

2  60 Ppeak  187.57 W

Ans.

(iv) Modulation efficiency, %  



EC2 (1  ma ) 2 2R

%

ma2  100 % 2  ma2

(2 / 3) 2  100 %  18.18% 2  (2 / 3) 2

Ans.

(v) Amplitude and phase of the additional carrier which must be added to the waveform to attain a modulation index of 80%. Additional carrier  AC cos(2f c t ) s(t )  EC 1  ma cos(2f m t )  cos(2f c t )  AC cos(2f c t )   EC  AC  1  ma 'cos(2f m t )  cos(2f c t )

New carrier amplitude,  EC  AC  90  AC Modulation index, ma ' 

0.8 

Vm Em  Vc EC  AC

60 90  AC



0.8  90  AC   60

Additional carrier amplitude, AC  15 mV  151800 mV

Ans.

Question 54 Derive the following relations.

Ans.

1.

Modulation index in terms of PT and PC

2.

Current relation of AM wave

3.

Modulating index in terms of IT and IC

4.

Voltage relation of AM wave

5.

Modulation index in terms of VT and VC

1.

Modulation index in terms of PT and PC

PC : Carrier power

PT : Transmitted power  m  PT  PC 1   2  



m2 PT  1 a PC 2

ma2 PT  1 2 PC



P  ma2  2  T  1 P  C 

2 a



P  ma  2  T  1 P  C 

Ans.

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2.

Current relation of AM wave

I C : Carrier current

IT : Total current

2 Carrier power, PC  I C R

Transmitted power, PT  I R 2 T

 m2  PT  PC 1  a  2  



 m2  IT2 R  I C2 R 1  a  2  

 m2  IT2  I C2 1  a  2  



 m2  IT  I C2 1  a  2  

ma2 2



IT  I C 1 

3.

Modulating index in terms of IT and IC

IT  I C 1 

4.

Ans.

ma2 2

m2  I  1 a   T  2  IC  

Amplitude Modulation

1 - 41



 m2  IT2  I C2 1  a  2  



 I  2  m  2  T   1  I C  

2

2 a

 I  2  ma  2  T   1  I C   Voltage relation of AM wave

AC  VC : Carrier voltage

AT  VT : Total voltage Transmitted power, PT 

Ans.

2 T

V R

Carrier power, PC 

 m2  PT  PC 1  a  2  



VT2 VC2  ma2   1   R R  2 

 m2  VT2  VC2 1  a  2  



 m2  VT  VC2 1  a  2  

 m2   VT  VC 1   a   2  5.

Ans.

Modulation index in terms of VT and VC

VT  VC 1 

ma2 2

m2  V  1 a   T  2  VC  

VC2 R



 m2  VT2  VC2 1  a  2  



 V  2  m  2  T   1  VC  

2

 V  2  ma  2  T   1  VC  

2 a

Ans.

Analog Communication

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Question 55 The antenna current of an AM transmitter is 8 A if any carrier is sent. It increases to 8.93 A, if carrier is modulated by a single sinusoidal wave. Determine the percentage of modulation. Also find the antenna current if percentage of modulation changes to 80%. Sol. Total current, IT 1  8.93 A Carrier current, I C  8 A

IT  I C 1 

ma2 2



8.93  8 1 



ma  0.701

ma2 2

2

ma2  8.93    2  8  Modulation index, ma  70% 1

Ans.

Percentage of modulation increases to 80% i.e. ma  0.8

ma2  2 Total antenna current, IT  9.19 A IT  I C 1 

IT  8 1 

0.82 2 Ans.

Question 56 The antenna current of an AM broadcast transmitter modulated to a depth of 40% by an audio sine wave is 11 A. It increases to 12 A as a result of sinusoidal modulation by another audio sine wave. What is the modulation index due to second wave ? Sol. Modulation index, ma1  0.4 Total current, IT 1  11 A

IT  I C 1 

ma2 2

Carrier current, IC 



11  I C 1 

0.42 2

11

 10.58 A 1  0.8 Total current is increased to 12 A as a result of simultaneous modulation. The current content of the carrier of an AM signal remains the same regardless of multitone modulation. 

IT  I C 1 



mat  ma21  ma22

IT 2  12 A ma2 2

mat  ma21  ma22  0.756

Modulation index, ma 2  0.6415

mat2 2



12  10.58 1 



0.7562  0.42  ma22 Ans.

Question 57 The output current of a 60 percent modulated AM generator is 1.5 A. (i) To what value will this current rise if the generator is modulated additionally by another audio wave, whose modulation index is 0.7 ? (ii) What will be the percentage power saving if the carrier and one of the sidebands are now suppressed ? [CSVTU May 2011]

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Ans.

Amplitude Modulation

1 - 43

(i) Current for multitone modulation Transmitted current at ma1  0.6 , IT  1.5 A

IT  I C 1 

ma21 2



1.5  I C 1 

0.62 2

 Carrier current, IC  1.38 A The current content of the carrier of an AM signal remains the same regardless of multitone modulation. Modulation index, ma1  0.6 and ma 2  0.7 2 2 2 2 Modulation index for multitone modulation, mat  ma1  ma 2  0.6  0.7  0.92

Transmitted current at mat  0.92 , IT  I C 1 

mat2 0.922  1.38 1  2 2

 Current for multitone modulation, IT  1.646 A

Ans.

(ii) Carrier and one of the sidebands are suppressed From the AM wave or DSB-FC wave if the carrier and one of the sidebands are suppressed then the remaining signal is called as SSB-SC (Single sideband suppressed carrier) signal.  m2  Total transmitted power in an AM (DSB-FC) wave is, PT  PC 1  a  2   m A m A s AM (t )  Ac cos(2f c t )  a c cos 2( f c  f m )t  a c cos 2( f c  f m )t 2 2 2 A Suppressed carrier power  PC  c 2 m2 A2 m2 P Suppressed single sideband power  a c  a C 8 4

Percentage power saving in SSB-SC  % Power saving 

4  0.922

2  2  0.922 

PC  PSSB 4  ma2  100 %   100 % PT 2  2  ma2 

 100 %  85.13%

Ans.

Question 58 The AM signal is given by :



















QAM  10cos 2.106 t  5cos 2.106 t cos 2.10 3 t  2cos 2.10 6 t cos 4.10 3 t

Find : (i) Net modulation index (ii) Total modulated power (iii) Lower sideband power (iv) Transmission efficiency for a 50% net modulation (v) rms value of the modulated signal



[CSVTU Dec 2008]

Analog Communication Sol.

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QAM  10 cos  2.106 t   5cos  2.106 t  cos  2.103 t   2 cos  2.106 t  cos  4.103 t 

2cos A cos B  cos( A  B)  cos( A  B) QAM  10cos  2.106 t   2.5cos 2(106  103 )t   2.5cos 2(106  103 )t   cos  2(106  2 103 )t   cos  2(106  2 103 )t  Ac ma1  2.5 2 Am Amplitude of second sideband, c a 2  1 2

Amplitude of first sideband,

QAM  10 1  0.5cos  2.103 t   0.2cos  4.103 t  cos  2.106 t  Time domain equation of multitone AM signal is given by,

sAM (t )  Ac 1  ma1 cos(2 f m1t )  ma 2 cos(2 f m2t ) cos  2 fct 

Carrier amplitude, Ac  10 V Modulation index, ma1  0.5 and ma 2  0.2 2 2 (i) Net modulation index, mat  ma1  ma 2



mat  0.52  0.22  0.538

Ans.

 m2  A2  m2  (ii) Total modulated power, PT  PC 1  at   c 1  at  2  2  2  



PT 

102 2

 0.5382  1    57.25 W 2  

(iii) Lower sideband power, PLSB 



PLSB 

Ans.

Ac2 ma21 Ac2 ma22 Ac2 mat2   8 8 8

102  0.5382  3.618 W 8

(iv) Transmission efficiency, %  

Ans.

ma2  100 % 2  ma2

Modulation index, ma  0.5



% 

0.52 100%  11.11% 2  0.52

(v) Total modulated signal voltage, VT  Vc 1  VT  10 1 

 Vrms 

VT 2

Ans.

mat2 2

0.5382  10.7 V 2

 7.56 V

Ans.

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Amplitude Modulation

Question 59 The Fourier transform of m(t ) is F  m(t )  M ( f ) .

Sol.

1 Show that F m( t )cos 2f c t    M ( f  f c )  M ( f  f c ) 2 Or State modulation theorem. [CSVTU Dec 2007] The signals like m(t ) cos 2f c t represents carrier modulation. Here, m(t ) is the

modulating signal and cos 2f c t is the carrier. The process of the modulation translates the signal up and down by frequency f c . Fourier transform of a signal m(t ) is given by,

M ( f )  F  m(t )  



 m(t ) e

 j 2 ft

dt



From frequency shifting property of Fourier transform F  m(t )e j 2 fc t   M ( f  f c ) F  m(t )e  j 2 fc t   M ( f  f c )   e j 2 fc t  e j 2 fc t    m(t )e j 2 fc t F  m(t ) cos 2f c t   F  m(t )    F  2 2     1  F  m(t ) cos 2f c t    M ( f  f c )  M ( f  f c ) 2 This relation is called modulation theorem. Question 60 A baseband signal f (t )  Sa(t ) is used to amplitude

Sol.

Draw the waveform and spectrum of AM signal. sin t Modulating or baseband signal, f (t )  Sa(t )  t c ( t )  cos 20 t Carrier signal, Modulated AM signal, s(t )  1  f (t ) cos c t Waveform of AM signal :

  m(t )e  j 2 fc t  F  2   

Proved

modulated a carrier cos 20t . [CSVTU May 2012]

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Spectrum of AM signal : Fourier transform of modulating signal, FT  f (t )  FT Sa(t )  t  FT    A rect    A Sa     2   t  FT Cancel A and put   2 , rect    2 Sa() 2 Apply duality property, FT FT X ()  x(t )  2.x(t ) X ()

 t  FT  rect    2 Sa() 2 Rectangular function is an even function.   FT  Sa(t )  .rect   2

   FT 2Sa(t )  2.rect    2 

Fourier transform of carrier signal, FT c(t )  FT cos20(t ) FT cos 0t  ( 0 )  ( 0 )



FT cos 20t  ( 20)  ( 20)

Spectrum of AM signal :

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Amplitude Modulation

Modulation and Demodulation of DSB-FC Question 61 Explain square law modulator for generation of AM wave. Ans. The circuit that generates AM wave is called amplitude modulator. Square law modulator or non linear modulator is used for generation of AM wave or DSB-FC wave. An AM wave can be obtained by combining the modulating signal and the carrier through a non linear device. A non linear device is the device with a non linear relation between its current and voltage. The non linear device could be a diode or a transistor. The transistor has an advantage that it also provides amplification. When the current through the diode is passed through a resistive load, the output voltage v0 (t ) is related to the input voltage vi (t ) by the following polynomial approximation v0 (t )  a1vi (t )  a2 vi2 (t )  a3vi3 (t )..... Where the ai s are constant. If | vi (t ) | is small enough, the above expression can be approximated by the first two terms and we have a square law device (SLD) satisfying the relationship v0 (t )  a1vi (t )  a2 vi2 (t ) The square law modulator circuit consists of the following (1) A carrier signal and a modulating signal (2) A non linear device (3) A bandpass signal

Block diagram of square law modulator :

vi (t )  m(t )  c(t ) where c(t )  Vc cos(2fc t ) vi (t )  m(t )  Vc cos(2f c t )

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2 Square law device, v0 (t )  a1vi (t )  a2 vi (t )



v0 (t )  a1  m(t )  Vc cos(2f c t )   a2  m(t )  Vc cos(2f c t ) 

2

v0 (t )  a1  m(t )  Vc cos(2f c t )   a2 m 2 (t )  2m(t )Vc cos(2f c t )  Vc2 cos 2 (2f c t ) 

1  cos(4f c t )  v0 (t )  a1m(t )  a1Vc cos(2f c t )  a2 m 2 (t )  2a2 m(t )Vc cos(2f c t )  a2Vc2   2   v0 (t ) 

a2Vc2 a V2  a1m(t )  a2 m 2 (t )  a1Vc cos(2f c t )  2a2 m(t )Vc cos(2f c t )  2 c cos(4f c t ) 2 2

v0 (t ) 

 2a  a2Vc2 a V2  a1m(t )  a2 m2 (t )  a1Vc 1  2 m(t )  cos(2f c t )  2 c cos(4f c t ) 2 a1 2  

The desired AM wave is obtained by passing v0 (t ) through a band pass filter centred at f c with a bandwidth corresponding to the AM bandwidth 2 f m .



 2a  S AM (t )  a1Vc 1  2 m(t )  cos(2f c t ) a1  

Which is of the form of standard AM wave, S AM (t )  Ac 1  Ka m(t ) cos(2f ct ) BPF produces desire AM wave in frequency range f c  f m to f c  f m and remove undesired components out of this frequency range. The main condition for generation of DSB-FC or AM wave is that the carrier frequency should be atleast three times the maximum frequency of the modulating signal.

fc  f m  2 f m



fc  3 f m

Question 62 Explain square law detector ?

[CSVTU May 2011] Or

Explain square law demodulator for AM signal. Ans.

[CSVTU May 2008]

A square law demodulator is used for the recovery of modulated signal from DSB-FC or AM wave. It uses a non linear square law device like diode which satisfies square law relationship. An AM signal can be demodulated by squaring it and then passing the squared signal through a low pass filter (LPF).

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1 - 49

Block diagram of square law demodulator :

The characteristics of a non-linear device is given by

v0 (t )  a1vi (t )  a2 vi2 (t ) Where, vi (t )  Input voltage, v0 (t )  output voltage, a1 and a2 are constants The input voltage of the AM wave is given by :

vi (t )  S AM (t )  AC 1  Ka m(t ) cos(2fct ) 

v0 (t )  a1  Ac 1  K a m(t ) cos(2f c t )   a2  Ac 1  K a m(t )  cos(2f c t ) 

2

v0 (t )  a1 Ac 1  K a m(t )  cos(2f c t )  a2 Ac2 1  K a m(t )  cos 2 (2f c t ) 2

1  cos(4f c t )  v0 (t )  a1 Ac 1  K a m(t )  cos(2f c t )  a2 Ac2 1  K a2 m 2 (t )  2 K a m(t )    2  

v0 (t ) 

a2 Ac2 a A2 K 2  a2 Ac2 K a m(t )  2 c a m 2 (t )  a1 Ac 1  K a m(t )  cos(2f c t )  2 2 a2 Ac2 2 1  K a m(t ) cos(4fc t ) 2

The low pass filter passes only the first three terms on the right hand side to the output. Hence the low pass filter output is  a A2  a A2 K 2 y(t )  LPF  2 c  a2 Ac2 K a m(t )  2 c a m2 (t )  2  2  y (t ) 

a2 Ac2 a A2 K 2  a2 Ac2 K a m(t )  2 c a LPF m 2 (t )  2 2

m 2 (t ) has a spectrum extending from dc to 2 f m . When it is low pass filtered, only the

spectrum portion 0 to f m  gets through and this is indicated in the right hand side of the above equation LPF  m 2 (t )  .

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For successful demodulation, the output should be proportional to the modulating signal, m(t ) alone. However, there is an unwanted signal which gives rise to signal distortion. Desired component : a2 Ac2 K a m(t ) Distortion component :

a2 Ac2 K a2 LPF  m 2 (t )  0, f m  2

The ratio of desired component to the undesired distortion component is

a2 Ac2 K a m(t ) 1 a2 Ac2 K a2 m2 (t ) 2



1 2  1 K a m(t ) K a m(t ) 2

Ideally, this ratio should be infinite; practically, as large as possible; for which | Ka m(t ) |  2 . This implies that the percentage modulation should be small. Thus, the square law demodulator allows distortion less recovery of the baseband signal only if the AM wave has a low percentage modulation. This is a major limitation of the square law demodulator since small percentage modulation implies that the transmitted AM wave has low power efficiency and more susceptibility to channel noise. Question 63 In a square law demodulator the output signal is the signal average over many carrier cycles, with a neat diagram and derivation clarify the above statement. [CSVTU May 2009] Ans.

An alternative method of recovering the baseband signal which has been superimposed as an amplitude modulation on a carrier is to pass the AM signal through a nonlinear device. Such demodulation is illustrated in below figure. We assume here for simplicity that the device has a square law relationship between input 2 signal x (current or voltage) and output signal y (current or voltage). Thus y  kx ,

where k is a constant.

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Amplitude Modulation

Because of the nonlinearity of the transfer characteristic of the device, the output response is different for positive and for negative excursions of the carrier away from the quiescent operating point O of the device. As a result, the output, when averaged over a time which encompasses many carrier cycles but only a very small part of the modulation cycle, have the wave shape of the envelope. The applied signal is, x  A0  Ac 1  m(t ) cos ct Thus the output of the squaring circuit is, y  kx 2  k  A0  Ac 1  m(t ) cos c t 

2

2 y  k  A02  Ac2 1  m(t ) cos2 c t  2 A0 Ac 1  m(t ) cos ct    2  1  cos 2c t  y  kA02  kAc2 1  m(t )     2kA0 Ac 1  m(t )  cos c t 2  

 1  cos 2c t  y  kA02  kAc2 1  2m(t )  m 2 (t )     2kA0 Ac 1  m(t )  cos c t 2   k k k y  kA02  Ac2  kAc2 m(t )  Ac2 m 2 (t )  Ac2 1  2m(t )  m 2 (t )  cos 2c t  2 2 2 2kA0 Ac 1  m(t )  cos c t

Dropping dc terms as well as terms whose spectral components are located near c and 2c , we find that the output signal s0 (t ) , that is, the signal output of a low pass filter located after the squaring circuit, is k 1   s0 (t )  kAc2 m(t )  Ac2 m2 (t )  kAc2 m(t )  m2 (t )  2 2  

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Observe that the modulation m(t ) is indeed recovered but that m 2 (t ) appears as well. Thus the total recovered signal is a distorted version of the original modulation. The 1 distortion is small, however, if m2 (t )  m(t ) or if | m(t ) |  2 . 2 Question 64

    The signal v(t )  1  0.2cos  m  t  cos c t is demodulated using a square law  3    2 demodulator having the characteristics v0  ( v  2) . The output v0 ( t ) is then filtered by an ideal low pass filter having cutoff frequency at f m Hz. Sketch the amplitude frequency characteristic of the output waveform in the frequency range 0  f  fm . [CSVTU May 2012] Sol.

  v(t )  1  0.2 cos  m  3 

   t  cos c t  

v0  (v  2) 2  v 2  4  4v    v0  1  0.2cos  m  3  

  v0  1  0.22 cos 2  m  3 

cos 2 A 

2

     m   t  cos c t   4  4 1  0.2cos  3      

  m  t  0.4 cos  3  

   t  cos c t  

    m 2  t  cos c t  4   4  0.8cos  3    

   t  cos c t  

1  cos 2 A 2

 0.04   2m        1  cos 2c t  v0  1  t   0.4cos  m  t   1  cos   4 2  2  3    3       m    4  0.8cos   t  cos c t  3      2 v0  0.5 1  0.02  0.02 cos  m  3 

  m  t  0.4 cos  3  

   t  1  cos 2c t         4   4  0.8cos  m  t  cos c t  3   

 2 v0  0.5  0.01  0.01cos  m  3

  m   t  0.2 cos  3  t  0.5cos 2c t  0.01cos 2c t      2    0.01cos 2c t.cos  m  t  0.2 cos 2c t.cos  m  t   3   3    4  4 cos c t  0.8cos c t.cos  m  t  3 

2cos A cos B  cos( A  B)  cos( A  B)

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Amplitude Modulation

2  2     v0  4.51  0.01cos  m  t  0.2 cos  m  t  0.51cos 2c t  0.005cos  2c  m 3  3   3   2         0.005cos  2c  m  t  0.1cos  2c  m  t  0.1cos  2c  m  t  3 3 3             4 cos c t  0.4 cos  c  m  t  0.4 cos  c  m  t 3  3   

 t  

After passing through the low pass filter with cutoff frequency f m we get

   2  v0  4.51  0.2cos  m  t  0.01cos  m  t  3   3  Single sided spectrum of amplitude frequency characteristics :

Ans.

Double sided spectrum of amplitude frequency characteristics :

At f  0 , V0 (0) is same in both single sided and double sided spectrum. Question 65 The signal v(t )  (1  0.1cos 1 t  0.1cos 21 t )cos c t is detected by a square-law 2 detector, v0  2v . Plot the amplitude-frequency characteristic of v0 ( t ) .

[CSVTU Dec 2010] Sol.

v(t )  (1  0.1cos 1t  0.1cos 21t )cos c t v0  2v 2  2 1  0.1cos 1t  0.1cos 21t  cos 2 c t 2

2 2 1  cos 2c t  v0  2 1  0.1cos 1t    0.1cos 21t   2(1  0.1cos 1t )(0.1cos 21t )      2 

v0  1  0.01cos 2 1t  0.2 cos 1t  0.01cos 2 21t  0.2 cos 21t  0.02 cos 21t.cos 1t  1  cos 2c t  cos 2 A 

1  cos 2 A and 2cos A cos B  cos( A  B)  cos( A  B) 2

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v0  1  0.005(1  cos 21t )  0.2cos 1t  0.005(1  cos 41t )  0.2cos 21t  0.01cos 31t  0.01cos 1t  1  cos 2c t 

v0  1  0.005  0.005cos 21t  0.2cos 1t  0.005  0.005cos 41t  0.2cos 21t  0.01cos 31t  0.01cos 1t  1  cos 2c t 

v0  1.01  0.21cos 1t  0.205cos 21t  0.01cos31t  0.005cos 41t] 1  cos 2c t 

v0  1.01  0.21cos 1t  0.205cos 21t  0.01cos31t  0.005cos 41t  1.01cos 2c t  0.21cos 1t.cos 2c t  0.205cos 21t.cos 2c t  0.01cos31t.cos 2c t  0.005cos 41t.cos 2c t

2cos A cos B  cos( A  B)  cos( A  B) v0  1.01  0.21cos 1t  0.205cos 21t  0.01cos 31t  0.005cos 41t  1.01cos 2c t  0.105cos(2c  1 )t  0.105cos(2c  1 )t  0.1025cos(2c  21 )t  0.1025cos(2c  21 )t  0.005cos(2c  31 )t  0.005cos(2c  31 )t  0.0025cos(2c  41 )t  0.0025cos(2c  41 )t Amplitude frequency characteristics :

Question 66 Derive an expression for the signal v3 ( t ) in figure for v1 (t )  10cos(2000t )  2 4sin(200t ) . Assume that v2 ( t )  v1 ( t )  0.1v1 ( t ) and that the BPF is an ideal unity

gain filter with pass-band from 800 Hz to 1200 Hz.

Ans.

v1 (t )  10cos(2000t )  4sin(200t )

v2 (t )  v1 (t )  0.1v12 (t )

v2 (t )  10 cos(2000t )  4sin(200t )   10 cos(2000 t )  4sin(200t )  0.1 2

v2 (t )  10 cos(2000t )  4sin(200t )  0.1 100 cos 2 (2000 t )  16sin 2 (200t )  80 cos(2000t )sin(200t ) 

2cos A sin B  sin( A  B)  sin( A  B) , cos 2 A 

1  cos 2 A 1  cos 2 A and sin 2 A  2 2

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  1  cos(4000t )   1  cos(400t )   v2 (t )  10 cos(2000t )  4sin(200t )  0.1 100    16    2 2      0.1 40 sin(2200t )  sin(1800t )  v2 (t )  10 cos(2000t )  4sin(200t )  5  5cos(4000t )  0.8  0.8cos(400t )  4sin(2200t )  4sin(1800t )

1600 2400 rad/sec ,

BPF has a pass-band from 800 Hz-1200 Hz

therefore the

output after filtering will be

v3 (t )  10cos(2000t )  4sin(2200t )  4sin(1800t )

Ans.

Question 67 The signal v(t )  1  m(t ) cos c t is square-law detected by a detector having the 2 characteristic v 0  v . If the Fourier Transform of m(t ) is a constant M 0 extending

from  f M to  f M , sketch the Fourier transform of v0 ( t ) in the frequency range

 fM  f  fM . Sol.

Fourier Transform of m(t ) is a constant M 0 extending from  f M to  f M 

F m(t )  M ( f )

v(t )  1  m(t ) cos c t  v0  v 2  1  m(t ) cos 2 c t  2

1 1  m2 (t )  2m(t )  1  cos 2c t  2

1 1  m2 (t )  2m(t )  cos 2c t  m2 (t ) cos 2c t  2m(t ) cos 2c t  2 Taking Fourier transform, v0 

F 1  ( f )

F m(t )  M ( f )

F  m 2 (t )   M ( f )  M ( f )  M c ( f )

F cos 2c t  

1  M ( f  fc )  M ( f  fc ) 2 1 F  m2 (t ) cos c t    M c ( f  f c )  M c ( f  f c )  2 F  m(t ) cos c t  

1 ( f  2 fc )  ( f  2 fc ) 2

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 ( f  2 f c ) ( f  2 f c ) 1 ( f )  M ( f )  M ( f )  2 M ( f )     2 2 2 M c ( f  2 fc ) M c ( f  2 fc )    M ( f  2 fc )  M ( f  2 fc ) 2 2 

In the frequency range,  f m  f  f m , frequency spectrum is given by Vm ( f ) 

1 ( f ) M c ( f )   M( f ) ( f )  M ( f )  M ( f )  2M ( f )  2 2 2

Question 68 Explain envelope detector with neat diagram. Ans. Envelope Detector : 1. The envelope detector is a simple and very efficient device which is suitable for the detection of a narrowband AM signal. A narrowband AM wave is the one in which the carrier frequency is much higher as compared to the bandwidth of the modulating signal. 2. Envelope detector is used for recovery of modulating signal form DSB-FC signal. 3. An envelope detector produces output signal that follows the envelope of the input AM signal exactly. 4. Envelope detector is used for recovery of modulating signal from under modulation ( ma  1 ) system. 5. The envelope detector is used in all the commercial AM radio receivers. Circuit diagram : It is assumed that the diode is ideal which presents a zero resistance when it is ON in forward biased and infinite resistance when it is OFF in reverse biased.

Operation of envelope detector : 1. The standard AM is applied at the input of the detector. 2. In every positive half cycle of the input the detector diode is forward biased. It will charge the filter capacitor C connected across the load resistance R to almost the peak value of the input voltage.

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Amplitude Modulation

3.

As soon as the capacitor charges to the peak value, the diode stops conducting. The capacitor C will discharge through R between the positive peaks as shown in the input output waveform. 4. The discharging process continues until the next positive half cycle. When the input signal becomes greater than the capacitor voltage, the diode conducts again and the process repeats itself. Input output waveforms for an envelope detector : 1. The input output waveform shows the charging discharging of the filter capacitor and the approximate output voltage. 2. It can be seed from these waveforms, that the envelope of an AM wave is being recovered successfully.

Selection of the RC time constants : 1. The capacitor charges through diode D and Rs when the diode is ON and it 2.

discharges through R when the diode is OFF. The charging time constant Rs C should be short as compared to the carrier period 1/ f c . 

3.

Rs C 

1 fc

The discharging time constant RC should be long enough so that the capacitor discharges slowly through the load resistance R. But this time constant should not be too long which will not allow the capacitor voltage to discharge at the maximum rate of change of envelope. 1 1  RC   fc fm

Tc  RC  Tm

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Where f m is maximum modulating frequency, Tc is the time period of carrier signal and Tm is the time period of modulating signal. 4.

For proper operation of envelope detector, the magnitude of the slope of the dvc (t ) capacitor discharge curve should be larger than the magnitude of the dt slope of the envelope 

dA(t ) . dt dvc (t ) dA(t )  dt dt

Drawbacks of envelope detector : There are two types of distortions which can occur in the detector output. They are : 1.

Diagonal Clipping

2.

Negative Peak Clipping

1. Diagonal clipping : This type of distortions occurs when the RC time constant of the load current is too long. Due to this the RC circuit cannot follow the fast change in the modulating envelope. Condition to avoid diagonal clipping :

1 1  RC  fc fm

2. Negative peak clipping : This distortion occurs due to a fact that the modulation index on the output side of the detector is higher than that on its input side. So at higher depths of modulation of the transmitted signal, the over modulation may take place at the output of the detector. The negative peak clipping will take place as a result of this over modulation as shown in figure.

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Question 69 The input

to

an

Amplitude Modulation

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envelope

detector

is

a

single-tone

AM

signal

x AM (t )  A(1  ma cos m t )cos c t where ma is a constant, 0  ma  1 and c  m . (a) Show that if the detector output is to follow the envelope of x AM ( t ) , it is  ma sin m t0  1  m   RC  1  ma cos m t0  (b) Show that if the detector output is to follow the envelope at all time, it is

required that at any time t  t0 ,

required that RC 

1 m

1  ma2 ma

Or The input of the envelop detector of an tone modulated signal is given as v(t )  Ac 1  m cos m t cos c t . Find the maximum value of the time constant RC of Sol.

the detector that can always follow the message envelop. Circuit diagram of envelope detector :

[CSVTU Dec 2011]

Input output waveform of envelope detector :

(a) Assume

that

the

capacitor discharges from the peak value E0  A(1  ma cos mt ) at t . Then the voltage vc (t ) across the capacitor is given by

vc (t )  E0 et /( RC ) . t   vc (t ) can be approximated by using taylor series vc (t )  E0 1    RC 

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where RC = discharging time constant dv (t ) E The slope of discharge is given by, c   0 dt RC The amplitude ‘ E0 ’ at any instant is given by E0  A 1  ma cos mt 

…..(i) …..(ii)

dE0 …..(iii)  ma Am sin m t dt In order for the capacitor to follow the envelope E0 (t ) , the magnitude of the slope

The slope of this envelope is given by,

of the RC (discharging time) must be greater than the magnitude of the slope of the envelope E0 (t ) . dvc (t ) dE0  dt dt

…..(iv)

Substituting equation (1) and (3) in equation (4), E0  ma Am sin m t RC Substituting equation (2) in equation (5), A1  ma cos mt   ma Am sin mt RC m  sin m t 1  a m RC 1  ma cos m t

…..(v)

 If the detector output is to follow the envelope of xAM (t ) , it is required that at

any time t  t0 ,

m  sin m t0 1  a m RC 1  ma cos m t0

Proved

(b) On cross multiplication, m 1  a cos m t0  ma m sin m t0 RC RC 1   1 ma  m sin mt0  cos mt0   RC   RC  1  1  1  1 ma 2m     sin  m t0  tan m RC  RC  RC   This inequality must hold for every t0 , 2

2



1  1  ma m2     RC RC  

On squaring,

2 2   1    1  ma2 m2         RC    RC  

 If the detector output is to follow the envelope at all time, it is required that

time constant, RC  Question 70

1 m

1  ma2 ma

Proved

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Sol.

Amplitude Modulation

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An amplitude modulated wave 10 1 + 0.6cos(2π.103 t )  cos(2π.106 t ) is to be detected by a linear diode detector. Find : (a) The time constant τ (b) The value of resistance R if the capacitor used is 100 pF. Time domain equation of AM wave, S AM (t )  Ac 1  ma cos(2f mt ) cos(2fct ) S AM (t )  10 1 + 0.6cos(2π.103t )  cos(2π.106 t )

On comparing, m  2π  103 and ma  0.6 (a) Time constant If the detector output is to follow the envelope at all time, it is required that (a)

RC 

1 m

1  ma2 ma

1 2103  0.6   4.712 103 RC 1  0.62 

 m 1  m a RC 1  ma2

 

  RC 

1 4.712  103

Time constant, τ  0.212 ms

(b) Value of resistance R if the capacitor used is 100 pF 1 1  R  100  1012  RC  3 4.712  10 4.712  103 6  Resistance, R  2.122 10 

Ans.

Ans.

Question 71 The signal v(t )  (1  m cos m t )cos c t is detected using a diode envelop detector. Sketch the detector output when m  2 . Sol.

Time domain equation of AM wave, S AM (t )  Ac 1  ma cos(2f mt ) cos(2fct )

v(t )  (1  m cos mt )cos c t where modulation index m  2 Diode envelope detector is only valid for under modulation system (ma  1) . For modulation index ma  2 demodulator cannot detect the modulating signal m(t ) properly since it can follow only the positive cycles. Negative peak clipping : This distortion occurs due to a fact that the modulation index on the output side of the detector is higher than that on its input side. So at higher depths of modulation of the transmitted signal, the over modulation may take place at the output of the detector. The negative peak clipping will take place as a result of this over modulation as shown in figure.

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Question 72 Consider

a

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resultant

wave

obtained

by

adding

a

non-coherent

carrier

Ac cos(2fc t  ) to a DSB-SC wave cos(2fc t ).m(t ) . This composite wave is applied to an ideal envelop detector. Find the resulting detector output. Evaluate this output for   0 . Sol.

The input to the envelop detector is the sum of non-coherent carrier and a DSB-SC wave :

v(t )  Ac cos(2fc t  )  m(t )cos 2f ct cos( A  B)  cos A cos B  sin A sin B

v(t )  Ac cos(2fc t )cos   Ac sin(2fc t )sin   m(t )cos(2fc t ) v(t )  cos(2fct )  Ac cos   m(t )  Ac sin(2fct )sin 

The output of the envelop detector is :

v0 (t )  (Inphase component)2  (Quadrature component)2 

v0 (t ) 

 Ac cos   m(t )   Ac sin  2

When   0, v0 (t ) 

v0 (t )  

2

Ans.

 Ac cos 0  m(t )   Ac sin 0 2

2

 Ac 1  m(t )   Ac  0 2

2

v0 (t )  Ac  m(t )

Hence, the output of the envelope detector contains the message signal m(t ) . Question 73

Ans.

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Amplitude Modulation

The envelop detector shown in below figure is used to recover the signal m(t ) form

the AM signal v(t )  1  m(t ) cos c t , where m(t ) is a square wave taking on the values 0 and  0.5 volt and having a period T  RC 

Sol.

1 . Sketch the recovered signal if fc

T and 4T. 20

[CSVTU Dec 2010]

Time domain equation of AM wave, S AM (t )  Ac 1  ma cos(2f mt ) cos(2fct ) Given : v(t )  1  m(t ) cos c t Peak amplitude of AM signal is 1 V.

On comparing given figure with actual envelope detector, Rs  0 The capacitor charges through diode D and Rs when the diode is ON and it discharges through R when the diode is OFF. Charging time constant, c  Rs C  0 The discharging time constant RC should be long enough so that the capacitor discharges slowly through the load resistance R. Discharging time constant, d  RC (i) If RC 

T (Low discharging time constant) 20

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(ii) If RC  4T (High discharging time constant)

Modulation and Demodulation of DSB-SC Question 74 With suitable diagram explain the generation of DSB-SC signal using balanced modulator. Or Draw and explain the basic circuit of balanced modulator. [CSVTU May 2011] Ans. Balanced modulator is used for generation of DSB-SC wave. Using two ideal AM generators it is possible to generate a DSB-SC signal. It consists of two amplitude modulators that are inter connected in such a way as to suppress the unwanted carrier in an AM wave and the output consists of upper and lower sidebands only.

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One input to the amplitude modulator is from an oscillator that generates a carrier wave. The second input to the amplitude modulator in the top path is modulating signal m(t) while in the bottom path is – m(t). The output of the two AM modulators are as follows :

s2 (t )  Ac 1  Ka m(t ) cos(2fct )

s1 (t )  Ac 1  Ka m(t ) cos(2f ct )

The output of the summer is : s(t )  s1 (t )  s2 (t )

s(t )  Ac 1  Ka m(t ) cos(2fct )  Ac 1  Ka m(t ) cos(2f ct )

s(t )  Ac cos(2fc t )  Ac Ka m(t )cos(2f ct )  Ac cos(2f ct )  Ac K a m(t )cos(2f ct ) 

s(t )  2 Ac Ka m(t )cos(2f c t )

The balanced modulator output is equal to the product of the modulating signal m(t) and the carrier signal c(t) except the scaling factor 2 K a , which is of the form of standard time domain DSB-SC signal. M ( f  fc )  M ( f  fc ) 2 Taking Fourier transform on both sides of equation, FT From modulation theorem, m(t ) cos(2f c t ) 



F  s(t )  F  2 Ac K a m(t ) cos(2f c t ) 

2 Ac K a  M ( f  f c )  M ( f  f c ) 2

S ( f )  Ac Ka  M ( f  fc )  M ( f  fc ) Message Spectrum :

DSB-SC Spectrum :

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Since the carrier component is eliminated, the output is called DSB-SC signal. Question 75 A signal vm ( t ) is bandlimited to the frequency range 0 to f m . It is frequencytranslated by multiplying it by the signal vc (t )  cos( 2fc t ) . Find f c so that the bandwidth of the translated signal is 1 percent of the frequency f c . Sol.

Baseband signal  vm (t )

Bandwidth of baseband signal  f m

Carrier signal  vc (t )  cos(2f c t )

Translated signal vm (t ) vc (t ) represents the DSB-SC signal whose bandwidth is 2 f m Bandwidth of translated signal  1% of f c i.e. 2 f m  0.01 f c 

fc  200 f m

Ans.

Question 76 The signals v1 (t )  2 cos 1t  cos 21t and v2 (t )  cos 2 t  2 cos 22 t are multiplied. Plot the resultant amplitude frequency characteristic, assuming that 2  21 , but is not a harmonic of 1 . Repeat for 2  21 . Sol.

v1 (t )  2cos 1t  cos 21t

v2 (t )  cos 2t  2cos 22t

(a) 2  21

x(t )  v1 (t )  v2 (t )  (2cos 1t  cos 21t )  (cos 2t  2cos 22t ) x(t )  2cos 2t cos 1t  cos 2t cos 21t  4cos 22t cos 1t  2cos 22t cos 21t

2cos A cos B  cos( A  B)  cos( A  B) 

x(t )  cos(2  1 )t  cos(2  1 )t  0.5cos(2  21 )t  0.5cos(2  21 )t  2 cos(22  1 )t  2 cos(22  1 )t  cos(22  21 )t  cos(22  21 )t

Amplitude frequency characteristics :

…..(i)

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(b) 2  21 Put 2  21 in equation (i)

x(t )  cos31t  cos 1t  0.5cos 41t  0.5  2cos51t  2cos31t  cos61t  cos 21t

x(t )  0.5  cos 1t  cos 21t  3cos31t  0.5cos 41t  2cos51t  cos61t

Question 77 Explain ring modulator. Ans.

Ring modulator is another product modulator, which is used to generate DSB-SC signal. In a ring modulator circuit, four diodes are connected in the form of a ring in which all four diodes point in the same manner. All the four diodes in ring are controlled by a square wave signal c(t ) of frequency f c applied through a centre tapped transformer.

In case, when diodes are ideal and transformer are perfectly balanced, the two outer diodes are switched on if the carrier signal is positive whereas the two inner diodes are switched off and thus presenting very high impedance as shown in figure (a). Under this condition, the modulator multiplies the modulating signal x(t ) by 1 . Now, in case when carrier signal is negative, the situation becomes reversed as shown in figure (b). In this case, the modulator multiplies the modulating signal by 1 .

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Figure (a) illustrates the condition when the diodes (1) and (3) are switched ON and diodes (2) and (4) are switched OFF. Figure (b) illustrates the condition when the diodes (2) and (4) are switched ON and diodes (1) and (3) switched OFF. Hence, the ring modulator is a product modulator for a square wave carrier and modulating signal.

c(t ) 

4  (1) n 1 cos  2fc t (2n  1)   n 1 (2n  1)

s(t )  x(t ).c(t )  x(t ) 

4  (1) n 1 cos  2fc t (2n  1)   n 1 (2n  1)

Thus, with the help of above mathematical analysis, it may be verified that the output from ring modulator does not have any components at carrier frequency. Hence, the modulated output does not have any components at carrier frequency. Hence the modulated output contains only product terms. A ring modulator is also known as a double balanced modulator since it is balanced with respect to the baseband signal as well as the square wave signal. The frequency spectrum of the ring modulator output contains sidebands around each of the odd harmonics of the square wave carrier signal.

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Question 78 What is synchronous detection? Explain with a block diagram. Ans. Synchronous detector or coherent detector Synchronous detector is used for over modulated system (ma  1) . It can recover modulating signal from DSB-FC, DSB-SC or SSB-SC. The frequency and phase of the locally generated carrier signal and the carrier signal at the transmitter carrier must be identical. This means that the local oscillator signal must be exactly coherent or synchronized with the carrier signal at the transmitter, both in frequency and phase, otherwise the detector signal would get distorted. Therefore, this method of recovery is called synchronous detection or coherent detection. Synchronous detection of DSB-SC : In synchronous detection method, the received modulated or DSB-SC signal is first multiplied with a locally generated carrier signal cos(2f c t ) and then passed through a low-pass filer (LPF). At the output of a low-pass filter (LPF), the original modulating signal is recovered.

The balanced modulator (or multiplier or product modulator) output can be given as v(t )  m(t ) cos(2f c t ).cos(2f c t )  m(t ) cos 2 (2f c t )

1  cos 2 A 2 1  cos(2.2 f c t )  m(t ) m(t ) cos(2.2 f c t ) v(t )  m(t )   2  2 2   cos 2 A 



The product waveform is transmitted through a low-pass filter which rejects the double-frequency signal 2 f c . Output signal, s (t ) 

m(t ) , is proportional to m(t ) and hence the modulating signal is 2

recovered. Frequency spectrum :

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v(t ) 

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m(t ) m(t ) cos(2.2 f c t ) FT M ( f ) 1  M ( f  2 f c )  M ( f  2 f c )       2 2 2 2 2 

FT m(t ) cos(2f c t ) 

M ( f  fc )  M ( f  fc ) 2

Effects of asynchrony between the carriers at the modulator and demodulator : (1) Phase drift in carrier oscillator

Product waveform, v(t )  m(t )cos(2fc t ).cos(2fc t  )

2cos A cos B  cos( A  B)  cos( A  B) m(t ) cos(2.2 fc t  )  cos  2 The product waveform is transmitted through a low-pass filter which rejects the double-frequency signal 2 f c . 

p(t ) 

Output signal, s(t ) 

m(t ) cos  2

For   0 , output signal s (t ) 

m(t ) and the modulating signal is recovered. 2

  900 , output signal s(t )  0 . This effect is known as Quadrature null effect 2 where the output of the receiver becomes zero in spite of the fact that there is a transmitted signal. (2) Frequency drift in carrier oscillator For  

Product waveform, v(t )  m(t )cos 2f c t.cos 2( f c  f )t

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2cos A cos B  cos( A  B)  cos( A  B) m(t ) cos 2(2 fc  f )t  cos 2(f )t  2 The product waveform is transmitted through a low-pass filter which rejects the double-frequency signal 2 f c . 

v(t ) 

m(t ) cos 2(f )t 2 The above result is known as beat effect. Disadvantage : The demerit of the synchronous detection is that it requires an additional system at the receiver to ensure that the locally generated carrier is synchronized with the transmitter carrier to avoid phase and frequency errors making receiver complex and costly. Question 79 The baseband signal m(t ) in the frequency translated v(t )  m(t )cos 2fc t is Output signal, s(t ) 

recovered by multiplying v(t ) by the waveform cos 2( fc  f )t . The product

Sol.

waveform is transmitted through a low-pass filter which rejects the doublefrequency signal. Find the output signal of the filter. [CSVTU May 2012] Given : v(t )  m(t )cos 2f c t It is multiplied by cos 2( f c  f )t .

Product waveform, p(t )  v(t ).cos 2( f c  f )t  m(t )cos 2f c t.cos 2( f c  f )t

2cos A cos B  cos( A  B)  cos( A  B) m(t ) cos 2(2 fc  f )t  cos 2(f )t  2 The product waveform is transmitted through a low-pass filter which rejects the double-frequency signal 2 f c . 

p(t ) 

Output signal, s(t ) 

m(t ) cos 2(f )t 2

Ans.

Question 80 The baseband signal m(t ) in the frequency translated signal v(t )  m(t )cos 2fc t is recovered by multiplying v(t ) by the waveform cos(2fc t  ) . (a) The product waveform is transmitted through a low-pass filter which rejects the double-frequency signal. What is the output signal of the filter ? (b) What is the maximum allowable value for the phase  if the recovered signal is to be 90 percent of the maximum possible value ?

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(c) If the baseband signal m(t ) is band limited to 10 kHz, what is the minimum value of f c for which it is possible to recover m(t ) by filtering the product waveform v( t )cos(2f c t  ) ? Sol.

Given : v(t )  m(t )cos 2f c t It is multiplied by cos(2fc t  ) .

(a) Product waveform, p(t )  v(t ).cos(2f c t  )  m(t )cos(2f c t ).cos(2f c t  )

2cos A cos B  cos( A  B)  cos( A  B) m(t ) cos(2.2 fc t  )  cos  2 The product waveform is transmitted through a low-pass filter which rejects the double-frequency signal 2 f c . 

p(t ) 

m(t ) cos  Ans. 2 (b) The maximum allowable value for the phase  if the recovered signal is to be 90 percent of the maximum possible value 1 1 cos   0.9 0.9  m(t )  m(t ) cos   2 2 Ans.    25.840 (c) The baseband signal m(t ) is band limited to 10 kHz. Output signal, s(t ) 

p(t ) 

m(t ) 1 1 cos(2.2 f c t  )  cos   m(t ) cos(2.2 f c t  )  m(t ) cos  2 2 2 Term 2 Term 1

From above figure, to recover m(t ) by using low pass filter.

2 fc  f m  f m



2 fc  2 f m

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The minimum value of f c for which it is possible to recover m(t ) by filtering the product waveform, is f c  f m or fc  10 kHz .

Ans.

Question 81 The baseband signal em (t ) is recovered from the DSB-SC signal by multiplying

s(t )  em (t )cos 2fc t by the locally generated carrier cos(2fc t  ) . The product is passed through a low-pass filter which rejects the double-frequency signal. What is the maximum allowable value for the phase angle if the recovered signal is to be 95% of the maximum possible output ? If the baseband signal m(t ) is band limited to 10 kHz, what is the minimum value of f c for which em (t ) can be recovered by filtering ? Sol.

Given : s(t )  em (t )cos 2f c t It is multiplied by cos(2fc t  ) .

Product waveform, p(t )  s(t ).cos(2f c t  )  em (t )cos(2f ct ).cos(2f ct  )

2cos A cos B  cos( A  B)  cos( A  B) em (t ) cos(2.2 fc t  )  cos  2 The product waveform is transmitted through a low-pass filter which rejects the double-frequency signal 2 f c . 

p(t ) 

em (t ) cos  2 The maximum allowable value for the phase  if the recovered signal is to be 95% of the maximum possible value 1 1 0.95  em (t )  em (t ) cos   cos   0.95 2 2 Ans.    18.20 Output signal, y (t ) 

The baseband signal m(t ) is band limited to 10 kHz. p(t ) 

em (t ) 1 1 cos(2.2 f c t  )  cos   em (t ) cos(2.2 f c t  )  em (t ) cos  2 2 2 Term 1

Term 2

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From above figure, to recover m(t ) by using low pass filter.



2 fc  f m  f m

2 fc  2 f m

The minimum value of f c for which it is possible to recover m(t ) by filtering the product waveform, is f c  f m or fc  10 kHz .

Ans.

Question 82 The signal m(t ) in the DSB-SC signal v(t )  m(t )cos(c t  ) is to be reconstructed 2 by multiplying v(t ) by a signal derived from v ( t ) . 2 (a) Show that v ( t ) has a component at the frequency 2 f c . Find its amplitude.

(b) If m(t ) is bandlimited to f m and has a probability density

f (m) 

1 2

em

2

/2

  m  

2 Find the value of the amplitude of the component of v ( t ) at 2 f c .

Sol.

(a) v(t )  m(t )cos(c t  )

1  cos(2c t  2)  m 2 (t ) m 2 (t )   2  2 cos  2c t  2 2  

v 2 (t )  m 2 (t ) cos 2 (c t  )  m 2 (t )  cos 2 A 

1  cos 2 A 2

v 2 (t ) consist two frequency, one at f  0 and other at f  2 fc 

2 Amplitude of v (t ) at f  2 fc is

(b) f (m ) 

1 2

em

2

/2

m 2 (t ) 2

Ans.

  m  

Standard gaussian density function is given by f (m)  Where  is mean value and  is standard deviation. On comparing (i) and (ii),   0 and   1

2  MSV  (mean) 2 where MSV is mean square value

…..(i) 1  2



e

( m  )2 2 2

…..(ii)

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MSV  2  mean 2  1  0  1 2 Amplitude of v (t ) at f  2 fc is

m 2 (t ) 2

 m2 (t )  1 1 1 2 2 Expected value, E    E  m (t )    MSV   (1 ) 2 2  2  2 

2 Amplitude of v (t ) at f  2 fc is

1 2

Ans.

Question 83 Two signals m1 ( t ) and m2 ( t ) , both band limited to 5000 rad/s, are to be transmitted simultaneously over a channel by the multiplexing scheme shown in given figure. The signal at point b is the multiplexed signal, which now modulates a carrier of frequency 20,000 rad/s. The modulated signal point is transmitted over a channel. (i) Sketch signal spectra at points a, b, c (ii) What must be the bandwidth of the channel? (iii) Design a receiver to recover signal m1 ( t ) and m2 ( t ) from the modulated signal at point c.

Sol.

[CSVTU Dec 2007]

At point a, a(t )  m2 (t )  2cos(10000t )

FT x(t )cos 0t   X ( 0 )  X ( 0 )

Taking Fourier transform, a()  2 M 2 (10000)  M 2 ( 10000) Spectrum of a ()

A point b, b(t )  m1 (t )  a(t )

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Taking Fourier transform, b()  m1 ()  a() Spectrum of b()

At point c, c(t )  b(t )  2cos(20000t ) Taking Fourier transform, c()  2b( 20000)  b( 20000) Spectrum of c()

(2) Bandwidth of the channel 30000 30000 2 fm  2   Hz 2  (3) Receiver At point c, c(t )  m1 (t )  2m2 (t )cos(10000t )  2cos(20000t )



c(t )  2m1 (t ) cos(20000t )  4m2 (t ) cos(10000t ) cos(20000t ) 2cos A cos B  cos( A  B)  cos( A  B) c(t )  2m1 (t ) cos(20000t )  2m2 (t ) cos(30000t )  2m2 (t ) cos(10000t )

From the figure, x(t )  c(t ) cos(20000t )

x(t )  2m1 (t ) cos 2 (20000t )  2m2 (t ) cos(30000t ) cos(20000t )  2m2 (t ) cos(10000t ) cos(20000t )

1  cos 2 A and 2cos A cos B  cos( A  B)  cos( A  B) 2 x(t )  m1 (t ) 1  cos(40000t )   m2 (t ) cos(50000t )  m2 (t ) cos(10000t ) 

cos 2 A 

m2 (t ) cos(30000t )  m2 (t ) cos(10000t )

x(t )  m1 (t )  m1 (t )cos(40000t )  m2 (t ) cos(50000t )  cos(30000t )  2cos(10000t )

Ans.

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When x(t ) is passed through low pass filter the resultant is m1 (t ) .

LPF  x(t )  m1 (t )

Also, y(t )  x(t )  2cos(10000t ) y (t )  m1 (t ) cos(10000t )  m1 (t ) cos(40000t ) cos(10000t )  m2 (t ) cos(50000t ) cos(10000t ) 

m2 (t ) cos(30000t ) cos(10000t )  m2 (t ).2 cos 2 (10000t )

1  cos 2 A and 2cos A cos B  cos( A  B)  cos( A  B) 2 y(t )  2m1 (t ) cos(10000t )  0.5m1 (t ) cos(50000t )  0.5m1 (t ) cos(30000t ) 

cos 2 A 

0.5m2 (t ) cos(60000t )  0.5m2 (t ) cos(40000t )  0.5m2 (t ) cos(40000t )  0.5m2 (t ) cos(20000t )  m2 (t ) 1  cos(20000t )

When y (t ) is passed through low pass filter the resultant is m2 (t ) .

LPF  y(t )  m2 (t )

Question 84 Define QAM.

Ans.

[CSVTU Dec 2011]

Or Explain Quadrature Amplitude Modulation (QAM). [CSVTU Dec 2007] The transmission bandwidth required by a DSB-SC signal is twice the bandwidth of the message signal. Using Quadrature Carrier Multiplexing or Quadrature Amplitude Modulation (QAM), it is possible to send two DSB-SC signals using carriers of the same frequency but with quadrature phase (i.e. 900 out of phase). Both the modulated signals occupy the same frequency band and this, in effect, increases the bandwidth efficiency. In fact, this modulation scheme enables two DSB-SC modulated signals to occupy the same transmission bandwidth and therefore it allows for the separation of the two message signals at the receiver output. It is therefore known as a bandwidth conservation scheme. Block diagram of the QAM transmitter :

Block diagram of the QAM receiver :

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The QAM transmitter consists of two separate product modulators or balanced modulators which are supplied with two carrier waves of the same frequency but differing in phase by 900 . The output of the two balanced modulators are added in the adder and transmitted. The transmitted signal is thus given by

s(t )  m1 (t ). A cos(2fc t )  m2 (t ). A sin(2f ct ) Where, m1 (t ) and m2 (t ) are the two different message signals applied to the product modulators. Both x1 (t ) and x2 (t ) are band-limited in the interval  f m  f  f m , then s(t) will occupy a bandwidth of 2 f m . This bandwidth 2 f m is centred at the carrier frequency f c , where f m is the bandwidth of message signal m1 (t ) or m2 (t ) . Hence, the multiplexed signal consists of the in-phase component Am1 (t ) and the quadrature phase component Am2 (t ) . The multiplexed signal s(t) from QAM transmitter is applied simultaneously to two separate coherent detectors that are supplied with two local carriers of the same frequency, but differing in phase by 900 . Output at receiver before low pass filter,

x1 (t )  s(t ).cos(2f c t )  m1 (t ). A cos 2 (2f c t )  m2 (t ). A cos(2f ct )sin(2f ct ) sin(4f c t ) 1  cos(4f c t )   Am1 (t ).   m2 (t ).  2 2   x2 (t )  s(t ).sin(2f c t )  m1 (t ). A cos(2f c t )sin(2f c t )  m2 (t ). A sin 2 (2f c t )  Am1 (t ).

cos 2 A 

sin(4f c t ) 1  cos(4f c t )   m2 (t ).   2 2  

1  cos 2 A 1  cos 2 A sin 2 A , sin 2 A  and sin A cos A  2 2 2

Output at receiver after low pass filter (0 to f m ) ,

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y1 (t ) 

Am1 (t ) 2

Amplitude Modulation

1 - 79 Am2 (t ) 2

y2 (t ) 

The output of one detector is (1/ 2) Am1 (t ) , whereas the output of the second detector is (1/ 2) Am2 (t ) . For satisfactory operation of the coherent detector, it is essential to maintain coherent phase and frequency relationship between the oscillators used in the QAM transmitter and receiver parts of the system. QAM has the problem of co-channel interference i.e. modulated signals having the same carrier frequency interfere with each other. The problem gets worse with frequency error. QAM finds application in colour television transmission for multiplexing colour information signals. Question 85 (a) The

baseband

signal

m( t )

in

the

frequency

translated

signal

v(t )  m(t )cos(2fc t ) is to be recovered. There is available a waveform p(t ) which is periodic with period 1 / f c . Show that m(t ) may be recovered by appropriately filtering the product waveform p(t )v(t ) . (b) Show that m(t ) may be recovered as well if the periodic waveform has a period

n / fc , where n is as integer. Assume m(t ) band limited to the frequency range from 0 to 5 kHz and let fc  1 MHz. Find the largest n which will allow m(t ) to Sol.

be recovered. Are all periodic waveforms acceptable? (a) The periodic waveform can be expanded in exponential Fourier series as under : p (t ) 



Ce

n 

( j 2 nf c t )

…..(i)

n

Where Cn ’s are complex Fourier series coefficient. Thus, C n  Cn * The product s (t ) is expressed as s(t )  p(t )v(t )

…..(ii)

Substituting equation (i) and v(t )  m(t )cos(2f c t ) in equation (ii), we get s(t )  p(t )v(t ) 



Ce

n 

n

( j 2 nf c t )

m(t )cos(2f ct ) 



C

n 

n

m(t )cos(2f ct )e( j 2 nf c t )

   e j 2 fc t  e  j 2 f c t  ( j 2 nfc t )  s (t )  m(t )  Cn   e 2 n      

s(t ) 

m(t )  m(t )  Cne j 2 ( n 1) fc t     Cne j 2( n 1) fct  2 n  2 n 

…..(iii)

The signal m(t ) can be recovered from product s(t )  p(t )v(t ) by passing it through a low-pass filter (LPF) which rejects double frequency signal. The output of the filter is then given by s0 (t ) 

m(t ) m(t ) (C1  C1 )   2 Re(C1 )  m(t ) Re(C1 ) 2 2

Proved

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(b) Substituting fc / n in place of f c in Fourier series expansion of p(t ) , we obtain the product signal s (t ) as 

s(t ) 

1







1





 j 2   1 f c t   j 2   1 f c t  m(t )  m(t )  Cne   n    Cne   n     2 n  2 n 

The original modulating signal m(t ) can be recovered from s (t ) only when the largest frequency contained in m(t ) is less than fc / n . Under this condition, we have s0 (t )  Re(Cn ) m(t )

Proved

(c) For the given band-limited modulation signal m(t ) , highest frequency component contained will be f m  5 kHz . The carrier frequency is f c  1 MHz . The largest value of n can be found from f c / nmax  f m nmax 

f c 1  106 Hz   200 f m 5  103 Hz

Ans.

Modulation and Demodulation of SSB-SC Question 86 Define Hilbert transform. Also write the properties and applications. Or Write a comment on Hilbert transform. [CSVTU Dec 2008] Ans. Hilbert transform is a linear, non-causal and phase shifter transform which gives 900 phase shift for positive frequency and 900 phase shift for negative frequency.

The Hilbert transform of a function m(t ) is defined by : 

mˆ (t )  m(t )  h(t )  m(t ) 

1 1 m()   d t   t 

1 t Frequency response of Hilbert transform is given by : H ( f )   j sgn( f )

Hilbert transform, h(t ) 

Magnitude spectrum, H ( f )  1,    f  

900  Phase spectrum, H ( f )   0  900 

f 0 f 0 f 0

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Amplitude Modulation

Properties of Hilbert Transform : (1) A signal m(t ) and its Hilbert transform mˆ (t ) have the same energy density spectrum. (2) A signal m(t ) and its Hilbert transform mˆ (t ) have the same autocorrelation function. (3) A signal m(t ) and its Hilbert transform mˆ (t ) are mutually orthogonal. 

 m(t ). mˆ (t ) dt  0



(4) If mˆ (t ) is a Hilbert transform of m(t ) , then the Hilbert transform of mˆ (t ) is m(t )

ˆ (t )  m(t ) ˆ (t ) then H m If H  m(t )  m Here ‘H’ denotes the Hilbert transform. m(t )  cos(2f c t ) Example :

ˆ (t )  cos(2fct  900 )  sin(2fct ) H m(t )  m

ˆ (t )  sin(2fct  900 )   cos(2fct )  m(t ) H m Application of Hilbert Transform : (1) For generation of SSB signals (2) For designing of minimum phase type filters (3) For representation of band-pass signals Question 87 What are the different methods of generating SSB? Ans. The different methods of SSB generation are 1. Phase shift method 2. Filter method Question 88 Draw the block diagram of phase cancellation SSB generation and explain how the carrier and the unwanted sideband are suppressed. Or Describe the Quadrature phase shift method to generate SSB-SC AM signal. Or Explain phase shift method of generation of SSB-SC signal. [CSVTU May 2008] Ans. Phase shift method or Phase discrimination method or Phase cancellation method

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It is the most powerful method for SSB generation. The block diagram for the phase shift method of SSB generation is shown. This system uses two balanced modulators 0 or product modulators 1 and 2 and two  90 phase shifting networks.

Operation of the phase shift method : 1. The SSB modulator uses two balanced modulators 1 and 2. 2. The message signal m(t ) and a carrier signal Ac cos(2fc t ) is directly applied to the 3.

balanced modulator 1, producing a DSB-SC wave. 0 The Hilbert transform mˆ (t ) ( 90 phaseshift) of m(t ) and carrier signal shifted by

 900 are applied to balanced modulator 2, producing a DSB-SC wave. 4.

The output of balanced modulator 1 is s1 (t )  m(t ) Ac cos (2f c t ) .

5.

ˆ (t ) Ac sin (2f c t ) . The output of balanced modulator 2 is s2 (t )  m

6.

These signals s1 (t ) and s2 (t ) are fed to a summer. The output of the summer is

s(t )  s1 (t )  s2 (t ) s(t )  Ac m(t )cos(2fc t )  Ac mˆ (t )sin (2f c t ) 7.

The plus sign at the summing junction yields an SSB with only the LSB i.e.

sLSB (t )  Ac m(t )cos (2f c t )  Ac mˆ (t )sin (2f ct ) Let m(t )  Am cos(2f mt ) then its Hilbert transform is

m(t )  Am cos(2f m t  900 )  Am sin(2f m t )

sLSB (t )  Ac Am cos(2f mt )cos(2fc t )  Ac Am sin(2f mt )sin (2f ct ) sLSB (t )  Ac Am cos(2f mt )cos(2fct )  sin(2f mt )sin (2fct )

cos( A  B)  cos A cos B  sin A sin B 

sLSB (t )  2 Ac Am cos 2( fc  f m )t

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Amplitude Modulation

This signal contains only lower sideband frequency component fc  f m . 8.

The minus sign at the summing junction yields an SSB with only the USB i.e.

sUSB (t )  Ac m(t )cos (2f c t )  Ac mˆ (t )sin (2f ct ) sUSB (t )  Ac Am cos(2f mt )cos(2fc t )  Ac Am sin(2f mt )sin (2f ct ) sUSB (t )  Ac Am cos(2fmt )cos(2fct )  sin(2f mt )sin (2fct )

cos( A  B)  cos A cos B  sin A sin B 

sUSB (t )  2 Ac Am cos 2( fc  f m )t This signal contains only upper sideband frequency component f c  f m .

Merits of phase shift method : 1. Band pass filters are not required. 2. Low audio frequencies may be used for modulation. 3. It can generate SSB at any frequency. 4. Easy switching from one sideband to other sideband is possible. 5. To generate SSB at high radio frequencies, up conversion and hence repetitive mixing is not necessary. Demerits of phase shift method : 1. The output of two balanced modulators must be exactly same, otherwise cancellation will be incomplete. 2.

0 The design of the  90 phase shifting networks for the carrier signal and

modulating signal is extremely critical. These networks have to provide a correct  900 phase shift at all the frequencies which are practically difficult to achieve. Question 89 In the SSB generating system by using phase discrimination method, the carrier phase shift network produces a phase shift which differs from 900 by a small angle  . Calculate the output waveform and point out the respect in which the output no longer meets the requirement for an SSB waveform. Assume that the input is a single spectral component cos ω m t . [CSVTU Dec 2011]

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Sol.

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The single-tone sinusoidal modulating signal is given by m(t )  cos(2f mt ) 0 Hilbert transform of m(t ) , m(t )  cos(2f m t  90 )  sin(2f m t )

The carrier phase shift network produces a phase shift which differs from 900 by a small angle  . The output of the balanced modulator 1 is

s1 (t )  m(t )cos(2fc t )  cos(2f mt )cos(2f ct ) The output of the balanced modulator 2 is

s2 (t )  m(t )sin (2f c t  )  sin (2f mt )sin (2f ct  ) The output of the SSB generating system will be

s(t )  s1 (t )  s2 (t ) s(t )  cos (2f mt )cos (2fc t )  sin (2f mt )sin (2fc t  ) 2cos A cos B  cos( A  B)  cos( A  B) 2sin A sin B  cos( A  B)  cos( A  B) 1 cos{2( fc  f m )t}  cos{2( fc  f m )t}  2 1 cos{2( fc  f m )t  }  cos{2( fc  f m )t  } 2 The SSB waveform contains either only upper sideband ( f c  f m ) or lower sideband



s(t ) 

( fc  f m ) . Here, the output no longer meets the requirement of SSB signal. Hence, the output waveform is no longer an SSB-SC waveform.

Proved

Question 90 In the SSB generating system by using phase discrimination method, the baseband phase shift network produces a phase shift which differs from 900 by a small angle  . Calculate the output waveform and point out the respect in which the output no longer meets the requirement for an SSB waveform. Assume that the input is a single spectral component cos ω m t .

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Sol.

1 - 85

Amplitude Modulation

The single-tone sinusoidal modulating signal is given by m(t )  cos(mt )  cos(2f mt ) 0 Hilbert transform of m(t ) , m(t )  cos(2f m t  90  )  sin(2f m t  )

The output of the balanced modulator 1 is

s1 (t )  m(t )cos(2fc t )  cos(2f mt )cos(2f ct ) The output of the balanced modulator 2 is

s2 (t )  m(t )sin (2fc t )  sin (2f mt  )sin (2f c t ) The output of the SSB generating system will be

s(t )  s1 (t )  s2 (t ) s(t )  cos (2f mt )cos (2fc t )  sin (2f mt  )sin (2f c t ) 2cos A cos B  cos( A  B)  cos( A  B) 2sin A sin B  cos( A  B)  cos( A  B) 

s(t ) 

1 cos{2( fc  f m )t}  cos{2( fc  f m )t}  2 1 cos{2( fc  f m )t  }  cos{2( fc  f m )t  } 2

The SSB waveform contains either only upper sideband ( f c  f m ) or lower sideband

( fc  f m ) . Here, the output no longer meets the requirement of SSB signal. Hence, the output waveform is no longer an SSB-SC waveform.

Proved

Question 91 Explain the filter method of SSB generation. What are limitations of this scheme? [CSVTU May 2012] Ans.

Filter method or Frequency discrimination method

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Filter method is used for generation of SSB-SC signal. In filter method, a DSB-SC signal is generated first using a balanced modulator or product modulator and then one sideband either upper sideband or lower sideband is removed using a sideband filter. Accordingly, the output is either LSB SSB-SC or USB SSB-SC signal.

Operation of the filter method : Filter method can be used for generating the SSB from modulated wave if the message signal does not have any low frequency content. 1.

The modulating signal is amplified using an audio amplifier and applied to the balanced modulator. The other input to the balanced modulator is the carrier signal generated by the crystal oscillator 1. This carrier frequency is much less as compared to the carrier which is to be actually transmitted.

2.

The balanced modulator will suppress the carrier and will produce upper and lower sidebands at its output which is a DSB-SC signal.

3.

Out of the sidebands the unwanted sideband is heavily attenuated by the sideband suppression filter and the other sideband is passed through without much attenuation. In general, the sideband filtering process is not perfect and a small amount of the unwanted sideband gets through. Therefore, the requirements of sideband filter are flat pass band, high attenuation stop band and sharp cut off frequencies.

4.

The frequency of the sideband is very low. So the frequency is boosted up to the transmitter frequency by the combination of the balanced mixer and second crystal oscillator. This is called as frequency up conversion.

5.

This signal is then amplified using linear amplifiers which are used to avoid any waveform distortion.

Frequency spectrum at various points :

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Amplitude Modulation

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Advantages of filter method : 1.

The filter method gives the adequate sideband suppression.

2.

The sideband filter also helps to attenuate carrier if present in the output of balanced modulator.

3.

The band width is sufficiently flat and wide.

4. This method is simpler as compared to other methods. Disadvantages of filter method : 1.

Low audio frequencies cannot be used as the filter becomes bulky.

2. 3.

Due to the inability of the system to generate SSB at high radio frequencies, the frequency up conversion is necessary. Two expensive filters are to be used one for each sideband.

4.

Switching of side band is not possible.

5.

We cannot generate SSB-SC at any frequency.

Question 92 Give comparison of Filter method and Phase shift method of SSB generation. Ans.

Comparison of Filter method and Phase shift method of SSB generation :

S.

Parameter

Filter method

Phase shift method

1.

Method to cancel the unwanted sideband

Using filter

Using phase shifting techniques

2.

900 phase shift

Not required

Requires complex phase shift network

3.

Possible frequency range of SSB generation

Not possible to generate SSB at any frequency

Possible to generate SSB at any frequency

4.

Need of up conversion

Needed

Not needed

5.

Use of low modulating frequencies

Not possible

Possible

6.

Need of linear amplifiers

Needed

Needed

7.

Switching ability

Extra filter and switching network is necessary

Easily possible

8.

Complexity

Less

Medium

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Critical points in system design

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Filter characteristics, its size and weight, cutoff frequency

Design of 900 phase shifter for modulating frequency, symmetry of balanced modulators

Question 93 Explain diode detector circuit for SSB-PC transmission. Explain the phenomenon of diagonal clipping. Ans. Single Side Band-Plus Carrier transmission : Let us consider an SSB signal in which a large carrier is also present. This carrier may be introduced at the transmitter as in AM, or may be added at the receiver. When a large carrier is added to the SSB-SC signal the detection becomes easy. When a large carrier is present, the SSB waveform shows amplitude variations, even for a single tone modulating frequency. The expression for such a wave is obtained by adding a large carrier term in the expression of SSB-SC, given as 1 SSB (t )  ma Ac cos(c  m )t  Ac cos c t ; Ac  Am 2 The phasor diagram is shown in figure. The SSB-SC waveform is formed by superposition of two waveforms (lower sideband and large carrier term).

Demodulation of the SSB Signal with Large Carrier using diode detector : When a large carrier is introduced, a synchronous detector is not necessary for recovering the baseband signal. An envelope detector or diode detector provides the approximate baseband signal. An envelope detector is simpler and cheaper than synchronous detector. An expression for an SSB-signal with a large carrier Ac cos c t can be written as

SSB (t )  f (t )cos ct  f h (t )sin ct  Ac cos c t

Here it should be kept in mind that the carrier is added to an SSB-SC signal (and not multiplied with it as in synchronous detection). This technique is called carrier reinsertion technique. SSB (t )   Ac  f (t ) cos ct  fh (t )sin ct  e(t )cos(ct  ) Where,

e(t ) 

 Ac  f (t )2  fh2 (t ) is

the

envelope

of

the

wave,

and

 f (t )     tan 1  h  is the phase of the wave. When this SSB signal is applied to an  Ac  f (t )  envelope detector, the output of the detector will be the envelope e(t ) :

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e(t )   tan

1

1 - 89 1/2

 A  f (t ) 2  f 2 (t )  h  c 

Amplitude Modulation 1/2

 2 f (t ) f 2 (t ) f h2 (t )   Ac 1     Ac Ac2 Ac2  

Which can be expanded in the form of a Binomial series, given by  2 f (t )  e(t )  Ac 1   higher terms  2 Ac  

Since Ac  f (t ) and f h (t ) , higher terms may be neglected, then

e(t )

Ac  f (t )

Thus, the output of the envelope detector is close to the desired baseband signal f (t ) . The carrier re-insertion technique is also useful for detection of DSB-SC signals. The large carrier can be added either at the transmitter or at the receiver. When the carrier is added at the receiver, the synchronization problem remains as it is, and this technique has no advantage. When carrier is added at the transmitter, this mode has the advantage of both SSB and AM systems. It needs only half the bandwidth, at the same time, detection becomes simple. Diagonal clipping : This type of distortions occurs in an envelope detector when the RC time constant of the load current is too long. Due to this the RC circuit cannot follow the fast change in the modulating envelope. Condition to avoid diagonal clipping :

1 1  RC  fm fc

Question 94 Consider a two-stage SSB modulator where the message signal occupies a band 0.3 KHz to 4 KHz and the two carrier frequencies are f1  10 kHz and f 2  100 kHz . Evaluate the following :

Sol.

1.

Sideband of SSB-SC modulated waves at the output of the product modulators.

2.

The sidebands of SSB modulated waves at the outputs of BPF.

3.

The pass bands and the guard bands of the two BPFs.

Given : m(t )  0.3 kHz to 4 kHz

f1  10 kHz , f 2  100 kHz

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fc  f1  10 kHz, f m  0.3 kHz to 4 kHz

LSB  fc  f m  10 kHz  (0.3 kHz to 4 kHz) LSB  6 kHz to 9.7 kHz

USB  fc  f m  10 kHz  (0.3 kHz to 4 kHz) USB  10.3 kHz to 14 kHz Output of BPF 1 : Assume that BPF 1 passes only the USB

s1 (t )  10.3 kHz to 14 kHz (2) Output of PM 2 :

Ans.

fc  f 2  100 kHz, f m  10.3 kHz to 14 kHz LSB  fc  f m  100 kHz  (10.3 kHz to 14 kHz) LSB  86 kHz to 89.7 kHz USB  fc  f m  100 kHz  (10.3 kHz to 14 kHz)

USB  110.3 kHz to 114 kHz Output of BPF 2 : Assume that BPF 2 passes only the USB

s2 (t )  110.3 KHz to 114 KHz

Ans.

(3) Guard Band of BPF’s : Guard band represents the separation between USB and LSB. Guard band of BPF 1 = 9.7 kHz to 10.3 kHz Guard band of BPF 2 = 89.7 kHz to 110.3 kHz

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Amplitude Modulation

Question 95 N

(a) Show that the signal s(t )   cos(c t )cos(i t  i )  sin(c t )sin(i t  i ) is an i 1

SSB-SC signal (c   N ) . Is it the upper or lower sideband ? (b) Write an expression for the missing sideband. (c) Obtain an expression for the total DSB-SC signal. Ans.

(a) Upper or lower sideband N

s(t )  cos(c t ) cos(i t  i )  sin(c t )sin(i t  i )

..…(i)

i 1

The given signal can be modified using trigonometric expression as N 1 1  s(t )    cos{(c  i )t  i }  cos{(c  i )t  i }  2  i 1  2 N 1 1    2 cos{(c  i )t  i }  2 cos{(c  i )t  i }  i 1 

2cos A cos B  cos( A  B)  cos( A  B) 2sin A sin B  cos( A  B)  cos( A  B) 

N

s(t )   cos{(c  i )t  i }  sUSB (t )

..…(ii)

i 1

Hence, s (t ) contains frequency components f c  f i higher than the carrier frequency

f c and thus it represents SSB signal with upper sideband (USB).

Ans.

(b) Expression for the missing lower sideband The lower sideband (missing sideband) can be expressed as N

s(t )  cos(c t )cos(i t  i )  sin(c t )sin(i t  i )

..…(iii)

i 1

The given signal can be modified using trigonometric expression as N 1 1  s(t )    cos{(c  i )t  i }  cos{(c  i )t  i }  2  i 1  2 N 1 1    2 cos{(c  i )t  i }  2 cos{(c  i )t  i }  i 1 

2cos A cos B  cos( A  B)  cos( A  B) 2sin A sin B  cos( A  B)  cos( A  B) 

N

s(t )   cos{(c  i )t  i }  sLSB (t )

..…(iv)

i 1

N

sLSB (t )   cos{(c  i )t  i } i 1

(c) Total DSB-SC signal The total DSB-SC signal is given by sDSB  SC (t )  sUSB (t )  sLSB (t )

Ans.

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(i) = (ii) and (iii) = (iv) N

sDSB  SC (t )   cos(c t ) cos(i t  i )  sin(c t )sin(i t  i )  i 1 N

 cos( t ) cos( t   )  sin( t )sin( t   ) i 1



c

i

i

c

i

i

N

sDSB  SC (t )  2 cos(c t ) cos(c t  i )

Ans.

i 1

Question 96 A synchronous detection of SSB-SC signal shows phase and frequency discrepancy. Consider the synchronous detection of SSB signal given by N

s(t )   cos(c t )cos(i t  i )  sin(c t )sin(i t  i ) i 1

The signal is multiplied by the locally generated carrier cos c t and then low-pass filtered to recover the modulating signal. (i) Prove that the modulating signal can be completely recovered if the cut-off frequency of the filter is f N  f0  2 fc . (ii) Determine the recovered signal when the multiplying signal is cos(c t  ) . (iii) Determine the recovered signal when the multiplying signal is cos(c  )t . Ans.

N

(i)

s(t )  cos(c t ) cos(i t  i )  sin(c t )sin(i t  i ) i 1

The given signal can be modified using trigonometric expression as N 1 1  s(t )    cos{(c  i )t  i }  cos{(c  i )t  i }  2  i 1  2 N 1 1    2 cos{(c  i )t  i }  2 cos{(c  i )t  i }   i 1

2cos A cos B  cos( A  B)  cos( A  B) 2sin A sin B  cos( A  B)  cos( A  B) 

N

s(t )   cos{(c  i )t  i }

..…(i)

i 1

The product signal is given by : v(t )  s(t )cos(c t ) Substituting equation (i) in equation (ii), N

v(t )   cos{(c  i )t  i }cos(c t ) i 1

N 1 1  v(t )    cos{(2c  i )t  i }  cos(i t  i )  2  i 1  2 N N 1 1 v(t )   cos{(2c  i )t  i }   cos(i t  i ) 2 i 1 2 i 1 The output waveform of the low-pass filter (LPF) is thus given by

..…(ii)

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1 N  cos(i t  i ) 2 i 1 When the cut-off frequency, f c of the LPF is in the range f N  fo  2 fc . v0 (t ) 

Ans.

(ii) The recovered signal when the multiplying signal is cos(c t  ) When the multiplying signal is cos(c t  ) , the product signal will be given by

v(t )  s(t )cos(c t  )

..…(iii)

Substituting equation (i) in equation (iii), N

v(t )   cos{(c  i )  i }cos(c t  ) i 1

Modifying equation (vi), we get N 1 1  v(t )    cos{(2c  i )t  (i  )}  cos{i t  (i  )} 2  i 1  2

1 N 1 N cos{(2c  i )t  (i  )}   cos{i t  (i  )}  2 i 1 2 i 1 The output of the low-pass filter (LPF) will be 1 N v0 (t )   cos{i t  (i  )} 2 i 1 Therefore, the demodulated signal shows phase error. (iii) The recovered signal when the multiplying signal is cos(c  )t v(t ) 

Ans.

When the multiplying signal is cos(c  )t , the product signal will be given by

v(t )  s(t )cos(c  )t

..…(iv)

Substituting equation (i) in equation (iv), N

v(t )   cos{(c  i )t  i }cos{(c  )t} i 1

N 1 1  v(t )    cos{(2c  i  )t  i }  cos{(i  )t  i } 2  i 1  2 N N 1 1 v(t )   cos{(2c  i  )t  i }   cos{(i  )t  i } 2 i 1 2 i 1 The output of the low-pass filter (LPF) will be 1 N v0 (t )   cos{(i  )t  i } Ans. 2 i 1 Therefore, the demodulated signal shows frequency error. Question 97 A received SSB signal in which the modulation is a single spectral component has a 2 normalized power of 0.5 volt . A carrier is added to the signal, and the carrier plus signal are applied to a diode demodulator. The carrier amplitude is to be adjusted so that at the demodulator output 90 percent of the normalized power is in the

Analog Communication

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recovered modulating waveform. Neglect de components. Find the carrier amplitude required. [CSVTU Dec 2010] For a single-tone SSB-SC signal, the waveform after carrier insertion becomes

s '(t )  s(t )  c(t ) s '(t )  cos{(c  m )t}  A cos(c t ) Using trigonometric relation, cos( A  B)  cos A cos B  sin A sin B

s '(t )  cos(c t )cos(mt )  sin(c t )sin(mt )  A cos(c t ) s '(t )   A  cos(mt ) cos(ct )  sin(ct )sin(mt ) This signal is applied to a diode demodulator. The output of the demodulator is given by

v(t )  (Inphase component)2  (Quadrature component)2

v(t )  [ A  cos(mt )]2  [sin(mt )]2 v(t )  A2  cos2 (mt )  2 A cos(mt )  sin 2 (mt ) 1/ 2

v(t )  A2  1  2 A cos(mt )   A2  1  2 A cos(mt )  1/ 2

2A   v(t )  A2  1 1  2 cos(m t )  A  1  

A   cos(mt )  Using binomial expansion, v(t )  A2  1 1  2 A  1   A v(t )  A2  1  cos(m t ) A2  1 Neglecting the dc component, the normalized power of the detected signal will be 2

Pd 

1 A  1  A2    2    2  A2  1  2  A 1

For the recovered signal at the demodulator output to be 90% of this normalized power, detector output = 90% of normalized power of SSB modulated waveform

Pd  90%  Pn Normalized power, Pn  0.5 volt 2 1  A2     0.90  0.5 2  A2  1 



A2  0.9( A2  1)

A2  0.9 A2  0.9



0.1A2  0.9

 A3 Thus, amplitude of the reinserted carrier must be almost 3 Volt. Ans. Question 98 An SSB signal is demodulated using a synchronous demodulator. However, the locally generated carrier has a phase error  . Determine the effect of the error in

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Amplitude Modulation

demodulation. What will be the effect of this error on demodulation if input is DSB-SC instead of SSB-SC. [CSVTU May 2010] Sol.

Phase error in carrier oscillator in SSB-SC

SSB signal, SSB  m(t )cos(2f c t )  m(t )sin(2f c t ) Product waveform, v(t )  SSB.cos(2f c t  )

v(t )  m(t )cos(2fc t )  m(t )sin(2fct ) cos(2fct  )

v(t )  m(t )cos(2fc t )cos(2f c t  )  m(t )sin(2f ct )cos(2f ct  )

2cos A cos B  cos( A  B)  cos( A  B) 2cos A sin B  sin( A  B)  sin( A  B) m(t ) m(t ) cos(2.2 fc t  )  cos   sin(2.2 fc t  )  sin  2 2 The product waveform is transmitted through a low-pass filter which rejects the double-frequency signal 2 f c . v(t ) 

m(t ) m(t ) cos  sin  2 2 m(t ) For   0 , output signal s (t )  and the modulating signal is recovered. 2 For   0 , the second term can cause serious distortion. Phase error in carrier oscillator in DSB-SC Output signal, s(t ) 

Product waveform, v(t )  m(t )cos(2fc t ).cos(2fc t  )

2cos A cos B  cos( A  B)  cos( A  B) m(t ) cos(2.2 fc t  )  cos  2 The product waveform is transmitted through a low-pass filter which rejects the double-frequency signal 2 f c . 

v(t ) 

Output signal, s(t ) 

m(t ) cos  2

For   0 , output signal s (t ) 

m(t ) and the modulating signal is recovered. 2

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  900 , output signal s(t )  0 . This effect is known as Quadrature null effect 2 where the output of the receiver becomes zero in spite of the fact that there is a transmitted signal. Question 99 For  

A baseband signal bandwidth to the frequency range 300 to 3000 Hz, is to be superimposed on a carrier of frequency 40 MHz as a single-sideband modulation using the filter method. Assume that bandpass filters are available which will provide 40 dB of attenuation in a frequency interval which is about 1% of the filter center frequency. Draw a block diagram of a suitable system. At each point in the system draw plots indicating the spectral range occupied by the signal present there. [CSVTU May 2010, Dec 2008, Dec 2007] Sol.

Frequency range of baseband signal = 300 Hz to 3000 Hz Baseband spectrum

It is desired to generate an SSB signal with a carrier of 40 MHz. Then we require a passband filter with a selectivity that provides 40 dB of attention within 600 Hz (twice of 300 Hz or twice of lower frequency) at a frequency of 40 MHz, a percentage frequency change of 1 percent. Filters with such sharp selectivity are very elaborate and difficult to construct. For this reason, it is customary to perform the translation of the baseband signal to the final carrier frequency in several stages. First stage of SSB generation : 1% of filter center frequency, 0.01 f01  2  300 Hz

0.01 f01  600 Hz



Consider f 01  20 kHz

Second stage of SSB signal generation :

f01  60 kHz

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1 % of filter center frequency, 0.01 f02  2  20.3 kHz

0.01 f02  40.6 kHz



f02  4.06 MHz

Consider f 02  1 MHz

Third stage of SSB signal generation : 1 % of filter center frequency, 0.01 f03  2 1.0203 MHz

0.01 f03  2.0406 MHz Consider f 03  40 MHz

Block diagram of system :



f 03  204.06 MHz

Amplitude Modulation

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Vestigial Sideband Question 100 How much bandwidth required to access a TV channel in Indian scenario ? [CSVTU Dec 2009] Ans. The television video signal has a bandwidth of about 5 MHz. Question 101 Explain applications of VSB in television Broadcasting. [CSVTU May 2010] Ans. The principle application of VSB is found in commercial television broadcasting. The television picture signal, i.e. the video signal, occupies a bandwidth of nominally 5 MHz. A carrier amplitude modulated with such a signal would give rise to a signal extending over 10 MHz. since this bandwidth is about 9 times the width of the entire medium wave AM broadcast band (550-1650 kHz), some means of conserving bandwidth is surely needed. SSB-SC is ruled out because of the complexity that it would introduce into the many millions of receivers. Thus, a compromise is made between DSB-SC and SSB-SC in the form of VSB for television broadcasting. Question 102 Explain the characteristics of Vestigial side band filter. [CSVTU May 2010] Or Give mathematical proof of Vestigial sideband Modulation and Demodulation along with waveforms. [CSVTU May 2009] Ans. SSB-SC is bandwidth efficient compared to DSB-SC and conventional AM. It is possible to simply filter out the unwanted sideband out of a DSB-SC signal, but this requires an excellent and hence an expensive ideal band pass filter. Vestigial sideband

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(VSB) modulation relaxes the sharp cutoff filter requirement of the SSB signal by retaining a vestige or trace of the unwanted sideband in the transmitted signal. In VSB modulation instead of rejecting one sideband completely as in SSB modulation scheme, a gradual cut-off of one sideband is allowed. This gradual cut is compensated by a vestige or portion of the other sideband. In VSB, one sideband and a part of the other sideband called as vestige is transmitted. So, the band width required for VSB transmission is somewhat higher than that of SSB modulation. Generation of VSB : To generate a VSB signal, a DSB-SC signal is generated first and is then passed through a sideband filter as shown in below figure. This filter will pass the wanted sideband as it is along with a part of unwanted sideband.

The VSB signal obtained at the output of filter is applied to a chain of linear amplifiers and raised in power by the power amplifiers. The amplified signal is then applied to the transmitting antenna for transmission of signal. Frequency spectrum of baseband signal x(t ) :

Frequency spectrum of DSB-SC signal :

Frequency spectrum of SSB-SC signal :

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Frequency spectrum of VSB signal :

In the frequency spectrum it is assumed that the upper sideband is transmitted as it is and the lower sideband is modified into vestigial sideband. Transmission bandwidth of VSB is given by, BW  ( f c  f m )  ( f c  f v )  f m  f v Where f m is the bandwidth of the message signal and f v is the width of the vestigial sideband. Modulation of VSB signal : VSB signal can be generated by passing a DSB-SC signal through an appropriate VSB filter having transfer function H ( f ) .

DSB-SC signal  x(t ).c(t )  x(t )cos(2f c t ) VSB-SC signal, SVSB (t )  x(t )cos(2fc t )  h(t ) The frequency spectrum of VSB signal, SVSB ( f ) is therefore, given by

SVSB ( f )  Fourier Transform SVSB (t ) FT x(t )  h(t )  X ( f ).H ( f )

SVSB ( f )  FT  x(t )cos(2fct ).H ( f ) FT x(t ) cos(2f c t ) 

SVSB ( f ) 

1  X ( f  f c )  X ( f  f c ) 2

1  X ( f  fc )  X ( f  f c ) H ( f ) 2

Where X ( f ) is the Fourier transform of modulating signal x(t ) .

…..(i)

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Demodulation of VSB signal : VSB signal can be demodulated by passing a VSB signal through a product modulator along with a carrier signal and then passing through a low pass filter.

The output of the product modulator is given by, v(t )  SVSB (t )cos(2fc t ) 1 …..(ii)  SVSB ( f  f c )  SVSB ( f  f c ) 2 Substituting equation (i) in equation (ii), 1 1 V ( f )   X ( f  2 f c )  X ( f ) H ( f  f c )   X ( f )  X ( f  2 f c ) H ( f  f c ) 4 4 1 1 V ( f )   H ( f  f c )  H ( f  f c ) X ( f )   X ( f  2 fc ) H ( f  f c )  X ( f  2 f c ) H ( f  f c ) 4 4 The first term of above equation corresponds to the frequency spectrum of the modulating signal. The second terms of V ( f ) corresponds to the frequency spectrum

Taking Fourier Transform, V ( f ) 

of the VSB signal having carrier frequency 2 f c . The second term can be removed by using low-pass filter (LPF). The frequency spectrum of the signal v0 (t ) at the output of the LPF is given by V0 ( f ) 

1  H ( f  f c )  H ( f  f c ) X ( f ) 4

For a distortion less reproduction of the original modulating signal x(t ) , V0 ( f ) must be a scaled version of X ( f ) . Hence, for perfect demodulation, the required condition is

H ( f  fc )  H ( f  fc )  constant for | f |  f m This is called vestigial symmetry condition. Advantages of VSB : 1. The main advantage of VSB modulation is the reduction in bandwidth. It is almost as efficient as the SSB. 2. Due to allowance of transmitting a part of lower sideband, the constraint on the filters has been relaxed. So, practically easy to design filters can be used. 3. It possesses good phase characteristics and makes the transmission of low frequency components possible. 4. Practical filters can be used for partial suppression of unwanted sideband. Disadvantages of VSB : 1. During the removal of part of lower sideband, some power is wasted in the filters (VSB filters). 2. Amplitude and phase distortion is introduced at low frequencies.

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3.

Because of narrow lower sideband used for VSB transmission, critical tuning is required at the receiver. Applications of VSB : VSB modulation has become standard for the transmission of Television signals, because the video signals need a larger transmission bandwidth if transmitted using DSB-FC or DSB-SC techniques. Question 103 Explain the amplitude response of VSB filter. Ans. In order to preserve the frequency spectrum of x(t ) properly in VSB, the phase spectrum of the signal should be linear in the interval fc  fv  f  fc  f m and its value should be equal to zero at f c , or an integral multiple of 2 .

Question 104 Compare various AM system. Or Write the comparisons between DSB-FC, DSB-SC, SSB-SC and VSB. [CSVTU Dec 2008] Ans. S.

Parameter

DSB-FC Standard AM

DSB-SC

SSB-SC

VSB

1

Power

High

Medium

Less

Less than DSBSC but greater than SSB

2

Bandwidth

2 fm

2 fm

fm

f m  BW  2 f m

3

Carrier suppression

No

Yes

Yes

No

4

Sideband suppression

No

No

One sideband completely

One sideband suppressed partially

5

Transmission efficiency

Minimum

Moderate

Maximum

Moderate

6

Receiver complexity

Simple

Complex

Complex

Simple

7

Modulation

Non-linear

Linear

Linear

Linear

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Amplitude Modulation

type 8

Applications

Radio communication

Linear Radio communication

Linear point to point mobile communication

Television broadcasting

Question 105 Explain main block diagram of communication system. Ans. Elements of Communication Systems Any electronic communication system can be represented in its basic form, as shown in the figure.

The basic components of communication systems are transmitter, a communication channel or medium, and a receiver. Noise is inherently present in the channel or medium. It gets added to the information being communicated. 1. Information : The communication systems communicate messages. The message comes from the information sources. The amount of information contained in any given message is measured in bits. To have a better communication system, selective, but all information must be communicated with no redundancy since no real information can be conveyed by a redundant message. 2. Transmitter : The transmitter is a collection of electronic circuits designed to convert the information into a signal suitable for transmission over a given communication medium. Most of the times message that comes from information source is non-electrical and therefore it is not suitable for immediate transmission. Such messages need to be coded or processed before transmission and also require suitable transducers to convert them into electrical signals. The built-in circuitry such as decoders, encoders, transducers, etc. in the transmission makes incoming information suitable for transmission and subsequent reception. The most of the transmitters have built-in amplifier circuits. These circuits amplify the incoming signals (information) before transmission which help in faithful reception of the transmitted information at the receiver end. 3. Communication Channel : The communication channel is the medium by which the electronic signal is transmitted from one place to another. The communication medium can be a pair of conducting wire, coaxial cable, optical fibre cable or free space. Depending on the type of communication medium the communication system can be classified as, 1. Wire communication or Line communication. 2. Wireless communication or Radio communication.

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4.

Noise : Noise is random, undesirable electric energy that enters the communication system via the medium and interferes with the transmitted message. Some noise is also produced in the receiver. Noise can be either natural or man-made. Noise is one of the serious problems of electronic communication. It cannot be completely eliminated. However, there are ways to deal with noise, and reduce the possibility of degradation of signal due to noise. 5. Receiver : A receiver is a collection of electronic circuits designed to convert the signal back to the original information. It consists of amplifier, detector, mixer, oscillator, transducers and so on. Question 106 What is meant by term Radio? [CSVTU Dec 2010] Ans. Radio is the transmission of signals through free space by electromagnetic waves with frequencies significantly below visible light, in the radio frequency range, from about 3 kHz to 300 GHz. These waves are called radio waves. Electromagnetic radiation travels by means of oscillating electromagnetic fields that pass through the air and the vacuum of space. Information, such as sound, is carried by systematically changing (modulating) some property of the radiated waves, such as their amplitude, frequency, phase, or pulse width. When radio waves strike an electrical conductor, the oscillating fields induce an alternating current in the conductor. The information in the waves can be extracted and transformed back into its original form. Question 107 What is Radio Communication? Ans. Radio is the broad general term applied to any form of wireless communication between two points. Radio communication is a wireless communication, requiring no physical wires between transmitter and receiver to carry the signal; on the contrary, the signal is sent through free space or air. Radio communication essentially requires two antennas, one at transmitting end and the other at the receiving end. Using transmitting antenna, the transmitter transmits the signal, over a carrier wave, into the free space. The receiver picks up the signal by means of receiving aerial and separates the signal from the carrier. The medium attenuates the signal and causes it to appear much lower in amplitude at the receiver. Considerable amplification of the signal, both at the transmitter and the receiver, becomes essential for successful communication. Radio communication makes possible communication over very long distances, even from Earth to moon. Question 108 Define radio transmitter and receiver. [CSVTU Dec 2011] Ans.

Radio Transmitter : 1. A radio transmitter takes the information to be communicated and converts it into an electronic signal compatible with the communication medium. 2. This process involves carrier generation, modulation and amplification.

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3.

The transmitter is an electronic unit which accepts the information signal to be transmitted, and converts it into a radio frequency (RF) signal capable of being transmitted over long distances. Radio Receiver : 1. A radio receiver is a collection of electronic circuits designed to convert the signal back to the original information. 2. This process involves demodulation and amplification.

3.

A receiver intercepts some of the radio wave and extracts the information bearing electronic signal, which is converted back to its original form by a transducer such as a speaker. Question 109 What are basic functions of transmitter? Ans.

Every transmitter has three basic functions as follows : 1. The transmitter must generate a signal of correct frequency at a desired point in the spectrum. 2. Secondly it must provide some form of modulation to modulate the carrier. 3. Third it must provide sufficient power amplification in order to carry the modulated signal to a long distance. Question 110 Write the main functions of a radio receiver. Ans. The main functions of a radio receiver are : 1. Select the desired signal and reject the unwanted signals. 2. Amplify this selected R.F. signal. 3. Demodulate the amplified signal. 4. After demodulation, the original modulating signal is obtained. 5. Apply the amplified demodulated signal to the loudspeaker. Question 111 Define antenna? Ans. An antenna (or aerial) is an electrical device which converts electric currents into radio waves, and vice versa. It is usually used with a radio transmitter or radio receiver. In transmission, a radio transmitter applies an oscillating radio frequency electric current to the antenna's terminals, and the antenna radiates the energy from the current as electromagnetic waves (radio waves). In reception, an antenna

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intercepts some of the power of an electromagnetic wave in order to produce a tiny voltage at its terminals that is applied to a receiver to be amplified. An antenna can be used for both transmitting and receiving.

AM Receiver Question 112 Explain the AM receiver characteristics. Or What is image frequency? [CSVTU May 2011] Or Define sensitivity & selectivity. [CSVTU May 2008] Or Define selectivity in case of radio receivers. [CSVTU May 2009] Ans. The performance of the radio receiver can be measured in terms of following receiver characteristics : (1) Sensitivity (2) Selectivity (3) Fidelity (4) Double spotting (5) Image frequency & its rejection (1) Sensitivity : (a) The sensitivity of a communication receiver refers to the receiver’s ability to pick up weak signals, and amplify it. (b) It is often defined in terms of the voltage that must be applied to the receiver input terminals to give a standard output power, measured at the output terminals. (c) A receiver with a good sensitivity will produce more output for a similar input signal as against a receiver with poor sensitivity. The sensitivity of a receiver is decided by the bandwidth of audio amplifier which amplifies the baseband signal. (d) The more gain that a receiver has, the smaller is the input signal necessary to produce desired output power. Therefore, sensitivity is a primary function of the overall receiver gain. It is often expressed in microvolt or in decibels. (e) The sensitivity of receiver mostly depends on the gain of the IF amplifiers. (f) Figure shows the typical sensitivity curve. As shown in the figure the sensitivity varies over the tuning band.

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(2) Selectivity : (a) Selectivity refers to the ability of a receiver to select a signal of a desired frequency while reject all others. Selectivity is receiver’s ability to distinguish two adjacent carrier frequencies. (b) By this feature it is decided how perfectly the receiver is able to select the desired carrier frequency and reject the others. (c) Selectivity in a receiver is obtained by using tuned circuits. These are LC circuits tuned to resonate at a desired signal frequency. (d) Selectivity depends on the sharpness of the resonance curve of tuned circuits involved in the receiver. The sharper the resonance curve, the better the selectivity. Better selectivity means that the receiver has greater capability to reject undesired signals.

(e) The sharpness of the resonance curve depends on the quality factor Q of the resonant circuit. The higher the Q, the more is the selectivity. (f) The selectivity of a receiver goes on increasing as the selectivity curve becomes more and more narrow. Thus, curve 1 is sharper than curve 2, and hence has better selectivity.

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Selectivity depends upon the following factors : (i) Selectivity varies with receiving frequency and becomes somewhat worse when the receiving frequency is raised. (ii) In general, selectivity is mainly determined by the response of the IF section, with the mixer and RF amplifier input circuits playing a small but significant part. (iii) Selectivity is the main factor which determines the adjacent channel rejection of a receiver. Higher the selectivity better is the adjacent channel rejection and less is the adjacent channel interference. (3) Fidelity : (a) The ability of a receiver to reproduce faithfully all frequency components present in the baseband signal is called fidelity. (b) If any component is missed, or attenuated considerably, fidelity suffers and the reproduced signal is distorted. This feature is mainly decided by the bandwidth of audio amplifier which amplifies the baseband signal. (c) Figure shows the typical fidelity curve for radio receiver.

(d) The fidelity at the lower modulating frequencies is determined by the low frequency response of the IF amplifier and the fidelity at the higher modulating frequencies is determined by the high frequency response of the IF amplifier. (e) Fidelity is difficult to obtain in AM receiver because good fidelity requires more bandwidth of IF amplifier resulting in poor selectivity. (4) Double spotting : (a) When a receiver picks up the same short wave station at two nearby points on the receiver dial, the double spotting phenomenon takes place.

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(b) The main cause for double spotting is poor front-end selectivity, i.e., inadequate image frequency rejection. The front-end of the receiver does not select different adjacent signal very well. (c) The adverse effect of double spotting is that a weak station may be marked by the reception of a nearly strong station at the undesired point on the dial. (d) When the receiver is tuned across the band, a strong signal appears to be at two different frequencies, once at the desired frequency and again when the receiver is tuned to 2 times IF below the desired frequency. In this second case, the signal becomes the image, reduced in strength by the image rejection, thus making it appear that the signal is located at two frequencies in the band. (e) Double spotting may be used to calculate the IF of an unknown receiver. The undesired point on the dial is precisely 2 fi below the correct frequency. If imagefrequency rejection is improved, then certainly there will be a corresponding decrease in the double spotting occurrence. (5) Image frequency and its rejection : (a) In standard broadcast receiver the local oscillator frequency f 0 is made higher than the signal frequency f s by an amount equal to intermediate frequency (IF). Therefore, f0  f s  fi . (b) When f 0 and f s are mixed, the difference frequency f 0  f s , which is one of the products, is equal to f i , only f i is passed and amplified by the IF stage. (c) If a frequency f si  f 0  fi , i.e. f si  f s  2 fi , appears at the input of the mixer then it will produce the sum and difference frequencies regardless of the inputs. Therefore, the mixer output will be difference frequency at the IF value. The term f si is called the image frequency and is defined as the signal frequency plus twice the intermediate frequency. (d) Unfortunately, this image frequency signal is also amplified by the IF amplifiers resulting in interference. This has the effect of two stations being received simultaneously and is naturally undesirable.

(e) The rejection of an image frequency by a signal tuned circuit is the ratio of the gain at the signal frequency to the gain at the image frequency. It is given by f f   1  Q 22 where   si  s and Q  loaded Q of tuned circuit f s f si

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(f) If the receiver has an RF stage, then there are two tuned circuits, both tuned to f s ; the rejection of each will be calculated by the same formula, and the total rejection will be the product of two. (g) The image rejection depends on the selectivity of the RF amplifier and tuned circuits and must be achieved before the IF stage. Once the undesired frequency enters the first IF amplifier, it becomes impossible to remove it from the wanted signal. Question 113 Define RF amplifier. Ans. An RF amplifier is a tuned amplifier with high gain and low noise. It is used for improving the selectivity, sensitivity and to provide amplification to weak signals. Below figures show the RF amplifier circuits. It is a tuned circuit followed by an amplifier. The RF amplifier is usually a simple class A circuit.

The values of resistors R1 and R2 in the bipolar circuit are adjusted such that the amplifier works as class A amplifier. The antenna is connected through coupling capacitor to the base of the transistor. This makes the circuit very broad band as the transistor will amplify virtually any signal picked up by the antenna. However the collector is tuned with a parallel resonant circuit to provide the initial selectivity for the mixer input. The FET circuit is more effective than the transistor circuit. Their high input impedance minimizes the loading on tuned circuits, thereby permitting the Q of the circuit to be higher and selectivity to be sharper. Question 114 Write the advantages of RF amplifier. Ans. The advantages of RF amplifier are : 1. Greater gain i.e. better sensitivity. 2. Improved rejection of adjacent undesired signals, i.e. better selectivity. 3. Improved signal to noise ratio. 4. Improved image frequency rejection. 5. Improved coupling of the receiver to the antenna due to impedance matching.

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Prevention of re-radiation of the local oscillator voltage through the antenna. Prevention of undesired frequencies from entering the mixer and heterodyne to produce interfering frequency equal to IF.

Question 115 What is mixer? Ans.

Mixer or Frequency changer : 1.

The mixer is basically a nonlinear resistance which has two inputs at different frequencies ( f s and f 0 ) .

2.

Its output has several frequency components including the difference between the two input frequencies. The difference frequency ( f 0  f s ) is called as intermediate frequency IF and it is

3.

selected for further amplification by the tuned circuit at the output of the mixer.

Question 116 What is local oscillator? Ans.

In shortwave broadcasting, the operating limit for receivers is 36 MHz. For such operating limit local oscillators such as Armstrong, Hartley, Colpitts, Clapp or ultraaudion are used. The Colpitts, Clapp and ultra-audion oscillators are used at the top of the operating limit, whereas Hartley oscillator is used for frequencies below 120 MHz. All these oscillators are LC oscillators and each employs only one tuned circuit to determine its frequency. When higher frequency stability of local oscillator is required, the circuits like AFC (Automatic Frequency Control) are used.

Question 117 Why should local oscillator frequency f 0 be greater than signal frequency f s ? Ans.

For the medium wave (MW) band, the signal frequency ranges from 540 kHz to 1650 kHz. Assuming IF = 455 kHz, the local oscillator frequency will be in the range of 995 kHz to 2105 kHz. This gives the frequency ratio, f 0 max  2.2 f 0 min If the local oscillator frequency is designed to be lower than the signal frequency, then f 0 will be in the range 85 kHz to 1195 kHz. This gives the frequency ratio of,

f 0 max 1195   14 f 0 min 85

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The normal tunable capacitor has a capacitance ratio,

Cs max  10 . Cs min

This capacitance ratio will easily give a frequency ratio of 3.2 which is greater than the frequency ratio of 2.2. The frequency ratio of 14 just cannot be obtained using the practically available tunable capacitance. That is why the local oscillator frequency should be higher than the signal frequency. There is one more reason for it. With f 0 greater than f s the tracking errors are reduced to a great extent. Question 118 Define IF amplifiers. Ans. IF amplifiers are tuned voltage amplifiers tuned for the fixed frequency. It’s important function is to amplify only tuned frequency signal and reject all others. Most of the receiver gain is provided by the IF amplifiers, to obtain required gain, usually two or more stages of IF amplifiers are required. The IF amplifier is a fixed frequency amplifier. It is supposed to select the desired frequency and reject the unwanted frequencies. The IF amplifiers will decide the sensitivity and selectivity of the receiver. Therefore, its frequency response must have steep skirts. Question 119 What are the conditions for the selection of IF amplifier? Ans. Intermediate Frequency selector : The choice of the IF depends on the following factors : 1. IF should not be too high as it will result in poor selectivity and therefore poor adjacent channel rejection. This is due to difficulty in obtaining a high quality factor Q at high frequencies. 2. If the IF is too high the tracking problems increase. 3. If the IF is lowered then the image frequency rejection is poorer. 4. A very low IF can make the selectivity too sharp cutting the sidebands. 5. For very low IF, the frequency stability of the local oscillator must be very high. Because a small drift in the local oscillator frequency will result in a larger error. 6. The IF must not fall within the tuning range of the receiver. Otherwise instability will occur and heterodyne whistles will be heard. This will make the tuning impossible. 7. In the standard broadcast AM receivers, IF is within the range of 438 to 465 kHz with 455 kHz being the most popular frequency. Question 120 Explain tuned radio frequency (TRF) receiver. Ans. Tuned radio frequency (TRF) receiver : Tuned radio frequency (TRF) receiver is the simplest radio receiver. The very first block of this receiver is an RF stage. This stage generally contains two or three RF amplifiers. Actually, these RF (radio frequency) amplifiers are tuned RF amplifiers i.e. they have variable tuned circuit at the input and output sides.

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At the input of the receiver, there is a receiving antenna as shown in block diagram. At this antenna signals from different sources (i.e. stations) are present. However, with the help of input variable tuned circuit of RF amplifier the desired signal (i.e. station) is selected. But this selected signal is usually very weak of the order of V . This selected weak signal is amplifier by the RF amplifier (i.e. RF

stage). 3. Thus the function of RF stage is to select the desired signal and amplify it. 4. After this, the amplified incoming modulated signal is applied to the demodulator. The demodulator or detector demodulates the modulated signal and thus at the output of the demodulator, we get modulating or baseband signal (i.e. audio signal). 5. This audio signal is amplified by audio amplifier. After that, this audio signal is further amplified by a power amplifier upto desired power level to drive the loudspeaker. 6. The last stage of this receiver is the loudspeaker. A loudspeaker is a transducer which converts electrical signal into sound signal. Question 121 Write advantages and disadvantages of TRF receiver. [CSVTU Dec 2008] Ans.

The main advantages of a TRF receiver are : 1. Simplest type of receiver since it does not involve mixing and IF operation. 2. Very much suitable to receive single frequency. 3. TRF receivers have good sensitivity. Disadvantages of TRF receiver are : (1) Instablility of the receiver. (2) Insufficient selectivity at high frequencies and poor adjacent channel rejection. (3) Bandwidth variation over the tuning range. Question 122 Explain the drawbacks of TRF Receiver. Ans.

Although TRF receiver is cheaper and the simplest one, it has certain drawbacks as under : (1) Instablility of the receiver : The TRF receiver suffers from a tendency to oscillate at higher frequencies from the multistage RF amplifiers with high gain and operating at same frequency. If such an amplifier has a gain of 20,000 then if a small portion of the output leaked back to the input of the RF stage, then positive feedback and oscillation will result. This type of leakage could result from power supply coupling, stray capacitance coupling, radiation coupling or coupling

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through any other element common to the input and output stages. Definitely, this type of condition is undesirable for a good receiver. This problem is also termed as instability of the receiver. (2) Insufficient selectivity at high frequencies and poor adjacent channel rejection : The selectivity of a receiver is its ability to distinguish between a desired signal and an undesired signal. The selectivity of TRF receiver is poor. In fact, it is difficult to achieve sufficient selectivity at high frequencies due to the enforced use of single-tuned circuits. (3) Bandwidth variation over the tuning range : Another problem associated with the TRF receiver is the bandwidth variation over the tuning range. For example, in AM broadcast system; let us consider that a tuned circuit is required to have a bandwidth of 10 kHz at a frequency of 540 kHz. According to the definition, the quality factor Q of this tuned circuit must be

Q

Resonance frequency 540   54 Bandwidth 10

Now, at the other end of this AM broadcast band (i.e. 1640 kHz), the quality factor Q of the coil, according to above equation must increase by a factor of 1640/535 (i.e. 3) to a value of 164. However, in practice due to several losses dependent upon frequency would prevent such a large increase. Thus, practically, the quality factor Q of this tuned circuit is unlikely to exceed 120 and hence providing a bandwidth of the tuned circuit equal to f 1640 f  r   13.8 kHz Q 120 Therefore, due to this increased bandwidth of 13.8 kHz in place of a fixed bandwidth of 10 kHz, the receiver would pick up or select adjacent frequencies (i.e. stations) with the desired frequency or station. This means that the bandwidth of the TRF receiver varies with the incoming frequency. Question 123 Draw a block schematic of super heterodyne radio receiver and explain why it is called so. Explain the function of each block. [CSVTU Dec 2011, Dec 2010] Or Explain the working of superheterodyne receiver with neat block diagram. [CSVTU May 2011] Or Draw block diagram of superhetrodyne AM receiver. [CSVTU Dec 2007] Ans.

Superheterodyne receiver or Superhet receiver : The principle of operation of the superheterodyne receiver depends on the use of heterodyning or frequency mixing. The process of mixing two signals having different frequencies to produce a new frequency is called as heterodyning. In a superheterodyne receiver, the incoming RF signal frequency is combined with the local oscillator signal frequency through a mixer and is converted into a signal of

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lower fixed frequency. This lower fixed frequency is known as intermediate frequency (IF). However, the IF signal contains the same modulation as the original signal. This IF signal is now amplified and demodulated to reproduce the original signal. The word heterodyne stands for mixing. Here, the incoming signal frequency is mixed with the local oscillator frequency. Therefore, the receiver is called superheterodyne receiver. The superheterodyne receiver is used in all types of receivers like television receiver, radar receiver, satellite receiver etc.

Operation of superheterodyne receiver : (1) Antenna : The DSBFC or AM signal transmitted by the transmitter travels through the air and reaches the receiving antenna. This signal is in the form of electronic waves. It induces a very small voltage (few μV ) into the receiving antenna. (2) RF stage : The RF stage consists of a pre-selector which is used to select the wanted signal and reject other out of many, present at the antenna. It also reduces the effect of noise. It also consists of an RF amplifier; at the output of the RF amplifier we get the desired amplified signal at frequency f s . (3) Mixer : The mixer receives signals from the RF amplifier at frequency f s and from the local oscillator at frequency f 0 such that ( f 0  f s ) . (4) Ganged tuning : In order to maintain a constant difference between the local oscillator frequency and the incoming frequency, ganged tuning is used. This is simultaneous tuning of RF amplifier, mixer and local oscillator and it is achieved by using ganged capacitors (tuning control in radio set). (5) Intermediate frequency (IF): The mixer will mix these signals to produce signals having frequencies f s , f 0 , ( f 0  f s ) and ( f 0  f s ) . Out of these the difference of frequency component i.e. ( f 0  f s ) is selected and all others are rejected. This frequency is called as the intermediate frequency (IF).  I .F .  ( f0  f s ) This frequency contains the same modulation as the original signal f s . (6) IF amplifier : The IF amplifier provide most of the gain (and hence sensitivity) and the bandwidth requirements (and hence selectivity) of the receiver. Therefore, the IF amplifier determines the sensitivity and the selectivity of the superheterodyne receiver.

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(7) Demodulator : The amplifier IF signal is detected by the detector or demodulator to recover the original modulating signal. (8) AGC means automatic gain control. This circuit controls the gains of the RF and IF amplifier to maintain a constant output voltage level even when the signal level at the receiver input is fluctuating. This is done by feeding a controlling dc voltage to the RF and IF amplifiers. The amplitude of this dc voltage is proportional to the detector output. (9) Audio and power amplifier : The modulating signal (audio signal) is then amplified by an audio amplifier to get a particular voltage level. This amplified audio signal is further amplified by a power amplifier to get a specified power level so that it may activate the loudspeaker. (10) Loudspeaker : The loudspeaker is a transducer which converts this audio electrical signal into audio sound signal and thus the original signal is reproduced i.e. the original transmission is received. Question 124 Draw waveforms and frequency spectrums at various points of a superheterodyne receiver. Ans. Waveforms at various points of a superheterodyne receiver :

Frequency spectrum at various points of a superheterodyne receiver :

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Question 125 Write advantages and disadvantages of superheterodyne receiver. [CSVTU Dec 2008] Ans. The main advantages of a superheterodyne receiver are : 1. No variation in bandwidth. The BW remains constant over the entire operating range. 2. High sensitivity and selectivity as a result of the use of high Q filters and the double RF down conversion scheme. 3. High adjacent channel rejection. 4. High gain can be obtained without instability. Disadvantages of superheterodyne receiver are : 1. Image frequency problem; to reject the image frequency an additional filter is often applied in front of the mixer. 2. Large in size and weight. 3. It consumes more dc power. 4. Only a fixed data bandwidth. 5. Higher cost for higher number of components. Question 126 What do you understand by heterodyning? How does heterodyning helps in image frequency rejection? [CSVTU May 2009] Ans. Heterodyning or frequency mixing : The principle of operation of the superheterodyne receiver depends on the use of heterodyning or frequency mixing. The process of mixing two signals having different frequencies to produce a new frequency is called as heterodyning. Heterodyning is useful for frequency shifting signals into a new frequency range. The two frequencies are combined in a non linear signal processing device such as a mixer. In the most common application, two signals at frequencies f1 and f 2 are mixed, creating two new signals, one at the sum

f1  f 2 of the two frequencies, and the other at the

difference f1  f 2 . These new frequencies are called heterodynes. Typically only one of the new frequencies is desired, and the other signal is filtered out of the output of the mixer. Image frequency rejection : A superheterodyne receiver is a better receiver than a TRF receiver. However, a superheterodyne receiver suffers from a major drawback known as image frequency problem. This problem of image frequency is inherent to a superheterodyne receiver and arises because of the use of heterodyne principle. In fact, the frequency conversion process carried out by the local oscillator and the mixer often allows an undesired frequency in addition to the desired incoming frequency. In a standard broadcast receiver, the local oscillator frequency f 0 is always made higher than the incoming signal frequency f s . The local oscillator frequency f 0 is kept equal to the signal frequency plus the intermediate frequency f i .

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fi  f 0  f s

Hence, the intermediate frequency is the difference between the local oscillator frequency and the incoming signal frequency. Now, if a frequency f si manages to reach the mixer, such that f si  f 0  fi , then this frequency f si would also produce f i when it is mixed with f 0 . This undesired IF signal will also be amplified by the stage and thus would cause interference. This has the effect of two sources or stations being received simultaneously. This situation is obviously undesirable. Now, f si  f s  fi  fi  f s  2 fi The term f si is known as the image frequency and is defined as the signal frequency plus twice the intermediate frequency.

For example, an AM broadcast station at f s  580 kHz is tuned on a receiver with an IF f i  455 kHz. The local oscillator is tuned to f0  580  455  1035 kHz. But a signal at f si  f s  fi  fi  580 + 455 + 455 = 1490 kHz is also 455 kHz away from the local oscillator; so both the desired signal and the image, when mixed with the local oscillator, will also appear at the intermediate frequency. This image frequency is within the AM broadcast band. Thus the undesired frequency signal cannot be distinguished by the IF stage and hence would be treated in the same manner as the desired frequency signal. The rejection of an image frequency signal by a single tuned circuit may be defined as the ratio of the gain at the signal frequency to the gain at the image frequency. This is given as f f   1  Q 22 where   si  s and Q  loaded quality factor of tuned circuit f s f si If the receiver has an RF stage, then there are two tuned circuits, both tuned to f s ; the rejection of each will be calculated by the same formula, and the total rejection will be the product of two. The image rejection depends on the selectivity of the RF amplifier and tuned circuits and must be achieved before the IF stage. Once the undesired frequency enters the first IF amplifier, it becomes impossible to remove it from the wanted signal. Question 127 Explain how the features of an AM receiver are improved by heterodyning process. [CSVTU May 2010] Ans.

The principle of operation of the superheterodyne receiver depends on the use of heterodyning or frequency mixing. The process of mixing two signals having different

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frequencies to produce a new frequency is called as heterodyning. A superheterodyne receiver suffers from a major drawback known as image frequency problem. Heterodyning helps in image frequency rejection. The rejection of an image frequency signal by a single tuned circuit may be defined as the ratio of the gain at the signal frequency to the gain at the image frequency. This is given as f f   1  Q 22 where   si  s and Q  loaded quality factor of tuned circuit f s f si Question 128 Write down the important features of AM receiver. Compare TRF & Superhetrodyne receiver. [CSVTU Dec 2008] Ans. Important features of AM receiver : The performance of the radio receiver can be measured in terms of following receiver characteristics : (1) Sensitivity (2) Selectivity (3) Fidelity (4) Double spotting (5) Image frequency & its rejection Comparison of TRF and Superhetrodyne receiver : S. TRF receiver Superhetrodyne receiver 1. Local oscillator stage is absent. Local oscillator stage is present. 2. IF is not used. All stages till detector IF is used. Amplifiers after mixer operate at incoming RF. stage operate at a common IF. 3. Tuning is required over a wide range Tuning is required over a fixed range of frequencies. of frequencies except RF stage. 4. They have good sensitivity but poor They have good sensitivity and selectivity. selectivity. 5. RF stages can make receiver unstable. Instability problems are absent. 6. Used for single channel low Used for multichannel and high frequency applications. frequency applications Question 129 What do you understand by double conversion used in communication receiver? Ans. The receiver has two mixers and local oscillators and therefore has two intermediate frequencies. That is why it is called as the double conversion receiver. The double conversion receiver allows the receiver to have good image rejection and also good adjacent channel selectivity. When choosing the intermediate frequency for a double conversion receiver there is a trade-off to be made between the advantages of using a low frequency IF or a high frequency one. High frequency IF : The use of a high frequency IF means that the difference between the wanted frequency and the unwanted image is much greater and it is easier to achieve high levels of performance because the front end filtering is able to provide high levels of rejection.

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Low frequency IF : The advantage of choosing a lower frequency IF is that the filters that provide the adjacent channel rejection are lower in frequency. The use of a low frequency IF enables the performance to be high, while keeping the cost low. Thus, the first IF is high, in the MHz range in order to have better image frequency rejection. Whereas the second IF is low, in the kHz range in order to have all the advantages of low IF. Question 130 Explain the method of double super heterodyne receiver and the state how it rejects image frequency of incoming signal also. [CSVTU Dec 2009] Ans.

The double superheterodyne radio receiver improves the performance in a number of areas including stability, image rejection and adjacent channel filter performance. The double superheterodyne radio receiver is widely used, especially at high frequencies where factors such as image rejection and filter performance are important. The receiver has two mixers and local oscillators and therefore has two intermediate frequencies. That is why it is called as the double conversion receiver. The double conversion receiver allows the receiver to have good image rejection and also good adjacent channel selectivity. Reason for using double superheterodyne radio receiver : When choosing the intermediate frequency for a superheterodyne radio receiver there is a trade-off to be made between the advantages of using a low frequency IF or a high frequency one. High frequency IF : The use of a high frequency IF means that the difference between the wanted frequency and the unwanted image is much greater and it is easier to achieve high levels of performance because the front end filtering is able to provide high levels of rejection. Low frequency IF : The advantage of choosing a lower frequency IF is that the filters that provide the adjacent channel rejection are lower in frequency. The use of a low frequency IF enables the performance to be high, while keeping the cost low. Accordingly there are two conflicting requirements which cannot be easily satisfied using a single intermediate frequency. The solution is to use a double conversion superheterodyne topology to provide a means of satisfying both requirements. Basic double superheterodyne receiver concept : The basic concept behind the double superheterodyne radio receiver is the use of a high intermediate frequency to achieve the high levels of image rejection that are required, and a further low intermediate frequency to provide the levels of performance required for the adjacent channel selectivity. Typically the receiver will convert the incoming signal down to a relatively high first intermediate frequency (IF) stage. This enables the high levels of image rejection to be achieved. As the image frequency lies at a frequency twice that of the IF away from the main or wanted signals, the higher the IF, the further away the image is and the easier it is to reject at the front end.

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Once the signal has passed through the first IF at the higher frequency, it is then passed through a second mixer to convert it down to a lower intermediate frequency where the narrow band filtering is accomplished so that the adjacent channel signals can be removed. As the lower frequency, filters are cheaper and the performance is often higher. (Although it must be said that filter technology now allows effective filters to be made at much higher frequencies than was previously possible.) Image frequency rejection : The front end selectivity of any receiver must reject the first image frequency at 2 fi above the desired signal frequency, and for this to occur, the intermediate frequency f i should be high enough. However, as the f i increases, its band-pass gets larger. Beyond the HF band (above 30 MHz), it becomes impossible to obtain the required band-pass with good image rejection using ordinary circuits. To avoid this problem two intermediate frequencies are used. The first f i is high, several megahertz or even higher. The second f i is quite low, of the order of 1 MHz or even less. As the receiver has two intermediate frequencies, such receiver is called double conversion receiver. The double conversion receiver allows the receiver to have good image rejection and also good adjacent channel selectivity. The image rejection is assured in the first conversion as image frequency is well outside the fixed-tuned RF amplifier band-pass. The narrow band-pass required for good adjacent channel rejection is obtained in the second intermediate frequency, which may be as low as 100 kHz. Let us consider the receiver is tuned to receive signal of 150 MHz, first intermediate frequency is 10.7 MHz with the bandwidth of 150 kHz and second intermediate frequency is 465 kHz with bandwidth of 15 kHz. With these values, the image frequency of the first intermediate frequency can be given as f si  f s  2 fi  150  2 10.7  171.4 MHz This mage frequency is beyond the band width limit of 150 kHz and hence it is rejected. But at the same time bandwidth of 150 kHz allows several 15 kHz wide adjacent channels to pass to the second mixer. The band-pass for the second intermediate frequency rejects these adjacent channels since its pass band is only of 15 kHz.

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Question 131 Determine the image frequency for a standard broadcast band AM receiver using a 455 kHz IF and tuned to a station at 640 kHz. Sol.

Intermediate frequency, fi  455 kHz

Signal frequency, f s  640 kHz

The image frequency is given by, f si  f s  2 fi

f si  640  2  455  640  910  1550 kHz

Ans.

Question 132 In a superheterodyne receiver having no RF amplifier, the loaded Q of the antenna coupling circuit (at the input of the mixer) is 90. If the intermediate frequency is 455 kHz, calculate the following : (i) The image frequency and image frequency rejection ratio at 950 kHz and (ii) The image frequency and its rejection ratio at 10 MHz. Sol.

(i) Intermediate frequency, fi  455 kHz

Signal frequency, f s  950 kHz

Quality factor, Q  90 The image frequency is given by, f si  f s  2 fi

f si  950  2  455  950  910  1860 kHz Image frequency rejection ratio is given by,   1  Q 22 where  



Ans.

f si f s  f s f si

1860 950   1.958  0.510  1.448 950 1860

Since the given receiver has no RF amplifier, therefore there is only single tuned circuit. The rejection ratio is given as   1  Q 22   1  (90) 2  (1.448) 2  130.32

(ii) Intermediate frequency, fi  0.455 MHz

Ans. Signal frequency, f s  10 MHz

The image frequency is given by, f si  f s  2 fi

f si  10  2  0.455  10.91 MHz

Ans.

f f Image frequency rejection ratio is given by,   1  Q 22 where   si  s f s f si

10.91 10   1.091  0.916  0.175 10 10.91 Since the given receiver has no RF amplifier, therefore there is only single tuned 

circuit. The rejection ratio is given as   1  Q 22   1  (90)2  (0.175)2  15.78

Ans.

Question 133 For a broadcast superheterodyne AM receiver having no RF amplifier, the loaded quality factor Q of the antenna coupling circuit is 100. Now if the intermediate frequency is 455 kHz, then determine the following :

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(i) The image frequency and its rejection ratio at an incoming frequency of 1000 kHz. (ii) The image frequency and its rejection ratio at an incoming frequency of 25 MHz. (i) Intermediate frequency, fi  455 kHz Signal frequency, f s  1000 kHz Quality factor, Q  100 The image frequency is given by, f si  f s  2 fi

f si  1000  2  455  1000  910  1910 kHz Image frequency rejection ratio is given by,   1  Q 22 where  

Ans.

f si f s  f s f si

1910 1000   1.91  0.524  1.386 1000 1910 Since the given receiver has no RF amplifier, therefore there is only single tuned 

circuit. The rejection ratio is given as   1  Q 22   1  (100) 2  (1.386) 2  138.6

(ii) Intermediate frequency, fi  0.455 MHz

Ans. Signal frequency, f s  25 MHz

The image frequency is given by, f si  f s  2 fi

f si  25  2  0.455  25.91 MHz Image frequency rejection ratio is given by,   1  Q 22 where  

Ans.

f si f s  f s f si

25.91 25   1.0364  0.9649  0.0715 25 25.91 Since the given receiver has no RF amplifier, therefore there is only single tuned 

circuit. The rejection ratio is given as   1  Q 22   1  (100)2  (0.0715)2  7.22

Ans.

Question 134 In a broadcast superheterodyne receiver having no RF amplifier, the loaded Q of the antenna coupling circuit is 100. If the intermediate frequency is 455 kHz, calculate : (i) The image frequency and its rejection ratio for tuning at 1.1 MHz station. (ii) The image frequency and its rejection ratio for tuning at 25 MHz. [CSVTU May 2010, Dec 2010, May 2008] Or In a broadcast superheterodyne receiver having no RF amplifier, the loaded Q of antenna coupling circuit is 100. If the intermediate frequency is 455 kHz, (i) Calculate rejection ratio at 1100 kHz. (ii) Calculate rejection ratio at 25 MHz. (iii) What change must be brought in IF to improve rejection ratio at 25 MHz to that at 1100 kHz? [CSVTU May 2012]

Analog Communication Sol.

1 - 124

(i) Intermediate frequency, fi  455 kHz

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Signal frequency, f s  1100 kHz

Quality factor, Q  100 The image frequency is given by, f si  f s  2 fi

f si  1100  2  455  1100  910  2010 kHz Image frequency rejection ratio is given by,   1  Q 22 where  

Ans.

f si f s  f s f si

2010 1100   1.827  0.547  1.28 1100 2010 Since the given receiver has no RF amplifier, therefore there is only single tuned 

circuit. The rejection ratio is given as   1  Q 22   1  (100)2  (1.28)2  128

(ii) Intermediate frequency, fi  0.455 MHz

Ans. Signal frequency, f s  25 MHz

The image frequency is given by, f si  f s  2 fi

f si  25  2  0.455  25.91 MHz Image frequency rejection ratio is given by,   1  Q 22 where  

Ans.

f si f s  f s f si

25.91 25   1.0364  0.9649  0.0715 25 25.91 Since the given receiver has no RF amplifier, therefore there is only single tuned 

circuit. The rejection ratio is given as   1  Q 22   1  (100)2  (0.0715)2  7.22

Ans.

Question 135 For a receiver with IF and RF frequencies of 455 kHz and 900 kHz respectively, determine the following : (i) The local oscillator frequency. (ii) Image frequency. (iii) Image frequency rejection ratio for a pre-selector Q of 80. Sol. Intermediate frequency, fi  455 kHz Signal frequency, f s  950 kHz (i) The local oscillator frequency is given by, f0  f s  fi

f0  900  455  1355 kHz

Ans.

(ii) The image frequency is given by, f si  f s  2 fi

f si  900  2  455  1810 kHz (iii) Quality factor, Q  80 Image frequency rejection ratio is given by,   1  Q 22 where  



1810 900   2.011  0.497  1.514 900 1810

f si f s  f s f si

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Amplitude Modulation

Since the given receiver has no RF amplifier, therefore there is only single tuned circuit. The rejection ratio is given as   1  Q 22   1  (80) 2  (1.514) 2  121.12

Ans.

Question 136 In a broadcast superheterodyne AM receiver having no RF stage, the loaded Q of the aerial coupling circuit (at the input of the mixer) is 125. If the intermediate frequency is 465 kHz calculate : (i) The image frequency and its rejection at 1 MHz and 30 MHz. (ii) The IF required to make the image rejection ratio as good at 30 MHz as it is at 1 MHz. Sol. (i) Intermediate frequency, fi  0.465 MHz Quality factor, Q  125 (A) Signal frequency, f s  1 MHz The image frequency is given by, f si  f s  2 fi

f si  1  2  0.465  1  0.93  1.93 MHz Image frequency rejection ratio is given by,   1  Q 22 where  

Ans.

f si f s  f s f si

1.93 1   1.93  0.518  1.412 1 1.93 Since the given receiver has no RF amplifier, therefore there is only single tuned 

circuit. The rejection ratio is given as   1  Q 22   1  (125) 2  (1.412) 2  176.5

Ans.

(B) Signal frequency, f s  30 MHz The image frequency is given by, f si  f s  2 fi

f si  30  2  0.465  30  0.93  30.93 MHz Image frequency rejection ratio is given by,   1  Q 22 where  

Ans.

f si f s  f s f si

30.93 30   1.031  0.97  0.061 30 30.93 Since the given receiver has no RF amplifier, therefore there is only single tuned 

circuit. The rejection ratio is given as   1  Q 22   1  (125) 2  (0.061) 2  7.69

Ans.

Thus, the image frequency rejection is adequately large at low frequency but it degrades as the signal frequency increases. (ii) The image rejection ratio   176.5 at f s  30 MHz . The corresponding value of   1.412 at   176.5 But,  

f si f s  f s f si

Analog Communication Therefore, 1.412 

GATE ACADEMY PUBLICATIONS ®

1 - 126 f si 30 MHz  30 MHz f si

On crosss multiplication,

f si2  900  1.412 30 f si

On arranging, f si2  42.36 f si  900  0 New image frequency, f si  57.9 MHz But, f si  f s  2 fi

f si  f s 57.9  30  2 2 Image frequency, fi  13.95 MHz Hence, fi 

Ans.

Question 137 A superheterodyne receiver has been designed to receive radio station in the frequency band 108 MHz to 157 MHz. (i) If intermediate frequency is chosen to be 12 MHz, show that the image frequency band overlaps the RF band. (ii) Work out the minimum required intermediate frequency such that the image frequency falls outside the 108 to 157 MHz region. Assume that local oscillator frequency is less than the carrier frequency. [CSVTU Dec 2009] Sol. Range of carrier signal frequency, f s  108 MHz to 157 MHz Intermediate frequency, fi  12 MHz (1) Local oscillator frequency f 0 is less than the carrier frequency f s . Therefore, fi  f s  f 0



f 0  f s  fi

So, f01  f s1  fi  108  12  96 MHz and f02  f s 2  fi  157  12  145 MHz Image frequency, f si  f 0  fi  f s  2 fi So, f si1  96  12  84 MHz and f si 2  145  12  133 MHz Therefore, the image frequency band overlaps the RF band. (2) Since, f si  f s  2 fi

Proved

f si1  108  2 fi , then for fi  0 , f si1 is always less than 108 MHz, so it never

overlaps the RF band. f si 2  157  2 fi , then to avoid overlap, f si 2  108 MHz .  2 fi  49 157  2 fi  108 

fi  24.5 MHz

Ans.

Question 138 A tuned circuit of the oscillator in a simple AM transmitter employs a 50 μH coil

Sol.

and 1 nF capacitor. If the oscillator output is modulated audio frequency upto 10 kHz, what is the frequency range occupied by the sidebands? [CSVTU Dec 2012] Tuned oscillator circuit : L  50 H and C  1 nF Carrier Frequency : f c 

1 2 LC



1 2 50  106  109

 712 kHz

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Amplitude Modulation

1 - 127

Since, the highest modulating frequency is f m  10 kHz the frequency range occupied by the sideband will range from 10 kHz belowto 10 kHz above the carrier, extending from 702 to 722 kHz.

Frequency range of sideband =  712 10 kHz to  712  10 kHz

 702kHz to 722 kHz

Q.1

Ans.

A message signal m(t )  cos 2000t  4cos 4000 t modulates the carrier c(t )  cos 2f c t where f c  1MHz to produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of the detector circuit should satisfy [GATE 2011, IIT-Madras]

Sol.

(A) 0.5ms < RC< 1 ms

(B) 1s << RC< 0.5 ms

(C) RC  1s

(D) RC  0.5ms

Given : m(t )  cos 2000t  4cos 4000 t , f c  1MHz The maximum frequency of message signal is f m  2 kHz

Tm 

1 1   0.5msec f m 2 kHz

Tc 

1 1   1 sec f c 1MHz

For proper operation of envelop detector time constant RC should lie between Tm and

Tc . Tc  RC  Tm 1 1  RC  fc fm Where

1  Tc  Time period of carrier signal fc 1  Tm  Time period of message signal fm

Analog Communication Note : (i)

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1  RC to avoid message signal fluctuations at recovered output. fc

(ii) RC 

1 to avoid diagonal clipping. fm 1μs  RC  0.5ms

Ans.

(B)

Q.2

Suppose that the modulating signal is m(t )  2cos  2f mt  and the carrier signal is xc (t )  Ac cos  2fc t  . Which one of the following is a conventional AM signal without

over-modulation ?

[GATE 2010, IIT-Guwahati]

(A) x(t )  Ac m(t )cos  2fct  (B) x(t )  Ac 1  m(t ) cos  2fc t 

Ac m(t ) cos  2f c t  4 (D) x(t )  Ac cos  2f mt  cos  2fc t   Ac sin  2f mt  sin  2fct  (C) x(t )  Ac cos  2f c t  

Sol.

Option (A) represents DSB-SC which is over modulated waveform  ma  1 . Option (B) represents DSB-FC. Option (C) represents DSB-FC. Option (D) represents SSB-SC which is over modulated waveform  ma  1 . So check for option (B) and Option (C). From option (B) x(t )  Ac 1  m(t ) cos 2fc t x(t )  Ac 1  2cos 2f mt  cos 2fc t

…..(i)

Standard AM wave is given by, x(t )  Ac 1  ma cos 2f mt  cos 2fct

…..(ii)

Comparing equation (i) and (ii), we get

ma  2 (Over-modulated waveform) Ac m(t ) cos 2f c t 4 A x(t )  Ac cos 2f c t  c  2cos 2f m cos f c t 4 A A x(t )  Ac cos 2f c t  c cos 2  f c  f m  t  c cos 2  f c  f m  t 4 4 …..(iii) Standard AM wave in terms of modulation index is given by, From option (C)

x(t )  Ac cos 2f c t 

x(t )  Ac cos 2f c t 

Ac ma Am cos 2  f c  f m  t  c a cos 2  f c  f m  t 2 2 …..(iv)

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Amplitude Modulation

1 - 129

Comparing equation (iii) and (iv), we get

ma 1  2 4

ma 

1 2

[Under-modulated waveform]

Option (C) satisfied. Ans.

(C)

Q.3

For a message signal m(t )  cos  2f mt  and carrier of frequency f c , , which of the following represents a single, side-band (SSB) signal ? (A) cos  2fmt  cos  2fc t 

(B) cos  2fc t 

(C) cos  2  f c  f m  t 

(D) 1  cos  2f m t  cos  2f c t  

Sol.

cos 2  fc  fm  t represents USB is i.e. SSB-SC signal.

Ans.

(C)

Q.4

[GATE 2009, IIT-Roorkee]

1 1 A message signal given by m(t )    cos 1t    sin 2t is amplitude-modulated with 2  2  a carrier of frequency c to generate s(t )  1  m(t ) cos c t

What is the power efficiency achieved by this modulation scheme? [GATE 2009, IIT-Roorkee] (A) 8.33 % Sol.

(B) 11.11 %

(C) 20 %

Given s(t )  1  m(t ) cos c t

……(i)

1 1 cos 1t  sin 2t 2 2 Power efficiency is given by, Where m(t ) 

% 

ka2 m 2 (t )  100 1  ka2 m 2 (t )

% 

m 2 (t )  100 1  m 2 (t )

From equation (i), ka  1

Where m 2 (t )  mean square value = Power

m (t )  2

1 2  2

2



1 2 

0.25 100 1  0.25 %   20%

% 

Ans.

(C)

(D) 25 %

2

2



1 4

Analog Communication

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Ac cos c t  2cos mt cos c t . For demodulating the signal using envelope detector, the minimum value of Ac should

Q.5

Consider the amplitude modulated (AM) signal

Sol.

be: (A) 2 (B) 1 Given : S AM  Ac cos c t  2cos mt cos c t

[GATE 2008, IISc-Bangalore] (C) 0.5 (D) 0

  2 S AM  Ac cos c t 1  cos m t   Ac  Standard AM signal is given by, X AM  Ac cos ct 1  ma cos mt 

……(i)

……(ii)

Comparing equation (i) and (ii), we get 2 ma  Ac For an envelope detection, ma  1

2  1 Ac Ac (min)  2 Ans. Q.6

Ac  2

(A) In the following scheme, if the spectrum M(f) of m(t) is as shown, then the spectrum Y(f) of y(t) will be : [GATE 2007, IIT-Kanpur]

(A)

(B)

(C)

(D)

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1 - 131

Sol.

F m(t )  M( f )

F m(t ) cos 2Bt  

M ( f  B)  M ( f  B) 2

m(t )  m(t )  h(t )  m(t ) 

1 t

M ( f )  M ( f ) H ( f )  M ( f )[ j sgn( f )]

Amplitude Modulation

Analog Communication

1 - 132

F m(t )  M ( f ) F m(t ) cos 2Bt  

F m(t )sin 2Bt  

Ans.

(B)

M ( f  B)  M ( f  B ) 2

M ( f  B)  M ( f  B ) 2j

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GATE ACADEMY PUBLICATIONS ®

Q.7

Amplitude Modulation

1 - 133

The diagonal clipping in amplitude demodulation (using envelope detector) can be avoided if RC time-constant of the envelope detector satisfies the following condition, (here W is message bandwidth and

c is carrier frequency both in rad/sec) [GATE 2006, IIT-Kharagpur]

(A) RC  Sol.

1 W

(B) RC 

1 W

(C) RC 

1 c

(D) RC 

1 c

For proper operation of envelop detector time constant RC should lie between Tm and

Tc . Tc  RC  Tm 1 1  RC  c W Note : (i)

1  RC to avoid message signal fluctuations at recovered output. c

(ii) RC 

1 to avoid diagonal clipping. W

Ans.

(A)

Q.8

Consider a system shown in figure. Let X(f) and Y(f) denote the Fourier transforms of x(t ) and y (t ) respectively. The ideal HPF has the cutoff frequency 10 KHz. [GATE 2004, IIT-Delhi]

The positive frequencies where Y(f) has spectral peaks are : (A) 1 KHz and 24 KHz

(B) 2 KHz and 24 KHz

(C) 1 KHz and 14 KHz

(D) 2 KHz and 14 KHz

Analog Communication

1 - 134

GATE ACADEMY PUBLICATIONS ®

Sol.

A  x(t ) cos(210 103 t )

A  0.5[ X ( f  10)  X ( f  10)]

y (t )  B cos(213 103 t )

Y ( f )  0.5[ B( f  13)  B( f  13)]

Ans. Q.9

Y(f) has spectral peaks at positive frequency (2 kHz and 24 kHz). (B) A 100 MHz carrier of 1V amplitude and a 1 MHz modulating signal of 1 V amplitude are fed to a balanced modulator. The output of the modulator is passed through an ideal high-pass filter with cut-off frequency of 100 MHz. The output of the filter is 0

added with 100 MHz signal of 1V amplitude and 90 phase shift as shown in below figure. The envelope of the resultant signal is : [GATE 2004, IIT-Delhi]

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Amplitude Modulation

1 - 135

(A) Constant (C) Sol.

5 / 4  sin  2106 t 

(B)

1  sin  2106 t 

(D)

5 / 4  sin  2 106 t 

Given : Vc  1 V , fc  100 MHz

Vm  1 V , f m  1 MHz m(t )  Vm cos 2f m t  cos(2106 t ) c(t )  Vc cos 2f c t  cos(2108 t )

Output of balanced modulator = m(t).c(t) m(t).c(t)  0.5[cos(2101 106 t )  cos(2 99 106 t )] High pass filter characteristics is given by below figure,

Output of HPF  0.5cos(2101106 t ) y (t )  0.5cos(2101106 t )  cos(2100 106 t  900 ) y (t )  0.5cos(2 101 106 t ) sin(2 100  106 t ) y (t )  0.5cos(2 (100  1) 106 t ) sin(2 100  106 t ) y (t )  0.5[cos(2100 106 t ) cos(2106 t )  sin(2100  106 t ) sin(2106 t )] sin(2100 106 t ) y (t )  cos(2100 106 t )[0.5cos(2106 t )]

Note : Envelope of A cos 2f c t  B sin 2f c t is

sin(2100 106 t )[1  0.5sin(2106 t )] A2  B2

Envelope of y (t ) is given by, y (t )  {0.5cos(2 106 t )}2  {1  0.5sin(2 106 t )}2 E  0.52  1  sin(2 106 t ) E  1.25  sin(2 106 t )  5 / 4  sin(2106 t )

If we consider 900 phase shift then,

E

5  sin 2106 t 4

Ans.

(C)

Q.10

An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value of the time constant of the envelope detector is : [GATE 2003, IIT-Madras] (A) 1μsec

(B) 0.2 μsec

(C) 20μsec

(D) 500μsec

Analog Communication Sol.

1 - 136

GATE ACADEMY PUBLICATIONS ®

Given : f c  1MHz , f m  2 kHz

Tm 

1 1   0.5msec = 500 sec f m 2 kHz

Tc 

1 1   1 sec f c 1MHz

For proper operation of envelope detector time constant RC should lie between Tm and Tc .

Tc  RC  Tm 1 1  RC  fc fm 1μsec  RC  500 sec Ans. Q.11

(C) A DSB-SC signal is to be generated with a carrier frequency f c  1MHz using a nonlinear device with the input-output characteristic v0  a0 vi  a1vi3 where a0 and a1 are constants. The output of the nonlinear device can be filtered by an appropriate band-pass filter. Let vi  Ac' cos  2f c't   m(t ) where m(t) is the message signal. Then, '

the value of f c (in MHz) is : Sol.

[GATE 2003, IIT-Madras]

(A) 1.0 (B) 0.333 (C) 0.5 (D) 3.0 Given : Carrier frequency of output, fc  1 MHz , vi  Ac' cos  2f c't   m(t ) v0  a0 vi  a1vi3

Second term of v0 will generate the DSB-SC.

a1vi3  a1  Ac' cos 2f c't  m(t ) 

a  b

3

3

 a 3  b3  3ab  a  b   a 3  b 3  3a 2b  3ab 2

3a2b represents DSB-SC. 2

DSB-SC = 3  Ac' cos 2f c't  m(t ) ' 2 1  cos 4f t   c DSB-SC = 3  Ac'   m(t ) 2   3 ' 2 3 ' 2 DSB-SC = Ac m(t )  Ac m(t ) cos 4f c't 2 2 f c  2 f c'  1 MHz

 

 

f c'  0.5 MHz

Ans. Q.12

(C) A1 MHz sinusoidal carrier is amplitude modulated by a symmetrical square wave of period 100μsec sec. Which of the following frequencies will NOT be present in the modulated signal ? (A) 990 kHz (C) 1020 kHz

[GATE 2000, IIT-Kharagpur] (B) 1010 kHz (D) 1030 kHz

GATE ACADEMY PUBLICATIONS ®

Sol.

Amplitude Modulation

1 - 137

Given : c(t )  cos 2106 t , m(t )  symmetricalsquare wave of period 100 sec

f c  1MHz, f m 

1  10 kHz 100μsec

Symmetrical square wave in terms of Fourier series is given by, 1 2 1 1  m(t )   cos mt  cos3m  cos5mt  ...... 2  3 5 

s(t )  m(t )c(t ) 1 2 2 c(t )  c(t ) cos m t  c(t ) cos 3mt  ....... 2  3 Frequency of modulated signal is given by, s(t ) 

fc , fc  f m , fc  3 f m , f c  5 f m ,........ 1000,1000  10,1000  30,1000  50,.......... 1000, 1010, 990, 1030, 970, ………. Ans. Q.13

(C)

The amplitude modulated wave form s(t )  Ac 1  Ka m(t ) cos ct is fed to an ideal envelope detector. The maximum magnitude of K a m(t ) is greater than 1. Which of the following could be the detector output ?

Sol.

[GATE 2000, IIT-Kharagpur]

(A) Ac m(t )

(B) A 1  K a m(t ) 

(C) Ac 1  K a m(t ) 

(D) Ac 1  K a m(t ) 

2

2 c

2

Given : s(t )  Ac [1  ka m(t )]cos c t As

ka m(t )  1 that means modulation index ma is greater than 1 i.e. m a  1 . The

detector output is | Ac [1  ka m(t )] | . Ans.

(C)

Q.14

A message m(t) band-limited to the frequency f m has a power of Pm .The power of the output signal in given figure is :

Pm cos  2 P sin 2  (C) m 4 Let x(t )  m(t )cos 0t  cos(0t  ) (A)

Sol.

x(t ) 

m(t ) [cos(20t  )  cos ] 2

[GATE 2000, IIT-Kharagpur]

Pm 4 P cos 2  (D) m 4 (B)

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x(t ) is passed through a low pass filter. 0  2f m

Output signal 

m(t ) cos  2 2

m 2 (t )  m(t )  Power   cos    cos 2  2 4  

Given : m 2 (t )  Pm Power  Ans.

Pm cos 2  4

(D)



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1 - 139 Space for Notes

Amplitude Modulation

Analog Communication

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