COMMUNICATIONS...Fig. Communication system 1.1.1 TRANSMITTER . In electronic and telecommunication, a radio transmitter is an electronic device which, with the aid of an antenna, produces

COMMUNICATIONS

For INSTRUMENTATION ENGINEERING

ELECTRONICS & COMMUNICATION ENGINEERING

SYLLABUS Random signals and noise: probability, random variables, probability density function, autocorrelation, power spectral density. Analog communication systems: amplitude and angle modulation and demodulation systems, spectral analysis of these operations, super – heterodyne receivers: elements of hardware, realizations of analog communication systems; signal to noise ratio (SNR) calculations for amplitude modulation (AM) and frequency modulation (FM) for low noise conditions. Fundamentals of information theory and channel capacity theorem. Digital communication systems: pulse code modulation (PCM), differential pulse code modulation (DPCM) Digital modulation schemes: amplitude, phase and frequency shift keying schemes (ASK, PSK, FSK), matched filter receivers, bandwidth consideration and probability of error calculations for these schemes. Basics of TDMA, FDMA, CDMA and GSM.

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COMMUNICATIONS

Topics Page No

1. AMPLITUDE MODULATION

1.1 Introduction 01 1.2 Modulation 02 1.3 Types of Modulation 03 1.4 Amplitude Modulation 03 1.5 Power in AM Wave 07 1.6 Efficiency of AM Transmission 07 1.7 Types of AM Modulation 10 1.8 Generation of Amplitude Modulation Wave 16 1.9 Demodulation of Amplitude Modulated Signal 13 1.10 QAM (Quadrature Amplitude Modulation) 15

2. ANGLE MODULATION

2.1 Introduction 16 2.2 Frequency Modulation (FM) 16 2.3 Power in Frequency Modulated Signal 17 2.4 Types of FM 18 2.5 Bandwidth of Signal 19 2.6 FM Generated Method 19 2.7 FM Modulation 20 2.8 Phase Modulation 22 2.9 Capture Effect 23 2.10 Threshold Effect 23 2.11 Pre- Emphasis 23 2.12 De- Emphasis 24

Gate Questions

3. RADIO RECEIVER

3.1 Radio Receiver 43 3.2 Tuned Radio Frequency (TRF) Receiver 44 3.3 Super Heterodyne Receiver 44 3.4 Receiver Characteristics 47 3.5 Image Frequency & Its Rejection 48 3.6 Receiver ratio of CAP. in AM Receiver 49

4. NOISE

4.1 Introduction 51 4.2 Classification of Noise 51 4.3 Noise Temperature 534.4 Noise Factor of Amplifiers In Cascade 54

Gate Questions

25

56

5. PULSE DIGITAL MODULATION

5.1 Introduction 80 5.2 Pulse Code Modulation 80 5.3 Delta Modulation 85 5.4 Differential PCM 87

6. COMMUNICATION SYSTEM

6.1 Random Single And Noise 88 6.2 Types of Probability Density Function 896.3 Joint Moment About the Origin 92

7. DIGITAL CARRIER MODULATION

7.1 Introduction 99 7.2 Line Coding 99 7.3 Binary Amplitude-Shift Keying 99 7.4 Binary Phase-Shift Keying (BPSK) 100 7.5 Binary Frequency Shift Keying (BFSK) 101 7.6 Noise Analysis of Digital Communication 102 7.7 Probability of Error Calculation 103

8. DIGITAL MODULATION SCHEMES107 108

8.1 Digital Modulation Schemes 8.2 Binary Phase Shift Keying 8.3 Differential PSK(DPSK)

Gate Questions 109

9. INFORMATION THEORY9.1 Introduction 138 9.2 Information Content of a Symbol 138 9.3 Entropy 140 9.4 Information Rate 1409.5 Channel Capacity 141

Gate Questions

10. MULTIPLEXING

10.1 Introduction 149 10.2 Classification of Multiplexing 149 10.3 Frequency Division Multiplexing 149 10.4 Time Division Multiplexing 150

11. ASSIGNMENT 153

114

142

1.1 INTRODUCTION

Communication is the field of study concerned with the transmission of information through various means. It can also be defined as technology employed in transmitting messages. It can also be defined as the inter-transmitting the content of data (speech, signals, pulses etc.) from one node to another. Communication is a process, it consists of main three blocks Transmitter, Receiver, channel, through which exchange of information between two or more systems

Inputmessage Input

Transducer

Inputsignal

Transmitter

Transmitted signal

Channel

Distortion and noise

Receivedsignal

Receiver

Outputsignal

Outputtransducer

Outputmessage

Fig. Communication system

1.1.1 TRANSMITTER

In electronic and telecommunication, a radio transmitter is an electronic device which, with the aid of an antenna, produces radio waves. The transmitter itself generates a radio frequency alternating current, which is applied to the antenna. When excited by this alternating current, the antenna radiates radio waves. In addition to their use in broadcasting, transmitters are necessary component parts of many electronic devices that communicate by radio, such as cell phones, wireless computer networks; Bluetooth enabled devices, garage door opener, two-way radios in aircraft, ships and spacecraft, radar sets and navigational beacons.

1.1.2 CHANNEL

In telecommunications and computer networking, a communication channel or channel, refers either to a physical transmission medium such as a wire or to a logical connection over a multiplexed medium such as a radio. A channel is used to convey an information signal for example a digital bit stream from one or several senders (or transmitters) to one or several receivers.

Unbounded Air, Satellite

Channel

Bounded Transmission Line Wave Guide

Optical Fibre

The communication channels use two types of media: cable (twisted-pair wire, cable and fiber-optic cable) and broadcast (microwave, satellite, radio and infrared).

Note: A channel has a certain capacity for transmitting information, often measured by its bandwidth in Hz or its data in bits per second.

1.1.3 RECEIVER

A radio receiver is an electronic device that receives radio waves and converts the information carried by them to a usable form. It is used with an antenna. The antenna intercepts radio waves (electromagnetic waves) and converts them to tiny alternating currents which are applied to the receiver and the receiver extracts the desired information.

1 AMPLITUDE MODULATION

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1.2 MODULATION Modulation is defined as a process by virtue of which, some characteristic of a high frequency sinusoidal wave is varied in accordance with the instantaneous amplitude of the baseband signal. Two signals are involved in the modulation process. 1) The baseband signal which is to be transmitted to the receiver. The frequency of this signal is generally low. In the modulation process, this baseband signal is called the modulating signal. The waveform of this signal is unpredictable. For example, the waveform of a speech signal is random in nature and cannot be predicted. In this case, the speech signal is the modulating signal. 2) The other signal involved in the modulation is a high frequency sinusoidal wave. This signal is called the carrier signal or carrier. The frequency of the carrier signal is always much higher than that of the baseband signal. After modulation, the baseband signal of low frequency is transferred to the high frequency carrier, which carries the information in the form of some variations. After the completion of the modulation process, some characteristic of the carrier is varied such that the resultant variations carry the information. 1.2.1 NEED OF MODULATION Modulation is needed for following reasons: 1) Practical antenna length: Theory shows that in order to transmit a wave effectively, the length of the transmitting antenna should be approximately equal to the wavelength of the wave. Now,

8velocity 3×10wavelength = = metersfrequency frequency(Hz)

As the audio frequencies range from 20 Hz to 20 kHz, therefore, if they are transmitted

directly into space, the length of the transmitting antenna required would be extremely large. For instance, to radiate a frequency of 20 kHz directly into space, we would need an antenna length of

8 33×10 /20×10 =15000meters . This is too long antenna to be constructed practically. For this reason, it is impractical to radiate audio signal directly into space. On the other hand, if a carrier wave say of 1000 kHz is used to carry the signal, we need an antenna length of 300 meters only and this size can be easily constructed. 2) Operating range: The energy of a wave depends upon its frequency. The greater the frequency of the wave, the greater the energy possessed by it. As the audio signal frequencies are small, therefore, these cannot be transmitted over large distances if radiated directly into space. The only practical solution is to modulate a high frequency carrier wave with audio signal and permit the transmission to occur at this high frequency (i.e. carrier frequency). 3) Wireless communication: One desirable feature of radio transmission is that it should be carried without wires i.e. radiated into space. At audio frequencies, radiation is not practicable because the efficiency of radiation is poor. However, efficient radiation of electrical energy is possible at high frequencies (> 20 kHz). For this reason, modulation is always done in communication systems. 4) To multiplex data using FDM: Several message signals can be transmitted on a given channel, by assigning to each message signal an appropriate slot in the pass band of the channel. Take the example of AM broadcast, used for voice and medium quality music broadcast. The pass band of the channel used to 550 kHz to 1650 kHz. That is, the width of the pass band of the channel that is being used is 1100 kHz. If the required transmission bandwidth is taken as 10 kHz, then it is

Amplitude Modulation


possible for us to multiplex, at least theoretically, 110 distinct message signals on the channel and still be able to separate them individually as and when we desire because the identity of each message is preserved in the frequency domain. 1.3 TYPES OF MODULATION 1.3.1 ANALOG MODULATION

1) Amplitude Modulation (AM), 2) Frequency Modulation (FM), 3) Phase Modulation (PM), 4) Quadrature Amplitude Modulation (QAM) 1.3.2 DIGITAL MODULATION 1) Pulse Code Modulation (PCM), 2) Differential Pulse Code Modulation (DPCM), 3) Delta modulation (DM), 4) Sigma Delta Modulation

1.3.3 PULSE MODULATION 1) Pulse Amplitude Modulation (PAM), 2) Pulse Position Modulation (PPM), 3) Pulse Duration/Width Modulation (PDM/PWM) 1.3.4 DATA TRANSMISSION

1) Amplitude Shift Keying (ASK), 2) Frequency Shift Keying (FSK), 3) Phase Shift Keying (PSK), 4) Differential Phase Shift Keying (DPSK), 5) Quadrature Phase Shift Keying (QPSK), 6) Minimum Shift Keying (MSK), 7) Quadrature Amplitude Phase Shift Keying (QAPSK), 8) Gaussian Minimum Shift Keying (GMSK), 9) M-ray Amplitude Shift Keying, 10) M-ray Phase Shift Keying, 11) M-ray Frequency Shift Keying

1.4 AMPLITUDE MODULATION

When the amplitude of high frequency carrier wave is changed in accordance with the intensity of the message or information signal, it is called amplitude modulation. In amplitude modulation, only the amplitude of the carrier wave is changed in accordance with the intensity of the signal. However, the frequency of the modulated wave remains the same i.e. carrier frequency. Figure shows the principle of amplitude modulation. Fig. (i) shows the audio electrical signal. Fig. (ii) shows a carrier wave of constant amplitude. Fig. (iii) shows the amplitude modulated (AM) wave.

Note: The amplitudes of both positive and negative half – cycles of carrier wave are changed in accordance with the signal. For instance, when the signal is increasing in the positive sense, the amplitude of carrier wave also increases. On the other hand, during negative half-cycle of the signal, the amplitude of carrier wave decreases.



The following points are worth noting in amplitude modulation: 1) The amplitude of the carrier wave

changes according to the intensity ofthe signal.

2) The amplitude variations of the carrierwave are at the signal frequency fm.

3) The frequency of the amplitudemodulated wave remains the same i.e.carrier frequency fc.

1.4.1 MATHEMATICAL EXPRESSION FOR AMPLITUDE MODULATION

Let the modulating signal m(t) be expressed as:

m mm(t) = V cos tω Where, m(t) is instantaneous value of information signal, Vm is peak amplitude of information signal, m mω = 2πf is angular velocity at frequency fm & the carrier wave is represented as:

c c cv (t)=V cos ω tWhere, cv (t) is instantaneous voltage of

carrier cV is amplitude of carrier c cω = 2πf is

angular velocity at carrier frequency cf

In amplitude modulation, the amplitude cVof the carrier wave is varied in accordance with the intensity of the signal as shown in Fig. below. The peak amplitude of carrier after modulation at any instant is given by

c m[V kv (t)]+ . The carrier signal after modulation or the modulated signal is represented by the equation:

am c cv (t)=[V +k m(t)]cos ω t

⇒ m mam c c

c

kV cos v (t)=V [1+ ]cos ω tV

tω

⇒ am c a m cv (t)=V [1+m cos ]cos ω tω t

Where ma

c

Vm kV

= is called modulation

index or modulation factor. Generally k = 1

∴ ma

c

VmV

=

Now expanding the equation of AM wave we get,

a cam c c c m

m Vv (t) V cos ω t cos (ω )2

= + + +ω t

a cc m

m V cos (ω )2

−ω t

1.4.2 SPECTRUM OF AM MODULATED WAVES

The following points may be noted from the above equation of amplitude modulated wave: 1) The AM wave is equivalent to the

summation of three sinusoidal waves;one having amplitude cV and frequency

cf , the second having amplitude a cm V2

and frequency c m(f f )+ and the third

having amplitude a cm V2

and frequency

c m(f f )− . 2) The AM wave contains three

frequencies• cf the carrier frequency• c m(f f )+ the upper sideband frequency• c m(f f )− the lower sideband frequency

Thus, the process of modulation doesnot change the original carrierfrequency but produces two newfrequencies c m(f f )+ and c m(f f )− which are called sideband frequencies.

1.4.3 MODULATION FACTOR

An important consideration in amplitude modulation is to describe the depth of modulation i.e. the extent to which the



amplitude of carrier wave is changed by the signal. This is described by a factor called modulation factor or modulation index which may be defined as the ratio of change of amplitude of carrier wave to the amplitude of normal carrier wave i.e.

aamplitude change of carrier wavemodulation index m =

normal carrier amplitude (unmodulated)

The value of modulation factor depends on signal. The change in carrier amplitude is equal to the amplitude of modulating signal therefore

ma

c

VmV

=

Fig. below shows amplitude modulation for different values of modulation factor m. i) When signal amplitude is zero, the

carrier wave is not modulated; the amplitude of carrier wave remains unchanged.

∴ ac

0m = = 0 or 0%V

ii) When signal amplitude is equal to the carrier amplitude i.e. m cV V= , the amplitude of carrier varies between

c2V and zero & amplitude change of carrier c c c=2V V V− = .

∴ ca

c

Vm = =1 or 100%V

In this case, the carrier is said to be 100% modulated.

iii) When the signal amplitude is one-half

the carrier amplitude i.e. m cV 0.5V= , the amplitude of carrier wave varies between c1.5V & c0.5V & amplitude change of carrier c c c= 1.5V V 0.5V− = .

∴ ca

c

0.5Vm = =0.5 or 50%V

In this case, the carrier is said to be 50% modulated.

iv) When the signal amplitude is 1.5 times

the carrier amplitude i.e. m cV 1.5V= , the maximum value of carrier wave becomes c2.5V & amplitude change of carrier wave c c c2.5V V 1.5V= − = .



∴ ca

c

1.5Vm = =1.5 or 150%V

In this case, the carrier is said to be 150% modulated i.e. over-modulated.

Note: 1) When the carrier is modulated to a

small degree (i.e. small m), the amount of carrier amplitude variation is small. Consequently, the audio signal being transmitted will not be very strong. The greater the degree of modulation (i.e. m), the stronger and clearer will be the audio signal. It may be emphasized here that if the carrier is over modulated( )i.e. m > 1 , distortion will occur during reception. The AM waveform is clipped and the envelope is discontinuous. Therefore, degree of modulation should never exceed 100%.

2) The equation for modulation index can

be derived in another for as:

max minm

V VV =2−

And max minc

V VV2+

=

∴

max min

ma

max minc

V VV 2m V VV

2

−

= =+

∴ max mina

max min

V VmV V

−=

+

Example A carrier of 100 V and 1200 kHz is modulated by a 50 V, 1000 Hz sine wave signal. Find the modulation factor. Solution Modulation factor,

ma

c

V 50m 0.5V 100

= = =

Example The maximum peak-to-peak voltage of an AM wave is 16 mV and the minimum peak-to-peak voltage is 4 mV. Calculate the modulation factor. Solution From the figure below

Maximum voltage of AM wave is

max16V = = 8mV2

Minimum voltage of AM wave is

min4V 2mV2

= =

∴ a8 2m 0.68 2−

= =+



1.4.4 MODULATION INDEX FOR MULTI-TONE AMPLITUDE MODULATION When more than one modulating signals are used for modulating the carrier wave, it is called multi tone modulation. Let

1 2 2 3 4 nm (t), m (t), m (t), m (t), m (t) - - - - - - - -m (t)are the modulating signal, then the modulated signal is given by

am c 1 2 2 3

4 n c

v (t)=[V m (t) m (t) m (t) m (t)m (t) - - - - - m (t))]cos ω t

+ + + + ++

The modulation index corresponding to each modulating signal is given by

m1a1

c

m2a2

c

m3 mna3 an

c c

Am ,AAm ,A

A Am - - - - - mA A

=

=

= =

The overall or effective modulation index is given by

2 2 2 2a a1 a2 a3 anm m m m .... m= + + +

1.5 POWER IN AM WAVE The equation for AM wave is

am c c

a cc m

a cc m

v (t) V cos ω tm V cos (ω )

2m V cos (ω )

2

=

+ +

+ −

t

t

ω

ω

Equation of AM wave reveals that it has three components of amplitude cV ,

a cm V2

And a cm V2

. Clearly, power output

must be distributed among these components.

1) Power of carrier, 2c

cVP2

=

2) Power in upper sideband, 2 2a c

USBm VP

8= >

3) Power in lower sideband, 2 2a

LSBm VP

8= c

Therefore the total power in an AM signal is

T c USB LSBP P P P= + +

⇒2 2 2 2 2c a c a c

TV m V m VP2 8 8

= + +

⇒2 2 2c a c

TV m VP2 4

= +

⇒2 2 2c a a

T cV m mP 1 P 12 2 2

= + = +

Note: 1) Out of the total power, the power

carried by the sideband is the only useful component.

2) From the equation of total power, 2aT

c

mP 1P 2

= +

If cI and TI are the r.m.s values of un modulated current and total modulated current and R is the resistance through which these current flow, then substituting

2T TP I R= & 2

c cP I R= , we get

T

2 2a

2c

I m1I 2

= +

⇒2a

T cmI I 1+2

=

1.6 EFFICIENCY OF AM TRANSMISSION The efficiency of transmission is the ratio of useful power transmitted (power in sidebands) to total power transmitted i.e.

useful power transmittedefficiency ( ) 100%total power transmitted

= ×η

⇒ SB USB LSB

T T

P P Pη 100%P P

+= = ×

⇒

2 2a c

2a

22 2ac a

m Vm4η 100%

2 mV m12 2

= = ×+

+



Note: In AM transmission, the efficiency is maximum when

am 1=2

max 2

1η 100% 33.33%2 1

= × =+

Example A carrier wave of 500 watts is subjected to 100% amplitude modulation. Determine: (i) power in sidebands (ii) power of modulated wave. Solution For 100% modulation am 1= (i)Sideband power,

2a

SB USB LSB cmP P P P2

= + =

∴

2

SB c1P P2

1 500 250watts2

= =

× =

Thus there are 125W in upper sideband and 125W in lower sideband. (ii) Power of AM wave,

2a

T cmP P 12

= +

21500 1 750watts2

= + =

1.7 TYPES OF AM MODULATION

1.7.1 DOUBLE SIDEBAND FULL CARRIER In double sideband full carrier AM (AM-DSBFC) both the sidebands are transmitted along with the carrier. It is also called as conventional AM signal. The mathematical expression for AM-DSB/FC is

a cam c c c m

m Vv (t) V cos ω t cos (ω )t2

= + +ω

a cc m

m V cos (ω )t2

+ −ω

Power saving: In AM-DSB/FC both the sidebands are transmitted along with carrier no power saved during transmission. ∴ Power saving 0%=

Bandwidth: The bandwidth required for transmission of AM-DSB/FC is twice the maximum frequency of modulating signal (In case if modulating signal contains many frequency components, then the bandwidth will be determined by the maximum frequency). ∴ BW 2 mf=

1.7.2 DOUBLE SIDEBAND SUPPRESSED CARRIER

We know that changing the degree of modulation of a particular carrier does not change the amplitude of the carrier component itself. Instead, the amplitude of the sidebands changes, thus altering the amplitude of the composite wave. Since the amplitude of the carrier component remains constant, all the transmitted information is contained in the sidebands. This means that the considerable power transmitted in the carrier is essentially wasted. For improved power efficiency, the carrier component may be suppressed (usually by the use of a balanced modulator circuit), so that the transmitted wave consists only of the upper and lower sidebands. This type of modulation is called double sideband suppressed carrier, or DSB-SC. The carrier must be reinserted at the receiver, however, to recover this modulation. In the time and frequency domains, DSB-SC modulation appears as shown in figure. The mathematical expression for AM-DSB/SC is

am cv (t) v (t).m(t)=

a c a cc m c m

m V m Vcos (ω )t cos (ω )t2 2

= +ω + −ω

Note: An AM-DSB/SC signal can be generated through a Product modulator as shown in fig.



Power saving: In AM-DSB/SC both the sidebands are transmitted & carrier is suppressed i.e. carrier power is saved.

( )c

2 2a a

c

P 2Power saving 100%m 2 mP 12

= = × ++

Power saving when am 1= is

( )2Power saving 100% 66.67%

2 1= × =

+ Bandwidth: The bandwidth requirement for AM-DSB/SC is equal to AM-DSB/FC as both the sidebands are transmitted. ∴

BW 2 mf=

1.7.3 SINGLE SIDEBAND FULL CARRIER In single sideband full carrier one of the sidebands is suppressed, written as AM-SSB/FC. Since each sideband is displaced from the carrier by the same frequency, and since the two sidebands have equal amplitudes, it follows that any information contained in one must also be in the other. And, more importantly, halves the transmission bandwidth (frequency spectrum width). Power Saving: Eliminating one of the sidebands cuts the power requirement.

2a

c

2a

c

mP4Power savingmP 12

= +

2a

2a

m4 100%m12

= × +

Power saving when am 1= is

14Power saving 100% 16.67%

112

= × = +

Bandwidth: The bandwidth requirement for AM-SSB/FC is half of double sideband modulated signal as only one sideband is transmitted. ∴

mBW=f

1.7.4 SINGLE SIDEBAND SUPPRESSED CARRIER In SSB-SC (single side band suppressed carrier) one of the side bands & carrier are suppressed. This type of transmission is power efficient & bandwidth efficient. Power Saving: Eliminating one of the sidebands & carrier cuts the power requirement to a large extent.

∴

2 2a a

c

2 2a a

c

m mP 1 14 4

Power saving 100%m mP 1 12 2

+ +

= = × + +

Power saving when am 1= is 54Power saving 100% 83.33%32

= × =

Bandwidth: The bandwidth requirement for AM-SSB/SC is half of double sideband modulated signal as only one sideband is transmitted. ∴

BW mf=

System Transmitted Power

Ps (saved power)

AMDBS/FC

2a

t cmP P 12

= +

-

AM-DSB/SC

2a

cmP2

PC



AM-SSB/FC

2a

cmP 14

+

2a

cmP4

AM-SSB/SC

2a

cmP4

2a

cmP 14

+

BW % power Saving Complexity

2ωm - Min. 2ωm 67% ↓ ωm 16% ↓ ωm 83% Max.

1.8 GENERATION OF AMPLITUTE MODULATION WAVE The circuit that generates the AM waves is called as amplitude modulator and we will discuss two modulator circuits namely: i) Square law modulator ii) Switching modulator Both of them use a non-linear element such as a diode for their implementation. A non-linear device is the device with a non-linear relation between its current and voltage. Both these modulators are low power modulator circuits. 1.8.1 SQUARE LAW MODULATOR The square law modulator circuit has been shown in figure. It consists of the following: i) A non-linear device ii) A band pass filter iii) A carrier source and modulating signal

The modulating signal and carrier are connected in series with each other and their sum v1(t) is applied at the input of the non-linear device, such as diode, transistor etc

Thus, ( ) ( ) ( )1 c cv t x t E cos 2 f t= + π … (i)

The input output relation for non-linear device is as under:

( ) ( ) ( )22 1 1v t av t bv t= + … (ii)

Where a and b are constants Substituting the expression for vt(t)we get,

( ) ( ) ( )( ) ( )

2 c c

2c c

v t a[x t E cos 2 f t

b[x t E cos 2 f t ]

= + π

+ + π

Or ( ) ( )

( ) ( )

( ) ( ) ( )

22 c c

1 2 32 2

c c c c

4 5

v t ax t aE cos 2 f t bx t

2bx t E cos 2 f t bE cos 2 f t

= + π +

+ π + π

The five terms in the expression for v2(t) are as under: Term1: ( )ax t Modulating signal→ Term2: ( )c caE cos cos 2πf t Carrier signal→ Term b x2(t) →Squared modulating signal Term4: 2b x(t)Ec cos(2πfct) → AM wave with only sidebands Term5:bEc2cos2(2πfct) → Squared carrier Out of these five terms, terms 2 and 4 are useful whereas the remaining terms are not useful. Let us club terms 2, 4 and 1, 3, 5 as follows to get,

( ) ( ) ( ) ( )

( ) ( ) ( )

2 2 22 c c

unuseful terms

c c c c

useful terms

v t ax t bx t bE cos 2 f t

aE cos 2 f t 2bx t E cos 2 f t

= + + π

+ π + π

The LC tuned circuit acts as a bandpass filter. Its frequency response is shown in figure which shows that the circuit is tuned to frequency fc and its bandwidth is equal to 2fm. This bandpass filter eliminates the unuseful terms from the equation of v2(t).



Hence, the output votage v0(t) contains only the useful terms.

( ) ( ) ( ) ( )0 c c c cV t aE cos 2 f t 2bx t E cos 2 f t= π + π

Or ( ) ( ) ( )0 c c cV t aE 2bx t E cos 2 f t = + π Therefore,

( ) ( ) ( )0 c c2bV t aE 1 x t cos 2 f ta

= + π … (iii)

Comparing this with the expression for standard AM wave i.e. ( ) ( ) ( )c cs t E 1 mx t cos 2 f t = + π , we find that the expression for V0(t) of equation (iii) represents an Am wave with m=(2b/a). Hence the square law modulato produces an AM wave. 1.8.2 SWITCHING MODULATOR The switching modulator using a diode has been shown in figure.This didoe is assumed to be operating as a switch. The modulating signal x(t) and the sinusoidal carrier signal c(t)are connected in series with each other. Therefore, the input voltage to the diode is given by,

( ) ( ) ( ) ( ) ( )1 c cv t c t x t E cos 2 f t x t= + = π + … (i) The amplitude of carrier is much larger than that of x(t) and c(t) decides the status of the diode (ON or OFF).

Working Operation and Analysis:

We assume that the diode acts as an ideal switch. It acts as a closed switch when it is forward biased in the positive half cycle of the carrier and offers zero impedance.Whereas ‘D’ acts as an open switch when it is reverse biased in the negative half cycles of the carrier and offers an infinite impedance.The equivalent circuit in the positive and negative half cycles of the carrier have been shown in (a) and (b) respectively.

Therefore, the output votage ( ) ( )2 1v t v t= in the positive jhalf cycle of c(t) and ( )2 0v t = in the negative half cycle of c(t),

Therefore, ( ) ( ) ( )( )

12

v t for c t 0v t

0 for c t 0 >= <

… (ii)

In other words, the load votage ( )2v t varies

periodically between the values ( )1v t and zero at the rate equal to carrier frequecny

cf the approximate transfere characteristics of the diode-loadresistor combbination has been shown in fig. (b) We can expres ( )2v t mathmatically as under:

( ) ( ) ( )( ) ( ) ( )

2 1 p

c c p

v t v t .g t

x t E cos 2 f t g t

=

= + π … (iii)

Where, ( )pg t is a periodic pulse train of duty cycle equal to one hald cycle period i.e., 0 / 2T (where 0 1/ cT f= )



Let us express ( )pg t with othe help of Foureier series as under:

( ) ( ) ( )n 1n

p cn 1

11 2g t cos 2 f t 2n 12 2n 1

−=∞

=

− = + π − π −∑

…(iv)

( ) ( )p c1 2g t cos 2 f t odd harmonic components2

= + π +π

… (v) Substituting gp(t) into equation (vi), we get

( ) ( ) ( )2 c cv t x t E cos cos 2 f t = + π

( ) ( )n 1n

cn 1

11 2 cos 2 f t 2n 12 2n 1

−=∞

=

− + π − π − ∑

Therefore, ( ) ( ) ( )2 c cv t x t E cos cos 2 f t = + π

( )c1 2 cos 2 f t odd harmonics2 + π + π

… (vi)

The odd harmonics in this expression are unwanted, and therfore, are assumed to be eliminated.

( ) ( )2

modulating signal

1v t x t2

=

( ) ( ) ( )c c c

AM wave

1 2 E cos 2 f t cos 2 f t x t2

+ π + ππ

( )nd

2c c

2 harmonic of carrier

2 E cos 2 f t+ ππ

In this expression, the firstand the fourth terms are unwanted terms whereas the second and third terms together represent the AM wave. Clubing the second and third terms together , we obtain

( ) ( ) ( )c2 c

c

E 4v t 1 x t cos 2 f t2 E

unwanted terms

= + π π

+

This is the required expression for the AM

wave withc

4mE

= π

. The unwanted terms

can be eliminatd using a band-pass filter(BPF). 1.8.3 RING MODULATOR Ring modulator is the most widely used product modulator for generating DSBSC wave and is shown below. The four diodes form a ring in which they all point in the same direction. The diodes are controlled by square wave carrier c(t) of frequency fc , which is applied longitudinally by means of two center-tapped transformers.

Ring Modulator

Assuming the diodes are ideal, when the carrier is positive, the outer diodes D1 and



D2 are forward biased where as the inner diodes D3 and D4 are reverse biased, so that the modulator multiplies the base band signal m(t) by c(t). When the carrier is negative, the diodes D1 and D2 are reverse biased and D3 and D4 are forward, and the modulator multiplies the base band signal –m (t) by c (t). Thus the ring modulator in its ideal form is a product modulator for square wave carrier and the base signal m (t). The square wave carrier can be expanded using Fourier series as

( ) ( ) ( )n 1

cn 1

14c t cos(2πf t 2n 1 )π 2n 1

−∞

=

−= −

−∑

Therefore the ring modulator output is given by s(t) = m(t)c(t)

( ) ( ) ( )( )n 1

cn 1

14s t m(t) cos cos 2πf t 2n 1π 2n 1

−∞

=

−= −

− ∑

From the above equation it is clear that output from the modulator consists entirely of modulation products. If the message signal m (t) is band limited to the frequency band– w < 𝑓𝑓 < 𝑤𝑤 , the output spectrum consists of side bands centered at fc . 1.8.4 GENERATION OF SSB 1) Filter method (frequency selectivity

filtering) 2) Phase shift method 3) Third Method 1.9 DEMODULATION OF AMPLITUDE MODULATED SIGNAL The process of detection or demodulation is the process of recovering the message signal from the received modulated signal. This means that the process of detection is exactly opposite to that of modulation.

1.9.1 SQUARE LAW DETECTOR

Fig. shows the block diagram ofa square law detector Working Operation and Analysis: The input output characteristics i.e., the characteristics of a square law device is non linear and it is expreassed matematically as under:

( ) ( )22 1 1v t av bv t= + … (i)

Where ( )1v t = input votage to the detector =AM

wave Therefore, ( ) ( ) ( )1 c cv t E 1 mx t cos 2 f t = + π

Substituting for ( )1v t in equation (i) , we get ,

( ) ( ) ( )2 c cv t aE 1 mx t cos 2 f t = + π +

( ) ( )22 2c cbE 1 mx t cos 2 f t + π …

But, [ ]2 1cos 1 cos 22

θ = + θ

Therefore, ( ) ( )2c c

1cos 2 f t 1 cos 4 f t2 π = + π

Substituitng this, we get,

( ) ( ) ( )2c

2 c cbEv t aE 1 mx t cos 2 f t

2 = + π +

( ) ( ) ( )2 2c1 2mx t m x t 1 cos 4 f t + + + π

Out of these terms, the only desired term is 2cbE mx(t) which is due to the 2

1bv term. Hence, the name of this detector is square law detector. This desired term is extracted by using a lowpasss filter (LPF) after the diode as shown in fig.Thus, after the low pass filter,we get,



( ) ( ) ( )20 cv t bE m x t= … (iv)

This mean that we have recovered the message signal x(t) at the output of the detector. 1.9.2 ENVELOPE DETECTOR The envelop detecor is a simple and very effiecent device which is suitable for the detection of a narrowband AM signal. A narrowband AM wave is the one in which the carrier frequecy fc is much higher as compared to the bandwidth of the modulating signal. An envelop detector produces an output signal that follows the envelop of the input AM signal exactly. The envelope detector is used in all the commercial AM radio receivers. Circuit diagram: The circuit diagram of the envelope etector is shown in figure.It consists ofa diode and RC filter.

Working Operation: The standard AM wave is applied at the input of the detector. In every positive half cycle of the input, the detector diode is froward baised.It will charge the filter capacitor C connected across the load resistance R to almost the peak value of the input voltage. As soon as the capacitor charges to the peak value, the diode stops conducting. The capacitor will then discharge through R between the positive

peaks as shown in the figure. 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.

The input-output waveforms for the envelop detector are shown in figure. It shows the charging & discharging of the filter capacitor and the approximate output voltage. It may be observed from these waveforms that the envelope of the AM wave is being recovered successfully. We assume that the diode is ideal which presents a zero resistance when it is ON and infinite resistance when it is OFF. We also assume that the AM wave aplied to the input of the detector is supplied by a source having internal resistance Rs . Selection of the RC time constant : The capacitor charges through D and Rs when thediode is ON and it discharges through R when the diode is OFF. The charging time constant RsC should be short compared to the carrier period 1/fc

Thus , sc

1R C=f

On the other hand, the

discharging time constant RC should 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 the envelope.

Therefore,c

1 1=RC=f W

Where, W = maximum modulating frequency



1.9.2.1 DISTORTIONS IN THE ENVELOPE DETECTOR OUTPUT

There are two types of distortions which can occur in the detector output.They are as under: i) Diagonal clipping ii) Negative peak clipping 1) Diagonal cliping: This type of distrotion occurs when the RC time constant of the load circuit is too long. Due to this the RC circuit cannot follow the fast changes in the modulating envelope. The diagonal clipping has been shown in figure.

2) Negative peak clippping: 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, Hence, at higher depths of modulation of the transmitted signal, the over modulation (more than 100% moudlation) 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.

Remedy:The only to reduce or eliminate the distrtions is to choose the RC time constants properly as discussed earlier. 1.10 QAM (QUADRATURE AMPLITUDE MODULATION)

Quadrature amplitude modulation (QAM) is both an analog and a digital modulation scheme. It conveys two analog message signals, or two digital bit streams, by changing (modulating) the amplitudes of two carrier waves, using the amplitude-shift keying (ASK) digital modulation scheme or amplitude modulation (AM) analog modulation scheme. The two carrier waves, usually sinusoids, are out of phase with each other by 90° and are thus called quadrature carriers or quadrature components, hence the name of the scheme. The modulated waves are summed, and the final waveform is a Combination of both phase-shift keying (PSK) and amplitude-shift Keying (ASK), or (in the analog case) of phase modulation (PM) and amplitude modulation.

- /2

~

Low-pass filter

- /2

~

Low-pass filterm (t)2

m (t)1

cos Ct

2sin Ct

2cos Ct

m (t)1

m (t)2

x (t)1

x (t)2

QAM(t)

sin Ct

Quadrature amplitude multiplexing

MATHEMETICAL EXPRESSION FOR QAM When transmitting two signals by modulating them with QAM, the transmitted signal will be of the form.



Q.1 In a FM system, a carrier of 100MHz is modulated by a sinusoidal signal of 5 kHz. The bandwidth by Carson’s approximation is 1MHz. If y(t) = (modulated waveform)3 then by using Carson’s approximation, the bandwidth of 𝑦𝑦(𝑡𝑡) around 300MHz and the spacing of spectral components are respectively. a)3MHz,5kHz b)1MHz,15kHz c) 3MHz, 15kHz d)1 MHz,5 kHz

[GATE-2000]

Q.2 The Hilbert transform of 1 2cos ω t sin ω t+ is

a) 1 2sinω t-cosω t b) 1 2sin ω t cos ω t+c) 1 2cos ω t sin ω t− d) 1 2sin ω t sin ω t+

[GATE-2000]

Q.3 A message m(t) band limited to the frequency fm has a power of Pm The power of the output signal in the figure is.

a) mP cosθ2

b) mP4

c)2

mP sin θ4

d)2

mP cos θ4

[GATE-2000]

Q.4 The amplitude modulated wave form ( ) [ ]C a cs t =A 1+K m(t) cosω t is fedto an ideal envelop detector. The maximum magnitude of Kam(t) is greater than1. Which of the following could be the detector output? a) CA m(t)

b) [ ]22c aA 1 K m(t)+

c) C aA 1 K m(t) +

d) 2C aA 1 K m(t)+

[GATE-2000]

Q.5 A band limited signal is sampled at the Nyquist rate. The signal can be recovered by passing the samples through a) an RC filterb) an envelope detectorc) a PLLd) an ideal low –pass filter with the

appropriate bandwidth[GATE-2001]

Q.6 In the figure m(t) = 2sin 2πtt

( )s t cos 200πt= and n(t) = sin199πtt

The output y(t) will be

a) sin 2πtt

b) sin 2πt sin sin πt cos3πtt t

+

c) sin 2πt sin 0.5sin 0.5πt cos1.5πtt t

+

d) sin 2πt sin sin πt cos 0.75πtt t

+

[GATE-2002]

Q.7 An angle –modulated signal is given by

( )62×10 t+30sin150t+

s t cos2π40cos150t

The maximum frequency and phase deviations of s(t) are

GATE QUESTIONS


a) 10.5 kHz, 140π rad b) 6 kHz, 80 π rad

c) 10.5kHz, 1000π rad d) 7.5kHz, 100 π rad

[GATE-2002]

Q.8 A 1 MHz sinusoidal carrier is amplitude modulated by a symmetrical square wave of period 100μsec.Which of the following frequencies will NOT be present in the modulated signal?

a) 990 kHz b) 1010kHz c) 1020 kHz d) 1030 kHz

[GATE-2002]

Common Data for Questions 110 & 111 Let ( ) ( )3m t =cos 4π×10 t be the message

signal & ( ) ( )6c t =5cos 2π×10 t be the

carrier. Q.9 c(t) and m(t) are used to generate

an AM signal. The modulation index of the generated AM signal is 0.5 then the quantity Total sidebands power

Carrier power is

a) 1/2 b) 1/4 c) 1/3 d) 1/8

[GATE-2003]

Q.10 c(t) and m(t) are used to generate an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term

( )3cos 2π 1008×10 t in the FM signal

(in terms of the Bessel coefficients) is

a) 45J (3) b) 85 J (3)2

c) 85 J (4)2

d) 45J (3)

[GATE-2003]

Q.11 A DSB-SC signal is to be generated with a carrier frequency fc = 1MHz using anon-linear device with the input-output characteristic

30 0 i 1 iV a v a v= + Where a0 and a1 are

constants. The output of the non-linear device can be filtered by an appropriate band –pass filter. Let ( )i i

i c cV A cos 2πf t m(t)= + where m(t) is the message signal .Then the value of i

cf (in MHz) is a) 1.0 b) 0.333 c) 0.5 d) 3.0

[GATE-2003] Q.12 The input to a coherent detector is

DSB-SC signal plus noise. The noise at the detector output is

a) the in-phase component b) the quadrature-component c) zero d) the envelope

[GATE-2003] Q.13 A super heterodyne receiver is to

operate in the frequency range 550 kHz-1650 kHz with the intermediate

frequency of 450 kHz. Let max

min

CRC

=

denote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then

a) R=4.41, I=1600 b) R=2.10, I =1150 c) R=3.0, I=1600

d) R=9.0, I=1150 [GATE-2003]

Q.14 Choose the correct one from among

the alternative a, b, c, d after matching an item in Group 1 with the most appropriate item is Group 2. Group1

P. Ring modulator

Analog Communication System


Q. VCO R. Foster-Seeley discriminator S. Mixer Group 2 1. Clock recovery 2. Demodulation of FM 3. Frequency conversion 4. Summing the two inputs 5. Generation of FM 6. Generation of DSB-SC a) P-1; Q-3; R-2; S-4

b) P-6; Q-5; R-2; S-3 c) P-6; Q-1; R-3; S-2

d) P-5; Q-6; R-1; S-3 [GATE-2003]

Q.15 Consider a system shown in the

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

The positive frequency where Y(f) has spectral peaks are a) 1kHz and 24kHz

b) 2kHz and 24 kHz c) 1 kHz and 14 kHz

d) 2 kHz and 14 kHz [GATE-2004]

Q.16 Two sinusoidal signals of same

amplitude and frequencies 10 kHz and 10.1 kHz are added together. The combined signal is given to an ideal frequency detector. The output of the detector is

a) 0.1kHzsiniusoid b) 20.1 kHz sinusoid c) a linear function of time d) a constant

[GATE-2004]

Q.17 A 100 MHz carrier of w v amplitude and a 1MHz modulation 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 cutoff frequency of 100MHz. The output of the filter is added with 100 MHz signal of 1 V amplitude and 90° phase shift as shown in the figure. The envelope of the resultant signal is

a) constant b) ( )61+sin 2π×10 t

c) ( )65/4-sin 2π×10 t

d) ( )65/4+cos 2π×10 t

[GATE-2004]

Q.18 An AM signal and a narrow-band FM signal with identical carriers, modulating signals and modulation indices of 0.1 are added together. The resultant signal can be closely approximated by

a) broadband FM b) SSB with FM c) DSB-SC d) SSC without carrier

[GATE-2004]

Q.19 An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 1MHz and 2kHz respectively. An appropriate value for the time constant of the envelope detector is

a) 500μsec b) 0.5μsec c) 0.2μsec d) 1μsec

[GATE-2004]



Q.20 A carrier is phase modulated (PM) with frequency deviation of 10kHz by a single tone frequency of 1 kHZ. If the single tone frequency is increased to 2kHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is

a) 21kHz b) 22kHz c) 42kHz d) 44kHz

[GATE-2005]

Q.21 A device with input x (t) and output y (t) is characterized by 𝑦𝑦(𝑡𝑡) =𝑥𝑥2(𝑡𝑡) An FM signal with frequency deviation of 90 kHz and modulating signal bandwidth of 5 kHz is applied to this device. The bandwidth of the output signal is

a) 370kHz b) 190kHz c) 380 kHz d) 95 kHz

[GATE-2005]

Q.22 Which of the following analog modulation scheme requires the minimum transmitted power and minimum channel bandwidth?

a) VSB b) DSB-SC c) SSB d) AM

[GATE-2005]

Q.23 Find the correct match between group 1 and group 2

Group 1 P. ( ){ } c1+km t A sin(ω t)

Q. ( ) ckm t A sin(ω t) R. ( ){ }cA sin ω t+km t

S. ( )t

cA sin ω t k m t dt−∞

+

∫

Group 2 W. Phase modulation

X. Frequency modulation Y. Amplitude modulation Z.DSB-SC modulation a)P-Z,Q-Y,R-X,S-W b)P-W,Q-X,R-Y,S-Z c)P-X,Q-W,R-Z,S-Y d)P-Y,Q-Z,R-W,S-X

[GATE-2005]

Common Data for Questions 24 & 25

Consider the following Amplitude Modulated (AM) signal where

( ) ( )m AM m cf <Bx t =10 1+0.5sin2πf t cos2πf t Q.24 The average side –band power for

the AM signal given above is a) 25 b) 12.5 c) 6.25 d) 3.125

[GATE-2006]

Q.25 The AM signal gets added to a noise with Power Spectral Density nS (f) given in the figure below. The ratio of average sideband power to mean noise power would be:

a)

0

258N B

b)0

254N B

c)0

252N B

d) 0

25N B

[GATE-2006]

Common Data for Questions 26 & 27 Let g(t)=p(t)*p(t) where * denotes convolution & ( ) ( )

xp t u t u(t 1) lim

→∞= − − with u(t) Being the

unit step function Q.26 The impulse response of filter

matched to the signal ( ) ( ) ( )s t =g t -δ t-2 *g(t) is given as:

a) s(1-t) b) -s(1-t) c) -s(t) d) s(t)

[GATE-2006]

Q.27 An Amplitude Modulated signal is given as

( ) ( ) ( )AM cx t =100 p t +0.5g t cosω t in the interval 0 t 1≤ ≤ one set of possible values of the modulating



signal and modulation index would be

a) t, 0.5 b)t, 1.0 c) t, 2.0 d)t2, 0.5

[GATE-2006]

Q.28 A massage signal with bandwidth 10kHz is Lower-Side Band SSB modulated with carrier frequency

6c1f 10 Hz= The resulting signal is

then passed though a Narrow- Band Frequency Modulator with carrier frequency 9

c2f 10 Hz= The bandwidth of the output would be a) 4×104Hz b) 2×106Hz c) 2×109Hz d) 2×1010Hz

[GATE-2006]

Q.29 The diagonal clipping in Amplitude Demodulation (using envelop detector) can be avoided if RC time – constant of the envelope detector satisfies the following condition, (here W is message bandwidth and ω is carrier frequency both in rad/sec)

a) 1RCW

< b) 1RCW

>

c) 1RCω

< d) 1RCω

>

[GATE-2006] Q.30 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

a)

b)

c)

d)

[GATE-2007]

Q.31 The signal c m ccosω t+0.5cosω t sinω t is a) FM only

b) AM only c) Both AM and FM

d) Neither Am nor FM [GATE-2008]

Q.32 Consider the frequency modulated

signal

( )

( )

52π×10 t+5sinsin 2π×1500t10cos

+7.5sin( 2π×1000t



With carrier frequency of 105 Hz.The modulation index is

a) 12.5 b) 7.5 c) 10 d) 5

[GATE-2008]

Q.33 Consider the amplitude modulated (AM) signal c c m cA cosω t+2cosω tcosω t. For demodulating the signal using envelope detector, the minimum value of 𝐴𝐴𝑐𝑐 should be

a) 2 b) 1 c) 0.5 d) 0

[GATE-2008]

Q.34 A message signal given by

( ) 1 21 1m t cos ω t sin ω t2 2

= −

is

amplitudemodulated with a carrier of frequency ωc to generate

( ) ( ) cs t =[1+m t ]cosω t What is the power efficiency achieved by this modulation scheme?

a) 8.33% b) 11.11% c) 20% d) 25% [GATE-2009]

Q.35 For a message signal

( ) ( )mm t =cos 2πf t and carrier of frequency fc which of the following represents a single side-band (SSB) signal?

a) ( ) ( )m ccos 2πf t cos 2πf t b) ( )ccos 2πf t

c) ( )c mcos 2π f +f t

d) ( ) ( )m c1+cos 2πf t cos 2πf t [GATE-2009]

Q.36 Consider an angle modulated signal

( ) ( )62π×10 t+2sinsin 8000πtx t =6cos

+4cos(8000πt)

V. The average power of x(t) is

a) 10W b) 18W c) 20W d) 28W

[GATE-2010] Q.37 Suppose that the modulating signal

is ( ) ( )mm t =2cos 2πf t and the carrier signal is ( ) ( )C C cx t =A cos 2πf t.Which one of the following is a conventional AM signal without over- modulation?

a) ( ) ( ) ( )C cx t =A m t cos 2πf t b) ( ) [ ] ( )C cx t =A 1+m(t) cos 2πf t

c) ( ) ( ) ( )cC c c

Ax t =A cos 2πf t + cos 2πf t4

d) ( ) ( ) ( )C m cx t A cos 2πf t cos 2πf t=

( ) ( )C m cA sin 2πf t sin 2πf t+ [GATE-2010]

Q.38 A message signal ( )m t cos 2000πt 4cos 4000πt= +

modulates the carrier c(t)= cos 2πfc t where fc = 1 MHZ 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

a) 0.5ms<RC<1ms b)1μs=RC<0.5ms c) RC 1μs= d) RC 0.5ms?

[GATE-2011]

Q.39 The List-1 (lists the attributes) and the List – II (lists of the modulation systems.) Match the attribute to the modulation system that best meets it.

List-I A. Power efficient transmission of

signals B. Most bandwidth efficient

transmission of voice signals. C. Simplest receiver structure D. Bandwidth efficient transmission

of signals with significant dc component

List-II 1. Conventional AM 2. FM



3. VSB 4. SSB-SC

A B C D a) 4 2 1 3 b) 2 4 1 3 c) 3 2 1 4 d) 2 4 3 1

[GATE-2011] Q.40 The signal m(t) as shown is applied

both to a phase modulator (with 𝑘𝑘𝑝𝑝 as the phase constant) and a frequency modulator with (𝑘𝑘𝑓𝑓 as the frequency constant ) having the same carrier frequency

The ratio kp/kf (in rad/ Hz) for the same maximum phase deviation is

a) 8π b) 4π c) 2π d) π

[GATE-2012]

Q.41 Consider sinusoidal modulation in an AM system. Assuming no over modulation, the modulation index (μ) when the maximum and minimum values of the envelope, respectively, are 3 V & 1 V, is _____.

[GATE-2014]

Q.42. In the figure, M(f) is the Fourier transform of the message signal .m(t) where A = 100 Hz and B = 40 Hz. Given v(t) = cos (2πfct) and w(t) = cos(2π (fc + A) t) , where fc > A The cutoff frequencies of both the filters are fc

The bandwidth of the signal at the output of the modulator (in Hz) is __.

[GATE-2014]

Q.43 A modulated signal is y(t)= m.(t)cos(40000 πt ), where the baseband signal m(t) has frequency components less than 5 kHz only. The minimum required rate (in kHz) at which y(t) should be sampled to recover m(t) is__________.

[GATE-2014] Q.44 Consider an FM signal f(t) = cos[2

πfct +β1 sin 2 πf1t + β2 sin 2 πf2t.] . The maximum deviation of the instantaneous frequency from the carrier frequency fc is a) β1f1+ β2f2 b) β1f2+ β2f1 c) β1+ β2 d) f1+ f2

[GATE-2014] Q.45 In a double side-band (DSB) full

carrier AM transmission system, if the modulation index is doubled, then the ratio of total sideband power to the carrier power increases by a factor of______.

[GATE-2014]

Q.46 Consider the signal ( ) ( ) ( )c cs t =m t cos(2πf t t π+m (2 f) t)

where m(t) denotes the Hilbert transform of m(t) and the bandwidth of m(t) is very small compared to fc . The signal s(t) is a a) high-pass signal b) low-pass signal c) band-pass signal



d) double sideband suppressed carrier signal

[GATE-2015]

Q.47 In the system shown in figure (a), m(t) is a low-pass signal with bandwidth W Hz. The frequency response of the band-pass filter H(f) is shown in figure (b). If it is desired that the output signal z(t) = 10x(t), the maximum value of W (in Hz) should be strictly less than ___.

[GATE-2015]

Q.48 A message signal ( ) m mm t = sin(2A πf t)

is used to modulate the phase of a carrier Ac cos(2πfct) to get the modulated signal

( ) ( )c cy t = cos(2 + A πf t m t ). The bandwidth of y(t) a) depends on Am but not on fm b) depends on fm but not on Am c) depends on both Am and fm d) does not depends on Am or fm

[GATE-2015]

Q.49 The block diagram of a frequency synthesizer consisting of a Phase Locked Loop (PLL) and a divide-by-N counter (comprising ÷2, ÷4, ÷8, ÷16 outputs) is sketched below. The synthesizer is excited with a 5 kHz signal (Input 1). The free-running frequency of the PLL is set to20 kHz. Assume that the commutator switch makes contacts repeatedly in the order 1-2-3-4.

The corresponding frequencies synthesized are: a) 10 kHz, 20 kHz, 40 kHz, 80 kHz b) 20 kHz, 40 kHz, 80 kHz, 160 kHz c) 80 kHz, 40 kHz, 20 kHz, 10 kHz d) 160 kHz, 80 kHz, 40 kHz, 20 kHz

[GATE-2016] Q.50 A super heterodyne receiver

operates in the frequency range of 58 MHz – 68MHz. The intermediate frequency f1F and local oscillator frequency fL0 are chosen such that.

lF L0f f≤ It is required that the image frequencies fall outside the 58MHz –68MHz band. The minimum required f1F (in MHz) is _________.

[GATE-2016] Q.51 The amplitude of a sinusoidal

carrier is modulated by a single sinusoid to obtain the amplitude modulated signal s(t) = 5cos1600πt + 20cos1800πt + 5cos2000πt. The value of the modulation index is ___.

[GATE-2016] Q.52 For a super heterodyne receiver, the

intermediate frequency is 15 MHz and the local oscillator frequency is 3.5 GHz. If the frequency of the received signal is greater than the local oscillator frequency, then the image frequency (in MHz) is ____.

[GATE-2016]



Q.53 A modulating signal given by ( ) ( )3 3x t 5sin 4 10 t 10 cos 2 10 t V= π − π π is fed

to a phase modulator with phase deviation constant kp = 0.5rad/V. If the carrier frequency is 20 kHz, the instantaneous frequency (in kHz) at t = 0.5 ms is __________ .

[GATE-2017, Set-2]

Q.54 The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________

[GATE-2017, Set-2]

Q.55 Consider the following amplitude modulated signal: ( ) ( ) ( ) ( )s t cos 2000 t 4cos 2400 t cos 2800 t= π + π + π

The ratio (accurate to three decimal places) of the power of the message signal to the power of the carrier signal is____.

[GATE-2018]



Q.1 (a) [ ]C c mx(t)=A cos ω t+βsinω t

[ ]c my(t)=Kcos 3ω t+3βsinω t

( ) ( )3After passing from y t x t = 1β 3β∴ =

i.e. 1mf 3 f f f∆ = ×∆ = ∆ ?

1BW 2 f 3 3MHz= ∆ × =fm Remains same ∴ fm=5kHz

Q.2 (a) H.T

1 1cos ω t sin ω t←→H.T

2 2sin ω t cos ω t←→−

Q.3 (d)

Output signal m(t) cosθ2

=

m(t) has power Pm

Power of a 2mm(t) a P→

Here 1a cosθ2

=

∴ Power of output signal 2mP cos θ

4=

Q.4 (c)

When the modulation index of AM wave is less than unity the output of the envelope detector is envelope of the AM wave but when the modulation index is greater than unity then the output of the envelope detector is not envelope but mode of the envelope of the AM wave. Thus the detector output in given case would be AC[1 + Kam(t)]

Q.5 (d)

Q.6 ( )

( ) ( ) 2sin2πtm t ×s t ×cos200πtt

( )1= sin202πt-sin198πtt( ) ( ) ( )m t ×s t +n t

( )1 sin199πt= sin202πt-sin198πtt t

+

( )1 sin 202πt sin198πt sin199πtt

= − +

( ) ( ) ( ) ( )y t m t s t n t s(t)= × +

[ ]1= sin202πt-sin198πt+sin199πt cos200πtt

Freq. present

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (a) (d) (c) (d) * (d) (c) (d) (d) (c) (a) (a) (b)15 16 17 18 19 20 21 22 23 24 25 26 27 28 (b) (a) (c) (b) (b) (d) (a) (c) (d) (c) (b) (c) (a) (b) 29 30 31 32 33 34 35 36 37 38 39 40 41 42 (a) (a) (c) (c) (a) (c) (c) (b) (c) (b) (b) (b) 0.5 60 43 44 45 46 47 48 49 50 51 52 53 54 55 10 (a) 4 (c) 350 (c) (a) 5 0.5 3485 70 5.22 *

ANSWER KEY:

EXPLANATIONS



402 2 2 398 1 399, , , , ,2 2 2 2 2 2

=

After passing from LPF with cf =1Hz

∴ ( ) [ ]1y t = sin2πt+sin2πt-sinπt2t

( ) [ ]sin2πt 1y t = + 2sin0.5πt.cos1.5πt2t 2t

⇒

( ) sin2t sin0.5πt.cos1.5πty t = +2t t

⇒

Hence none of the option matches with answer.

Q.7 (d)

( )62π 2 10 t 2π 30sin150t

s t cos2π 40cos150t

× × + × + ×

6idΦ 2π 2 10 t 4500cos1 2 6000(sin150t)

dt= × × + + ×

( )1/22 2

ω 1 cω ω 2π 4500 6000∴∆ = − = +

2π 7.5k rad / sec= ×

ωΔf = =7.5kHz2πΔ

∴

2 2Φ 2 30 40π∆ = + ΔΦ=100π

Q.8 (c) C1MHz 1000kHz f⇒ =

m100μs 10kHz f⇒ = ∴ Frequencies present C m1000 10(f f )= ± ± =990,1010 = 970,1030

Half wave symmetric wave, so only odd harmonics are present

Q.9 (d)

2a

t cmP P 12

= +

2c

cAP2

=

Sideband power 2 2c aA m.

2 2=

∴

2 2c a

2S a

2cc

A m4P m 1

AP 2 82

= = =

Q.10 (d)

( ) ( ) ( )FM c n c mn

x t A J β cos ω nω t∞

=−∞

= +∑

mω 3 BW 6ω∆ = × =

m

ωβ 6ω∆

∴ = =

3c mω +nω =2π×1008×10

(given in question) 32π×n.4π×10∴ 3=2π×1008×10

∴ n=4 ∴ Besssel coefficient 45J (6)

Q.11 (c) 3

0 0 i 1 iV =a v +a v

( ) ( ) ( )31 i i 3 i 30 c c 0 1 c c 1= a A cosω +a m t +a A cos ω t+a m t

( ) 32 i i i 3 ii c c 1 c c+3a m t A osω t+3a A cos ω tm(t)

For DSB-SC, we are concerned with only last term. ( ) c cm t cos ω t f 1MHz→ =

For 2 iccos term2ω 2π 1MHz= ×

2 iccos term2 2 1MHzω = π×

Q.12 (a)

The coherent detector rejects the quadrature component of noise and therefore noise at the output has in phase component only. The in phase component of noise and output are additive at output of detector.

Q.13 (a)

2

max max

min min

C fRC f

= =

21650 450550 450

+ = + = 4.41



s ifI f 2f 700 2 450 1600= + = + × = Q.14 (b) Q.15 (b)

Peaks of X(f)are at 1 k and -1 k initially Output of balanced modulator C= f ±1k and

C C Cf ±-1kf ±1k=11k,9kf ±-1k=9k,11k Output of HPF with Cf =1ok will be 11k Frequency component

( )y f =13k±11k=24k and 2k∴ Q.16 (a)

( ) [ ] [ ]s t =Acos 2π×10Kt +Acos 2π×10.1Kt

11T = =100μs

10K

21T = =99μs

10.1K

1

2

T rationalT

∴ s(t) Will be periodic with period LCM of 1 2T &T =9900μs;10,000μs ∴ Frequency of detector = 0.1 kHz

Q.17 (c)

( ) { }61y t = cos 2π×101×10 t2

6sin 2 100 10+ × ×π t

( )( )

6 6

6

cos 2π100 10 tcos10 t.2π12 sin 2π100 10 tsin2π

× = − ×

6 610 t+sin2π×100×10 t

61= cos2π×100×10 t2

6 61cos2π×10 t + sin2π×100×10 t2

6-sin2π×10 t+2 6 6=Acos2π×10 t+Bsincos2π×10 t

Envelope 2 2= A +B 2 6 61 1= cos 2π×10 t+ sin2π×10 t+2

4 4

Envelope 6= 5/4-sin2π×10 t Q.18 (b)

( ) ( )

( )

cc c c m

cc m

A βs t =A cosω t+ coscos ω +ω t2

A β cos cos ω ω2

− −

( ) c ac cAM

A ms t =A cosω t+2

( ) ( )c ac m c m

A mcos w + w t+ cos w - w2

( ) ( )AMs t s t carrier USB+ = + = SSB With carrier Note:- ( ) ( )AM

-s t +s t =SSB+SC Q.19 (b)

c m

1 1RCf f< ≤

(where 1/ cf =to avoid fluctuations at recovered output and 1/ mf=to avoid diagonal clipping)

11μsec<RC2kHz

≤

1μsec<RC 0.5ms≤

Q.20 (d) fΔΦ = m

fm Remains same for both cases, f 10

2 1∆

⇒ =

f 20∴∆ =



( ) ( )mBW 2 f f 2 20 2 44K∴ = ∆ + = + =

Q.21 (a)

( ) ( )mBW 2 f f 2 180 5 370K= ∆ + = + =

Note:- When FM signal is applied to doublers frequency deviation doubles but fm remains the same.

Q.22 (c) Q.23 (d) Q.24 (c)

2 2 2a c a

s Cm A mP P2 2 2

= = ×

s100 0.25P 6.25Watts

2 2= × =

Q.25 (b)

S25P Watts4

=

N 0P N B=

S

N

PSNRP

∴ =

0

254N B

=

Q.26 (c)

( ) ( )p t u t u(t 1)= − −

( ) ( ) ( )g t =p t *p t

( ) ( ) ( ) ( ) ( )s t g t δ t 2 * g t g t

g(t 2)= − − = =

− −

( ) ( )s t =g t -g(t-2)

Impulse response of match filter ( )h t =s(T-t)

( )h t =-s(T)

Q.27 (a)

( ) ( ) ( )AM cx t 100 p t 0.5g t cos ω t= +

( ) ( )( ) ( ) ( ) ( )cs t 100 1+m t cosω tp t =1m t =0.5g t

Comparing AM equation with (i) =0.5t ⇒Modulating signal=t (from equation of line)



Modulation index = 0.5 Q.28 (b)

6

mf =10 Hz∴ (as 10k is small in comparison to 610)

6mB.W. 2f 2 10 Hz∴ = = ×

Q.29 (a) Q.30 (a)

Hilbert transformer is used for SSB generation. Sum of quadrature components gives LSB.

Q.31 (c) The signal c m ccos ω t 0.5cos ω t cosω t+ is either AM or narrow –band FM signal.

Q.32 (c)

Modulation index fm

δmf

=

Where δ=Maximum frequency deviation

mf = Maximum frequency Component Given that

mf =1500Hz Deviation,

( ) ( )θ 5sin 2π 1500t 7.5sin 2π 1000t∆ = × + ×

( )

( )

dθω 5 2π 1500cos 2π 500tdt

7.5 2π 1000 2π 1000t cos

∆ = = × × ×

+ × × ×

maxω 2π(7500 7500)∆ = +

maxωδ2π

∆= = 1500 Hz

fm

δ 15000mf 1500

= = = 10

Q.33 (a) Modulated signal,

( )AM c c m cΦ t A cos ω t 2cos ω tcosω t= +

[ ]c m c= A +2cosω t cosω t Condition for envelop detection of an AM signal is mA 2cos ω t 0= + ≥ For mcosω t=-1

c cA -2 0 A 2≥ ⇒ ≥ Therefore minimum value of

cA should be 2.

Q.34 (c) 2a

2a

mη= ×100%2+m

1 2 1 2

2 2ma a a a a

c

V 1m = = =m m = m +mV 2

0.25 2η 100%2 0.25 2

×= ×

+ ×

=20%

Q.35 (c) ( )c mcos 2π f +f t

Represents only USB of AM-SSB/SC signal

Q.36 (b) Average power of angle modulated AM signal x(t) is

2cA /2 Here cA =6

Therefore,

Average power 26= =18W

2

Q.37 (c)

Conventional AM signal is ( ) [ ] ( )C cx t A 1 m(t) cos 2πf t= +

( ) ( )C c C cA cos 2πf t A m(t)cos 2πf t= + In option (b)

Modulation index 2 21

= =



So, it is conventional AM signal but with over-modulation. In options (c)

( ) ( ) ( )cC c c

Ax t A cos 2πf t cos 2πf t4

= +

Here, Modulation index 2 14 2

= =

Therefore, it is conventional AM signal without over-modulation.

Q.38 (b)

c m

1 1RCf f

= =

I 1RC1MHz 2kHz

= =

⇒ 1μs ≪ RC ≪ 0.5ms Q.39 (b)

AM has simple receiver and VSB is bandwidth efficient transmission of signals with significant dc component. SSB–SC is the most bandwidth efficient transmission of voice signals.

Q.40 (b) For phase modulator ( ) c pΦ t =2πf t+k m(t) Maximum phase deviation is ( ) ( )D p pmax

Φ =k maxmax m t =2k (1) For frequency modulator

( ) ( )c

t

0fΦ t =2πf t+2πk m t dt∫

( ) ( )t

D fmax0 max

Φ' =2πk t m t dt ∫

( )2

D fmax0

Φ' =2πk 2dt ∫

( )D fmaxΦ' =8πk (2)

Given ( ) ( )D Dmax max

Φ' Φ= f p8πk =2k

p

f

k=4π

k

Q.41 0.5

max min

max min

A(t) -A(t)μ=A(t) +A(t)

3-1 1μ= = =0.53+1 2

Q.42 60

Exp: m(t) ↔M(f)

After multiplication with V(t) =cos (2πfct) Let ( ) ( )1w t =m t .V(t)

( ) ( )( )1 1W f specturm of w t is⇒

After high pass filter

After multiplication with

( )c(2π fcos +A t) and low pass filter of cut off 𝑓𝑓𝑐𝑐

Band width =A-B =100-40=60 Q.43 (10)

Since m(t) is a base band signal with maximum frequency 5 KHz, assumed spreads as follows:



∴y(t) = m(t) cos(40000πt)

( ) ( )7 f -20k f 201( k*f )2

m +→ δ +δ

y(f) = 12

[M(f ‒ 20k) + M(f+ 20k)]

Thus the spectrum of the modulated signal is as follows:

If y(t) is sampled with a sampling frequency `fs’ then the resultant signal is a periodic extension of successive replica of y(f) with a period ‘fs'. It is observed that 10 KHz and 20 KHz are the two sampling frequencies which causes a replica of M(f) which can be filtered out by a LPF. Thus the minimum sampling frequency (fs) which extracts m(t) from g(f) is 10 KHz.

Q.44 A

Exp: Instantaneous phase( )1 c 1 1 2 2f t =2πf t+β sin2πf +β sin2πf t

Instantaneous frequency fi(t) =

1d 1(t)×dt 2πφ

c 1 1 1 2 2 2=f +β f cos2πf t+β f 2os πfc tInstantaneous frequency deviation

1 1 1 2 2 2=β f cos2πf t+β f 2os πfc t Maximum 1 1 2 2Δf = β f +β f

Q.45 4

2Ratio of total side band power αμCarrier power

If it in doubled, this ratio will be come 4 time

Q.46 (C) Given s(t) is the equation for single side band modulation (lower side band). Thus it is a Band pass signal.

Q.47 350 x(t) = m(t).cos(2400𝜋𝜋𝜋𝜋) Spectrum of x(t)

Y(t) =10x(t) + 2x (t) Let us drawn the spectrum of positive frequencies of y(t) Y(t)=10m(t)

( ) 2 2cos 2400πt m (t)cos (2400πt) =10m(t)

( ) 2 1+cos4800πtcos 2400πt +m (t)2

Y(t)

( ) ( )2 2m (t) m (t)= + cos 2400πt + cos(4800πt)2 2

10m t

⇒2w <700

2400-2w>1700 ⇒w<350

Q.48 (C)



c cy(t) A cos[2πf t m(t)]= + m mm(t) A sin(2πf t)=

Since y(t) is phase modulated signal, c(t) 2πf t m(t)∅ = +

Bandwidth = n2[Δf+f ] 1 dΔf= m(t)

2π dtm mΔ f depends on A as well as f .⇒

Thus Bandwidth deeen s on both Am and fm.

Q.49 (a) Q.50 5 Q.51 0.5

S(t) = 20 cos 1800 πt + 5cos 1600 πt + 5cos 2000 πt

mc

c

A=A 1+ cos200πt cos(1800πt)A

Ac = 20 cA μ 5×2=5Þμ= =0.52 20

μ = 0.5

Q.52 3485 fimage = ?

S LO Si L0f f =±(f -f= )

Sif = Image freq

L0f =Signal freq S L0 Si L0f -f =-f +f

Si L0 Sf =2f -f S L0 IFf -f =f

Si L0 IF L0 L0 IFf =2f -f -f =f -f=3500-15=3485mHz

Q.53 70 (69.9-70.1)

( ) ( )( )3 3x t 5sin 4 x10 t 10 cos 2 x10 t= π − π π

Transform theorem frequency

( ) ( )( )( ) ( )

( ) ( )( )( )( )

( ) ( ) ( )( )

( )

3 3

i c p

3 3

3 3 3

3

t 0.53 3

i

x t 5sin 4 x10 t 10 cos 2 x10 t

1 df t f k . m t2 dt

d m t 5cos 4 x10 t 10 cos 2 x10 tdt 4 x10 t 10 cos 2 x10 t 2 x10

d m t 5cos 2 10 x 4 x10 0dt =20 x10 cos 12 20 x10

1f t 202

=

= π − π π

= +π

= π − π π

π − π π π

= π+ π π +

π π = π

= +π

x5x20 7kHzπ =

Q.54 5.22 (5.19-5.23)

Total power when µ = 50% is

( )

[ ][ ]

2

T C

2

T

T

P P 12

0.5P 5 1

2

5 1 0.125

=5 1.125P 5.625

µ= +

= +

= +

=

When µ = 40%

Total power remains 5.625

( )

[ ]

2

C

C

C

0.45.625 P 1

2

5.625 P 1 0.08P 5.22

⇒ = +

⇒ = +

=

Q.55 * Given,



( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

cos 2000 t 4cos 2400 t cos 2800 ts t

side band carrier side band4cos 2400 t 2cos 400 t cos 2400 t

14 1 cos 400 t cos 2400 t2

π π π= + +

= π + π π

= + π π Now at this point, some data is missing, if we compare this with standard AM signal equation

( ) ( ) ( )( ) ( )

( ) ( )

( ) ( )

c a

a

a

a

s t A 1 k m t cos 2400 t

if k =1/2, m t cos 400 t , PM 1/ 21if k =1, m t cos 400 t , PM=1/821if k =2, m t cos 400 t , PM=1/324

= + π = π =

= π

= π

Like this infinite possibilities exist Hence many answers are possible.



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