Multiple Choice Questions and Answers on Analog Communication (Part-3)

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1)   FM is advantageous over AM as

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a. The amplitude of FM is constant. So transmitter power remains unchanged in FM but it changes in AM
b. The depth of modulation in FM can be changed to any value by changing the frequency deviation. So the signal is not distorted
c. There is less possibility of adjacent channel interference due to presence of guard bands
d. All of the above

ANSWER: All of the above

2)   VCO is used to generate

a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC

ANSWER:  (a) Direct FM

3)   Change in instantaneous phase of the carrier with change in amplitude of the modulating signal generates

a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC

ANSWER: (b) Indirect FM

4)   Phase-locked loop can be used as

a. FM demodulator
b. AM demodulator
c. FM receiver
d. AM receiver

ANSWER:  (a) FM demodulator

5)   The increase or decrease in the frequency around the carrier frequency is termed as

a. Figure factor
b. Frequency deviation
c. Modulation index
d. Frequency pectrum

ANSWER:  (b) Frequency deviation

6)   Carson’s rule is used to calculate

a. Bandwidth of FM signal
b. Signal to noise ratio
c. Modulation index
d. Noise figure

ANSWER:  (a) Bandwidth of FM signal

7)   The ratio of maximum peak frequency deviation and the maximum modulating signal frequency is termed as

a. Frequency deviation
b. Deviation ratio
c. Signal to noise ratio
d. Frequency spectrum

ANSWER:  (b) Deviation ratio

8)   The equation v(t) = A cos [ωct + kp Φ(t)] represents the signal as

a. Phase modulated signal
b. SSBSC signal
c. DSB SC signal
d. None of the above

ANSWER:  (a) Phase modulated signal

9)   Calculate the maximum frequency deviation for the FM signal
v(t) = 10 cos (6000t+ 5sin2200t)

a. 2200 Hz
b. 6000 Hz
c. 1750 Hz
d. 11000 Hz

ANSWER:  (c) 1750 Hz

Explanation:
A standard FM signal is represented by
v(t) = Ac cos(2πfct + kfsin2πfmt)
Ac = carrier amplitude
fc = carrier frequency
kf = modulation index
fm = modulating frequency = 2200/2π = 350 Hz
kf = frequency deviation/modulating frequency
5 = freq deviation/ 350
Therefore, deviation = 5 * 350
= 1750Hz

10)   Calculate the dissipation in power across 20Ω resistor for the FM signal
v(t)= 20 cos(6600t+ 10sin2100t)

a. 5W
b. 20W
c. 10W
d. 400W

ANSWER:  (a) 5W

Explanation:
A standard FM signal is represented by
v(t) = Ac cos(2πfct + kfsin2πfmt)
Ac = carrier amplitude
fc = carrier frequency
kf = modulation index
fm = modulating frequency
kf = frequency deviation/modulating frequency
the power dissipated across 20Ω resistor is given by
Vrms2/R
=(20/√2)2/R
= 5W

11)   What is the value of carrier frequency in the following equation for the FM signal?
v(t)= 5 cos(6600t+ 12sin2500t)

a. 1150 Hz
b. 6600 Hz
c. 2500 Hz
d. 1050 Hz

ANSWER:  (d) 1050 Hz

Explanation:
A standard FM signal is represented by
v(t) = Ac cos(2πfct + kfsin2πfmt)
Ac = carrier amplitude
fc = carrier frequency
kf = modulation index
fm = modulating frequency
kf = frequency deviation/modulating frequency
therefore, fc = 6600/2π
= 1050Hz

12)   Calculate the modulation index in an FM signal when fm (modulating frequency) is 250Hz and Δf (frequency deviation) is 5KHz.

a. 20
b. 35
c. 50
d. 75

ANSWER:  (a) 20

Explanation:
Modulation index is the measure of how much the modulation parameter changes from its un modulated value. The modulation index of FM is given by
μ = frequency deviation/ modulating frequency
= Δf/ fm
Where Δf is the peak frequency deviation i.e. the deviation in the instantaneous value of the frequency with modulating signal.
fm is the value of modulating frequency
μ = 5000/250
= 20

13)   After passing the FM signal through mixer, what is the change in the frequency deviation Δ when the modulating frequency is doubled?

a. Becomes 2 Δ
b. Becomes Δ /2
c. Becomes Δ2
d. Remains unchanged

ANSWER:  (d) Remains unchanged

14)   In frequency modulation,

a. Armstrong method is used for generation
b. Multiple side bands are generated
c. The FM signal has infinite bandwidth
d. All of the above

ANSWER:  (d) All of the above

15)   Maximum frequency deviation and the maximum bandwidth allowed for commercial FM broadcast is

a. 80KHz, 160Khz
b. 75KHz, 200Khz
c. 60KHz, 170Khz
d. 75KHz, 250Khz

ANSWER: 7 (b) 5KHz, 200Khz

16)   Guard bands are provided in FM signal to

a. Prevent interference from adjacent channels
b. To increase the noise
c. To increase bandwidth
d. None of the above

ANSWER: (a)  Prevent interference from adjacent channels

17)   For a FM signal v(t) = 15 cos ( 10 * 108t + 10 sin 1220t), calculate
1. Carrier frequency
2. Modulating frequency

a. 159.1MHz, 194.1Hz
b. 185.5MHz, 200.15Hz
c. 350.1MHz, 200.1Hz
d. 159.1Hz, 194.1Hz

ANSWER: 1 (a) 59.1MHz, 194.1Hz

18)   For a FM signal v(t) = 25 cos (15 * 108t + 10 sin 1550t), calculate:

  1. Modulation index
  2. Maximum frequency deviation

a. 10, 3000.1Hz
b. 20, 1550.9Hz
c. 10, 2465.9Hz
d. 10, 2000.0Hz

ANSWER: (c)  10, 2465.9Hz

Explanation:
Standard expression for FM signal is given by
v(t) = A cos ( ωct + mf sin ωmt)
Comparing with the given equation,
Modulation index mf = 10
Maximum frequency deviation is given by
mf = Δf/fm
Δf = mf * fm
Here fm = 1550/2Π = 246.59 Hz
Δf = 10 * 246.59
= 2465.9Hz

19)   For a FM signal v(t) = 20 cos ( 10 * 108t + 30 sin 3000t), calculate the power dissipated by the FM wave in a 20Ω resistor.

a. 100 Watts
b. 10 Watts
c. 200 Watts
d. 20 Watts

ANSWER:  (b) 10 Watts

Explanation:
Standard expression for FM signal is given by
v(t) = A cos ( Ωct + mf sin Ωmt)
Comparing with the given equation,
A = 20
The dissipated power is given by P = V2rms/R
= (20/√2)2/ 20
= 10Watts

20)   A 100MHz carrier is frequency modulated by 5 KHz wave. For a frequency deviation of 100 KHz, calculate the carrier swing of the FM signal.

a. 2000 KHz
b. 100 KHz
c. 105 KHz
d. 200 KHz

ANSWER: (d) 200 KHz

Explanation:
Carrier frequency fc = 100MHz
Modulating frequency fm = 5 KHz
Frequency deviation Δf = 100 KHz
Carrier swing of the FM signal = 2 * Δf
= 2 * 100
= 200 KHz

21)   A 100MHz carrier is frequency modulated by 10 KHz wave. For a frequency deviation of 50 KHz, calculate the modulation index of the FM signal.

a. 100
b. 50
c. 70
d. 90

ANSWER:  (b) 50

Explanation:
Carrier frequency fc = 100MHz
Modulating frequency fm = 10 KHz
Frequency deviation Δf = 500 KHz
Modulation index of FM signal is given by
mf = Δf/fm
= 500 * 103/ 10 * 103
= 50

22)   Narrow band FM has the characteristics:

a. The frequency sensitivity kf is small
b. Bandwidth is narrow
c. Both a and b
d. None of the above

ANSWER: (c)  Both a and b

23)   Wide band FM has the characteristics:

a. The frequency sensitivity kf is large
b. Bandwidth is wide
c. Both a and b
d. None of the above

ANSWER:  (c) Both a and b

24)   Determine the Bandwidth of a FM wave when the maximum deviation allowed is 75KHz and the modulating signal has a frequency of 10KHz.

a. 170 KHz
b. 200 KHz
c. 100 KHz
d. 1000 KHz

ANSWER:  (a) 170 KHz

Explanation:
Modulating frequency fm = 10 KHz
Frequency deviation Δf = 75 KHz
According to Carson s rule, BW = 2(Δf + fm)
= 2 (75 + 10)
= 170 KHz

25)   FM signal is better than AM signal because

a. Less immune to noise
b. Less adjacent channel interference
c. Amplitude limiters are used to avoid amplitude variations
d. All of the above

ANSWER:  (d) All of the above

26)   FM is disadvantageous over AM signal because

a. much wider channel bandwidth is required
b. FM systems are more complex and costlier
c. Adjacent channel interference is more
d. Both a and b

ANSWER: (d) Both a and b

27)   For a three stage cascade amplifier, calculate the overall noise figure when each stage has a gain of 12 DB and noise figure of 8dB.

a. 12
b. 24
c. 13.55
d. 8

ANSWER:  (c) 13.55

Explanation:
As the signal passes through various stages of an amplifier, the output has the original signal and some noise that gets amplified at different stages of amplifiers. So the final noise figure of the cascaded amplifier is obtained by

FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ G1G2+……+(FN– 1)/ G1G2G3 GN
F1, F2, F3 .. FN, G1,G2, G3…. GN are the noise figures and the gains respectively of the amplifiers at different stages.
F1 = 12, F2 = 12, F3 = 12
G1 = 8, G2 = 8, G3 = 8
FN = 12 + (12- 1)/ 8+ (12- 1)/ 8 * 8
= 12 + 11/8 + 11/64
= 13.55

28)   The Hilbert transform of the signal sinω1t + sinω2t is

a. sinω1t + sinω2t
b. cosω1t + cosω2t
c. sinω2t + cosω2t
d. sinω1t + sinω1t

ANSWER:  (b) cosω1t + cosω2t

Explanation:
In Hilbert transform, the signal gets shifted by 900.

So the signal sinω1t+ sinω2t gets shifted by 900
sinω1(t+900)+ sinω2(t+900)
= cosω1t+ cosω2t

29)   The noise due to random behaviour of charge carriers is

a. Shot noise
b. Partition noise
c. Industrial noise
d. Flicker noise

ANSWER:  (a) Shot noise

30)   Transit time noise is

a. Low frequency noise
b. High frequency noise
c. Due to random behavior of carrier charges
d. Due to increase in reverse current in the device

ANSWER:  (b) High frequency noise

31)   Figure of merit γ is

a. Ratio of output signal to noise ratio to input signal to noise ratio
b. Ratio of input signal to noise ratio to output signal to noise ratio
c. Ratio of output signal to input signal to a system
d. Ratio of input signal to output signal to a system

ANSWER: (a) Ratio of output signal to noise ratio to input signal to noise ratio

32)   Signum function sgn(f), for f>0, f=0 and f<0, has the values:

a. -1 to +1
b. +1, 0, -1 respectively
c. -∞ to + ∞
d. 0 always

ANSWER: (b) +1, 0, -1 respectively

Explanation:
The sgn(f) is a signum function that is defined in the frequency domain as
sgn(f) = 1, f> 0
= 0, f = 0
= -1, f< 0
Mathematically, the sign function or signum function is an odd mathematical function which extracts the sign of a real number and is often represented as sgn

33)   In Hilbert transform of a signal, the phase angles of all components of a given signal are shifted by

a. +/- π
b. +/- π/4
c. +/- π/2
d. Any angle from 00 to 3600

ANSWER:  (c) +/- π/2

34)   The noise voltage (Vn) and the signal bandwidth (B) are related as

a. Vn is directly proportional to bandwidth
b. Vn is directly proportional to √bandwidth
c. Vn is inversely proportional to absolute temperature
d. Vn is inversely proportional to bandwidth

ANSWER:  (b) Vn is directly proportional to √bandwidth

35)   Noise factor for a system is defined as the ratio of

a. Input noise power (Pni) to output noise power (Pno)
b. Output noise power (Pno) to input noise power (Pni)
c. Output noise power (Pno) to input signal power (Psi)
d. Output signal power (Pso) to input noise power (Pni)

ANSWER:  (b) Output noise power (Pno) to input noise power (Pni)

36)   Noise Factor(F) and Noise Figure(NF) are related as

a. NF = 10 log10(F)
b. F = 10 log10(NF)
c. NF = 10 (F)
d. F = 10 (NF)

ANSWER: (a)  NF = 10 log10(F)

37)   The Noise Factor for cascaded amplifiers (FN) is given by (F1, F2, F3 .. FN, G1, G2, G3….GN) are the noise factors and the gains of the amplifiers at different stages:

a. FN = F1 + F2/ G1 + F3/ G1G2+ ..+ FN/ G1G2G3GN
b. FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ (G1+G2)+ ..+(FN – 1)/ (G1+G2+G3+…+GN)
c. FN = F1 + F2/ G1 + F3/ (G1+G2) +…+ FN/ (G1+G2+G3+…+GN)
d. FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ G1G2+…+(FN – 1)/ G1G2G3GN

ANSWER: (d) FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ G1G2+…+(FN – 1)/ G1G2G3GN

38)   For a two stage amplifier, first amplifier has Voltage gain = 20, Input Resistance Rin1=700Ω, equivalent Resistance Req1=1800Ω and Output Resistor Ro1 = 30KΩ. The corresponding values of second amplifier are : 25, 80 KΩ, 12 KΩ, 1.2 MΩ respectively. What is the value of equivalent input noise resistance of the given two stage amplifier?

a. 2609.1Ω
b. 2607.1Ω
c. 107.1Ω
d. 2107.1Ω

ANSWER:  (b) 2607.1Ω

Explanation:
R1 = Rin1 + Req1 = 700 + 1800 = 2500Ω
R2 = (Ro1 Rin2)/ (Ro1 + Rin2) + Req2 = 30 * 80/(30 + 80) + 12 = 40.92KΩ
R3 = Ro2 = 1.2MΩ
Equivalent input noise resistance of a two stage amplifier is given by
Req = R1 + R2/ A21 + R3/ (A21 A22)
= 2500 + 40.92 * 103/(20)2 + 1.2 * 106/(20)2(25)2
= 2607.1Ω

39)   The noise temperature at a resistor depends upon

a. Resistance value
b. Noise power
c. Both a and b
d. None of the above

ANSWER: (b)  Noise power

40)   Noise voltage Vn and absolute temperature T are related as

a. Vn = 1/ √(4RKTB)
b. Vn = √(4RK)/ (TB)
c. Vn = √(4RKTB)
d. Vn = √(4KTB)/R

ANSWER:  (c) Vn = √(4RKTB)