Multiple Choice Questions and Answers on Analog Communication (Part-1)

1)   Amplitude modulation is

a. Change in amplitude of the carrier according to modulating signal
b. Change in frequency of the carrier according to modulating signal
c. Change in amplitude of the modulating signal according to carrier signal
d. Change in amplitude of the carrier according to modulating signal frequency

ANSWER: (a) Change in amplitude of the carrier according to modulating signal

2)   The ability of the receiver to select the wanted signals among the various incoming signals is termed as

a. Sensitivity
b. Selectivity
c. Stability
d. None of the above

ANSWER:  (b) Selectivity

3)   Emitter modulator amplifier for Amplitude Modulation

a. Operates in class A mode
b. Has a low efficiency
c. Output power is small
d. All of the above

ANSWER: (d) All of the above

4)   Super heterodyne receivers

a. Have better sensitivity
b. Have high selectivity
c. Need extra circuitry for frequency conversion
d. All of the above

ANSWER: (d) All of the above

5)   The AM spectrum consists of

a. Carrier frequency
b. Upper side band frequency
c. Lower side band frequency
d. All of the above

ANSWER: (d) All of the above

6)   Standard intermediate frequency used for AM receiver is

a. 455 MHz
b. 455 KHz
c. 455 Hz
d. None of the above

ANSWER: (b) 455 KHz

7)   In the TV receivers, the device used for tuning the receiver to the incoming signal is

a. Varactor diode
b. High pass Filter
c. Zener diode
d. Low pass filter

ANSWER: (a) Varactor diode

8)   The modulation technique that uses the minimum channel bandwidth and transmitted power is

a. FM
b. DSB-SC
c. VSB
d. SSB

ANSWER: (d) SSB

9)   Calculate the bandwidth occupied by a DSB signal when the modulating frequency lies in the range from 100 Hz to 10KHz.

a. 28 KHz
b. 24.5 KHz
c. 38.6 KHz
d. 19.8 KHz

ANSWER: (d) 19.8 KHz

10)   In Amplitude Demodulation, the condition which the load resistor R must satisfy to discharge capacitor C slowly between the positive peaks of the carrier wave so that the capacitor voltage will not discharge at the maximum rate of change of the modulating wave (W is message bandwidth and ω is carrier frequency, in rad/sec) is

a. RC < 1/W
b. RC > 1/W
c. RC < 1/ω
d. RC > 1/ω

ANSWER: (a) RC < 1/W

11)   A modulation index of 0.5 would be same as

a. 0.5 of Modulation Depth
b. 1/2% of Modulation Depth
c. 5% of Modulation Depth
d. 50% of Modulation Depth

ANSWER: (d) 50% of Modulation Depth

12)   A 3 GHz carrier is DSB SC modulated by a signal with maximum frequency of 2 MHz. The minimum sampling frequency required for the signal so that the signal is ideally sampled is

a. 4 MHz
b. 6 MHz
c. 6.004 GHz
d. 6 GHz

ANSWER: (c) 6.004 GHz

13)   The function of multiplexing is

a. To reduce the bandwidth of the signal to be transmitted
b. To combine multiple data streams over a single data channel
c. To allow multiple data streams over multiple channels in a prescribed format
d. To match the frequencies of the signal at the transmitter as well as the receiver

ANSWER: (b) To combine multiple data streams over a single data channel

14)   Aliasing refers to

a. Sampling of signals less than at Nyquist rate
b. Sampling of signals greater than at Nyquist rate
c. Sampling of signals at Nyquist rate
d. None of the above

ANSWER: (a) Sampling of signals less than at Nyquist rate

15)   The amount of data transmitted for a given amount of time is called

a. Bandwidth
b. Frequency
c. Noise
d. Signal power

ANSWER: (a) Bandwidth

16)   An AM broadcast station transmits modulating frequencies up to 6 kHz. If the AM station is transmitting on a frequency of 894 kHz, the values for maximum and minimum upper and lower sidebands and the total bandwidth occupied by the AM station are:

a. 900 KHz, 888 KHz, 12 KHz
b. 894 KHz, 884 KHz, 12 KHz
c. 894 KHz, 888 KHz, 6 KHz
d. 900 KHz, 888 KHz, 6 KHz

ANSWER: (a) 900 KHz, 888 KHz, 12 KHz

Explanation:
Maximum Frequency fUSB = 894 + 6 = 900 kHz

Minimum Frequency fLSB = 894 – 6 = 888 kHz

Bandwidth BW = fUSB fLSB = 900 888 = 12 kHz OR = 2(6 kHz) = 12 kHz

17)   The total power in an Amplitude Modulated signal if the carrier of an AM transmitter is 800 W and it is modulated 50 percent.

a. 850 W
b. 1000.8 KW
c. 750 W
d. 900 W

ANSWER: (d) 900 W

Explanation:
The total power in an Amplitude Modulated wave is given by

PT = PC (1+ m22)

Here, PC = 800W,
m = 0.5

therefore,

PT = 800 (1+ (0.5)2/2) = 900 W

18)   An unmodulated AM signal produces a current of 5.4 A. If the modulation is 100 percent,
calculate:

(a) the carrier power,
(b) the total power,
(c) the sideband power when it is transmitted through an antenna having an impedance of 50Ω.

a. 1458 W, 2187.5 W, 729.25 W
b. 278 W, 2187.5 W, 1917.25 W
c. 1438 W, 2187.5 W, 759.25 W
d. 280 W, 2187.5 W, 750.25 W

ANSWER: (a) 1458 W, 2187.5 W, 729.25 W

Explanation:
a) PC=I2R = (5.4)2*50 = 1458W

b) IT = Ic√(1+m2/2) = 5.4√(1+12/2)
=6.614 A

PT = IT2R
= (6.614)2 * 50
= 2187.25 W

c) PSB = PT – PC
= 2187.25 – 1458 W
= 729.25W (for two bands)

For single band, PSB = 729.25/2
= 364. 625 W

19)   Calculate the depth of modulation when a transmitter radiates a signal of 9.8KW after modulation and 8KW without modulation of the signal.

a. 80%
b. 67%
c. 50%
d. 100%

ANSWER: (b) 67%

Explanation:
Ptotal = 9.8KW

Pc = 8KW

Power of the signal (Ptotal) transmitted by a transmitter after modulation is given by

Ptotal = Pc (1+ m2/2)

Where Pc is the power of carrier i.e., without modulation

M is the modulation index

Therefore,
9.8= 8 (1+ m2/2)
9.8/8=1+ m2/2
m=0.67 = 67%

20)   When AM signal is of 25KHz, calculate the number of channels required in Medium Frequency (MF) band of 300KHz-3000KHz.

a. 94
b. 69
c. 85
d. 54

ANSWER: (d) 54

Explanation:
Medium Frequency (MF) is the band of frequencies from 300 KHz to 3MHz. The lower portion of the MF band (300to 500 kilohertz) is used for ground-wave transmission for reasonably long distances. The upper and lower ends of the mf band are used for naval purpose.

Frequency available in MF band= 3000 – 300 = 2700 KHz

The bandwidth required by 25 KHz signal = 2 * 25= 50 KHz

Therefore the number of channels available = 2700/ 50 = 54

21)   Calculate the power in one of the side band in SSBSC modulation when the carrier power is 124W and there is 80% modulation depth in the amplitude modulated signal.

a. 89.33 W
b. 64.85 W
c. 79.36 W
d. 102 W

ANSWER: (c) 79.36 W

Explanation:
Modulation Index = 0.8
Pc = 124W
Power in sidebands may be calculated as = m2 Pc/4
= (0.8)2 * 124/4
= 79.36 W

22)   Calculate the total modulation Index when a carrier wave is being modulated by two modulating signals with modulation indices 0.8 and 0.3.

a. 0.8544
b. 0.6788
c. 0.9999
d. 0.5545

ANSWER: (a) 0.8544

Explanation:
Here, m1 = 0.8
m2 = 0.3
total modulation index mt = √( m12 + m22 )
= √( 0.82 + 0.32 )
= √ 0.73
= 0.8544

23)   Calculate the frequencies available in the frequency spectrum when a 2MHz carrier is modulated by two sinusoidal signals of 350Hz and 600Hz.

a. 2000.35, 1999.65 and 2000.6, 1999.4
b. 1999.35, 1999.65 and 2000.6, 2000.4
c. 2000.35, 2000.65 and 2000.6, 2000.4
d. 1999.35, 1999.65 and 1999.6, 1999.4

ANSWER: (a) 2000.35, 1999.65 and 2000.6, 1999.4

Explanation:
The frequencies obtained in the spectrum after the amplitude modulation are
fc + fm and fc + fm
therefore,

the available frequencies after modulation by 0.350 KHz are
2000KHz + 0.350 KHz = 2000.35 and 2000KHz – 0.350 KHz = 1999.65

the available frequencies after modulation by 0.6 KHz are
2000KHz + 0 .6 KHz = 2000.6 and 2000KHz – 0.6 KHz = 1999.4

24)   If an AM signal is represented by
v = ( 15 + 3 Sin( 2Π * 5 * 103 t) ) * Sin( 2Π * 0.5 * 106 t) volts

i) Calculate the values of the frequencies of carrier and modulating signals.
ii) Calculate the value of modulation index.
iii) Calculate the value of bandwidth of this signal.

a. 1.6 MHz and 8 KHz, 0.6, 16 MHz
b. 1.9 MHz and 18 KHz, 0.2, 16 KHz
c. 2.4 MHz and 18 KHz, 0.2, 16 KHz
d. 1.6 MHz and 8 KHz, 0.2, 16 KHz

ANSWER: (d) 1.6 MHz and 8 KHz, 0.2, 16 KHz

Explanation:
The amplitude modulated wave equation is
v = ( 10 + 2 Sin( 2Π * 8 * 103 t) ) * Sin (2Π * 1.6 * 106 t) volts

Instantaneous value of AM signal is represented by the equation
v = {Vc + Vm Sin ( ωm t )} * Sin (ωc t )
comparing it with the given equation,

Vc = 10 V
Vm = 2V
ωc (= 2Π fc) = 2Π * 1.6 * 106
ωm (= 2Π fm) = 2Π * 8 * 103

(i) The carrier frequency fc is = 1.6 * 106 = 1.6 MHz
The modulating frequency fm is = 8* 103 = 8 kHz

(ii) The modulation index m = Vm/Vc = 2/10 = 0.2

(iii) The bandwidth BW = 2 fm = 16 kHz

25)   An AM signal has a total power of 48 Watts with 45% modulation. Calculate the power in the carrier and the sidebands.

a. 39.59 W, 4.505 W
b. 40.59 W, 4.205 W
c. 43.59 W, 2.205 W
d. 31.59 W, 8.205 W

ANSWER: (c) 43.59 W, 2.205 W

Explanation:
Given that Pt = 48 W

Modulation index m = 0.45

The total power in an AM is given by
Pt = Pc ( 1 + m2/2)
= Pc ( 1 +0.452/2)
48 = Pc * 1.10125

Therefore, Pc = 48/ 1.10125
= 43.59 W
The total power in two sidebands is 4843.59 = 4.41 W
So the power in each sideband is 4.41/2 = 2.205 W

26)   Calculate the power saved in an Amplitude Modulated wave when it is transmitted with 45% modulation
– Without carrier
– Without carrier and a sideband

a. 90%, 95%
b. 82%, 91%
c. 82%, 18%
d. 68%, 16%

ANSWER: (a) 90%, 95%

Explanation:
i) The total power in an AM is given by
Pt = Pc ( 1 + m2/2)

Given: m = 0.45
Therefore Pt = Pc ( 1 + 0.452/2 )
Pt= Pc *1.10125
Pc/ Pt = 1/1.10125
= 0.908
= 90%

This shows that the carrier occupies 90% of total power. So 90% of total power may be saved if carrier is suppressed in the AM signal.

(ii) If one of the sidebands is also suppressed, half of the remaining power will be saved
i.e., 10/2 = 5 %. So a total of 95% (90% + 5% ) will be saved when carrier and a side band are suppressed.

27)   What is the carrier frequency in an AM wave when its highest frequency component is 850Hz and the bandwidth of the signal is 50Hz?

a. 80 Hz
b. 695 Hz
c. 625 Hz
d. 825 Hz

ANSWER: (d) 825 Hz

Explanation:
Upper frequency = 850Hz

Bandwidth = 50Hz

Therefore lower Frequency = 850 – 50= 800 Hz

Carrier Frequency = (850-800)/2
= 825 Hz

28)   Noise figure of merit in SSB modulated signal is

a. 1
b. Less than 1
c. Greater than 1
d. None of the above

ANSWER: (a) 1

29)   For low level modulation, amplifier used is

a. Class A
b. Class C
c. Class A & C
d. None of the above

ANSWER: (a) Class A

30)   The antenna current of the transmitter is 10A. Find the percentage of modulation when the antenna current increases to 10.4A.

a. 32%
b. 28.5%
c. 64%
d. 40%

ANSWER:(b) 28.5%

Explanation:
It = Ic √(1+ m2/2)
10.4= 10 √(1+ m2/2)
√ (1+ m2/2) = 1.04

Therefore m = 0.285
= 28.5%

31)   What is the change in the value of transmitted power when the modulation index changes from 0 to 1?

a. 100%
b. Remains unchanged
c. 50%
d. 80%

ANSWER: (c) 50%

Explanation:
Pt = Pc ( 1 + m2/2)

Pt= Pc ( 1 + 02/2) = Pc ..(1)

New total power Pt1= Pc ( 1 + 12/2)
= Pc *3/2 ..(2)
(2) / (1),

We get , Pt1/ Pt= 3/2= 1.5
Pt1= 1.5 Pt
i.e. there is increase in total power by 50%

32)   Function of RF mixer is

a. Addition of two signals
b. Multiplication of two signals
c. Rejection of noise
d. None of the above

ANSWER: (b) Multiplication of two signals

33)   If a receiver has poor capacity of blocking adjacent channel interference then the receiver has

a. Poor selectivity
b. Poor Signal to noise ratio
c. Poor sensitivity
d. None of the above

ANSWER: (a) Poor selectivity

34)   Advantage of using a high frequency carrier wave is

a. Signal can be transmitted over very long distances
b. Dissipates very small power
c. Antenna height of the transmitter is reduced
d. All of the above

ANSWER: (d) All of the above

35)   Advantage of using VSB transmission is

a. Higher bandwidth than SSB
b. Less power required as compared to DSBSC
c. Both a and b
d. None of the above

ANSWER: (c) Both a and b

36)   Modulation is required for

a. Reducing noise while transmission
b. Multiplexing the signals
c. Reduction of Antenna height
d. Reduction in the complexity of circuitry
e. All of the above

ANSWER: (e) All of the above

37)   Bandwidth required in SSB-SC signal is (fm is modulating frequency):

a. 2fm
b. < 2fm
c. > 2fm
d. fm

ANSWER: (d) fm

Explanation:
In an amplitude modulated wave, total bandwidth required is from fc + fm to fc – fm

i.e. BW = 2fm where fc is carrier frequency.

In SSB-SC transmission, as the carrier and one of the sidebands are suppressed, the bandwidth remains as fm.

38)   For over modulation, the value of modulation index m is

a. m < 1
b. m = 1
c. m > 1
d. Not predetermined

ANSWER: (c) m > 1

39)   Demodulation is:

a. Detection
b. Recovering information from modulated signal
c. Both a and b
d. None of the above

ANSWER: (c)Both a and b

40)   Calculate the side band power in an SSBSC signal when there is 50% modulation and the carrier power is 50W.

a. 50 W
b. 25 W
c. 6.25 W
d. 12.5 W

ANSWER: (c) 6.25 W

Explanation:
The side band power is given by
Pc m2/2
= 50 * (0.5) 2/2
= 6.25W

41)   TRF receiver and super heterodyne receiver are used for

a. Detection of modulating signal
b. Removal of unwanted signal
c. Both a and b
d. None of the above

ANSWER:  (c) Both a and b

42)   Disadvantage of using a DSB or SSB signal modulation is

a. Difficult to recover information at the receiver
b. Carrier has to be locally generated at receiver
c. Both a and b are correct
d. None of the above

ANSWER: (c) Both a and b are correct

43)   Calculate the modulation index when the un modulated carrier power is 15KW, and after modulation, carrier power is 17KW.

a. 68%
b. 51.63%
c. 82.58%
d. 34.66%

ANSWER:  (b) 51.63%

Explanation:
The total power in an AM is given by
Pt = Pc ( 1 + m2/2)
17 = 15(1 + m2/2)
m2/2 = 0.134
m = 0.5163
= 51.63%

44)   An AM transmitter has an antenna current changing from 5 A un modulated to 5.8 A. What is the percentage of modulation?

a. 38.8%
b. 83.14%
c. 46.8%
d. 25.2%

ANSWER: (b) 83.14%

Explanation:
Modulation index m is given by
m= √ (2{It/Ic}2-1)
= √ (2 (5.8/5)2 -1)
= √ (2 (5.8/5)2 -1)
= 0.8314
= 83.14%

45)   Calculate the power in a DSB SC signal when the modulation is 60% with a carrier power of 600W.

a. 600 W
b. 540 W
c. 108 W
d. 300 W

ANSWER:  (c) 108 W

Explanation:
The total power in an AM is given by
Pt = Pc (1 + m2/2)
Given: m = 0.6
Therefore DSB power = (m2/2)Pc
= 600* (0.6)2/2
= 108 W

46)   Analog communication indicates:

a. Continuous signal with varying amplitude or phase
b. No numerical coding
c. AM or FM signal
d. All of the above

ANSWER:(d) All of the above

47)   Types of analog modulation are:

a. Phase modulation
b. Frequency modulation
c. Amplitude modulation
d. All of the above

ANSWER: (d) All of the above

48)   What is the effect on the transmitted power of AM signal when the modulation index changes from 0.8 to 1?

a. 0.1364
b. 0.3856
c. 1.088
d. 0.5

ANSWER: (a) 0.1364

Explanation:
The total power in an AM is given by
Pt = Pc (1 + m2/2)
Where Pc is the carrier power and m is the modulation index.

Therefore,

Pt1 = Pc (1 + 0.82/2) = 1.32 Pc
Pt2 = Pc ( 1 + 12/2) = 1.5 Pc
Increase in power = (1.5 Pc – 1.32 Pc)/ 1.32 Pc
= 0.1364

49)   Synchronous detection means

a. Extracting week signal from noise
b. Need a reference signal with predetermined frequency and phase
c. Both a and b
d. None of the above

ANSWER: (c)  Both a and b

50)   Analog signal may be converted into digital signal by

a. Sampling
b. Amplitude modulation
c. Filtering
d. Mixing

ANSWER: (a) Sampling

51)   The minimum antenna height required for transmission in reference to wavelength λ is

a. λ
b. λ/4
c. λ/2
d. 4 λ

ANSWER:(b) λ/4

52)   Advantages of analog communication over digital communication are:

a. Data rate is low
b. Less transmission bandwidth is required
c. Synchronization is not needed
d. All of the above

ANSWER: (d) All of the above

53)   Radio waves travel through

a. Electromagnetic waves
b. Water
c. Wires
d. Fiber optic cable

ANSWER: (a) Electromagnetic waves

54)   AM wave may be represented as E(t) cos ωct where E(t) is

a. Envelope of the AM wave
b. Carrier signal
c. Amplitude of modulating signal
d. None of the above

ANSWER: (a) Envelope of the AM wave

55)   USB (Upper Side Band) is the band of frequency

a. Above the carrier frequency
b. Includes the carrier frequency
c. That lies in AM spectrum
d. Both a and c are correct

ANSWER:(d)  Both a and c are correct

56)   LSB (Lower Side Band) is the band of frequency

a. Below the carrier frequency
b. Includes the carrier frequency
c. That lies in AM spectrum
d. Both a and c are correct

ANSWER: (d) Both a and c are correct

57)   Bandwidth (B) of an AM signal is given by

a. B = 2 ωm
b. B = (ωc + ωm) – (ωc – ωm)
c. ωm
d. None of the above
e. Both a and b are correct

ANSWER: (e) Both a and b are correct

58)   An oscillator for an AM transmitter has a 100μH coil and a 10nF capacitor. If a modulating frequency of 10 KHz modulates the oscillator, find the frequency range of the side bands.

a. 149 KHz to 169 KHz
b. 184 KHz to 296 KHz
c. 238 KHz to 296 KHz
d. 155 KHz to 166 KHz

ANSWER: (a) 149 KHz to 169 KHz

Explanation:
Carrier frequency fc = 1/2Π√LC
= 1/ 2Π√100 * 10 – 6 * 10 * 10-9
= 1/2Π * 10-6
= 1.59 * 105 Hz
= 159 KHz
The modulating frequency fm is 10KHz
Therefore the range of AM spectrum is given by (fc fm ) to (fc + fm )
= (159 – 10) to (159 + 10)
= 149 KHz to 169 KHz

59)   In Low level Amplitude Modulation

a. Modulation is done at lower power of carrier and modulating signal
b. Output power is low
c. Power amplifiers are required to boost the signal
d. All of the above

ANSWER: (d) All of the above

60)   In High level Amplitude Modulation

a. Modulation is done at high power of carrier and modulating signal
b. Collector modulation method is High level Amplitude Modulation
c. Power amplifiers are used to boost the carrier and modulating signals before modulation
d. All of the above

ANSWER: (d) All of the above