Analog Communication Test Questions - Set 2

1)   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  Explanation  Related Ques

ANSWER: 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


2)   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  Explanation  Related Ques

ANSWER: 0.8544

Explanation:
Here, m1 = 0.8
m2 = 0.3

total modulation index mt = √( m12 + m22 )
= √( 0.82 + 0.32 )
= √ 0.73
= 0.8544 = 85.44%


3)   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  Explanation  Related Ques

ANSWER: 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


4)   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  Explanation  Related Ques

ANSWER: 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


5)   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  Explanation  Related Ques

ANSWER: 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


6)   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  Explanation  Related Ques

ANSWER: 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.


7)   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  Explanation  Related Ques

ANSWER: 825 Hz

Explanation:
Upper frequency = 850Hz

Bandwidth = 50Hz

Therefore lower Frequency = 850 - 50= 800 Hz

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


8)   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  Explanation  Related Ques

ANSWER: 1

Explanation:
A figure of merit used to describe the performance of a system. The figure of merit γ is the ratio of output signal to noise ratio to input signal to noise ratio of a receiver system. Figure of merit for SSB modulation is always 1.


9)   For low level modulation, amplifier used is

a. Class A
b. Class C
c. Class A & C
d. None of the above
Answer  Explanation  Related Ques

ANSWER: Class A

Explanation:
When the modulation takes place prior to the output element of the final stage of the amplifier, it is low level modulation. Class A amplifiers are used for this purpose.


10)   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  Explanation  Related Ques

ANSWER: 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%


11)   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  Explanation  Related Ques

ANSWER: 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%


12)   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  Explanation  Related Ques

ANSWER: Multiplication of two signals

Explanation:
RF mixer translates the frequencies of the two incoming signals from one band to another by multiplying them. When the two incoming signals are mixed, the new frequencies obtained are the sum and difference of incoming signal frequencies .i.e., if the two input signal frequencies are f1 and f2, then the output of the mixer will have signals at frequencies (f1 + f2) and (f1 - f2).


13)   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  Explanation  Related Ques

ANSWER: Poor selectivity

Explanation:
Frequency selectivity means to separate out closely lying components of frequency. If the receiver has poor selectivity, it has poor capacity of blocking adjacent channel interference.


14)   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  Explanation  Related Ques

ANSWER: All of the above

Explanation:
The main advantage of high frequency signals is that the signal may be transmitted over very long distances and thus dissipates very less power. The antenna height required for transmission is reduced at higher frequencies. Thus, the audio signals must be sent with the high frequency carrier signals for communication purpose.


15)   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  Explanation  Related Ques

ANSWER: Both a and b

Explanation:
In AM, the signal appears in the form of components at frequencies slightly higher and lower than that of the carrier frequency. These sidebands carry the information.

Both the sidebands are similar to each other, so one of the sidebands may be rejected. Vestigial Side Band transmission is similar to single-sideband (SSB) transmission, in which the second sideband is not completely removed, but is filtered to remove the range of frequencies that are not required.


16)   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  Explanation  Related Ques

ANSWER: All of the above

Explanation:
Modulation is required as with modulation, the signal is transmitted at higher frequency that reduces the probability of addition of noise while transmission and reduces the height of antenna at the transmitter. With modulation, there is reduction in the complexity of circuitry and more signals may be transmitted using multiplexing technique.


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

a. 2fm
b. < 2fm
c. > 2fm
d. fm
Answer  Explanation  Related Ques

ANSWER: 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.


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

a. m < 1
b. m = 1
c. m > 1
d. Not predetermined
Answer  Explanation  Related Ques

ANSWER: m > 1

Explanation:
In AM, the modulation index m is defined as the ratio of the amplitude modulation signal to the amplitude of carrier signal. When m > 1, it is called over-modulation and it generates modulation with distortions in the envelope of the modulated signal, from which it is difficult to recover the information.

For Under modulation m < 1

For Critical modulation m = 1

For Over modulation m > 1


19)   Demodulation is:

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

ANSWER: Both a and b

Explanation:
Demodulation is the process of recovering the original information from a modulated carrier wave. Systems are designed to be used as demodulators that detect the information signal from the carrier. The envelope detector and product detector are few of the AM detectors.


20)   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  Explanation  Related Ques

ANSWER: 6.25 W

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