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Analog Communications Multiple Choice Questions focuses on “AM Wave Representation”.
1. AM waves is represented by which equation?
a) [1 + m(t)].c(t)
b) [1 – m(t)].c(t)
c) [1 + m(t)].2c(t)
d) [1 + 2m(t)].c(t)
Answer: a
Clarification: Amplitude wave is represented by [1 + um(t)].c(t), where c(t) is carrier signal, m(t) is message signal, u is Modulation Index.
Generally, c(t) = Accos(wct), Ac = Amplitude of Carrier Signal.
2. For attenuation of high frequencies we can use ________
a) inductor
b) shunt capacitance
c) series capacitance
d) combination of inductor and resistor
Answer: b
Clarification: Capacitive reactance, Xc is low for high frequencies as Xc=(1/wC), where w is the frequency and C is the capacitance. Due to this, the high frequencies are shunted to ground and are not transmitted.
3. If each element of signal occupies 40ms, determine its speed?
a) 20 bauds
b) 25 bauds
c) 40 bauds
d) 0.05 bauds
Answer: b
Clarification:
4. A modem is classified as low if speed of data rate is ________
a) upto 600bps
b) upto 200bps
c) upto 400bps
d) upto 500bps
Answer: a
Clarification: Modem is device that can modulate carrier wave as well as demodulate transmitted modulated signal to decode the original information. When data rate in bps is upto 600, modem is considered having low speed.
5. Modulation is also called detection.
a) True
b) False
Answer: b
Clarification: Modulation is encoding the message signal for efficient transmission. Whereas, Demodulation is the process to extract or decode the original message signal from the transmitted modulated signal. Demodulation is also called detection.
6. It is suitable to connect woofer from the input through ________
a) band pass filter
b) band stop filter
c) low pass filter
d) high pass filter
Answer: c
Clarification: Woofer is considered as a low frequency loud speaker. It has a range from 10Hz to 500Hz. So it is better to use low pass filter for connection. Los Pass filter allows low frequencies to pass while high pass filter allows high frequencies to pass. Band pass filter allows a range of frequencies to pass whereas. Band stop filter does the opposite.
7. The radiation at right angles is zero means ________
a) l = ʎ
b) l = ʎ⁄4
c) l = 2 ʎ
d) l = ʎ⁄2
Answer: a
Clarification: Antenna is used for converting electromagnetic radiation into electric currents or vice-versa. If the length of antenna is equal to the whole wavelength then the radiation at right angles is zero.
8. Power of carrier wave is 500W and modulation index is 0.25. Find its total power?
a) 500W
b) 415W
c) 375W
d) 516W
Answer: d
Clarification: Total Power (Pt) = (1+m2/2)*Pc
So, here, m = 0.25, Pc = 500W
Pt = (1+(0.252/2))*500 = 516W.
9. Commercial frequency deviation is ________
a) 75 KHz
b) 80 KHz
c) 85 KHz
d) 65 KHz
Answer: a
Clarification: Frequency deviation is the difference between an FM modulated frequency and the carrier frequency. Commercial FM broadcasting uses a maximum frequency deviation of 75 KHz and maximum modulating frequency is approximately 15KHz.
10. Which of the following cannot be the Fourier series of a periodic signal?
a) x(t) = 2Cost + 4Cos7t
b) x(t) = 2Cosπt + 4Cos7t
c) x(t) = Cost + 0.7
d) x(t) = 2Cost + 4Sin7t
Answer: c
Clarification: x(t) = Cost + 0.7 does not follow Dirchlet’s conditions. Fourier series of a periodic signal will only be applicable if it satisifies Dirchlet’s conditions firstly. Dirichlet integral states that a finite number should have finite number of discontinuities and also have finite number of extrema. x(t) doesn’t satisfy any of the criteria, whereas other options satisfy Dirchlet’s conditions. Thus, x(t) = Cost + 0.7, cannot be the Fourier series of a periodic signal.
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