Analog Communications Questions and Answers – Coherent Detection of DSBSC Waves

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This set of Analog Communications Questions and Answers for Freshers focuses on “Coherent Detection of DSBSC Waves”.

1. Which among the following is the drawback of Pulse Position Modulation (PPM)?
a) Synchronization is required between receiver and transmitter
b) Synchronization is not required between receiver and transmitter
c) Amplitude is constant
d) Instantaneous power of PPM modulated signals is constant
View Answer

Answer: a
Explanation: Pulse Position Modulation (PPM) is a type of modulation process in which position of the pulse of the carrier wave is varied with respect to the instantaneous position values of the message signal. In PPM, there is minimum noise interference but the main disadvantage of the PPM modulation technique is that synchronization between transmitter and receiver must be needed.

2. Calculate the Nyquist rate for the signal x(t) = 12cos50πt + 7 cos75π t – 13 cos100πt.
a) 300 Hz
b) 600 Hz
c) 100 Hz
d) 150 Hz
View Answer

Answer: c
Explanation: On comparing with general Equation, Acosωt and substituting ω = 2×π× f. We have f1 = 25 Hz, f2 = 37.5Hz, f3 = 100π/2π = 50Hz.
Taking the maximum modulating frequency, fmax = 50Hz
Nyquist rate = 2 × fmax = 2 × 50 = 100 Hz.

3. What is the need of doing Pre emphasis?
a) For boosting of modulating signal voltage
b) For boosting of modulating signal frequency
c) For removing amplitude variations due to noise
d) For removing frequency variations
View Answer

Answer: a
Explanation: Pre-emphasis is vastly applied in communication systems to improve signal strength before transmission. It refers to boosting the amplitudes of the weak modulating voltages for high audio frequencies in the range of 2 to 15KHz.
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4. Amount of data transmitted for a given time is called _________
a) Noise
b) Frequency
c) Bandwidth
d) Power
View Answer

Answer: c
Explanation: Bandwidth is the maximum amount of data transmitted over a particular period of time. It is also defined as the difference of high frequency and low frequency, of a given band.

5. Consider an AM broadcast station that transmits modulating frequencies up to 10kHz. If it transmits a frequency of 1000 kHz. Find its maximum and minimum upper and lower sidebands frequencies and also the total bandwidth?
a) 900 KHz, 820 KHz, 1000 Hz
b) 720 KHz, 650 KHz, 1020 Hz
c) 1010 KHz, 880 KHz, 15000 Hz
d) 1010 KHz, 990 KHz, 20000 Hz
View Answer

Answer: d
Explanation: Maximum Frequency, fm = (1000 + 10) KHz = 1010 KHz.
Minimum Frequency, fl = (1000 – 10) KHz = 990 KHz and,
Bandwidth = 2 * Modulating frequency = Maximum Frequency – Minimum Frequency
= (1010 – 990) KHz = 20000 Hz.

6. A superheterodyne receiver receives signal within frequency range of 120 to 180 MHz. Then the required Intermediate frequency is _________
a) 30MHz
b) 60MHz
c) 90MHz
d) 50MHz
View Answer

Answer: a
Explanation:
The required Intermediate frequency is 30MHz if frequency range of 120 to 180 MHz
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7. Carson’s rule is used to calculate ________
a) Bandwidth of FM signal
b) SNR
c) Modulation index of FM signal
d) Figure of merit
View Answer

Answer: a
Explanation: Carson’s rule states that only (β+1) (where β = modulation index) upper and lower sidebands along with the carrier, have significant magnitude and contain 99% of total power. Thus, as per Carson’s rule, required bandwidth is equal to the twice of sum of the maximum frequency deviation (fd) and the maximum modulating frequency(fm), B = 2(fd + fm)Hz.
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8. What is the bandwidth of a FM wave when maximum allowed deviation is 50KHz and the modulating signal has a frequency of 15KHz?
a) 130 KHz
b) 260 KHz
c) 65 KHz
d) 50 KHz
View Answer

Answer: a
Explanation: According to Carson s rule, B = 2(fd +fm) = 2 (50 + 15) = 130 KHz.

9. For signal, m(t) = 50cos(10 × 90t + 30 sin100t), the power dissipated by the 20Ω resistor is ________
a) 100W
b) 65W
c) 74.7W
d) 62.5W
View Answer

Answer: d
Explanation: In angle modulation, the amplitude of carrier remains constant and thus power depends on amplitude only. On Comparing with the equation, m(t) = Acos(ωct + mf sinωmt). We have A = 50, dissipated power, The power dissipated by the 20Ω resistor is 62.5W if m(t) is 50cos(10 × 90t + 30 sin100t)
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10. Signal and its Hilbert transform have ________
a) same energy density spectrum
b) same power
c) a phase difference of 60°
d) a phase difference of 120°
View Answer

Answer: a
Explanation: Properties of Hilbert transform states that the signal and its Hilbert transform :
i) have same energy density spectrum
ii) are mutually orthogonal
iii) have same auto correlation function
iv) have same magnitude
v) have a phase difference of “-90” degree.

Sanfoundry Global Education & Learning Series – Analog Communications.

To practice all areas of Analog Communications for Freshers, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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