Analog Communications Questions and Answers – Heterodyne Receiver

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This set of Analog Communications Multiple Choice Questions & Answers (MCQs) focuses on “Heterodyne Receiver”.

1. What is the bandwidth required in SSB signal?
a) fm
b) 2fm
c) > 2fm
d) < 2fm
View Answer

Answer: a
Explanation: In an AM modulated system, total bandwidth required is from fc + fm to fc – fm i.e. bandwidth is equal to 2fm. In SSB-SC transmission, the carrier and one of the sideband gets suppressed, so the bandwidth becomes fm only.

2. One of the advantage of using a high frequency carrier wave is that it dissipates very small power.
a) True
b) False
View Answer

Answer: a
Explanation: The main advantage of using high frequency signals is that the signal gets transmitted over very long distances and thus dissipates very less power. The antenna height required for transmission also gets reduced at high frequencies. And also it allows less noise interference and enables multiplexing. This is the reason for sending the audio signals at high frequency carrier signals for communication purpose.

3. What is the function of RF mixer?
a) Addition of two signals
b) Multiplication of two signals
c) Subtraction of two signals
d) To reduce the amount of noise
View Answer

Answer: b
Explanation: RF mixer translates the frequencies of the two incoming signals by multiplying them and bringing them to a suitable band which can be processed.
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4. The antenna current is 10A. Find the percentage of modulation when the antenna current increases to 10.4A?
a) 50%
b) 30%
c) 28.5%
d) 23%
View Answer

Answer: c
Explanation:
The percentage of modulation is 28.5% when the antenna current increases to 10.4A which gives m = 0.285 or 28.5%.

5. Find the total power, if the carrier of an AM transmitter is 800W and it is modulated to 50%?
a) 100W
b) 800W
c) 500W
d) 900W
View Answer

Answer: d
Explanation: PT=PC (1+u22), according to the problem PC = 800W and m = 0.5. On substituting values in the equation we get PT=800(1+ 0.522) = 900W.

6. Aliasing refers to ________
a) Sampling of signals less than at Nyquist rate
b) Sampling of signals at Nyquist rate
c) Sampling of signals greater than at Nyquist rate
d) Unsampled the original signal
View Answer

Answer: a
Explanation: Aliasing refers to the sampling of signals less than at Nyquist rate. Nyquist rate states that the rate of sampling of signals should be greater than or equal to twice the bandwidth of modulating signal. It gets reduced if sampling is done at a higher rate than nyquist rate of sampling. Aliasing can be avoided by using anti-aliasing filters.

Sanfoundry Global Education & Learning Series – Analog Communications.

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To practice all areas of Analog Communications, 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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