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# GATE-2012 ECE Q39 (communication)

by on November 17, 2012

Question 39 on communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Q39. The signal  as shown is applied both to  a phase modulator (with  as the phase constant) and a frequency modulator (with as the frequency constant) having the same carrier frequency. ## (D) ## Solution

To answer this question, let us understand basics of Frequency modulation and Phase modulation. Am referring to the discussion in Section 10 of  Electronic Communication, 4th Edition, Dennis Roddy, John Coolen (buy from Amazon.com buy from Flipkart.com).

Frequency Modulation

In frequency modulation, the modulation signal  will cause a change in the instantaneous carrier frequency and is related as

where

is the frequency deviation constant expressed in hertz/volt (Hz/V) and  is the carrier frequency in Hz.

The angular velocity  radians per second.

Since the rate of change of phase  is the angular velocity,

and so, the instantaneous phase is

.

The frequency modulated carrier signal is

Phase modulation

In phase modulation, the modulation signal  will cause a change in phase and the instantaneous phase is

where

is the phase deviation constant expressed in radians/volt  and  is the carrier frequency in radians per second.

The phase modulated carrier signal is

.

Now, applying the above equations to solve the problem :

In frequency modulation case, the maximum phase deviation happens when the term hits the maximum.

In phase modulation case, the maximum phase deviation happens when the term  hits the maximum.

.

The ratio is,

and so,

Based on the above, the right choice is (B)

## References

 GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

 Electronic Communication, 4th Edition, Dennis Roddy, John Coolen (buy from Amazon.com buy from Flipkart.com).

D id you like this article? Make sure that you do not miss a new article by subscribing to RSS feed OR subscribing to e-mail newsletter. Note: Subscribing via e-mail entitles you to download the free e-Book on BER of BPSK/QPSK/16QAM/16PSK in AWGN. Raghava G D November 17, 2012 at 8:21 pm

Though the answer seems to right..mathematically there is a glitch i believe..why have integrated from 0 to t when the signal exists from “- infinity” ?..you should have integrated from “- infinity” to “t” right ?..basically i am just saying proof is not complete..though you will get the same answer when you make it complete ..

Sorry for being so critical on math rather than engineering!!  Krishna Sankar November 18, 2012 at 6:18 am
It feels nice to have you cross-checking the accuracy of the answers. Thanks much ! 