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.

## The ratio for the same maximum phase deviation is,

## (A)

## (B)

## (C)

## (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

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

[2] **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.*

{ 2 comments… read them below or add one }

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!!

@Raghava: Yes, this point briefly crossed my mind when writing up – however, since we are only concerned with the max() value, later forgot about it. Doing a re-read, seems the integral can be changed from -infinity to t with out affecting any other description.

It feels nice to have you cross-checking the accuracy of the answers. Thanks much !