GATE 2025 EC Q.14: solution and answer

GATE 2025 · ECQ.14 · 1 mark · MCQ

Consider a frequency-modulated (FM) signal $f (t) = A_{c} cos (2 π f_{c} t + 3 sin (2 π f_{1} t) + 4 sin (6 π f_{1} t))$ , where $A_{c}$ and $f_{c}$ are, respectively, the amplitude and frequency (in Hz) of the carrier waveform. The frequency $f_{1}$ is in Hz, and assume that $f_{c} > 100 f_{1}$ . The peak frequency deviation of the FM signal in Hz is ______.

(A)
$15 f_{1}$
(B)
$12 f_{1}$
(C)
$4 f_{1}$
(D)
$2 f_{1}$

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Answer and solution

ANSWER

The peak frequency deviation is $15 f_{1}$ Hz, option (A).

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

Write the total phase: $θ (t) = 2 π f_{c} t + 3 sin (2 π f_{1} t) + 4 sin (6 π f_{1} t)$ .
The instantaneous frequency in Hz is $f_{i} (t) = \frac{1}{2 π} \frac{d θ}{d t}$ .
Differentiate: $\frac{d θ}{d t} = 2 π f_{c} + 3 (2 π f_{1}) cos (2 π f_{1} t) + 4 (6 π f_{1}) cos (6 π f_{1} t)$ .
Divide by $2 π$ : $f_{i} (t) = f_{c} + 3 f_{1} cos (2 π f_{1} t) + 12 f_{1} cos (6 π f_{1} t)$ .
The frequency deviation is $Δ f (t) = f_{i} (t) - f_{c} = 3 f_{1} cos (2 π f_{1} t) + 12 f_{1} cos (2 π (3 f_{1}) t)$ .
The deviation is largest when both cosines equal $1$ . This happens together at $t = 0$ , since both arguments are zero there. So $Δ f_{m a x} = 3 f_{1} + 12 f_{1} = 15 f_{1}$ . No other time can give a larger value, because each term is bounded by its amplitude.
The condition $f_{c} > 100 f_{1}$ only keeps the instantaneous frequency positive and the carrier well above the message tones. It does not change the deviation. Answer: (A) $15 f_{1}$ .

KEY IDEA

Peak frequency deviation is the maximum of $\frac{1}{2 π} \frac{d}{d t}$ of the phase excess over the carrier. Each sinusoidal phase term $β sin (2 π f_{m} t)$ adds $β f_{m} cos (2 π f_{m} t)$ to the frequency, so its amplitude is $β f_{m}$ . The terms peak together at $t = 0$ and add.

WATCH OUT FOR

Reading the phase amplitudes 3 and 4 as the deviations, or taking only the larger tone. The second term has frequency $3 f_{1}$ , so its deviation is $4 \times 3 f_{1} = 12 f_{1}$ , not $4 f_{1}$ . Choosing $12 f_{1}$ ignores the first tone, which adds $3 f_{1}$ .

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 15 Analog modulation questions, with notes

GATE 2025 EC Q.14: solution and answer

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