GATE 2026 IN Q.52: solution and answer

GATE 2026 · INQ.52 · 2 marks · MCQ

An amplitude modulated voltage signal $x (t)$ drives a load of $1 Ω$ . $x (t) = K cos (300 π t) + L cos (240 π t) + L cos (360 π t)$ . If the efficiency is 60% and the carrier power is 50 W, the value of $L$ is ______ V.

(A)
8.66
(B)
1.22
(C)
0.81
(D)
17.32

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Answer and solution

ANSWER

$L = 75 \approx 8.66$ V, option (A).

Official answer: ASource: Official answer key, GATE 2026 IN (IIT Guwahati)

Read the spectrum: the term at $150$ Hz ( $300 π$ rad/s) is the carrier with amplitude $K$ . The terms at $120$ Hz and $180$ Hz ( $240 π$ and $360 π$ rad/s) are the lower and upper sidebands, each of amplitude $L$ , so $f_{m} = 30$ Hz.
Carrier power in the $1 Ω$ load: $P_{c} = \frac{K ^{2}}{2 R} = \frac{K ^{2}}{2} = 50 W$ , so $K = 10 V$ .
Each sideband is a sinusoid of amplitude $L$ , so it carries $\frac{L ^{2}}{2}$ W. The total sideband power is $P_{s} = 2 \cdot \frac{L ^{2}}{2} = L^{2} W$ .
Efficiency is the fraction of total power carried by the sidebands: $η = \frac{P _{s}}{P _{c} + P _{s}} = \frac{L ^{2}}{50 + L ^{2}}$ .
Set $η = 0.6$ : $L^{2} = 0.6 (50 + L^{2})$ , so $0.4 L^{2} = 30$ and $L^{2} = 75$ .
$L = 75 \approx 8.66 V$ , which is option (A).
Check: $L = μ K /2$ gives $μ = \frac{2 L}{K} \approx 1.732$ . Then $η = \frac{μ ^{2}}{2 + μ ^{2}} = \frac{3}{5} = 0.6$ , which is consistent.

KEY IDEA

In AM, efficiency is sideband power divided by total power. With a carrier of amplitude $K$ and two sidebands of amplitude $L$ each, $η = \frac{L ^{2}}{K ^{2} /2 + L ^{2}}$ in a $1 Ω$ load.

WATCH OUT FOR

Counting only one sideband ( $L^{2} /2$ instead of $L^{2}$ ) gives $L^{2} = 37.5$ and $L \approx 6.12$ V, which is not an option. Another slip is using $L^{2}$ instead of $L^{2} /2$ for the power of a sinusoid.

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GATE 2026 IN Q.52: solution and answer

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