GATE 2025 EC Q.15: solution and answer

GATE 2025 · ECQ.15 · 1 mark · MCQ

Consider an additive white Gaussian noise (AWGN) channel with bandwidth $W$ and noise power spectral density $\frac{N _{0}}{2}$ . Let $P_{a v}$ denote the average transmit power constraint.

Which one of the following plots illustrates the dependence of the channel capacity $C$ on the bandwidth $W$ (keeping $P_{a v}$ and $N_{0}$ fixed)?

GATE 2025 EC question 15: Four small plots, each with the vertical axis labelled C and the horizontal axis labelled W, both starting at the origin 0. — Figure from the GATE 2025 EC paper. It is part of the question.

(A)
Figure — C versus W: concave curve rising steeply from the origin and saturating at a horizontal asymptote.
(B)
Figure — C versus W: straight line through the origin with constant positive slope.
(C)
Figure — C versus W: convex curve starting with a small slope and rising increasingly steeply.
(D)
Figure — C versus W: curve rising from the origin to a peak and then decreasing.

Answer and solution

ANSWER

The correct plot is (A): capacity rises from zero, is concave, and saturates at $\frac{P _{a v}}{N _{0}} lo g_{2} e$ as $W \to \infty$ .

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

Use the two-sided noise PSD $\frac{N _{0}}{2}$ . In a bandwidth $W$ (band $- W$ to $W$ in the baseband picture, total width $2 W$ ) the noise power is $N = \frac{N _{0}}{2} \times 2 W = N_{0} W$ .
Shannon capacity of the AWGN channel: $C = W lo g_{2} (1 + \frac{P _{a v}}{N _{0} W}) bits/s .$ Write $a = \frac{P _{a v}}{N _{0}}$ (fixed), so $C = \frac{1}{ln 2} W ln (1 + \frac{a}{W})$ .
Small $W$ : $\frac{a}{W} ≫ 1$ , so $C \approx W lo g_{2} \frac{a}{W}$ . This is $0$ at $W = 0$ and rises steeply (slightly less than linearly).
Large $W$ : $x = \frac{a}{W} ≪ 1$ and $ln (1 + x) \approx x$ , so $C \to \frac{W}{ln 2} \cdot \frac{a}{W} = \frac{P _{a v}}{N _{0}} lo g_{2} e \approx 1.44 \frac{P _{a v}}{N _{0}},$ a finite limit. Capacity does not grow without bound.
Slope: $\frac{d C}{d W} = \frac{1}{ln 2} [ln (1 + \frac{a}{W}) - \frac{a}{W + a}] > 0,$ because $ln (1 + x) > \frac{x}{1 + x}$ for $x > 0$ . So $C$ increases with $W$ (no peak).
Curvature: $\frac{d ^{2} C}{d W ^{2}} = \frac{a}{ln 2 ( W + a )} [\frac{1}{W + a} - \frac{1}{W}] < 0,$ so $C$ is concave in $W$ .
So $C$ starts at $0$ , increases, is concave, and flattens to the horizontal asymptote $1.44 P_{a v} / N_{0}$ . A straight line (B), unbounded convex growth (C) and a curve with a peak (D) are all ruled out. This is plot (A).

KEY IDEA

With fixed $P_{a v}$ the noise power grows with bandwidth ( $N = N_{0} W$ ), so $C = W lo g_{2} (1 + P_{a v} / N_{0} W)$ is increasing and concave and tends to the finite limit $1.44 P_{a v} / N_{0}$ : a bandwidth-limited regime at small $W$ , a power-limited one at large $W$ .

WATCH OUT FOR

Reading $C = W lo g_{2} (1 + SNR)$ as linear in $W$ and choosing the straight line (B), forgetting that the SNR itself falls as $1/ W$ because noise power is $N_{0} W$ .

From the GATE 2025 EC paper. Every question of this paper · Download the full paper · All 9 Information theory questions, with notes

GATE 2025 EC Q.15: solution and answer

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