GATE 2026 EC Q.55: solution and answer

GATE 2026 · ECQ.55 · 2 marks · MSQ

Let the relevant bandwidth ( $B$ ) of a digital communication system be 1 MHz and $k T = - 174$ dBm/Hz, where $k$ is Boltzmann's constant and 'T' is equivalent noise temperature of the receiver. The power ( $S$ ) of signal received through an additive Gaussian channel is $- 80$ dBm. Which of the following options is/are TRUE about Shannon capacity ( $C$ ) of the channel?

(A)
$C = B$
(B)
$C = 2 B$
(C)
$C > 3 B$
(D)
$C < B$

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Answer and solution

ANSWER

Only option (C) is true: $C \approx 11.3 B > 3 B$ .

Official answer: CSource: Official answer key, GATE 2026 EC (IIT Guwahati)

Take the given $k T = - 174 dBm/Hz$ as the one-sided noise power spectral density $N_{0}$ (the question gives $k T$ directly, so no extra factor of 2 is applied). Then the noise power in the bandwidth $B = 1 MHz = 1 0^{6} Hz$ is $N = k T B$ .
In decibels: $N_{dBm} = - 174 + 10 lo g_{10} (1 0^{6}) = - 174 + 60 = - 114 dBm$ .
The received signal power is $S = - 80 dBm$ , so $SNR_{dB} = - 80 - (- 114) = 34 dB$ .
Convert to a ratio: $\frac{S}{N} = 1 0^{3.4} \approx 2512$ .
Shannon capacity of the band-limited AWGN channel: $C = B lo g_{2} (1 + \frac{S}{N}) = B lo g_{2} (2513)$ .
$lo g_{2} (2513) = \frac{ln 2513}{ln 2} \approx \frac{7.829}{0.693} \approx 11.3$ , so $C \approx 11.3 B \approx 11.3 Mbit/s$ .
Check the options. (A) $C = B$ : false, since $C \approx 11.3 B$ . (B) $C = 2 B$ : false. (C) $C > 3 B$ : true, since $11.3 B > 3 B$ . (D) $C < B$ : false.
Even if $k T$ were read as $N_{0} /2$ (so $N_{0} = 2 k T$ ), the SNR would drop by 3 dB to 31 dB and $C \approx 10.3 B$ , still above $3 B$ . The result does not depend on this convention.

KEY IDEA

Convert the noise PSD in dBm/Hz to noise power in the band by adding $10 lo g_{10} B$ , subtract in dB to get the SNR (34 dB), then apply $C = B lo g_{2} (1 + SNR)$ . At this high SNR the capacity is many times $B$ .

WATCH OUT FOR

Forgetting the bandwidth term $10 lo g_{10} (1 0^{6}) = 60$ dB, which gives an SNR of $94$ dB instead of $34$ dB. Another slip is treating the SNR in dB as a linear ratio, or taking $C = B lo g_{2} (SNR)$ and then confusing dB with ratio. Another is reading " $C > 3 B$ " as needing a precise value, when any SNR above 7 (8.5 dB) is enough.

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GATE 2026 EC Q.55: solution and answer

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