GATE EC Communications Previous Year Questions with Solutions

Full question on page 31 of the official paper (PDF)

GATE 2026, Q.32

Communications1 markNAT

16-QAM carries $4 \times 10^6$ bit/s over an AWGN channel. Find the minimum transmission bandwidth.

Answer format: MHz, rounded off to two decimal places

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Key idea: Minimum passband bandwidth equals the symbol rate.

Official answer: 0.95 to 1.05

16-QAM carries $\log_2 16 = 4$ bits per symbol, so the symbol rate is 1 Msymbol/s. A passband signal needs at least the symbol rate in bandwidth (Nyquist pulses, both sidebands):

B_{\min} = R_s = \frac{4 \times 10^6}{4} = 1.00\text{ MHz}

\begin{aligned} B_{\min} &= R_s \\ &= \frac{4 \times 10^6}{4} \\ &= 1.00\text{ MHz} \end{aligned}

Full question on page 41 of the official paper (PDF)

GATE 2026, Q.44

Communications2 marksMCQ

Coherent QPSK over AWGN is received at $E_b/N_0 = 8.4$ dB. Using $\mathrm{erfc}(u) \approx \dfrac{e^{-u^2}}{u\sqrt{\pi}}$ , find the nearest bit error rate.

Option A: $10^{-3}$
Option B: $10^{-4}$
Option C: $10^{-5}$
Option D: $10^{-6}$

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Key idea: QPSK has the same bit error rate as BPSK: $\tfrac{1}{2}\mathrm{erfc}\sqrt{E_b/N_0}$ .

Official answer: B

For Gray-coded coherent QPSK:

P_b = \tfrac{1}{2}\,\mathrm{erfc}\!\left(\sqrt{E_b/N_0}\right)

Convert the ratio: $E_b/N_0 = 10^{0.84} \approx 6.92$ , so $u = \sqrt{6.92} \approx 2.63$ . With the given approximation,

P_b \approx \frac{e^{-6.92}}{2 \times 2.63\sqrt{\pi}} \approx \frac{9.9 \times 10^{-4}}{9.32} \approx 1.06 \times 10^{-4}

\begin{aligned} P_b &\approx \frac{e^{-6.92}}{2 \times 2.63\sqrt{\pi}} \\ &\approx \frac{9.9 \times 10^{-4}}{9.32} \\ &\approx 1.06 \times 10^{-4} \end{aligned}

The nearest option is $10^{-4}$ . Option (B).

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Full question on page 52 of the official paper (PDF)

GATE 2026, Q.55

Communications2 marksMSQ

Bandwidth $B = 1$ MHz, $kT = -174$ dBm/Hz and received power $S = -80$ dBm over an additive Gaussian channel. What is true of the Shannon capacity $C$ ?

One or more options are correct.

Option A: $C = B$
Option B: $C = 2B$
Option C: $C > 3B$
Option D: $C < B$

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Key idea: Find the SNR in dB, then $C = B\log_2(1 + \text{SNR})$ .

Official answer: C

Noise power in 1 MHz (60 dB-Hz):

N = -174 + 10\log_{10}(10^6) = -114\text{ dBm}, \qquad \text{SNR} = -80 - (-114) = 34\text{ dB} \approx 2512

\begin{aligned} N &= -174 + 10\log_{10}(10^6) \\ &= -114\text{ dBm}, \\ \text{SNR} &= -80 - (-114) \\ &= 34\text{ dB} \\ &\approx 2512 \end{aligned}

C = B\log_2(1 + 2512) \approx 11.3\,B

\begin{aligned} C &= B\log_2(1 + 2512) \\ &\approx 11.3\,B \end{aligned}

So $C > 3B$ is the only true statement. Option (C).

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Full question on page 54 of the official paper (PDF)

GATE 2026, Q.57

Communications2 marksMSQ

A real narrowband signal is $x(t) = A(t)\cos\left[2\pi f_c t + \theta(t)\right]$ , where $A(t)$ and $\theta(t)$ vary slowly compared with $f_c = 1000 f_M$ . Which statements are correct?

One or more options are correct.

Option A: $x(t)$ can be a PSK signal.
Option B: $x(t)$ can be an AM signal.
Option C: $x(t)$ can be band-limited Gaussian noise.
Option D: $x(t)$ can never be a narrowband FM signal.

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Key idea: Any narrowband signal around a carrier can be written with a slowly varying envelope and phase.

Official answer: A, B and C

Check whether each signal fits the form $A(t)\cos[2\pi f_c t + \theta(t)]$ :

PSK: constant $A(t)$ , and $\theta(t)$ stepping between the symbol phases. True, (A).
AM: $A(t) = A_c[1 + k_a m(t)]$ and $\theta(t) = 0$ . True, (B).
Band-limited Gaussian noise around $f_c$ : its envelope (Rayleigh) and phase (uniform) are random processes, but the form is the same. True, (C).
Narrowband FM: constant $A(t)$ and $\theta(t) = 2\pi k_f \int m(\tau)\,d\tau$ with a small peak deviation. So x(t) can represent it, and (D), which says it never can, is false.

Options (A), (B) and (C).

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Full question on page 58 of the official paper (PDF)

GATE 2026, Q.63

Communications2 marksNAT

A (7, 4) Hamming decoder that corrects one error sees an average bit error rate of 0.10. Find the probability that it decodes a received word wrongly.

Answer format: A probability, rounded off to two decimal places

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Key idea: It fails only when two or more of the seven bits are wrong.

Official answer: 0.10 to 0.20

Subtract the chance of zero or one error in 7 bits from 1:

P_{\text{fail}} = 1 - (0.9)^7 - 7(0.1)(0.9)^6 = 1 - 0.4783 - 0.3720 \approx 0.15

\begin{aligned} P_{\text{fail}} &= 1 \\ &\quad - (0.9)^7 \\ &\quad - 7(0.1)(0.9)^6 \\ &= 1 - 0.4783 - 0.3720 \\ &\approx 0.15 \end{aligned}

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Full question on page 13 of the official paper (PDF)

GATE 2025, Q.14

Communications1 markMCQ

An FM signal is $f(t) = A_c \cos\!\big(2\pi f_c t + 3\sin(2\pi f_1 t) + 4\sin(6\pi f_1 t)\big)$ , with $f_c > 100 f_1$ . What is its peak frequency deviation in Hz?

Option A: $15 f_1$
Option B: $12 f_1$
Option C: $4 f_1$
Option D: $2 f_1$

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Key idea: Frequency deviation is the time derivative of the phase deviation divided by $2\pi$ , not the sum of the modulation indices.

Official answer: A

The phase deviation is $\phi(t) = 3\sin(2\pi f_1 t) + 4\sin(6\pi f_1 t)$ . Differentiating and dividing by $2\pi$ :

\Delta f(t) = \frac{1}{2\pi}\frac{d\phi}{dt} = 3f_1\cos(2\pi f_1 t) + 12 f_1\cos(6\pi f_1 t)

\begin{aligned} \Delta f(t) &= \frac{1}{2\pi}\frac{d\phi}{dt} \\ &= 3f_1\cos(2\pi f_1 t) \\ &\quad + 12 f_1\cos(6\pi f_1 t) \end{aligned}

Both cosines equal 1 at $t = 0$ , so the peak is $3f_1 + 12f_1 = 15f_1$ . Option (A).

The tone at $3f_1$ contributes four times its index, 12, because deviation scales with the modulating frequency. Adding the indices, 3 + 4, is the usual slip.

Full question on page 14 of the official paper (PDF)

GATE 2025, Q.15

Communications1 markMCQ

An AWGN channel has bandwidth $W$ , noise power spectral density $N_0/2$ and average power constraint $P_{av}$ . Which plot in the figure shows how the capacity $C$ depends on $W$ , with $P_{av}$ and $N_0$ fixed?

Four sketches of capacity C against bandwidth W, labelled A to D. A: C rises steeply from the origin and flattens toward a constant. B: a straight line through the origin. C: a curve that starts flat and bends sharply upward. D: a curve that rises to a peak and then falls. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Option A: Plot (A)
Option B: Plot (B)
Option C: Plot (C)
Option D: Plot (D)

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Key idea: More bandwidth also lets in more noise, so the signal-to-noise ratio falls as $W$ grows.

Official answer: A

By the Shannon-Hartley formula,

C = W \log_2\!\left(1 + \frac{P_{av}}{N_0 W}\right)

$C$ is 0 at $W = 0$ and always increases with $W$ , but ever more slowly. As $W \to \infty$ , using $\ln(1 + x) \approx x$ ,

C \to \frac{P_{av}}{N_0 \ln 2} \approx 1.44\,\frac{P_{av}}{N_0}

\begin{aligned} C &\to \frac{P_{av}}{N_0 \ln 2} \\ &\approx 1.44\,\frac{P_{av}}{N_0} \end{aligned}

So the curve rises and flattens toward a finite limit: plot (A). It never falls, which rules out (D), and it doesn't grow without limit as (B) and (C) do.

Full question on page 27 of the official paper (PDF)

GATE 2025, Q.28

Communications1 markMSQ

$m(t)$ is band-limited to $[-W, W]$ Hz. DSB-SC is $A_c\, m(t) \cos(2\pi f_c t)$ , and AM is $A_c\,(1 + \mu\, m(t)) \cos(2\pi f_c t)$ with modulation index $\mu$ . (i) An envelope detector can demodulate DSB-SC if $m(t) > 0$ for all $t$ . (ii) An envelope detector can demodulate AM only if $m(t) > 0$ for all $t$ . Which are correct?

One or more options are correct.

Option A: (i) is TRUE
Option B: (i) is FALSE
Option C: (ii) is TRUE
Option D: (ii) is FALSE

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Key idea: An envelope detector recovers whatever multiplies the carrier, provided that factor never goes negative.

Official answer: A and D

The envelope is $|A_c\, m(t)|$ for DSB-SC and $|A_c\,(1 + \mu\, m(t))|$ for AM.

(i): if $m(t) > 0$ always, $|A_c\, m(t)| = A_c\, m(t)$ , so the envelope is the message. (i) is TRUE.
(ii): AM needs $1 + \mu\, m(t) > 0$ , not $m(t) > 0$ . A tone $m(t) = \cos(2\pi W t)$ with $\mu = 0.5$ goes negative, yet $1 + 0.5\cos(2\pi W t)$ stays between 0.5 and 1.5, and the envelope detector recovers it. So the "only if" is FALSE.

Options (A) and (D).

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Full question on page 29 of the official paper (PDF)

GATE 2025, Q.33

Communications1 markNAT

A (6,3) binary linear block code has generator matrix $G = \begin{bmatrix} 1 & 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 0 \end{bmatrix}$ . What is the minimum Hamming distance $d_{min}$ between codewords?

Answer format: An integer

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Key idea: For a linear code, the minimum distance equals the smallest weight of a non-zero codeword.

Official answer: 3

List the 7 non-zero codewords, the sums of the rows of $G$ in every combination, with their weights:

Single rows: 100101, 010011, 001110 (weight 3 each)
Two rows: 110110, 101011, 011101 (weight 4 each)
All three: 111000 (weight 3)

The smallest weight is 3, so $d_{min} = 3$ .

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Full question on page 36 of the official paper (PDF)

GATE 2025, Q.41

Communications2 marksMCQ

A source sends $S$ , equally likely to be $-2$ , $0$ or $2$ . The receiver gets $Y = S + N$ , where $N$ is zero-mean Gaussian noise independent of $S$ , and uses a maximum-likelihood decoder. If the probability of a symbol error is $P_e = \alpha\, P(N > 1)$ , what is $\alpha$ ?

Option A: $\tfrac{1}{3}$
Option B: $1$
Option C: $\tfrac{2}{3}$
Option D: $\tfrac{4}{3}$

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Key idea: With equal priors and Gaussian noise, the ML thresholds sit halfway between neighbouring symbols, and an inner symbol can be pushed out either way.

Official answer: D

The decision thresholds are at $-1$ and $1$ . Using the symmetry of $N$ , with $P = P(N > 1)$ :

$S = -2$ is wrong only if $N > 1$ : probability $P$ .
$S = 0$ is wrong if $|N| > 1$ : probability $2P$ .
$S = 2$ is wrong only if $N < -1$ : probability $P$ .

P_e = \frac{1}{3}(P + 2P + P) = \frac{4}{3}\,P(N > 1)

\begin{aligned} P_e &= \frac{1}{3}(P + 2P + P) \\ &= \frac{4}{3}\,P(N > 1) \end{aligned}

So $\alpha = \tfrac{4}{3}$ . Option (D).

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Full question on page 37 of the official paper (PDF)

GATE 2025, Q.42

Communications2 marksMCQ

$f(t) = \sum_{n=1}^{N} a_n\, p(t - nT)$ , with $T > 0$ , $p(t) = 1$ for $t \in [0, 0.5T]$ and 0 otherwise, and the $a_n$ pairwise independent, zero-mean and unit-variance. (i) The mean of $f(t)$ is independent of $t$ . (ii) The autocorrelation $E[f(t)f(t + \tau)]$ is independent of $t$ for all $\tau$ . Which option is correct?

Option A: (i) is TRUE and (ii) is FALSE
Option B: Both (i) and (ii) are TRUE
Option C: Both (i) and (ii) are FALSE
Option D: (i) is FALSE and (ii) is TRUE

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Key idea: Zero-mean coefficients make the mean zero everywhere, but the pulses fill only half of each slot, so the power switches on and off with time.

Official answer: A

(i) $E[f(t)] = \sum_n E[a_n]\, p(t - nT) = 0$ for every $t$ : TRUE.

(ii) Since $E[a_n a_m]$ is 1 for $n = m$ and 0 otherwise,

E[f(t)f(t+\tau)] = \sum_{n=1}^{N} p(t - nT)\,p(t + \tau - nT)

\begin{aligned} &E[f(t)f(t+\tau)] \\ &\quad = \sum_{n=1}^{N} p(t - nT)\,p(t + \tau - nT) \end{aligned}

At $\tau = 0$ this is 1 when $t$ is in the first half of a slot and 0 in the second half, so it depends on $t$ : FALSE. Option (A).

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Full question on page 47 of the official paper (PDF)

GATE 2025, Q.54

Communications2 marksMSQ

$X$ takes values $-1$ , 0 and 1 with $P(X = -1) = P(X = 1) = \alpha$ and $P(X = 0) = 1 - 2\alpha$ , where $0 < \alpha < \tfrac{1}{2}$ . Let $g(\alpha)$ be the entropy of $X$ in bits. Which statements are TRUE?

One or more options are correct.

Option A: $g(0.4) > g(0.3)$
Option B: $g(0.3) > g(0.4)$
Option C: $g(0.3) > g(0.25)$
Option D: $g(0.25) > g(0.3)$

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Key idea: Entropy is largest when the three outcomes are equally likely, at $\alpha = \tfrac{1}{3}$ , and falls away on either side.

Official answer: B and C

g(\alpha) = -2\alpha\log_2\alpha - (1 - 2\alpha)\log_2(1 - 2\alpha)

\begin{aligned} g(\alpha) &= -2\alpha\log_2\alpha \\ &\quad - (1 - 2\alpha)\log_2(1 - 2\alpha) \end{aligned}

$g(0.25) = 1.5$ bits (probabilities 0.25, 0.5, 0.25)
$g(0.3) \approx 1.571$ bits (0.3, 0.4, 0.3)
$g(0.4) \approx 1.522$ bits (0.4, 0.2, 0.4)

So $g(0.3) > g(0.4)$ and $g(0.3) > g(0.25)$ : options (B) and (C). Both make sense: 0.3 is the closest of the three to $\tfrac{1}{3}$ .

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Full question on page 50 of the official paper (PDF)

GATE 2025, Q.60

Communications2 marksNAT

$X$ and $Y$ take values in $\{0, 1\}$ with $P(0, 0) = 0.06$ , $P(0, 1) = 0.14$ , $P(1, 0) = 0.24$ and $P(1, 1) = 0.56$ , where $P(x, y) = P(X = x, Y = y)$ . What is the mutual information $I(X; Y)$ ?

Answer format: Rounded off to two decimal places

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Key idea: Mutual information is zero exactly when the joint distribution is the product of the marginals.

Official answer: 0

The marginals are $P(X = 0) = 0.2$ , $P(X = 1) = 0.8$ , $P(Y = 0) = 0.3$ and $P(Y = 1) = 0.7$ . Every joint probability is the product of its marginals:

0.2 \times 0.3 = 0.06, \quad 0.2 \times 0.7 = 0.14, \quad 0.8 \times 0.3 = 0.24, \quad 0.8 \times 0.7 = 0.56

\begin{aligned} 0.2 \times 0.3 &= 0.06, \\ 0.2 \times 0.7 &= 0.14, \\ 0.8 \times 0.3 &= 0.24, \\ 0.8 \times 0.7 &= 0.56 \end{aligned}

So $X$ and $Y$ are independent and $I(X; Y) = 0.00$ .

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Full question on page 12 of the official paper (PDF)

GATE 2024, Q.14

Communications1 markMCQ

A digital system sends symbols $b[n]$ through a noiseless channel band-limited to $[-W, W]$ . The received signal is $z(t) = \sum_n b[n]\,x(t - nT)$ , where $x(t)$ is the overall response to one symbol and has Fourier transform $X(f)$ , and it is sampled at $t = mT$ . Which Nyquist condition must $X(f)$ satisfy for zero intersymbol interference?

Option A: $\sum_{m=-\infty}^{\infty} X\left(f + \frac{m}{T}\right) = T$
Option B: $\sum_{m=-\infty}^{\infty} X\left(f + \frac{m}{T}\right) = \frac{1}{T}$
Option C: $\sum_{m=-\infty}^{\infty} X(f + mT) = T$
Option D: $\sum_{m=-\infty}^{\infty} X(f + mT) = \frac{1}{T}$

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Key idea: A question given marks to all scores full marks for every candidate, whatever they answered.

Official answer: Marks to all

The final key awards marks to all for this question and doesn't say why, so we don't pick an option.

The idea behind it: zero ISI means $x(nT)$ is zero for every $n \neq 0$ . Sampling at $1/T$ repeats $X(f)$ every $1/T$ in frequency, so the condition is that the shifted copies add to a constant, $\sum_m X(f + m/T) = T\,x(0)$ . One likely reason for the ruling is that the question doesn't say what $x(0)$ is, so the constant on the right isn't fixed.

Ruling in the key: The final key gives marks to all (MTA) for this question: every candidate received its 1 mark.

Full question on page 17 of the official paper (PDF)

GATE 2024, Q.20

Communications1 markMCQ

Zero-mean white Gaussian noise $w(t)$ with power spectral density $N_0/2$ passes through a first-order RC low-pass filter, giving $n(t)$ . What is the variance of $n(t_k)$ at any time $t_k$ ?

Option A: $\dfrac{N_0}{4RC}$
Option B: $\dfrac{N_0}{2RC}$
Option C: $\dfrac{N_0}{RC}$
Option D: $\dfrac{2N_0}{RC}$

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Key idea: The output variance is the input PSD integrated against $|H(f)|^2$ over all frequencies.

Official answer: A

|H(f)|^2 = \frac{1}{1 + (2\pi f RC)^2}

\sigma^2 = \int_{-\infty}^{\infty} \frac{N_0}{2}\,\frac{df}{1 + (2\pi f RC)^2} = \frac{N_0}{2} \cdot \frac{1}{2RC} = \frac{N_0}{4RC}

\begin{aligned} \sigma^2 &= \int_{-\infty}^{\infty} \frac{N_0}{2}\,\frac{df}{1 + (2\pi f RC)^2} \\ &= \frac{N_0}{2} \cdot \frac{1}{2RC} \\ &= \frac{N_0}{4RC} \end{aligned}

The output is stationary, so this holds at every $t_k$ . Option (A).

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Full question on page 24 of the official paper (PDF)

GATE 2024, Q.28

Communications1 markNAT

An amplitude modulator's output is $s(t) = A\cos(400\pi t) + B\cos(360\pi t) + B\cos(440\pi t)$ V. The carrier power, normalised to 1 Ω, is 50 W, and the total sideband power is $\tfrac{1}{9}$ of the total power. What is $B$ , in volts?

Answer format: Volts, rounded off to two decimal places

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Key idea: Each cosine of amplitude $a$ carries $a^2/2$ into 1 Ω, so carrier and sidebands can be added as separate powers.

Official answer: 2.49 to 2.51

The carrier power is $A^2/2 = 50$ W. The two sidebands carry $B^2/2$ each, $B^2$ together:

\frac{B^2}{50 + B^2} = \frac{1}{9} \;\Rightarrow\; 8B^2 = 50 \;\Rightarrow\; B = 2.5\ \text{V}

\begin{aligned} \frac{B^2}{50 + B^2} &= \frac{1}{9} \\ \Rightarrow\; 8B^2 &= 50 \\ \Rightarrow\; B &= 2.5\ \text{V} \end{aligned}

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Full question on page 26 of the official paper (PDF)

GATE 2024, Q.34

Communications1 markNAT

A source sends symbols from an alphabet of 16. What is the maximum achievable entropy, in bits?

Answer format: Bits

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Key idea: Entropy is largest when every symbol is equally likely.

Official answer: 4

H_{max} = \log_2 16 = 4\ \text{bits}

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Full question on page 35 of the official paper (PDF)

GATE 2024, Q.45

Communications2 marksMCQ

A source sends $s$ from $\{-4, 0, 4\}$ with equal probability over an AWGN channel: $r = s + w$ , with $w$ zero-mean, variance 4 and independent of $s$ . With $Q(x) = \frac{1}{\sqrt{2\pi}}\int_x^\infty e^{-t^2/2}\,dt$ , what is the optimum symbol error probability?

Option A: $\tfrac{2}{3}Q(2)$
Option B: $\tfrac{4}{3}Q(1)$
Option C: $\tfrac{2}{3}Q(1)$
Option D: $\tfrac{4}{3}Q(2)$

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Key idea: With equal priors the optimum thresholds are midway between symbols, and the middle symbol can be pushed out on either side.

Official answer: B

The thresholds are at $\pm 2$ , a distance 2 from each symbol, and $\sigma = 2$ , so each one-sided error has probability $Q(2/2) = Q(1)$ :

$s = -4$ : $Q(1)$
$s = 0$ : $2Q(1)$
$s = 4$ : $Q(1)$

P_e = \frac{1}{3}\big(Q(1) + 2Q(1) + Q(1)\big) = \frac{4}{3}Q(1)

\begin{aligned} P_e &= \frac{1}{3}\big(Q(1) + 2Q(1) + Q(1)\big) \\ &= \frac{4}{3}Q(1) \end{aligned}

Option (B). Dividing the distance 2 by the variance 4 instead of $\sigma$ gives $Q(0.5)$ , and forgetting the standard deviation altogether gives $Q(2)$ .

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Full question on page 36 of the official paper (PDF)

GATE 2024, Q.47

Communications2 marksMCQ

The bits 1 1 1 0 1 0 1 0 1 are encoded with CRC-4, generator $C(x) = x^4 + x + 1$ . What is the encoded sequence?

Option A: 1 1 1 0 1 0 1 0 1 1 1 0 0
Option B: 1 1 1 0 1 0 1 0 1 1 1 0 1
Option C: 1 1 1 0 1 0 1 0 1 1 1 1 0
Option D: 1 1 1 0 1 0 1 0 1 0 1 0 0

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Key idea: Append four zeros, divide by the generator in modulo-2 arithmetic, and append the 4-bit remainder.

Official answer: A

The generator is 10011. Dividing 1110101010000 by it, with XOR in place of subtraction at each step, leaves the remainder 1100.

So the codeword is the message followed by 1100: 1 1 1 0 1 0 1 0 1 1 1 0 0. Option (A). A check: the codeword itself divides by 10011 with no remainder.

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Full question on page 44 of the official paper (PDF)

GATE 2024, Q.57

Communications2 marksNAT

$X(t) = A\cos(2\pi f_0 t + \theta)$ , with $A$ uniform on $[-2, 2]$ and $\theta$ uniform on $[0, 2\pi]$ , independent. It feeds an 8-bit uniform mid-rise quantiser, and its autocorrelation is $R_X(\tau) = \tfrac{2}{3}\cos(2\pi f_0\tau)$ . What is the signal-to-quantisation-noise ratio at the output, in dB?

Answer format: dB, rounded off to two decimal places

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Key idea: Signal power is $R_X(0)$ , and a uniform quantiser's noise power is $\Delta^2/12$ with $\Delta$ the full range over $2^n$ .

Official answer: 45.00 to 45.30

The signal swings over $[-2, 2]$ , so with 8 bits $\Delta = 4/256 = 1/64$ and the noise power is $\Delta^2/12 = 1/49152$ . The signal power is $R_X(0) = 2/3$ :

\text{SQNR} = \frac{2/3}{1/49152} = 32768, \qquad 10\log_{10} 32768 \approx 45.15\ \text{dB}

\begin{aligned} \text{SQNR} &= \frac{2/3}{1/49152} \\ &= 32768, \\ 10\log_{10} 32768 &\approx 45.15\ \text{dB} \end{aligned}

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Full question on page 19 of the official paper (PDF)

GATE 2023, Q.27

Communications1 markMCQ

Let $m(t)$ be a strictly band-limited signal with bandwidth $B$ and energy $E$ . Assuming $\omega_0 = 10B$ , what is the energy in the signal $m(t)\cos\omega_0 t$ ?

Option A: $\dfrac{E}{4}$
Option B: $\dfrac{E}{2}$
Option C: $E$
Option D: $2E$

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Key idea: Modulation splits the spectrum into two half-height copies; with $\omega_0 > B$ they don't overlap.

Official answer: B

The Fourier transform of $m(t)\cos\omega_0 t$ is

\tfrac{1}{2}\left[M(\omega - \omega_0) + M(\omega + \omega_0)\right]

Each copy has a quarter of the energy density, $|M|^2/4$ , and since $\omega_0 = 10B$ the copies don't overlap, so the energies add: $E/4 + E/4 = E/2$ . Option (B). In the time domain, $\cos^2\omega_0 t = \tfrac{1}{2}(1 + \cos 2\omega_0 t)$ , and the $\cos 2\omega_0 t$ part integrates to zero.

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Full question on page 21 of the official paper (PDF)

GATE 2023, Q.30

Communications1 markMSQ

For a real signal, which of the following is/are valid power spectral density/densities? Options (C) and (D) are the plots in the figure.

Two sketches of S_X(omega). (C): equal to 1 over a band of positive omega and equal to -1 over the matching band of negative omega. (D): equal to 1 over a band of positive omega and zero for all negative omega. — Figure from the official GATE 2023 EC paper, IIT Kanpur

One or more options are correct.

Option A: $S_X(\omega) = \dfrac{2}{9 + \omega^2}$
Option B: $S_X(\omega) = e^{-\omega^2}\cos^2\omega$
Option C: The plot labelled (C) in the figure
Option D: The plot labelled (D) in the figure

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Key idea: The PSD of a real signal is real, never negative, and even in $\omega$ .

Official answer: A and B

(A) $2/(9 + \omega^2)$ is positive and even: valid. It is the PSD of $e^{-3|\tau|}/3$ .
(B) $e^{-\omega^2}\cos^2\omega$ is never negative and is even: valid.
(C) is negative for $\omega < 0$ : not a PSD.
(D) is not even: for a real signal $S_X(-\omega) = S_X(\omega)$ , so a one-sided band isn't valid.

Options (A) and (B).

Full question on page 32 of the official paper (PDF)

GATE 2023, Q.45

Communications2 marksMCQ

An FM signal $x(t) = A\cos\left(\omega_c t + k_f \int_{-\infty}^{t} m(\lambda)\, d\lambda\right)$ , where $m(t)$ has bandwidth $W$ , is passed through a non-linear system with output $y(t) = 2x(t) + 5x^2(t)$ . Let $B_T$ denote the FM bandwidth. What is the minimum value of $\omega_c$ required to recover $x(t)$ from $y(t)$ ?

Option A: $B_T + W$
Option B: $\dfrac{3}{2}B_T$
Option C: $2B_T + W$
Option D: $\dfrac{5}{2}B_T$

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Key idea: Squaring an FM wave gives a DC term and an FM wave at $2\omega_c$ with twice the deviation, so twice the bandwidth.

Official answer: B

x^2(t) = \frac{A^2}{2} + \frac{A^2}{2}\cos\left(2\omega_c t + 2k_f \int m\right)

\begin{aligned} x^2(t) &= \frac{A^2}{2} \\ &\quad + \frac{A^2}{2}\cos\left(2\omega_c t + 2k_f \int m\right) \end{aligned}

The term at $2\omega_c$ occupies $2\omega_c \pm B_T$ (twice the deviation, so twice the bandwidth), while $x(t)$ occupies $\omega_c \pm B_T/2$ , and the DC term sits at 0. A bandpass filter can keep $x(t)$ alone if the bands don't overlap:

\omega_c + \frac{B_T}{2} \le 2\omega_c - B_T \quad\Rightarrow\quad \omega_c \ge \frac{3}{2} B_T

\begin{aligned} \omega_c + \frac{B_T}{2} &\le 2\omega_c - B_T \\ \Rightarrow\; \\ \omega_c &\ge \frac{3}{2} B_T \end{aligned}

The DC term needs only $\omega_c > B_T/2$ , which this already meets. Option (B).

Full question on page 42 of the official paper (PDF)

GATE 2023, Q.59

Communications2 marksNAT

$X(t)$ is white Gaussian noise with power spectral density $\tfrac{1}{2}$ W/Hz. It is input to an LTI system with impulse response $e^{-t}u(t)$ . What is the average power of the system output, in W?

Answer format: W, rounded off to two decimal places

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Key idea: For white noise of PSD $S_0$ , the output power is $S_0 \int |h(t)|^2 dt$ , by Parseval.

Official answer: 0.24 to 0.26

P = \frac{1}{2} \int_0^\infty e^{-2t}\, dt = \frac{1}{2} \times \frac{1}{2} = 0.25\ \text{W}

\begin{aligned} P &= \frac{1}{2} \int_0^\infty e^{-2t}\, dt \\ &= \frac{1}{2} \times \frac{1}{2} \\ &= 0.25\ \text{W} \end{aligned}

In the frequency domain, $|H(f)|^2 = 1/(1 + 4\pi^2 f^2)$ , whose integral over all $f$ is also $\tfrac{1}{2}$ . Answer: 0.25.

Full question on page 44 of the official paper (PDF)

GATE 2023, Q.65

Communications2 marksNAT

Eight symbols a to h occur with frequencies $\tfrac{1}{2}$ , $\tfrac{1}{4}$ , $\tfrac{1}{8}$ , $\tfrac{1}{16}$ , $\tfrac{1}{32}$ , $\tfrac{1}{64}$ , $\tfrac{1}{128}$ and $\tfrac{1}{128}$ . A symbol is chosen, and it is identified by a series of truthfully answered yes/no questions. What is the average number of questions when they are asked in the most efficient sequence?

Answer format: rounded off to two decimal places

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Key idea: The best question sequence is a Huffman code: each question is one bit.

Official answer: 1.97 to 1.99

Ask "is it a?", then "is it b?", and so on: a takes 1 question, b 2, ..., f 6, and g and h 7 each. This is the Huffman code for these probabilities, so no sequence does better on average:

\bar{L} = \frac{1}{2}(1) + \frac{1}{4}(2) + \frac{1}{8}(3) + \frac{1}{16}(4) + \frac{1}{32}(5) + \frac{1}{64}(6) + 2 \cdot \frac{1}{128}(7) = \frac{127}{64} = 1.984

\begin{aligned} \bar{L} &= \frac{1}{2}(1) \\ &\quad + \frac{1}{4}(2) \\ &\quad + \frac{1}{8}(3) \\ &\quad + \frac{1}{16}(4) \\ &\quad + \frac{1}{32}(5) \\ &\quad + \frac{1}{64}(6) \\ &\quad + 2 \cdot \frac{1}{128}(7) \\ &= \frac{127}{64} \\ &= 1.984 \end{aligned}

Every probability is a power of $\tfrac{1}{2}$ , so this equals the entropy exactly. Answer: 1.98.