GATE ECE · Communication Systems

Digital Communication Systems – GATE ECE Communication SystemsPractice Questions & PYQs

Generate GATE-level questions on Digital Communication. Focus on: 1. Pulse Modulation: PCM, DPCM, Delta Modulation, and Adaptive Delta Modulation. 2. Digital Modulation Techniques: ASK, PSK, FSK, QPSK, and QAM. 3. Pulse Shaping: Matched Filter, Inter-Symbol Interference (ISI), and Nyquist Criterion. 4. Probability of Error: Calculations for various digital modulation schemes in AWGN.

78 questions · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2024MCQ

A digital communication system transmits through a noiseless bandlimited channel $[-W W]$ . The received signal $z(t)$ at the output of the receiving filter is given by $z(t)=\sum_{n} b[n] x(t-n T)$ where $b[n]$ are the symbols and $x(t)$ is the overall system response to a single symbol. The received signal is sampled at $t=m T$ . The Fourier transform of $x(t)$ is $X(f)$ . The Nyquist condition that $X(f)$ must satisfy for zero intersymbol interference at the receiver is ______

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📖 Explanation

For zero ISI, $x(t)$ should have zero crossings at $\pm T, \pm 2 T, \pm 3 T \ldots$ $x(t) \sum_{m=-\infty }^{\infty } \delta(t-m T)=x(0) \delta(t)$ $\frac{1}{T} \sum_{m=-\infty }^{\infty } x\left(f-\frac{m}{T}\right)=x(0)$ $\sum_{m=-\infty }^{\infty } X\left(f-\frac{m}{T}\right)=x(0) T$ (or) $\sum_{m=-\infty }^{\infty } X\left(f+\frac{m}{T}\right)=x(0) T$ As per GATE official Answer Key-MTA

Explanation Figure 1

Q2GATE 2022NAT

A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).

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📖 Explanation

\begin{aligned} \text{BIT RATE}&=[\text{SYBOL RATE}] \times \log_{20}M\\ 1.\;\; QPSK, \; M&=4,n=2\\ R_{b1}&=1 \times 2 =2mbps\\ 2.\;\; 16 QAM_1\Rightarrow M=16,n=4\\ R_{b2}&=1 \times 4=4mbps\\ R_b&=\frac{2m+4m}{2}\\ R_b&=3 mbps \end{aligned}

Q3GATE 2022NAT

Consider a channel over which either symbol $x_A$ or symbol $x_B$ is transmitted. Let the output of the channel Y be the input to a maximum likelihood (ML) detector at the receiver. The conditional probability density functions for Y given $x_A$ and $x_B$ are: $f_{Y|x_A}(y)=e^{-(y+1)}u(y+1),$ $f_{Y|x_B}(y)=e^{(y-1)}(1-u(y-1)),$ where $u(\cdot )$ is the standard unit step function. The probability of symbol error for this system is _________ (rounded off to two decimal places).

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📖 Explanation

ML decoding $\rightarrow \; f_{Y/X_A}(y)\underset{\overset{ \lt }{X_B}}{\overset{X_A}{ \gt }} f_{Y/X_B}(y)$ $i.e. \;\; f_{Y/X_A}(y) \gt f_{Y/X_B}(y) \rightarrow \text{ Decision favor of }X_A$ $i.e. \;\; f_{Y/X_A}(y) \lt f_{Y/X_B}(y) \rightarrow \text{ Decision favor of }X_B$ For $-1 \lt y \lt 0$ and $1 \lt y \lt \infty \rightarrow f_{Y/X_A}(y) \gt f_{Y/X_B}(y)$ For above internal decision in favor of $X_A$ For $-\infty \lt y \lt -1$ and $0\lt y \lt 1 \rightarrow f_{Y/X_B}(y) \gt f_{Y/X_A}(y)$ For above internal decision in favor of $X_B$ $P_e=P(X_A)\cdot P_{eX_A}+P(X_B)\cdot P_{eX_A}$ $P_{eX_A}\rightarrow$ Probability of error $X_A$ transmitted $P_{eX_B}\rightarrow$ Probability of error $X_B$ transmitted $P_{eX_A}=\int_{0}^{1} f_{Y/X_A}(y)dy=\int_{0}^{1}e^{-(y+1)}u(y+1)dy= \int_{0}^{1}e^{-(y+1)} dy=e^{-1}-e^{-2}=0.23$ $P_{eX_B}=\int_{0}^{1} f_{Y/X_B}(y)dy=\int_{0}^{1}e^{(y-1)} [1-u(y-1)]dy= \int_{0}^{1}e^{(y-1)} dy=e^{-1}-e^{-2}=0.23$ $P_e=P(X_A) \times 0.23+P(X_B) \times 0.23 =0.23[P(X_A)+P(X_B)]=0.23$

Explanation Figure 1

Q4GATE 2022MSQ

Let $H(X)$ denote the entropy of a discrete random variable $X$ taking $K$ possible distinct real values. Which of the following statements is/are necessarily true?

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📖 Explanation

$H(X^2)=H(Y)=\Sigma P_Y(Y_i) \log _2 \frac{1}{P_Y(Y_i)}= \frac{1}{2} \log _2 2+\frac{1}{2} \log _2 2=1 bit/symbol$ $H(X)=\frac{1}{4} \log _2 4+\frac{1}{2} \log _2 2+\frac{1}{4} \log _2 4=1.5 bit/symbol$ $H(X) \gt H(X^2)$ Hence option (C) is not correct.

Explanation Figure 1

Q5GATE 2021NAT

A message signal having peak-to-peak value of $2\:V$ , root mean square value of $0.1\:V$ and bandwidth of $\text{5 kHz}$ is sampled and fed to a pulse code modulation $(\text{PCM})$ system that uses a uniform quantizer. The $\text{PCM}$ output is transmitted over a channel that can support a maximum transmission rate of $\text{50 kbps}$ . Assuming that the quantization error is uniformly distributed, the maximum signal to quantization noise ratio that can be obtained by the $\text{PCM}$ system (rounded off to two decimal places) is ______

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📖 Explanation

\begin{aligned} v_{p-p} &=2 \mathrm{~V} \\ \text { Root MSQ }[m(t)] &=0.1 \mathrm{~V} ; \quad f_{m}=5 \mathrm{kHz} \\ \text { Channel capacity, } \quad C &=50 \mathrm{kbps} \\ \text{Max} \frac{S}{N_{Q}}=?\\ & \text{Signal power},\\ S&=\text{MSQ}[m(t)]=(0.1)^{2}=0.01\\ C &\geq R_{b} \Rightarrow 50 \mathrm{kbps} \geq n f_{s}\\ \because \qquad f_{s}&=N R=2 f_{m}=10 \mathrm{kHz}\\ n &\leq 5 \Rightarrow n_{\max }=5\\ N_{Q} &=\frac{\Delta^{2}}{12} \\ \therefore\qquad \Delta &=\frac{V_{p-p}}{2^{n}} \\ \Delta_{\min } &=\frac{V_{p-p}}{2^{7} \max }=\frac{2 V}{2^{5}}=\frac{1}{16} \\ \left(N_{Q}\right)_{\min } &=\frac{\Delta_{\min }^{2}}{12}=3.25 \times 10^{-4} \\ \left(\frac{S}{N Q}\right)_{\max } &=\frac{S}{\left(N_{Q}\right)_{\min }}=\frac{0.01}{3.25 \times 10^{-4}}=30.72 \end{aligned}

Q6GATE 2019MCQ

A single bit, equally likely to be 0 and 1, is to be sent across an additive white Gaussian noise (AWGN) channel with power spectral density $N_0/2$ . Binary signaling, with $0\rightarrow p(t)$ and $1\rightarrow q(t)$ , is used for the transmission, along with an optimal receiver that minimizes the bit-error probability. Let $\varphi _1(t),\varphi _2(t)$ form an orthonormal signal set. If we choose $p(t)=\varphi _1(t)\; and \; q(t)=-\varphi _2(t)$ , we would obtain a certain bit-error probability $P_b$ . If we keep $p(t)=\varphi _1(t)$ , but take $q(t)=\sqrt{E}\varphi _2(t)$ , for what value of E would we obtain the same bit-error probability $P_b$ ?

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📖 Explanation

$\text { When } p(t)=\phi_{1}(t) \text { and } q(t)=-\phi_{1}(t)$ $d_{\min }=2$ $\text { When } p(t)=\phi_{1}(t) \text { and } q(t)=-\sqrt{E}\phi_{2}(t)$ $d_{\min }=\sqrt{(\sqrt{E})^{2}+1}=\sqrt{E+1}$ To obtain same bit-error probability, $\sigma_{\text {min }}$ should be same.

\begin{aligned} \text { So, } \quad \sqrt{E+1} &=2 \\ E &=3 \end{aligned}

Explanation Figure 1

Explanation Figure 2

Q7GATE 2019NAT

A random variable X takes values -1 and +1 with probabilities 0.2 and 0.8, respectively. It is transmitted across a channel which adds noise N, so that the random variable at the channel output is Y=X+N. The noise N is independent of X, and is uniformly distributed over the interval [-2,2]. The receiver makes a decision

\hat{X}=\left\{\begin{matrix} -1 & if & Y\leq \theta \\ +1& if & Y> \theta \end{matrix}\right.

where the threshold $\theta \in [-1,1]$ is chosen so as to minimize the probability of error $Pr[\hat{X}\neq X]$ . The minimum probability of error, rounded off to 1 decimal place, is ______.

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📖 Explanation

MAP criteria should be used to minimise the probability of error.

\begin{array}{c} f_{Y}(y \mid+1) P(+1) \sum_{-1}^{+1} f_{Y}(y \mid-1) F(-1) \\ P(+1)=0.80 \text { and } P(-1)=0.20 \end{array}

$P_{\theta(m \text { \in })}=$ Shaded area $=2 \times \frac{1}{20}=0.10$

Explanation Figure 1

Q8GATE 2019NAT

A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal $f(t)=A(1+m(t))cos 2\pi f_c t$ , where $f_c$ =600kHz. The AM signal f(t) is to be digitized and archived. This is done by first sampling f(t) at 1.2 times the Nyquist frequency, and then quantizing each sample using a 256-level quantizer. Finally, each quantized sample is binary coded using K bits, where K is the minimum number of bits required for the encoding. The rate, in Megabits per second (rounded off to 2 decimal places), of the resulting stream of coded bits is _______ Mbps.

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📖 Explanation

\begin{aligned} \text { Nyquist rate } &=2 \times 615 \mathrm{kHz} \\ &=1230 \mathrm{kHz}=1.23 \mathrm{MHz} \\ f_{s} &=1.2 \times 1.23=1.476 \mathrm{MHz} \\ \text { Bits/sample }, n &=\log _{2}(256)=8 \\ \text { So, } \quad A_{D} &=n f_{s}=8 \times 1.476 \\ &=11.808 \simeq 11.81 \mathrm{Mbps} \end{aligned}

Explanation Figure 1

Q9GATE 2018NAT

A random variable X takes values -0.5 and 0.5 with probabilities 1/4 and 3/4, respectively. The noisy observation of X is Y = X + Z, where Z has uniform probability density over the interval (-1, 1). X and Z are independent. If the MAP rule based detector outputs $\hat{X}$ as

\hat{X}=\left\{\begin{matrix} -0.5 \; \; \; Y \lt \alpha \\ 0.5 \; \; \; Y \geq \alpha \end{matrix}\right.

then the value of $\alpha$ (accurate to two decimal places) is _______.

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📖 Explanation

\begin{array}{c} P\left(x_{0}\right)=\frac{1}{4} ; P\left(x_{1}\right)=\frac{3}{4} \\ \text { MAP criteria, } f_{Y}\left(y \mid x_{0}\right) P\left(x_{0}\right) \overset{x_{2}}{\underset{x_{1}}{\gtrless}}\left(y \mid x_{1}\right) P\left(x_{1}\right) \end{array}

$\text{So}\quad \alpha=-0.5$

Explanation Figure 1

Explanation Figure 2

Q10GATE 2018NAT

A binary source generates symbols $X \in \{-1,1\}$ which are transmitted over a noisy channel. The probability of transmitting X= 1 is 0.5. Input to the threshold detector is R= X + N. The probability density function $f_{N}(n)$ of the noise N is shown below. If the detection threshold is zero, then the probability of error (correct to two decimal places) is __________.

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📖 Explanation

Let $s_{0}$ and $s_{1}$ be the transmitted symbols representing the transmitted value {-1 , 1} respectively and let $r_{0}$ and $r_{1}$ be the received symbols. Probability of error,

\begin{aligned} P_{e} &=P\left(s_{1}\right) P\left(r_{0} \mid s_{1}\right)+P\left(s_{0}\right) P\left(r_{1} \mid s_{0}\right) \\ P\left(r_{0} \mid s_{1}\right) &=P\left(r_{1} \mid s_{0}\right)=\frac{1}{2} \times 1 \times \frac{1}{4}=\frac{1}{8}\\ \text{Given that,}\\ P\left(s_{0}\right)&=P\left(s_{1}\right)=\frac{1}{2}\\ \text{So}\quad P_{e}&=\frac{1}{2}\left(\frac{1}{8}+\frac{1}{8}\right)=\frac{1}{8}=0.125 \end{aligned}

Explanation Figure 2

Q11GATE 2017NAT

A sinusoidal message signal is converted to a PCM signal using a uniform quantizer. The required signal-to-quantization noise ratio (SQNR) at the output of the quantizer is 40dB. The minimum number of bits per sample needed to achieve the desired SQNR is _______

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📖 Explanation

For sinusoidal input is applied to a PCM system, $\mathrm{SQNR}=6 n+1.8$ n= number of bits per sample Given that, the required $\mathrm{SQNR}=40 \mathrm{dB}$

\begin{aligned} \mathrm{so}, 6 n+1.8 &\geq 40 \\ n &\geq \frac{38.2}{6}\\ n_{\min }&=7 \\ \because \; &n \text{ is always an integer value} \end{aligned}

Q12GATE 2017MCQ

In a digital communication system, the overall pulse shape p(t) at the receiver before the sampler has the Fourier transform P(f). If the symbols are transmitted at the rate of 2000 symbols per second, for which of the following cases is inter symbol interference zero?

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📖 Explanation

Given, Symbol rate =2000 symbol/sec So $R_{b}=2 \mathrm{kHz}$ Nyquist's criterion for ISI free transmission is. $\sum_{n=-\infty }^{\infty } P\left(f-n R_{b}\right)=\text { Constant }$ So, only option (b) satisfies this condition for given $R_{b}$

Q13GATE 2017MCQ

In binary frequency shift keying (FSK), the given signal waveform are $u_{0}(t)=5cos(20000\pi t);o\leq t\leq T$ , and $u_{1}(t)=5cos(22000\pi t);o\leq t\leq T$ , Where T is the bit-duration interval and t is in seconds. Both $u_{0}(t)$ and $u_{1}(t)$ are zero outside the interval $0\leq t \leq T$ . With a matched filter (correlator) based receiver, the smallest positive value of T (in milliseconds) required to have $u_{0}(t)$ and $u_{1}(t)$ uncorrelated is

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📖 Explanation

\begin{aligned} u_{0}(t) &=5 \cos 20000 \pi t \\ u_{1}(t) &=5 \cos 22000 \pi t \\ f_{1} &=11000 \mathrm{Hz} \text { and } f_{2}=10000 \mathrm{Hz} \end{aligned}

For FSK waveforms to be uncorrelated,

\begin{aligned} f_{1}-f_{2} &=n \frac{R_{b}}{2} ; n=1,2,3, \ldots \\ R_{b} &=\frac{2\left(f_{1}-f_{2}\right)}{n}=\frac{2000}{n}-\text { bits } / \mathrm{sec} \\ R_{b(\max )} &=2000 \mathrm{bits} / \mathrm{sec}\\ &\quad \quad \because \text{minimum value of n=1 }\\ T_{b(\min )}&=\frac{1}{R_{b(\max )}}=0.5 \mathrm{ms} \end{aligned}

Q14GATE 2017MCQ

Which one of the following statements about differential pulse code modulation (DPCM) is true?

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Q15GATE 2016MCQ

A binary baseband digital communication system employs the signal

P(t)=\left\{\begin{matrix} \frac{1}{\sqrt{T_{s}}}, & 0\leq t\leq T_{s}\\ 0,& otherwise \end{matrix}\right.

for transmission of bits. The graphical representation of the matched filter output y(t) for this signal will be

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📖 Explanation

Impulse response of matched filter, $h(t)=P\left(T_{s}-t\right)$ Output of matched filter, $y(t) = P( t) * h(t)$

Explanation Figure 1

Explanation Figure 2

Explanation Figure 3

Q16GATE 2016NAT

An ideal band-pass channel 500Hz - 2000Hz is deployed for communication. A modem is designed to transmit bits at the rate of 4800 bits/s using 16-QAM. The roll-off factor of a pulse with a raised cosine spectrum that utilizes the entire frequency band is ________

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📖 Explanation

\begin{array}{l} \text { Channel spectrum } 500 \mathrm{Hz}-2000 \mathrm{Hz} \\ \text { hence } B W=1500 \mathrm{Hz} \\ B W=\frac{R_{b}}{\log _{2} M}(1+\alpha) \\ 1500=\frac{R_{b}}{\log _{2} 16}(1+\alpha) \\ 1500=\frac{4800}{\log _{2} 16}(1+\alpha) \\ 1500=1200(1+\alpha) \\ \Rightarrow \quad \alpha=0.25 \end{array}

Q17GATE 2016NAT

Consider binary data transmission at a rate of 56 kbps using baseband binary pulse amplitude modulation (PAM) that is designed to have a raised-cosine spectrum. The transmission bandwidth (in kHz) required for a roll-off factor of 0.25 is ________

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📖 Explanation

Bit rate, $\quad R_{b}=56 \mathrm{kbps}$ Roll-off factor, $\alpha=0.25$

\begin{aligned} \text { Transmission } \mathrm{BW} &=\frac{R_{b}}{2}(1+\alpha) \\ &=\frac{56}{2}(1.25)=28 \times 1.25 \\ &=35 \mathrm{kHz} \end{aligned}

Q18GATE 2016NAT

A speech signal is sampled at 8 kHz and encoded into PCM format using 8 bits/sample. The PCM data is transmitted through a baseband channel via 4-level PAM. The minimum bandwidth (in kHz) required for transmission is _______

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📖 Explanation

\begin{array}{l} f_{s}=8 \mathrm{kHz} \\ n=8 \text { bits/sample } \\ M=4 \\ B W_{\min }=\frac{R_{b}}{2 \log _{2} M}=\frac{R_{b}}{4}=\frac{8 \times 8}{4}=16 \mathrm{kHz} \end{array}

Q19GATE 2016NAT

The bit error probability of a memoryless binary symmetric channel is $10^{-5}$ . If $10^{5}$ bits are sent over this channel, then the probability that not more than one bit will be in error is __________

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📖 Explanation

bit error probability, $p=10^{-5}$ Number of bits transmitted, $n=10^{5}$ If n-bits are transmitted, probability of getting error in 'r' bits is given by $p_{e}={ }^{n} C_{r} \rho^{r}(1-p)^{n-r}$ The probability that not more than one bit will be in error is

\begin{array}{l} ={ }^{n} C_{0} p^{0}(1-p)^{-0}+{ }^{n} C_{\{(p)}^{1}(1-p)^{p-1} \\ =(1-p)^{n}+n p(1-p)^{n-1} \\ =\left(1-10^{-5}\right)^{10^{5}}+10^{5} \times 10^{-5}\left(1-10^{-5}\right)^{10^{5}-1} \\ =0.7357 \end{array}

Q20GATE 2016MCQ

An analog pulse s(t) is transmitted over an additive white Gaussian noise (AWGN) channel. The received signal is r(t) = s(t) + n(t), where n(t) is additive white Gaussian noise with power spectral density $\frac{N_{0}}{2}$ . The received signal is passed through a filter with impulse response h(t). Let $E_{s}$ and $E_{h}$ denote the energies of the pulse s(t) and the filter h(t), respectively. When the signal-to-noise ratio (SNR) is maximized at the output of the filter ( $SNR_{max}$ ), which of the following holds?

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📖 Explanation

When the signal to Noise ratio is maximum $h(t)=s(T-t)$ but shifting doesn't change the energy

\begin{aligned} \Rightarrow \quad E_{h}&=E_{s} \\ \text{and(SNR)}_{\max }&=\frac{2 E_{s}}{N_{0}} \end{aligned}

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