GATE EC · IN

PCM, digital modulation and detection

This topic covers how an analogue signal becomes bits (sampling, quantisation, PCM, DPCM, delta modulation), how pulses are shaped to avoid inter-symbol interference, how bits ride on a carrier (ASK, PSK, FSK, QAM), and how the receiver decides which symbol was sent (matched filter, MAP, ML). GATE questions reduce to a few formulas used with care about conventions: SQNR, bit rate and bandwidth, roll-off, Eb/N0 and Q-function error probabilities, and decision thresholds.

15 previous-year questions · 15 solved · 5 practice questions

Sampling, quantisation and PCM

Pulse code modulation (PCM) has three steps: sample the message at $f_{s} \geq 2 f_{m}$ , quantise each sample to one of $L = 2^{n}$ levels, and encode each level as an $n$ -bit word. Conventions used here: a uniform mid-riser quantiser covers $[- m_{p}, + m_{p}]$ , and the quantisation error is uniformly distributed over one step.

STEP SIZE

Δ = \frac{2 m _{p}}{L} = \frac{2 m _{p}}{2 ^{n}}

Uniform quantiser with peak-to-peak range $2 m_{p}$ and $L = 2^{n}$ levels.

QUANTISATION NOISE POWER

N_{q} = \overline{e^{2}} = \frac{Δ ^{2}}{12}

Error uniform on $(- Δ/2, Δ/2)$ ; valid when the signal is large compared with $Δ$ and does not overload the quantiser.

SQNR FOR A GENERAL SIGNAL

SQNR = \frac{S}{N _{q}} = \frac{3 L ^{2} m ^{2}}{m _{p}^{2}}

Here $S = \overline{m^{2}}$ is the average signal power and $m_{p}$ the quantiser's peak value.

SQNR, FULL-SCALE SINUSOID

SQNR_{dB} = 6.02 n + 1.76 dB

Sinusoid of amplitude $m_{p}$ so $S = m_{p}^{2} /2$ and SQNR $= 1.5 \cdot 4^{n}$ . Each extra bit adds about 6 dB. For a full-scale uniformly distributed signal ( $\overline{m^{2}} = m_{p}^{2} /3$ ) the SQNR is $4^{n}$ , i.e. $6.02 n$ dB.

PCM BIT RATE AND MINIMUM BANDWIDTH

R_{b} = n f_{s}, B_{m i n} = \frac{R _{b}}{2} = \frac{n f _{s}}{2}

Minimum (ideal Nyquist, sinc pulses) baseband bandwidth for binary PCM; with a raised-cosine pulse of roll-off $α$ it becomes $(1 + α) R_{b} /2$ . The first-null bandwidth of rectangular NRZ pulses is $R_{b}$ .

EXAM TIP

Time-division multiplexing $N$ channels, each sampled at $f_{s}$ with $n$ bits, gives $R_{b} = N n f_{s}$ (plus any framing bits), and a minimum bandwidth of $N n f_{s} /2$ .

EXAM TIP

$N_{q}$ depends only on $Δ$ , so if the signal power is $X$ dB below that of a full-scale sinusoid (amplitude $m_{p}$ ), the SQNR is $X$ dB below $6.02 n + 1.76$ dB. Always check whether the question says "full-scale" or gives the signal's own peak value.

Companding, DPCM and delta modulation

A uniform quantiser gives the same noise power to loud and soft signals, so soft signals have poor SQNR. Companding compresses the signal before a uniform quantiser and expands it after, which gives small amplitudes finer effective steps. With $x = m / m_{p}$ and $∣ x ∣ \leq 1$ :

Μ-LAW COMPRESSOR

y = sgn (x) \frac{ln ( 1 + μ ∣ x ∣ )}{ln ( 1 + μ )}

$μ = 255$ in the standard North American/Japanese system. In the limit $μ \to 0$ , $y = x$ (no compression).

SQNR WITH Μ-LAW COMPANDING

SQNR \approx \frac{3 L ^{2}}{[ ln ( 1 + μ ) ] ^{2}}

Valid when $μ^{2} \overline{m^{2}} ≫ m_{p}^{2}$ . The SQNR then hardly depends on the signal level, which is the point of companding.

A-LAW COMPRESSOR

y = ⎩ ⎨ ⎧ \frac{A ∣ x ∣}{1 + ln A}, \frac{1 + ln ( A ∣ x ∣ )}{1 + ln A}, 0 \leq ∣ x ∣ \leq \frac{1}{A} \frac{1}{A} \leq ∣ x ∣ \leq 1 (times sgn x)

$A = 87.6$ in the standard European system; linear for small inputs, logarithmic for large.

DPCM quantises the difference $d$ between the current sample and a prediction made from past (reconstructed) samples. The difference has a smaller variance than the signal, so fewer bits give the same SQNR, or the same bits give a better SQNR. Delta modulation (DM) is the one-bit extreme: the sampling rate is far above Nyquist and each bit says whether the staircase approximation steps up or down by $Δ$ .

DPCM PREDICTION GAIN

G_{p} = \frac{σ _{m}^{2}}{σ _{d}^{2}}, SQNR_{DPCM} = G_{p} \times \frac{σ _{d}^{2}}{σ _{q}^{2}}

$σ_{m}^{2}$ : signal variance, $σ_{d}^{2}$ : prediction-error variance, $σ_{q}^{2}$ : quantisation noise of the difference quantiser. In dB the gain adds $10 lo g_{10} G_{p}$ to the SQNR of the same quantiser.

NO SLOPE OVERLOAD (DM)

\frac{d m}{d t}_{m a x} \leq \frac{Δ}{T _{s}} = Δ f_{s}

For $m (t) = A sin 2 π f_{m} t$ this is $2 π f_{m} A \leq Δ f_{s}$ . If violated, the staircase cannot follow the signal (slope overload).

DM OUTPUT SNR, SINUSOID AT THE OVERLOAD LIMIT

SNR_{m a x} = \frac{3 f _{s}^{3}}{8 π ^{2} f _{m}^{2} B}

Assumes amplitude $A = Δ f_{s} / (2 π f_{m})$ (largest without slope overload), granular noise power $Δ^{2} /3$ spread evenly over $(0, f_{s})$ , and an output low-pass filter of bandwidth $B$ . Doubling $f_{s}$ improves the SNR by about 9 dB.

Delta modulation errors

Error	Cause	Cure
Slope overload	Step $Δ$ too small, or $f_{s}$ too low, for the signal slope	Increase $Δ$ or $f_{s}$ (adaptive DM)
Granular noise	Step $Δ$ too large when the signal is nearly flat; staircase hunts around it	Decrease $Δ$

EXAM TIP

Slope overload and granular noise pull $Δ$ in opposite directions. Fixing one worsens the other, which is why adaptive step size is used.

Line codes and inter-symbol interference

A line code maps bits to pulse amplitudes. Spectra are quoted for equally likely random bits with $T_{b} = 1/ R_{b}$ . The first-null bandwidth is the quick comparison at GATE level.

Line codes

Line code	Pulses	DC component	First-null bandwidth
Unipolar NRZ	1 = $+ A$ , 0 = 0, full bit width	Yes (and a discrete line at DC)	$R_{b}$
Polar NRZ	1 = $+ A$ , 0 = $- A$ , full bit width	None for equiprobable bits	$R_{b}$
Unipolar RZ (50% duty)	1 = pulse for $T_{b} /2$ , 0 = 0	Yes (discrete lines at DC and odd multiples of $R_{b}$ )	$2 R_{b}$
Manchester	Transition in the middle of every bit	None	$2 R_{b}$
Bipolar (AMI), full-width pulses	0 = 0, 1s alternate $+ A, - A$	None	$R_{b}$

Inter-symbol interference (ISI) is the smearing of one symbol's pulse into neighbouring symbol slots, caused by a band-limited channel. With symbol period $T$ and overall pulse $p (t)$ (transmit filter, channel and receive filter together), ISI is zero if the pulse is sampled at $t = k T$ and $p (k T) = 0$ for $k \neq = 0$ , with $p (0) \neq = 0$ .

NYQUIST CRITERION FOR ZERO ISI

k = - \infty \sum \infty P (f - \frac{k}{T}) = T

Equivalent to $p (k T) = δ_{k 0}$ (normalised so $p (0) = 1$ ). Smallest bandwidth satisfying it: $1/ (2 T)$ Hz, with $p (t) = sinc (t / T)$ , where $sinc x = sin (π x) / (π x)$ (ideal, not realisable).

RAISED-COSINE PULSE BANDWIDTH

B = \frac{1 + α}{2 T} = \frac{( 1 + α ) R _{s}}{2}, 0 \leq α \leq 1

Baseband (one-sided) bandwidth for symbol rate $R_{s} = 1/ T$ . $α$ is the roll-off factor. At $α = 0$ it is the ideal $R_{s} /2$ ; at $α = 1$ it is $R_{s}$ .

MAXIMUM ZERO-ISI SYMBOL RATE

R_{s} \leq \frac{2 B}{1 + α}

For a baseband channel of bandwidth $B$ . For a bandpass (double-sideband modulated) channel of bandwidth $B_{c}$ the limit is $R_{s} \leq B_{c} / (1 + α)$ .

EXAM TIP

Do not mix baseband and passband. A raised-cosine pulse of roll-off $α$ needs $(1 + α) R_{s} /2$ at baseband but $(1 + α) R_{s}$ after double-sideband modulation onto a carrier.

Digital modulation schemes

Digital modulation changes the amplitude, phase or frequency of a carrier $A cos (2 π f_{c} t + ϕ)$ according to the bits. With $M = 2^{k}$ symbols, $k$ bits go into each symbol, so $R_{s} = R_{b} / k$ and $E_{s} = k E_{b}$ . Bandwidths below are null-to-null for rectangular pulses, which is $2 R_{s}$ for a linearly modulated scheme.

Digital modulation schemes

Scheme	What varies	Bits per symbol	Null-to-null bandwidth
ASK / OOK	Amplitude (OOK: carrier on or off)	1	$2 R_{b}$
BPSK	Phase $0^{\circ}$ or $18 0^{\circ}$	1	$2 R_{b}$
BFSK	Frequency $f_{0}$ or $f_{1}$	1	$∣ f_{1} - f_{0} ∣ + 2 R_{b}$
QPSK	Phase, four values 90 degrees apart	2	$R_{b}$
$M$ -ary PSK	Phase, $M$ values	$lo g_{2} M$	$2 R_{b} / lo g_{2} M$
Square $M$ -QAM	Amplitude on I and Q	$lo g_{2} M$	$2 R_{b} / lo g_{2} M$

BANDWIDTH EFFICIENCY (NULL-TO-NULL, PSK AND QAM ONLY)

η = \frac{R _{b}}{B} = \frac{lo g _{2} M}{2} b/s/Hz

With a raised-cosine pulse the passband bandwidth is $(1 + α) R_{s}$ , so $η = lo g_{2} M / (1 + α)$ . Not for FSK, whose bandwidth also depends on the tone spacing.

AVERAGE ENERGY, SQUARE M-QAM

E_{s} = \frac{( M - 1 ) d ^{2}}{6}

$d$ is the minimum distance between adjacent points (per axis levels $\pm d /2, \pm 3 d /2, \dots$ ), equiprobable points. $M$ -PSK has constant energy $E_{s}$ on a circle of radius $E_{s}$ .

MINIMUM TONE SPACING FOR ORTHOGONAL BFSK

Δ f_{m i n} = \frac{1}{2 T _{b}} (coherent), Δ f_{m i n} = \frac{1}{T _{b}} (non-coherent)

The two tones must be orthogonal over one bit interval for the stated error probability to hold.

EXAM TIP

Gray-coded QPSK has the same bit-error probability as BPSK at the same $E_{b} / N_{0}$ , but needs half the bandwidth for the same bit rate. For PSK and QAM, going beyond $M = 4$ saves bandwidth but needs more $E_{b} / N_{0}$ for a given BER. Orthogonal $M$ -FSK goes the other way: larger $M$ needs less $E_{b} / N_{0}$ and more bandwidth.

Detection: matched filter, MAP and ML

The channel adds white Gaussian noise with two-sided PSD $N_{0} /2$ . The receiver filters, samples once per symbol and compares the sample with a threshold. The best linear filter for a known pulse $s (t)$ of duration $T$ is the matched filter.

MATCHED FILTER

h (t) = s (T - t), (\frac{S}{N})_{m a x} = \frac{2 E}{N _{0}}, E = \int_{0}^{T} s^{2} (t) d t

Output sampled at $t = T$ . In the frequency domain $H (f) = S^{*} (f) e^{- j 2 π f T}$ . The peak SNR depends only on the pulse energy, not its shape. A correlation receiver (multiply by $s (t)$ and integrate over $T$ ) gives the same output at the sampling instant $t = T$ .

For a decision between hypotheses $H_{0}$ (symbol $s_{0}$ ) and $H_{1}$ (symbol $s_{1}$ ) on the sample $r$ : MAP picks the hypothesis with the larger posterior probability, i.e. the larger $P (H_{i}) p (r ∣ H_{i})$ . ML is MAP with equal priors, so it picks the larger likelihood $p (r ∣ H_{i})$ . With Gaussian noise of the same variance under both hypotheses, ML is a minimum-distance rule: the threshold is the midpoint of the two signal points.

MAP THRESHOLD, TWO LEVELS IN GAUSSIAN NOISE

decide H_{1} if r > \frac{s _{0} + s _{1}}{2} + \frac{σ ^{2}}{s _{1} - s _{0}} ln \frac{P ( H _{0} )}{P ( H _{1} )}

Sample $r = s_{i} + n$ with $s_{1} > s_{0}$ and noise $n \sim N (0, σ^{2})$ under both hypotheses. For antipodal levels $s_{1} = + a$ , $s_{0} = - a$ it is $r_{t h} = \frac{σ ^{2}}{2 a} ln \frac{P ( H _{0} )}{P ( H _{1} )}$ . For equal priors the threshold is the midpoint (ML). A less likely $H_{1}$ moves the threshold toward $s_{1}$ .

Procedure for a detection question

Write the received sample distribution under each hypothesis (mean $μ_{i}$ and variance $σ^{2}$ ). For a correlator whose reference is $s (t)$ of energy $E$ , the signal part of the sample is $E$ and $σ^{2} = N_{0} E /2$ ; with a unit-energy reference the signal part is $E$ and $σ^{2} = N_{0} /2$ .
State the priors. Equal priors means ML; unequal means MAP.
Set the likelihoods times priors equal and solve for the threshold $r_{t h}$ .
Find each error probability as a Gaussian tail, with $μ_{1} > μ_{0}$ and $H_{1}$ decided above $r_{t h}$ : $P (err ∣ H_{0}) = Q (\frac{r _{t h} - μ _{0}}{σ})$ and $P (err ∣ H_{1}) = Q (\frac{μ _{1} - r _{t h}}{σ})$ .
Combine: $P_{e} = P (H_{0}) P (err ∣ H_{0}) + P (H_{1}) P (err ∣ H_{1})$ .

Error probability and BER

Convention: $Q (x) = \frac{1}{2} erfc (\frac{x}{2})$ is the tail probability of a standard normal variable. $E_{b}$ is the energy per bit, noise PSD is $N_{0} /2$ two-sided, and detection is coherent with an optimum (matched-filter) receiver and equal priors. For two equiprobable signals at distance $d$ in signal space, $P_{b} = Q (\frac{d}{2 N _{0}})$ .

BPSK AND QPSK (GRAY CODED) BIT ERROR PROBABILITY

P_{b}^{BPSK} = P_{b}^{QPSK} = Q (\frac{2 E _{b}}{N _{0}})

Here $d^{2} = 4 E_{b}$ for BPSK. QPSK is two independent BPSK streams on the I and Q carriers, so its per-bit error probability is the same.

COHERENT BFSK

P_{b}^{BFSK, coh.} = Q (\frac{E _{b}}{N _{0}})

Orthogonal tones, $d^{2} = 2 E_{b}$ . This needs 3 dB more $E_{b} / N_{0}$ than BPSK for the same BER.

COHERENT ASK (ON-OFF), AVERAGE EB

P_{b}^{OOK, coh.} = Q (\frac{E _{b}}{N _{0}})

Equal priors, threshold midway between the two received levels. Here $E_{b}$ is the average bit energy: the "1" has energy $2 E_{b}$ , the "0" none. If the same formula is written with the energy $E_{1}$ of the "1", use $Q (E_{1} / (2 N_{0}))$ .

NON-COHERENT BFSK AND DPSK

P_{b}^{non-coh. BFSK} = \frac{1}{2} e^{- E_{b} / (2 N_{0})}, P_{b}^{DPSK} = \frac{1}{2} e^{- E_{b} / N_{0}}

For comparison: both are closed-form exponentials rather than $Q$ functions.

LINK BETWEEN EB/N0 AND SNR

\frac{E _{b}}{N _{0}} = \frac{S}{N _{0} R _{b}} = \frac{S / N}{R _{b} / B}

$S$ is the received signal power and $N = N_{0} B$ the noise power in bandwidth $B$ . In dB: $(E_{b} / N_{0})_{dB} = SNR_{dB} - 10 lo g_{10} (R_{b} / B)$ .

QPSK SYMBOL ERROR PROBABILITY

P_{s} \approx 2 Q (\frac{E _{s}}{N _{0}}) (QPSK, high SNR)

Exactly $P_{s} = 1 - (1 - p)^{2} = 2 p - p^{2}$ , where $p = Q (E_{s} / N_{0}) = Q (2 E_{b} / N_{0})$ is the error probability of each of the two independent I and Q decisions ( $E_{s} = 2 E_{b}$ ).

M-PSK AND SQUARE M-QAM SYMBOL ERROR (HIGH SNR)

P_{s}^{M -PSK} \approx 2 Q (\frac{2 E _{s}}{N _{0}} sin \frac{π}{M}), P_{s}^{M -QAM} \approx 4 (1 - \frac{1}{M}) Q (\frac{3 E _{s}}{( M - 1 ) N _{0}})

For $M \geq 4$ ; both reduce to the QPSK result at $M = 4$ . With Gray coding, $P_{b} \approx P_{s} / lo g_{2} M$ .

WORKED EXAMPLE

A binary PCM system quantises a full-scale sinusoid, bandlimited to 3.4 kHz, using 8 bits per sample at a sampling rate of 8 kHz. Find the bit rate, the minimum Nyquist baseband bandwidth of the PCM stream and the SQNR in dB.

Solution

Bit rate: $R_{b} = n f_{s} = 8 \times 8000 = 64 kb/s$ .
Minimum bandwidth: $B_{m i n} = R_{b} /2 = 32 kHz$ .
SQNR for a full-scale sinusoid: $6.02 n + 1.76 = 6.02 \times 8 + 1.76 = 49.92 dB$ .

Answer: Bit rate 64 kb/s, minimum bandwidth 32 kHz, SQNR about 49.9 dB.

WORKED EXAMPLE

A BPSK signal $s (t) = \pm A cos (2 π f_{c} t)$ , $0 \leq t \leq T_{b}$ , with $A = 1$ V, $T_{b} = 1 μ$ s and $f_{c}$ an integer multiple of $1/ T_{b}$ , is received across a 1 ohm load in white noise of two-sided PSD $N_{0} /2$ with $N_{0} = 6.25 \times 1 0^{- 8}$ W/Hz. A coherent matched-filter receiver is used and the bits are equiprobable. Find $P_{b}$ and the matched filter's maximum output SNR for one bit.

Solution

Bit energy: $E_{b} = A^{2} T_{b} /2 = 5 \times 1 0^{- 7}$ J.
$E_{b} / N_{0} = 5 \times 1 0^{- 7} / (6.25 \times 1 0^{- 8}) = 8$ .
$P_{b} = Q (2 E_{b} / N_{0}) = Q (16) = Q (4) \approx 3.2 \times 1 0^{- 5}$ .
Matched filter maximum SNR $= 2 E_{b} / N_{0} = 16$ (about 12 dB).

Answer: $P_{b} \approx 3.2 \times 1 0^{- 5}$ and maximum SNR $= 16$ (about 12 dB).

PREVIOUS-YEAR QUESTIONS

15 questions from the EC and IN papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE ECE 2025 · GATE ECE 2024 · GATE ECE 2022 · GATE ECE 2020 · GATE ECE 2018 · GATE IN 2026 · GATE IN 2018

GATE 2026 EC Q.22

GATE 2026 · ECQ.22 · 1 mark · MCQ

Consider a discrete memoryless source with an alphabet of four source symbols. s(t) is a multi-level (-1, 0, +1, +2) signal representing a long sequence of random symbols from the above source which is generating 10^4 symbols per second. Which of the following options is the correct value of equivalent Nyquist bandwidth of s(t)?

(A)
10 kHz
(B)
64 kHz
(C)
5 kHz
(D)
20 kHz

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GATE 2026 EC Q.32

GATE 2026 · ECQ.32 · 1 mark · NAT

A wireless digital transmission scheme is using 16-QAM over an additive white Gaussian noise channel and a maximum-likelihood receiver. Consider the information bit rate from source to be $4 \times 1 0^{6}$ bits per second. The minimum transmission bandwidth (in MHz) of the modulated signal necessary for optimum recovery of information at the receiver is ____. (rounded off to two decimal places)

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GATE 2026 EC Q.44

GATE 2026 · ECQ.44 · 2 marks · MCQ

A QPSK modulated signal from an additive white Gaussian noise (AWGN) channel is received with an $E_{b} / N_{o} = 8.4$ dB at the input of a coherent QPSK demodulator. A maximum-likelihood reception method is used in the demodulator.

Assume the complimentary error function $erfc (u) ≅ [1/ (u π)] exp (- u^{2})$ .

Which is the nearest bit error rate (BER) at the output of the demodulator?

(A)
$1 0^{- 3}$
(B)
$1 0^{- 4}$
(C)
$1 0^{- 5}$
(D)
$1 0^{- 6}$

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GATE 2026 IN Q.27

GATE 2026 · INQ.27 · 1 mark · MCQ

An analog speech signal contains signals from 200 Hz to 2400 Hz. It is sampled at 6 kHz and quantized with 512 levels for pulse code modulation (PCM). The bit rate will be _____ kbps.

(A)
16
(B)
48
(C)
19.2
(D)
54

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GATE 2025 EC Q.41

GATE 2025 · ECQ.41 · 2 marks · MCQ

A source transmits symbol $S$ that takes values uniformly at random from the set ${- 2, 0, 2}$ . The receiver obtains $Y = S + N$ , where $N$ is a zero-mean Gaussian random variable independent of $S$ . The receiver uses the maximum likelihood decoder to estimate the transmitted symbol $S$ . Suppose the probability of symbol estimation error $P_{e}$ is expressed as $P_{e} = α P (N > 1)$ , where $P (N > 1)$ denotes the probability that $N$ exceeds 1. What is the value of $α$ ?

(A)
$1/3$
(B)
$1$
(C)
$2/3$
(D)
$4/3$

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GATE 2024 EC Q.14

GATE 2024 · ECQ.14 · 1 mark · MCQ

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 ______.

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

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GATE 2024 EC Q.45

GATE 2024 · ECQ.45 · 2 marks · MCQ

A source transmits a symbol $s$ , taken from ${- 4, 0, 4}$ with equal probability, over an additive white Gaussian noise channel. The received noisy symbol $r$ is given by $r = s + w$ , where the noise $w$ is zero mean with variance 4 and is independent of $s$ . Using $Q (x) = \frac{1}{2 π} \int_{x}^{\infty} e^{- t^{2} /2} d t$ , the optimum symbol error probability is ______.

(A)
$\frac{2}{3} Q (2)$
(B)
$\frac{4}{3} Q (1)$
(C)
$\frac{2}{3} Q (1)$
(D)
$\frac{4}{3} Q (2)$

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GATE 2024 EC Q.57

GATE 2024 · ECQ.57 · 2 marks · NAT

Let $X (t) = A cos (2 π f_{0} t + θ)$ be a random process, where amplitude $A$ and phase $θ$ are independent of each other, and are uniformly distributed in the intervals $[- 2, 2]$ and $[0, 2 π]$ , respectively. $X (t)$ is fed to an 8-bit uniform mid-rise type quantizer. Given that the autocorrelation of $X (t)$ is $R_{X} (τ) = \frac{2}{3} cos (2 π f_{0} τ)$ , the signal to quantization noise ratio (in dB, rounded off to two decimal places) at the output of the quantizer is ______.

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GATE 2022 EC Q.35

GATE 2022 · ECQ.35 · 1 mark · NAT

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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GATE 2022 EC Q.63

GATE 2022 · ECQ.63 · 2 marks · NAT

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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GATE 2022 EC Q.64

GATE 2022 · ECQ.64 · 2 marks · NAT

Consider a real valued source whose samples are independent and identically distributed random variables with the probability density function, f(x), as shown in the figure. Consider a 1 bit quantizer that maps positive samples to value α and others to value β. If α and β are the respective choices for α and β that minimize the mean square quantization error, then (α − β) = _________ (rounded off to two decimal places).

GATE 2022 EC question 64: Plot of f(x) versus x. — Figure from the GATE 2022 EC paper. It is part of the question.

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GATE 2020 EC Q.48

GATE 2020 · ECQ.48 · 2 marks · NAT

In a digital communication system, a symbol $S$ randomly chosen from the set ${s_{1}, s_{2}, s_{3}, s_{4}}$ is transmitted. It is given that $s_{1} = - 3$ , $s_{2} = - 1$ , $s_{3} = + 1$ and $s_{4} = + 2$ . The received symbol is $Y = S + W$ . $W$ is a zero-mean unit-variance Gaussian random variable and is independent of $S$ . $P_{i}$ is the conditional probability of symbol error for the maximum likelihood (ML) decoding when the transmitted symbol $S = s_{i}$ . The index $i$ for which the conditional symbol error probability $P_{i}$ is the highest is ______.

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GATE 2020 EC Q.48: answer and solution

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GATE 2020 EC Q.48: answer and solution

GATE 2018 EC Q.17

GATE 2018 · ECQ.17 · 1 mark · NAT

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 ______.

GATE 2018 EC question 17: Plot of f_N(n) versus n: a triangular pdf. — Figure from the GATE 2018 EC paper. It is part of the question.

Answer and solution

GATE 2018 EC Q.17: answer and solution

GATE 2018 EC Q.40

GATE 2018 · ECQ.40 · 2 marks · NAT

A random variable $X$ takes values $- 0.5$ and $0.5$ with probabilities $\frac{1}{4}$ and $\frac{3}{4}$ , 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 $\hat{X} = - 0.5$ for $Y < α$ and $\hat{X} = 0.5$ for $Y \geq α$ , then the value of $α$ (accurate to two decimal places) is ______.

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GATE 2018 EC Q.40: answer and solution

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GATE 2018 EC Q.40: answer and solution

GATE 2018 IN Q.53

GATE 2018 · INQ.53 · 2 marks · NAT

The sampling rate for Compact Discs (CDs) is 44,000 samples/s. If the samples are quantized to 256 levels and binary coded, the corresponding bit rate (in bits per second) is _____.

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GATE 2018 IN Q.53: answer and solution

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PRACTICE

5 questions to test yourself

PRACTICE 11 mark · NAT

A full-scale sinusoidal signal is quantised by a uniform quantiser with $n$ bits per sample. The SQNR must be at least 70 dB. Using SQNR $= 6.02 n + 1.76$ dB, find the minimum integer value of $n$ .

Answer and solution

Answer: $12$

Require $6.02 n + 1.76 \geq 70$ , so $n \geq 68.24/6.02 = 11.34$ .
The smallest integer is $n = 12$ (check: $n = 11$ gives 67.98 dB, $n = 12$ gives 74.0 dB).

PRACTICE 22 marks · NAT

A baseband binary-polar system sends symbols at 4800 baud through an ideal channel whose overall response is a raised-cosine pulse with zero ISI. The available baseband bandwidth is 3600 Hz. Find the largest roll-off factor $α$ that can be used.

Answer and solution

Answer: $0.5$ (accepted range $\pm 0.01$ )

Baseband raised-cosine bandwidth: $B = (1 + α) R_{s} /2$ .
$3600 = (1 + α) \times 2400$ , so $1 + α = 1.5$ .
$α = 0.5$ .

PRACTICE 32 marks · NAT

A rectangular pulse $s (t) = 2$ V for $0 \leq t \leq 3 μ$ s (and zero elsewhere) is received in white noise of two-sided PSD $N_{0} /2$ with $N_{0} = 1 0^{- 6}$ W/Hz (unit 1 ohm load). The pulse is passed through its matched filter. Find the maximum output SNR as a linear ratio.

Answer and solution

Answer: $24$ (accepted range $\pm 0.01$ )

Pulse energy: $E = \int_{0}^{T} s^{2} d t = 4 \times 3 \times 1 0^{- 6} = 1.2 \times 1 0^{- 5}$ J.
Maximum SNR $= 2 E / N_{0} = 2.4 \times 1 0^{- 5} /1 0^{- 6} = 24$ .

PRACTICE 42 marks · MSQ

Which of the following statements are correct for coherent detection in white Gaussian noise at the same $E_{b} / N_{0}$ and the same bit rate, with rectangular pulses and Gray coding where relevant?

(A)
Gray-coded QPSK and BPSK have the same bit error probability.
(B)
Coherent orthogonal BFSK has bit error probability $Q (E_{b} / N_{0})$ , which is 3 dB worse than BPSK.
(C)
QPSK needs the same null-to-null bandwidth as BPSK.
(D)
16-QAM carries 4 bits per symbol.

Answer and solution

Answer: (A), (B), (D)

A is true: QPSK is two independent BPSK streams, so $P_{b} = Q (2 E_{b} / N_{0})$ for both.
B is true: $d^{2} = 2 E_{b}$ for orthogonal BFSK gives $Q (E_{b} / N_{0})$ , i.e. 3 dB more $E_{b} / N_{0}$ for the same BER.
C is false: QPSK halves the symbol rate, so its null-to-null bandwidth $R_{b}$ is half that of BPSK ( $2 R_{b}$ ).
D is true: $lo g_{2} 16 = 4$ .

PRACTICE 52 marks · MCQ

A receiver sees $r = s + n$ where $s = + 1$ V for bit 1 and $s = - 1$ V for bit 0, and $n$ is zero-mean Gaussian with variance $σ^{2} = 0.5 V^{2}$ . The priors are $P (1) = 0.2$ and $P (0) = 0.8$ . The MAP receiver decides bit 1 if $r$ exceeds a threshold. The threshold is closest to:

(A)
0
(B)
0.347
(C)
-0.347
(D)
0.693

Answer and solution

Answer: (B)

Use $r_{t h} = \frac{σ ^{2}}{2 a} ln \frac{P ( 0 )}{P ( 1 )}$ with $a = 1$ .
$r_{t h} = \frac{0.5}{2} ln 4 = 0.25 \times 1.386 = 0.347$ V.
Positive, because bit 1 is less likely, so the decision boundary moves toward the +1 V point.

OFFICIAL GATE 2027 SYLLABUS WORDING

Copied from the GATE 2027 syllabus; see all 2027 syllabi.

GATE EC

Digital Communications: PCM, DPCM, digital modulation schemes (ASK, PSK, FSK, QAM), bandwidth, inter- symbol interference, MAP, ML detection, matched filter receiver, SNR and BER

GATE IN

pulse code modulation … amplitude, phase, frequency, quadrature amplitude, pulse shift keying for digital modulation and demodulation

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