GATE ECE Communications Formula Sheet 2027: Formulas and Traps

Q: Which Communications topics are asked most often in GATE ECE?

Across the five counted EC papers (2010, 2019, 2020, 2025 and 2026), angle modulation and MAP or ML detection with error probability each appear in all five. AM and DSB-SC and random processes through LTI filters each appear in four. Entropy, capacity and Hamming codes appear in three, all of them recent. Prepare those formulas first.

Q: What is the bit error probability of BPSK in GATE ECE?

For coherent BPSK in additive white Gaussian noise, the bit error probability is Q(√(2Eb/N₀)), where Eb is the energy per bit and N₀/2 is the two-sided noise PSD. Coherent orthogonal signalling such as BFSK gives Q(√(Eb/N₀)), so it needs twice the energy, or 3 dB more, for the same error rate.

Q: How do you find the bandwidth of an FM signal for GATE?

Use Carson's rule. The bandwidth is 2(Δ f + fₘ), where Δ f is the peak frequency deviation and fₘ is the highest message frequency. Written with the modulation index it is 2fₘ(β + 1). Check whether the deviation constant is given in hertz per volt or radians per second per volt before you compute Δ f.

Q: Does the power of an FM signal change with the modulation index?

No. An FM or PM signal has a constant envelope, so its average power is Ac²/2 whatever the message or the modulation index. Modulation only redistributes that power between the carrier and the sidebands. This single fact was the subject of 2010 Q21 and is one of the most asked facts in the section.

Q: How many errors can a Hamming code correct?

A Hamming code has a minimum distance of 3, so it corrects one error or detects two, but not both at once. In general a code corrects t errors if dₘᵢₙ ≥ 2t + 1 and detects s errors if dₘᵢₙ ≥ s + 1. A single-error-correcting decoder given two errors decodes to the wrong codeword.

By MD ANISH AHAMADUpdated 4 Oct 202610 min read

GATE ECE Communications Formula Sheet 2027: Formulas and Traps

A GATE ECE Communications formula sheet needs about thirty formulas, not three hundred. Communications averaged 10.0 marks across the five EC papers counted for the ECE book, and half of its questions asked for a typed number rather than an option. This sheet gives you the formulas those questions use, each with a one-line example and the trap that costs the mark.

In this guide

Key takeaways
The notation used on this sheet
Random processes: what a filter does to noise power
AM: modulation index, power and efficiency
DSB-SC and SSB: what a wrong local carrier does
FM and PM: instantaneous frequency and Carson's rule
Sampling, PCM and quantisation noise
M-ary signalling: symbols, bits and bandwidth
MAP and ML detection with the Q function
The N₀ conventions in one table
Entropy and channel capacity
Hamming codes and the dₘᵢₙ rules
Block error probability
How to use this sheet in the last weeks
Quick revision

Key takeaways

Most lost marks in Communications come from a factor of two in a convention, not from a wrong method.
Angle modulation and MAP or ML detection appeared in all five counted papers, so their formulas come first.
Noise power through a filter uses $\vert H(f)\vert^2$ , never $H(f)$ , and white noise has two-sided PSD $N_0/2$ .
Antipodal signalling gives $Q(\sqrt{2E_b/N_0})$ and orthogonal signalling gives $Q(\sqrt{E_b/N_0})$ ; the factor of 2 is the whole difference.
Bit rate, symbol rate and bandwidth are linked by $\log_2 M$ , and baseband needs half the minimum bandwidth of passband.
Entropy and capacity are in bits only when you use $\log_2$ .

The notation used on this sheet

$m(t)$ is the message, $A_m$ its peak amplitude and $f_m$ its frequency. $W$ is the message bandwidth.
$A_c$ and $f_c$ are the carrier amplitude and frequency, with $\omega_c = 2\pi f_c$ .
$\mu$ is the AM modulation index. $\beta$ is the angle modulation index, and $\Delta f$ the peak frequency deviation.
$N_0/2$ is the two-sided noise PSD in W/Hz, so $N_0$ is the one-sided value.
$E_b$ is energy per bit, $E_s$ energy per symbol, $M$ the number of symbols, $R_b$ the bit rate and $R_s$ the symbol rate.
$Q(x)$ is the probability that a standard Gaussian exceeds $x$ . You will need $Q(3) \approx 1.35 \times 10^{-3}$ .
A process is wide-sense stationary (WSS) when its mean is constant and its autocorrelation depends only on the lag $\tau$ .

Random processes: what a filter does to noise power

A filter shapes a random signal's power spectrum by the square of its gain. Integrate the output PSD and you have the output power.

Quantity	Formula	Watch out for
WSS test	$E[X(t)] = \text{const}$ , $R_X(t_1, t_2) = R_X(t_1 - t_2)$	Both conditions; a time-varying mean fails at once
Average power	$P = R_X(0) = \int_{-\infty}^{\infty} S_X(f)\,df$	Integrate over negative frequencies too
Output PSD	$S_Y(f) = \vert H(f)\vert^2 S_X(f)$	The square is the mark
Output mean	$m_Y = m_X H(0)$	Uses $H(0)$ , not $\vert H(0)\vert^2$
White noise in an ideal filter of bandwidth $B$	$\displaystyle P = \frac{N_0}{2} \times 2B = N_0 B$	Count the negative band if you use $N_0/2$

Example: two-sided PSD $N_0/2 = 10^{-6}$ W/Hz through an ideal low-pass filter of bandwidth 5 kHz gives

P = 10^{-6} \times 2 \times 5000 = 0.01 \text{ W}

A filter with gain 2 in that band would give 0.04 W, not 0.02 W.

Trap: integrating the two-sided $N_0/2$ over $0$ to $B$ only halves the answer, and in a numerical question half the answer scores zero.

AM: modulation index, power and efficiency

Conventional AM adds a carrier to the message, so most of the power sits in a carrier that carries no information.

Quantity	Formula	Watch out for
Signal	$s(t) = A_c[1 + \mu \cos \omega_m t]\cos \omega_c t$	Envelope detection needs $\mu \le 1$
Index from the envelope	$\displaystyle \mu = \frac{V_{\max} - V_{\min}}{V_{\max} + V_{\min}}$	Read peak values, not peak-to-peak
Total power	$\displaystyle P_t = P_c\left(1 + \frac{\mu^2}{2}\right)$	Current version: $I_t = I_c\sqrt{1 + \mu^2/2}$
Efficiency	$\displaystyle \eta = \frac{\mu^2}{2 + \mu^2}$	Maximum $1/3$ at $\mu = 1$
Several tones	$\mu_t = \sqrt{\mu_1^2 + \mu_2^2}$	Add squares, not indices
Bandwidth	$2W$	SSB needs only $W$

Example: $V_{\max} = 15$ V and $V_{\min} = 5$ V give $\mu = 0.5$ . With $P_c = 100$ W, $P_t = 112.5$ W and $\eta = 0.25/2.25 \approx 11.1$ per cent.

DSB-SC and SSB: what a wrong local carrier does

DSB-SC has no carrier, so the receiver must multiply by its own. The counted papers test what happens when that carrier is slightly wrong.

Case	Output after the low-pass filter	Watch out for
DSB-SC, phase error $\phi$	$\displaystyle \frac{A_c}{2} m(t)\cos\phi$	Zero output at $\phi = 90^\circ$
DSB-SC, frequency offset $\Delta\omega$	$\displaystyle \frac{A_c}{2} m(t)\cos(\Delta\omega\,t)$	A slow beat, not a constant loss
SSB, phase error $\phi$	proportional to $m(t)\cos\phi \pm \hat{m}(t)\sin\phi$	Phase distortion, not a null
SSB, frequency offset $\Delta f$	every message frequency shifted by $\Delta f$	A shift, not a beat

Example: a phase error of $60^\circ$ halves the DSB-SC output, because $\cos 60^\circ = 0.5$ .

FM and PM: instantaneous frequency and Carson's rule

Every angle-modulation question starts from one definition: instantaneous frequency is the carrier frequency plus the rate of change of the phase, divided by $2\pi$ .

f_i(t) = f_c + \frac{1}{2\pi}\frac{d\phi(t)}{dt}

The topic appeared in all five counted papers: 2010 Q21, 2019 Q32, 2020 Q57, 2025 Q14 and 2026 Q57.

Quantity	Formula	Watch out for
FM frequency	$f_i = f_c + k_f m(t)$ , $k_f$ in Hz/V	If $k_f$ is in rad/s/V, $\Delta f = k_f A_m/2\pi$
PM frequency	$\displaystyle f_i = f_c + \frac{k_p}{2\pi}\frac{dm}{dt}$	PM deviation follows the message slope
FM index	$\displaystyle \beta = \frac{\Delta f}{f_m} = \frac{k_f A_m}{f_m}$	Doubling $f_m$ halves $\beta$
PM index	$\beta = k_p A_m$	Doubling $f_m$ leaves $\beta$ unchanged
Carson's rule	$B = 2(\Delta f + f_m) = 2f_m(\beta + 1)$	Narrowband ( $\beta \ll 1$ ): $B \approx 2f_m$
Power	$\displaystyle P = \frac{A_c^2}{2}$	Independent of $\beta$

Example: $\Delta f = 75$ kHz and $f_m = 15$ kHz give $\beta = 5$ and $B = 2(75 + 15) = 180$ kHz.

Remember: an angle-modulated signal has a constant envelope, so its power is $A_c^2/2$ whatever the message. Modulation moves power from the carrier into sidebands; it never adds any.

Sampling, PCM and quantisation noise

PCM samples the message, rounds each sample to one of $L$ levels and sends $n$ bits for each sample.

Quantity	Formula	Watch out for
Nyquist rate	$f_s \ge 2W$	"1.2 times Nyquist" means $f_s = 2.4W$
Bits per sample	$n = \lceil \log_2 L \rceil$	100 levels need 7 bits, not 6.64
Bit rate	$R_b = n f_s$	Use the highest frequency actually present
Step size	$\displaystyle \Delta = \frac{V_{\max} - V_{\min}}{L}$	Range is peak to peak
Quantisation noise	$\displaystyle \frac{\Delta^2}{12}$	Uniform rounding error only
SQNR, full-scale sinusoid	$1.76 + 6.02n$ dB	Each extra bit adds about 6 dB

Example: a 5 kHz message sampled at 12 kHz with 128 levels needs $n = 7$ , so $R_b = 84$ kbit/s and the SQNR is about $1.76 + 42.14 = 43.9$ dB.

M-ary signalling: symbols, bits and bandwidth

One M-ary symbol carries $\log_2 M$ bits, and the bandwidth follows the symbol rate, not the bit rate.

Quantity	Formula	Watch out for
Symbol rate	$\displaystyle R_s = \frac{R_b}{\log_2 M}$	16-QAM carries 4 bits, not 16
Energy	$E_s = E_b \log_2 M$	Compare schemes at equal $E_b$
Minimum bandwidth, baseband	$\displaystyle \frac{R_s}{2}$	Nyquist pulses assumed
Minimum bandwidth, passband (PSK, QAM)	$R_s$	Double the baseband figure

Example: 64-QAM at 12 Mbit/s carries 6 bits a symbol, so $R_s = 2$ Msymbol/s and the minimum passband bandwidth is 2 MHz.

The last-minute revision sheet in the GATE ECE 2027 book covers all nine sections in this shape, and the book adds 925 questions with worked solutions and 10 full mock tests.

MAP and ML detection with the Q function

Detection was asked in all five counted papers: 2010 Q55, 2019 Q44 and Q57, 2020 Q58, 2025 Q41 and 2026 Q44. MAP detection picks the symbol with the largest $P(m)\,p(r \mid m)$ . ML detection drops the prior and picks the largest $p(r \mid m)$ . The two agree only when the priors are equal.

Quantity	Formula	Watch out for
Error between two points	$\displaystyle P_e = Q\!\left(\frac{d}{2\sigma}\right)$ , $\displaystyle \sigma^2 = \frac{N_0}{2}$	$d$ is the full distance between the points
BPSK, also QPSK bit error with Gray coding	$\displaystyle P_b = Q\!\left(\sqrt{\frac{2E_b}{N_0}}\right)$	QPSK symbol error is about twice this
Coherent orthogonal (BFSK)	$\displaystyle P_b = Q\!\left(\sqrt{\frac{E_b}{N_0}}\right)$	Needs 3 dB more than BPSK
MAP threshold, $\pm A$ in Gaussian noise	$\displaystyle \gamma = \frac{\sigma^2}{2A}\ln\frac{P_0}{P_1}$	Moves towards the less likely symbol
Gray-coded M-ary	$\displaystyle P_b \approx \frac{P_s}{\log_2 M}$	Symbol error is not bit error

Example: BPSK at $E_b/N_0 = 4.5$ gives $P_b = Q(3) \approx 1.35 \times 10^{-3}$ . Orthogonal signalling needs $E_b/N_0 = 9$ for the same figure.

In one line: antipodal points sit $2\sqrt{E_b}$ apart and orthogonal points $\sqrt{2E_b}$ apart, which is where the factor of 2 inside the root comes from. Convert $E_b/N_0$ from dB to a ratio before taking the root.

The $N_0$ conventions in one table

Quantity	Formula	Watch out for
White noise PSD	$\displaystyle S_n(f) = \frac{N_0}{2}$ , two-sided	Some questions state the one-sided $N_0$
Noise variance at a correlator, unit-energy basis	$\displaystyle \sigma^2 = \frac{N_0}{2}$	Not $N_0$
Matched filter peak SNR	$\displaystyle \frac{2E}{N_0}$	Independent of the pulse shape
Thermal noise	$N_0 = kT$ , about $-174$ dBm/Hz at 290 K	Add $10\log_{10}B$ for the power in $B$

Entropy and channel capacity

Entropy is the average information per symbol. Capacity is the highest rate a channel supports with an arbitrarily small error probability.

Quantity	Formula	Watch out for
Entropy	$H = -\sum_i p_i \log_2 p_i$ bits/symbol	Natural log gives nats
Maximum entropy	$\log_2 M$ , at equal probabilities	Binary: 1 bit at $p = 0.5$
Mutual information	$I(X;Y) = H(X) - H(X \mid Y)$	Never negative
Shannon capacity	$\displaystyle C = B\log_2\left(1 + \frac{S}{N}\right)$	Convert SNR from dB first
Infinite bandwidth limit	$\displaystyle C_\infty \approx 1.44\frac{P}{N_0}$	Capacity stays finite
BSC capacity	$C = 1 - H(p)$	$p = 0.5$ gives zero
Source coding	$H \le \bar{L} < H + 1$ , $\displaystyle \eta = \frac{H}{\bar{L}}$	New in 2027; no counted precedent

Examples: probabilities $\displaystyle \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}$ give $H = 1.75$ bits. A 3 kHz channel at 30 dB has $\displaystyle \frac{S}{N} = 1000$ , so $C \approx 3000 \times 9.97 \approx 29.9$ kbit/s. A BSC with $p = 0.11$ has $H(p) \approx 0.5$ , so $C \approx 0.5$ bit per use.

Hamming codes and the $d_{\min}$ rules

A Hamming code uses $m = n - k$ parity bits to protect a block of $n = 2^m - 1$ bits, and always has $d_{\min} = 3$ .

Quantity	Formula	Watch out for
Hamming bound	$2^{n-k} \ge n + 1$	$m = 3$ gives (7,4), $m = 4$ gives (15,11)
$d_{\min}$ of a linear code	smallest weight of a nonzero codeword	Exclude the all-zero codeword
Correct $t$ errors	$d_{\min} \ge 2t + 1$	$t = \lfloor (d_{\min} - 1)/2 \rfloor$
Detect $s$ errors	$d_{\min} \ge s + 1$	Detection alone, no correction
Correct $t$ and detect $s$	$d_{\min} \ge t + s + 1$ , $s \ge t$	Not the sum of the two rules

Example: a code with $d_{\min} = 5$ corrects 2 errors or detects 4.

Block error probability

With independent bit errors of probability $p$ , a block fails when more errors arrive than the code corrects.

P_{\text{uncoded}} = 1 - (1 - p)^n, \qquad P_{\text{fail}} = 1 - \sum_{i=0}^{t}\binom{n}{i}p^i(1 - p)^{n-i}

Example: an uncoded 8-bit block at $p = 0.01$ fails with probability $1 - 0.99^8 \approx 0.077$ . A (15,11) Hamming code at the same $p$ fails with about $0.0096$ .

The trap is the $i = 0$ term: it is the no-error case and belongs inside the sum.

How to use this sheet in the last weeks

Read the tables daily and say each trap aloud. The GATE ECE important topics guide shows where Communications sits among the eight technical sections, and the subject-wise weightage guide explains why its 10.0-mark mean is so steady. Half of the section's counted questions are numerical, so rehearse $Q$ values, logarithms and dB conversions on the GATE virtual calculator. Numerical answers carry no negative marks; the marking scheme, which is common to every GATE paper, sets out the rules for MCQ, MSQ and NAT.

More formula sheets: all of GATE ECE · Analog Circuits · Electromagnetics · Signals and Systems

Quick revision

Output PSD is $\vert H(f)\vert^2 S_X(f)$ , and white noise in bandwidth $B$ has power $N_0 B$ .
AM power is $P_c(1 + \mu^2/2)$ and efficiency peaks at one third when $\mu = 1$ .
A DSB-SC phase error scales the output by $\cos\phi$ ; a frequency offset makes it beat.
FM index is $\Delta f/f_m$ , PM index is $k_p A_m$ , Carson's bandwidth is $2(\Delta f + f_m)$ , and power stays $A_c^2/2$ .
PCM bit rate is $n f_s$ , and SQNR is $1.76 + 6.02n$ dB for a full-scale sinusoid.
BPSK error is $Q(\sqrt{2E_b/N_0})$ , orthogonal is $Q(\sqrt{E_b/N_0})$ , and MAP equals ML only for equal priors.
Use $\log_2$ for entropy, convert SNR from dB before $B\log_2(1 + S/N)$ , and remember $C = 1 - H(p)$ for a BSC.
Correct $t$ errors when $d_{\min} \ge 2t + 1$ ; Hamming codes have $d_{\min} = 3$ .

Frequently asked questions

Which Communications topics are asked most often in GATE ECE?

Across the five counted EC papers (2010, 2019, 2020, 2025 and 2026), angle modulation and MAP or ML detection with error probability each appear in all five. AM and DSB-SC and random processes through LTI filters each appear in four. Entropy, capacity and Hamming codes appear in three, all of them recent. Prepare those formulas first.

What is the bit error probability of BPSK in GATE ECE?

For coherent BPSK in additive white Gaussian noise, the bit error probability is $Q(\sqrt{2E_b/N_0})$ , where $E_b$ is the energy per bit and $N_0/2$ is the two-sided noise PSD. Coherent orthogonal signalling such as BFSK gives $Q(\sqrt{E_b/N_0})$ , so it needs twice the energy, or 3 dB more, for the same error rate.

How do you find the bandwidth of an FM signal for GATE?

Use Carson's rule. The bandwidth is $2(\Delta f + f_m)$ , where $\Delta f$ is the peak frequency deviation and $f_m$ is the highest message frequency. Written with the modulation index it is $2f_m(\beta + 1)$ . Check whether the deviation constant is given in hertz per volt or radians per second per volt before you compute $\Delta f$ .

Does the power of an FM signal change with the modulation index?

No. An FM or PM signal has a constant envelope, so its average power is $A_c^2/2$ whatever the message or the modulation index. Modulation only redistributes that power between the carrier and the sidebands. This single fact was the subject of 2010 Q21 and is one of the most asked facts in the section.

How many errors can a Hamming code correct?

A Hamming code has a minimum distance of 3, so it corrects one error or detects two, but not both at once. In general a code corrects $t$ errors if $d_{\min} \ge 2t + 1$ and detects $s$ errors if $d_{\min} \ge s + 1$ . A single-error-correcting decoder given two errors decodes to the wrong codeword.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.