GATE EC Signals and Systems Formula Sheet

Organised by the official GATE 2027 EC syllabus, Section 2: Networks, Signals and Systems (the signals and systems topics). Each worked example was checked in code. Practise these: GATE EC Networks, Signals and Systems previous year questions.

LTI systems

Syllabus: Definition and properties, causality, stability, impulse response, convolution, poles and zeroes, frequency response, group delay, phase delay.

Convolution

y(t) = \int_{-\infty}^{\infty} x(\tau)\, h(t - \tau)\, d\tau, \qquad y[n] = \sum_k x[k]\, h[n - k]

\begin{aligned} y(t) &= \int_{-\infty}^{\infty} x(\tau)\, h(t - \tau)\, d\tau, \\ y[n] &= \sum_k x[k]\, h[n - k] \end{aligned}

Symbols

$h$ : impulse response of the LTI system
$x, y$ : input and output

Applies when

Linear time-invariant systems only. The output starts at the sum of the start times and lasts for the sum of the durations.

Common mistake

Forgetting to flip h before sliding it: the integrand has h(t − τ), not h(τ − t).

Worked example

e^(−t)u(t) convolved with u(t), at t = 1: 0.632 (1 − e^(−t))

Causality and BIBO stability

\text{causal} \iff h(t) = 0 \text{ for } t < 0, \qquad \text{BIBO stable} \iff \int_{-\infty}^{\infty} |h(t)|\, dt < \infty

\begin{gathered} \text{causal} \iff h(t) = 0,\ t < 0 \\ \text{BIBO stable} \iff \\ \int_{-\infty}^{\infty} |h(t)|\, dt < \infty \end{gathered}

Symbols

$h(t)$ : impulse response

Applies when

LTI systems. For a causal rational H(s), stability means every pole is in the left half-plane; in discrete time, inside the unit circle.

Common mistake

Calling a system stable because its impulse response stays bounded: h(t) = u(t) is bounded, but its integral diverges, so it is not BIBO stable.

Worked example

h(t) = e^(−2t)u(t): the integral of |h|: 0.5 (finite, so stable)

Phase delay and group delay

\tau_p(\omega) = -\frac{\theta(\omega)}{\omega}, \qquad \tau_g(\omega) = -\frac{d\theta(\omega)}{d\omega}

\begin{aligned} \tau_p(\omega) &= -\frac{\theta(\omega)}{\omega}, \\ \tau_g(\omega) &= -\frac{d\theta(\omega)}{d\omega} \end{aligned}

Symbols

$\theta(\omega)$ : phase of H(jω), in radians
$\tau_p, \tau_g$ : phase delay and group delay (s)

Applies when

Group delay is the delay of a narrow-band envelope; for a linear phase, θ = −ωt₀, both equal t₀.

Common mistake

Computing group delay from a wrapped phase, which jumps by 2π. Unwrap it, or differentiate the formula.

Worked example

H(jω) = e^(−j2ω)/(1 + jω) at ω = 1 rad/s: group delay: 2.5 s (θ = −2ω − tan⁻¹ω; the phase delay is 2.785 s)

First-order frequency response

H(j\omega) = \frac{1}{1 + j\omega/\omega_c}, \qquad |H(j\omega_c)| = \frac{1}{\sqrt 2}\ (-3\text{ dB}), \qquad \angle H(j\omega_c) = -45^\circ

\begin{aligned} H(j\omega) &= \frac{1}{1 + j\omega/\omega_c}, \\ |H(j\omega_c)| &= \frac{1}{\sqrt 2}\ (-3\text{ dB}), \\ \angle H(j\omega_c) &= -45^\circ \end{aligned}

Symbols

$\omega_c$ : corner (cut-off) frequency, 1/RC for an RC low-pass (rad/s)

Applies when

A single real pole. Far above

ω_{c}

the magnitude falls at 20 dB per decade.

Common mistake

Mixing f and ω: the corner of an RC low-pass is 1/RC in rad/s, or 1/(2πRC) in Hz.

Worked example

R = 1 kΩ, C = 1 µF: |H| at 1000 rad/s: 0.7071

Continuous-time signals

Syllabus: Fourier series and Fourier transform, Nyquist sampling theorem, sampling and reconstruction.

Fourier series coefficients

x(t) = \sum_k c_k e^{jk\omega_0 t}, \qquad c_k = \frac{1}{T}\int_T x(t)\, e^{-jk\omega_0 t}\, dt, \qquad \omega_0 = \frac{2\pi}{T}

\begin{aligned} x(t) &= \sum_k c_k e^{jk\omega_0 t}, \\ c_k &= \frac{1}{T}\int_T x(t)\, e^{-jk\omega_0 t}\, dt, \\ \omega_0 &= \frac{2\pi}{T} \end{aligned}

Symbols

$T$ : period (s)
$c_k$ : complex coefficient of the k-th harmonic

Applies when

Periodic x(t) meeting the Dirichlet conditions. For a real x, c₋ₖ is the conjugate of cₖ.

Common mistake

Leaving out the 1/T, or integrating over something other than one full period.

Worked example

Square wave of ±1, odd symmetry: the amplitude of the fundamental sine: 1.273 (4/π; odd harmonics only, 4/(kπ))

Parseval's theorem

P = \frac{1}{T}\int_T |x(t)|^2\, dt = \sum_k |c_k|^2, \qquad E = \int |x(t)|^2\, dt = \frac{1}{2\pi}\int |X(j\omega)|^2\, d\omega

\begin{aligned} P &= \frac{1}{T}\int_T |x(t)|^2\, dt \\ &= \sum_k |c_k|^2, \\ E &= \int |x(t)|^2\, dt \\ &= \frac{1}{2\pi}\int |X(j\omega)|^2\, d\omega \end{aligned}

Symbols

$P$ : average power of a periodic signal
$E$ : energy of an aperiodic signal

Applies when

Power form for periodic signals, energy form for finite-energy signals.

Common mistake

Dropping the 1/(2π) when integrating over ω; with f in Hz it is ∫|X(f)|² df, no factor.

Worked example

x(t) = 3 + 4 cos ω₀t: average power: 17 (9 + 2 × 2²)

Fourier transform pairs

e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}, \qquad \operatorname{rect}\!\left(\frac{t}{\tau}\right) \leftrightarrow \tau\,\frac{\sin(\omega\tau/2)}{\omega\tau/2}, \qquad e^{-t^2} \leftrightarrow \sqrt{\pi}\, e^{-\omega^2/4}

\begin{aligned} e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}, \\ \operatorname{rect}\!\left(\frac{t}{\tau}\right) \leftrightarrow \tau\,\frac{\sin(\omega\tau/2)}{\omega\tau/2}, \\ e^{-t^2} \leftrightarrow \sqrt{\pi}\, e^{-\omega^2/4} \end{aligned}

Symbols

$a$ : a positive constant
$\tau$ : pulse width (s)

Applies when

X(jω) = ∫x(t)e^(−jωt) dt; these pairs need a > 0.

Common mistake

Getting the sinc's zeros wrong: the first zero of the rect's transform is at ω = 2π/τ, not π/τ.

Worked example

e^(−2t)u(t): |X(jω)| at ω = 2 rad/s: 0.3536 (1/√8)

Time shift and scaling

x(t - t_0) \leftrightarrow X(j\omega)\, e^{-j\omega t_0}, \qquad x(at) \leftrightarrow \frac{1}{|a|}\, X\!\left(\frac{j\omega}{a}\right)

\begin{aligned} x(t - t_0) \leftrightarrow X(j\omega)\, e^{-j\omega t_0}, \\ x(at) \leftrightarrow \frac{1}{|a|}\, X\!\left(\frac{j\omega}{a}\right) \end{aligned}

Symbols

$t_0$ : delay (s)
$a$ : time-scale factor, a ≠ 0

Applies when

Any transformable signal. A shift changes only the phase; compressing in time stretches in frequency.

Common mistake

Forgetting the 1/|a|, so the transform's height doesn't change when it should.

Worked example

x(t) = e^(−t)u(t): the value at ω = 0 of the transform of x(2t): 0.5

Nyquist sampling theorem

f_s \ge 2 f_{max}, \qquad \text{alias: } f_a = |f - k f_s|\ \text{(nearest multiple)}

\begin{gathered} f_s \ge 2 f_{max} \\ \text{alias: } f_a = |f - k f_s| \\ \text{(nearest multiple)} \end{gathered}

Symbols

$f_s$ : sampling rate (Hz)
$f_{max}$ : highest frequency in the signal (Hz)
$f_a$ : frequency the samples appear to have (Hz)

Applies when

A band-limited low-pass signal, reconstructed with an ideal low-pass filter.

2f_{max}

is the Nyquist rate.

Common mistake

Doubling the bandwidth of a product or a modulated signal from the wrong number: find its highest frequency first.

Worked example

A 700 Hz tone sampled at 1 kHz: the alias: 300 Hz

Discrete-time signals

Syllabus: DTFT, DFT, z-transform, FIR and IIR filter design.

DTFT

X(e^{j\Omega}) = \sum_{n=-\infty}^{\infty} x[n]\, e^{-j\Omega n}, \qquad a^n u[n] \leftrightarrow \frac{1}{1 - a e^{-j\Omega}}\ (|a| < 1)

\begin{gathered} X(e^{j\Omega}) = \sum_{n=-\infty}^{\infty} x[n]\, e^{-j\Omega n} \\ a^n u[n] \leftrightarrow \frac{1}{1 - a e^{-j\Omega}} \\ (|a| < 1) \end{gathered}

Symbols

$\Omega$ : digital frequency (rad/sample)

Applies when

Periodic in Ω with period 2π; Ω = π is the highest frequency a sequence can have.

Common mistake

Treating the DTFT as non-periodic and looking for content beyond Ω = π.

Worked example

x[n] = 0.5ⁿu[n]: X at Ω = 0: 2 (1/(1 − 0.5))

DFT

X[k] = \sum_{n=0}^{N-1} x[n]\, e^{-j2\pi kn/N}, \qquad X[0] = \sum_n x[n]

\begin{aligned} X[k] &= \sum_{n=0}^{N-1} x[n]\, e^{-j2\pi kn/N}, \\ X[0] &= \sum_n x[n] \end{aligned}

Symbols

$N$ : number of points
$k$ : frequency index, 0 to N − 1

Applies when

N samples of the DTFT, evenly spaced over one period. Multiplying DFTs gives circular, not linear, convolution.

Common mistake

Expecting the product of two N-point DFTs to give linear convolution: pad both to at least N₁ + N₂ − 1 points first.

Worked example

x = [1, 2, 3, 4]: X[2]: -2 (1 − 2 + 3 − 4)

z-transform and region of convergence

a^n u[n] \leftrightarrow \frac{z}{z - a},\ |z| > |a|; \qquad -a^n u[-n-1] \leftrightarrow \frac{z}{z - a},\ |z| < |a|

\begin{aligned} &a^n u[n] \leftrightarrow \frac{z}{z - a},\ |z| \\ &\quad > |a|; \\ &-a^n u[-n-1] \leftrightarrow \frac{z}{z - a},\ |z| \\ &\quad < |a| \end{aligned}

Symbols

$z$ : complex variable
$a$ : pole location

Applies when

The same expression has different inverses for different ROCs; a causal sequence has an ROC outside its outermost pole. Stable means the ROC includes |z| = 1.

Common mistake

Inverting without the ROC and always picking the causal sequence.

Worked example

X(z) = z/(z − 0.5), |z| > 0.5: x[3]: 0.125

Stability in discrete time

\text{causal LTI stable} \iff |p_i| < 1 \text{ for every pole } p_i \text{ of } H(z)

\begin{gathered} \text{causal LTI is stable} \\ \iff |p_i| < 1 \\ \text{for every pole } p_i \text{ of } H(z) \end{gathered}

Symbols

$p_i$ : poles of H(z)

Applies when

Causal systems with rational H(z). Equivalently, Σ|h[n]| is finite.

Common mistake

Checking the zeros. Zeros can sit anywhere; only the poles decide stability.

Worked example

y[n] = 1.2 y[n − 1] + x[n]: the pole magnitude: 1.2 (outside the unit circle: unstable)

Linear-phase FIR filter

h[n] = h[N - 1 - n] \;\Rightarrow\; \tau_g = \frac{N - 1}{2}\ \text{samples}

\begin{aligned} h[n] &= h[N - 1 - n] \\ \Rightarrow\; \tau_g &= \frac{N - 1}{2}\ \text{samples} \end{aligned}

Symbols

$N$ : filter length (number of taps)
$h[n]$ : impulse response, symmetric about its centre

Applies when

Symmetric (or antisymmetric) coefficients give exactly linear phase, so every frequency is delayed equally.

Common mistake

Using N/2 instead of (N − 1)/2 for the delay.

Worked example

h = [1, 2, 3, 2, 1]: group delay: 2 samples

Bilinear transformation (IIR design)

s = \frac{2}{T}\,\frac{1 - z^{-1}}{1 + z^{-1}}, \qquad \omega_a = \frac{2}{T}\tan\frac{\Omega_d}{2}

\begin{aligned} s &= \frac{2}{T}\,\frac{1 - z^{-1}}{1 + z^{-1}}, \\ \omega_a &= \frac{2}{T}\tan\frac{\Omega_d}{2} \end{aligned}

Symbols

$T$ : sampling period (s)
$\Omega_d$ : digital frequency wanted (rad/sample)
$\omega_a$ : analog frequency to design for (rad/s)

Applies when

Maps the whole jω axis onto the unit circle, so there is no aliasing, but frequencies are warped: pre-warp the critical ones.

Common mistake

Designing the analog prototype at

\Omega_d/T

without pre-warping, which shifts the digital cut-off.

Worked example

T = 1 s,

\Omega_d = \pi/2

: the pre-warped analog frequency: 2 rad/s

Practise these: GATE EC Networks, Signals and Systems previous year questions

Source for the section headings: the official GATE 2027 EC syllabus, IIT Madras. The formulas, examples and notes are Studyloaf's own.

GATE EC Signals and Systems Formula Sheet

LTI systems

Convolution

Causality and BIBO stability

Phase delay and group delay

First-order frequency response

Continuous-time signals

Fourier series coefficients

Parseval's theorem

Fourier transform pairs

Time shift and scaling

Nyquist sampling theorem

Discrete-time signals

DTFT

DFT

z-transform and region of convergence

Stability in discrete time

Linear-phase FIR filter

Bilinear transformation (IIR design)

Other GATE EC formula sheets