GATE EC · EE · IN

LTI systems

Properties, impulse response, convolution and correlation

A linear time-invariant (LTI) system is described completely by its impulse response $h (t)$ or $h [n]$ . The output for any input is the convolution of that input with $h$ , and causality, stability, memory and invertibility can all be read from $h$ . Questions on this topic ask you to test a system equation for linearity, time invariance, causality, memory, invertibility and stability, to evaluate a convolution (often at one instant, or with the width and area rules), to find the response to a sinusoid through the frequency response, and to compute or reason about an autocorrelation or a cross-correlation.

31 previous-year questions · 28 solved · 5 practice questions

1. System properties and the test for each

A system $T$ maps an input $x$ to an output $y = T {x}$ . Each property below is a statement about every admissible input and every instant. One counterexample disproves a property. A proof must hold in general. Everything here applies to continuous time ( $t$ ) and to discrete time ( $n$ , with integer shifts). Unless a question says otherwise, assume the system is at rest (zero initial conditions) before the input is applied.

LINEARITY (SUPERPOSITION)

T {a x_{1} + b x_{2}} = a T {x_{1}} + b T {x_{2}}

Must hold for all inputs $x_{1}, x_{2}$ and all constants $a, b$ (complex constants if the signals may be complex). It combines additivity ( $a = b = 1$ ) and homogeneity ( $b = 0$ ). Zero input giving zero output is necessary, not sufficient.

TIME INVARIANCE

x (t) \to y (t) \Rightarrow x (t - t_{0}) \to y (t - t_{0}) for every t_{0}

Discrete time: $x [n - n_{0}] \to y [n - n_{0}]$ for every integer $n_{0}$ . Also called shift invariance.

CAUSALITY AND MEMORY

causal: y (t_{0}) depends only on x (t), t \leq t_{0} memoryless: y (t_{0}) depends only on x (t_{0})

Both must hold for every $t_{0}$ . A memoryless (static) system is always causal. A causal system need not be memoryless.

BIBO STABILITY

∣ x (t) ∣ \leq M_{x} < \infty for all t \Rightarrow ∣ y (t) ∣ \leq M_{y} < \infty for all t

Every bounded input must give a bounded output. One bounded input with an unbounded output proves the system unstable.

How to run each test

Linearity. Find $y_{1} = T {x_{1}}$ and $y_{2} = T {x_{2}}$ . Then compute $T {a x_{1} + b x_{2}}$ and compare it with $a y_{1} + b y_{2}$ . As a quick first check, set $x = 0$ . A nonzero output (for example from $y = x + 2$ ) proves the system is nonlinear. Passing this check proves nothing, because $y = x^{2}$ passes it too.
Time invariance. (i) Delay only the input: replace $x (\cdot)$ by $x (\cdot - t_{0})$ inside the operator and leave every explicit $t$ unchanged. Call the result $y_{d} (t)$ . (ii) Delay the output: replace every $t$ in the output expression by $t - t_{0}$ to get $y (t - t_{0})$ . (iii) The system is time-invariant only if $y_{d} (t) = y (t - t_{0})$ for all $t$ and all $t_{0}$ .
Causality. Ask which input instants $y (t_{0})$ needs. Check negative $t_{0}$ as well as positive $t_{0}$ .
Memory. A delay, an advance, a scaled or reversed argument, an integral or a derivative gives the system memory. A system is memoryless only if its output at each instant is a function of the input at that same instant alone. The function may change with time, as in $t x (t)$ or $x^{2} (t) + t$ .
Invertibility. The system is invertible if distinct inputs always give distinct outputs. To disprove it, find two different inputs with the same output: $x$ and $- x$ for $y = x^{2}$ , or $x$ and $x + c$ for a differentiator. To prove it, write down the inverse system.
BIBO stability. To disprove it, apply a simple bounded input such as $x = 1$ , $u (t)$ or $cos ω_{0} t$ and show that the output grows without bound. To prove it, bound $∣ y ∣$ using $∣ x ∣ \leq M_{x}$ .

EXAM TIP

A straight line is not a linear system. $y = 2 x + 3$ gives $y = 3$ for $x = 0$ , so it is not linear. It is incrementally linear: a linear system plus a response that does not depend on the input. For the same reason, a system described by a linear constant-coefficient differential or difference equation is linear and time-invariant only when it starts at rest. A nonzero initial condition produces an output even with zero input. A related trap is $y (t) = Re {x (t)}$ . It is additive, but it is not homogeneous for complex constants: with $a = j$ and a real $x$ , $Re {j x} = 0$ while $j Re {x} = j x$ . It is linear only when inputs and constants are restricted to real values.

EXAM TIP

Delay the input correctly. For $y (t) = x (2 t)$ , the delayed input is the new signal $x_{1} (t) = x (t - t_{0})$ . Its output is $x_{1} (2 t) = x (2 t - t_{0})$ , not $x (2 (t - t_{0}))$ . The delayed output is $y (t - t_{0}) = x (2 t - 2 t_{0})$ . The two differ, so the system is time-varying. Quick screens, to be confirmed with the formal test: an explicit $t$ or $n$ as a coefficient ( $t x (t)$ , $x (t) cos t$ ), a scaled or reversed argument ( $x (2 t)$ , $x (- t)$ ), or an integration or summation limit that is neither infinite nor of the form $t - constant$ (such as $\int_{- \infty}^{2 t}$ or $\int_{0}^{t}$ ) each point to a time-varying system. Limits such as $\int_{- \infty}^{t}$ and $\int_{t - T}^{t}$ are consistent with time invariance.

2. Property tests on typical system equations

The table applies the six tests to standard system equations. Yes means the system has the property. The note after a colon is the counterexample or the argument. Learn the reasons rather than the table, because a question usually changes one detail of an equation like these. The differentiator is taken as causal: the derivative at $t_{0}$ can be found from the left, using only $x (t)$ for $t \leq t_{0}$ , and its impulse response $δ^{'} (t)$ is zero for $t < 0$ .

System	Linear	Time-invariant	Causal	Memoryless	BIBO stable	Invertible
$y (t) = x (2 t)$	Yes	No	No: $y (1) = x (2)$	No	Yes	Yes: $x (t) = y (t /2)$
$y [n] = n x [n]$	Yes	No	Yes	Yes	No: $x = 1$ gives $y = n$	No: $y [0] = 0$ whatever $x [0]$ is
$y (t) = x (t) cos t$	Yes	No	Yes	Yes	Yes: $∣ y ∣ \leq ∣ x ∣$	No: $y = 0$ at $t = π /2 + k π$ whatever $x$ is there
$y (t) = x^{2} (t)$	No	Yes	Yes	Yes	Yes: $∣ y ∣ \leq M_{x}^{2}$	No: $x$ and $- x$ give the same $y$
$y (t) = e^{x (t)}$	No	Yes	Yes	Yes	Yes: $y \leq e^{M_{x}}$	Yes, for real $x$ : $x = ln y$
$y (t) = x (t) + 2$	No: $x = 0$ gives $y = 2$	Yes	Yes	Yes	Yes	Yes
$y (t) = x (t - 2)$	Yes	Yes	Yes	No	Yes	Yes: $x (t) = y (t + 2)$
$y (t) = x (- t)$	Yes	No	No: $y (- 1) = x (1)$	No	Yes	Yes: $x (t) = y (- t)$
$y (t) = \int_{- \infty}^{t} x (τ) d τ$	Yes	Yes	Yes	No	No: $u (t)$ gives $t u (t)$	Yes: $x = d y / d t$
$y (t) = \int_{t - T}^{t} x (τ) d τ$ , $T > 0$	Yes	Yes	Yes	No	Yes: $∣ y ∣ \leq M_{x} T$	No: $cos (2 π t / T)$ gives $y = 0$
$y (t) = \frac{d x ( t )}{d t}$	Yes	Yes	Yes	No	No: $sin (t^{2})$ gives $2 t cos (t^{2})$	No: $x$ and $x + c$ give the same $y$

WORKED EXAMPLE

Test the system $y (t) = \int_{- \infty}^{2 t} x (τ) d τ$ for linearity, time invariance, causality, memory, BIBO stability and invertibility.

Solution

Linearity: $\int_{- \infty}^{2 t} [a x_{1} (τ) + b x_{2} (τ)] d τ = a y_{1} (t) + b y_{2} (t)$ . The system is linear.
Time invariance: the delayed input $x (t - t_{0})$ gives $y_{d} (t) = \int_{- \infty}^{2 t} x (τ - t_{0}) d τ = \int_{- \infty}^{2 t - t_{0}} x (λ) d λ$ , with $λ = τ - t_{0}$ .
The delayed output is $y (t - t_{0}) = \int_{- \infty}^{2 t - 2 t_{0}} x (λ) d λ$ . The upper limits differ for $t_{0} \neq = 0$ , so the system is time-varying.
Causality: $y (1) = \int_{- \infty}^{2} x (τ) d τ$ needs $x$ on $1 < τ \leq 2$ , which is in the future. The system is non-causal. (For $t < 0$ it happens to use only past values, but one counterexample is enough.)
Memory: the output depends on the input over an interval, so the system has memory.
Stability: $x (t) = u (t)$ is bounded by 1, but $y (t) = \int_{0}^{2 t} d τ = 2 t$ for $t > 0$ grows without bound. The system is BIBO unstable.
Invertibility: differentiating gives $\frac{d y}{d t} = 2 x (2 t)$ , so $x (t) = \frac{1}{2} \frac{d y}{d t}_{t /2}$ , meaning the derivative of $y$ evaluated at $t /2$ . The input can be recovered, so the system is invertible.

Answer: Linear, time-varying, non-causal, with memory, BIBO unstable and invertible.

EXAM TIP

Scaled arguments. A system $y (t) = x (a t)$ with $a \neq = 0$ and $a \neq = 1$ is time-varying and non-causal. For $a > 1$ it needs future input when $t > 0$ . For $0 < a < 1$ or $a < 0$ it needs future input when $t < 0$ . It is still linear, BIBO stable and invertible. Check negative time before calling any system with a scaled argument causal.

3. The impulse response describes an LTI system completely

Any signal can be written as a sum of weighted, shifted impulses: $x (t) = \int x (τ) δ (t - τ) d τ$ (the sifting property). If the system is time-invariant, its response to $δ (t - τ)$ is $h (t - τ)$ . If it is also linear, superposition gives the output for any input as a convolution with $h$ . So the impulse response $h (t) = T {δ (t)}$ , measured with the system at rest, describes an LTI system completely. This fails for a nonlinear or time-varying system, whose response to $δ (t)$ does not determine its response to other inputs. The same argument works in discrete time with $x [n] = \sum_{k} x [k] δ [n - k]$ and $h [n] = T {δ [n]}$ .

CONVOLUTION INTEGRAL

y (t) = x (t) * h (t) = \int_{- \infty}^{\infty} x (τ) h (t - τ) d τ = \int_{- \infty}^{\infty} h (τ) x (t - τ) d τ

$τ$ is the integration variable and $t$ is held fixed while integrating. In practice the limits shrink to the interval where both factors are nonzero.

CONVOLUTION SUM

y [n] = x [n] * h [n] = k = - \infty \sum \infty x [k] h [n - k] = k = - \infty \sum \infty h [k] x [n - k]

Each output sample is a sum of products of samples whose indices add to $n$ .

CAUSAL SYSTEM WITH A CAUSAL INPUT

y (t) = \int_{0}^{t} x (τ) h (t - τ) d τ (t \geq 0), y [n] = k = 0 \sum n x [k] h [n - k] (n \geq 0)

When $x$ and $h$ are both zero before the origin, $y$ is zero there too. If either signal has an impulse at the origin, take the lower limit as $0^{-}$ so that the impulse is included.

CAUSALITY OF AN LTI SYSTEM

h (t) = 0 for t < 0, h [n] = 0 for n < 0

Then $y (t) = \int_{0}^{\infty} h (τ) x (t - τ) d τ$ uses only present and past input. Necessary and sufficient for an LTI system.

BIBO STABILITY OF AN LTI SYSTEM

\int_{- \infty}^{\infty} ∣ h (t) ∣ d t < \infty, n = - \infty \sum \infty ∣ h [n] ∣ < \infty

Necessary and sufficient for an LTI system. The integral of $∣ h ∣$ must be finite, not the integral of $h$ . The output is then bounded by $∣ y ∣ \leq M_{x} \int ∣ h ∣ d t$ . If $h (t)$ contains impulses, each term $c δ (t - t_{i})$ adds $∣ c ∣$ to the integral, so $h (t) = δ (t)$ (a plain wire) is stable. A doublet term $δ^{'} (t)$ (a differentiator) makes the system unstable.

Property of the LTI system	Continuous time	Discrete time
Causal	$h (t) = 0$ for $t < 0$	$h [n] = 0$ for $n < 0$
BIBO stable	$\int_{- \infty}^{\infty} ∣ h (t) ∣ d t < \infty$	$\sum_{n = - \infty}^{\infty} ∣ h [n] ∣ < \infty$
Memoryless	$h (t) = K δ (t)$	$h [n] = K δ [n]$
Invertible	some $h_{i} (t)$ satisfies $h * h_{i} = δ (t)$	some $h_{i} [n]$ satisfies $h * h_{i} = δ [n]$
Stable, in transform terms	ROC of $H (s)$ contains the $j ω$ -axis, and $h (t)$ has no $δ^{'} (t)$ or higher-derivative terms (for a rational $H (s)$ : $H (s)$ proper)	ROC of $H (z)$ contains the unit circle $∣ z ∣ = 1$
Causal and stable, rational transfer function	$H (s)$ proper and every pole in $Re {s} < 0$	$H (z)$ proper in $z$ (numerator degree $\leq$ denominator degree, in powers of $z$ ) and every pole inside $∣ z ∣ < 1$

STEP RESPONSE (CONTINUOUS TIME)

s (t) = h (t) * u (t) = \int_{- \infty}^{t} h (τ) d τ, h (t) = \frac{d s ( t )}{d t}

Here $s (t)$ is the step response, not the Laplace variable. The step response is the running integral of $h$ . For a causal $h$ , $s (t) = \int_{0}^{t} h (τ) d τ$ . For a stable system the final value is $s (\infty) = \int_{- \infty}^{\infty} h (t) d t = H (0)$ , the DC gain. The ramp response is the running integral of the step response.

STEP RESPONSE (DISCRETE TIME)

s [n] = h [n] * u [n] = k = - \infty \sum n h [k], h [n] = s [n] - s [n - 1]

The step response is the running sum of $h$ , and $h$ is the first difference of the step response.

INTERCONNECTIONS OF LTI SYSTEMS

cascade: h = h_{1} * h_{2} = h_{2} * h_{1}, parallel: h = h_{1} + h_{2}

In any transform domain a cascade multiplies, $H = H_{1} H_{2}$ , and a parallel connection adds, $H = H_{1} + H_{2}$ . The order of LTI blocks in a cascade does not change the overall response.

INVERSE SYSTEM

h (t) * h_{i} (t) = δ (t), H (s) H_{i} (s) = 1

Examples: the delay $δ (t - t_{0})$ has inverse $δ (t + t_{0})$ . The accumulator $h [n] = u [n]$ has inverse $δ [n] - δ [n - 1]$ , because $u [n] - u [n - 1] = δ [n]$ . An inverse need not be causal or stable: for $t_{0} > 0$ the delay is causal, but its inverse, an advance, is not.

LTI SYSTEM FROM A LINEAR CONSTANT-COEFFICIENT EQUATION

k = 0 \sum N a_{k} \frac{d ^{k} y ( t )}{d t ^{k}} = k = 0 \sum M b_{k} \frac{d ^{k} x ( t )}{d t ^{k}} \Rightarrow H (s) = \frac{\sum _{k = 0}^{M} b _{k} s ^{k}}{\sum _{k = 0}^{N} a _{k} s ^{k}} k = 0 \sum N a_{k} y [n - k] = k = 0 \sum M b_{k} x [n - k] \Rightarrow H (z) = \frac{\sum _{k = 0}^{M} b _{k} z ^{- k}}{\sum _{k = 0}^{N} a _{k} z ^{- k}}

Holds when the system is at rest before the input starts (initial rest), which makes it LTI and causal. $h$ is the inverse transform with the ROC to the right of the rightmost pole of $H (s)$ , or outside the outermost pole of $H (z)$ . Examples: $\frac{d y}{d t} + a y = x (t)$ gives $H (s) = \frac{1}{s + a}$ and $h (t) = e^{- a t} u (t)$ , stable only for $a > 0$ . $y [n] - a y [n - 1] = x [n]$ gives $H (z) = \frac{1}{1 - a z ^{- 1}}$ and $h [n] = a^{n} u [n]$ , stable only for $∣ a ∣ < 1$ . An $h [n]$ of finite length is called FIR and one of infinite length IIR. A difference equation with no past outputs ( $N = 0$ ) is always FIR, with $h [n] = b_{n} / a_{0}$ .

EXAM TIP

Stability traps. (1) The ideal low-pass filter $h (t) = \frac{sin t}{π t}$ has $\int h d t = 1$ , which is finite, but $\int ∣ h ∣ d t$ diverges, so it is not BIBO stable. It is also non-causal. (2) $h [n] = \frac{1}{n} u [n - 1]$ is unstable because the harmonic series diverges, while $h [n] = \frac{1}{n ^{2}} u [n - 1]$ is stable. (3) $h (t) = e^{- a ∣ t ∣}$ with $a > 0$ is stable but non-causal. (4) The integrator $h (t) = u (t)$ and $h (t) = sin (ω_{0} t) u (t)$ are causal but unstable. (5) A finite-length $h [n]$ (FIR) with finite samples is always stable. (6) An improper $H (s)$ , such as $H (s) = s$ , contains $δ^{'} (t)$ terms and is not BIBO stable even though it has no finite poles.

EXAM TIP

Only LTI blocks can be reordered. For blocks that are not LTI, the order can matter. Multiplying by $t$ and then delaying by 1 gives $(t - 1) x (t - 1)$ . Delaying first and then multiplying by $t$ gives $t x (t - 1)$ .

4. Convolution: graphical method, properties and standard results

Graphical convolution (flip and slide)

Write both signals against the integration variable: $x (τ)$ and $h (τ)$ . Convolution is commutative, so flip whichever signal is simpler.
Flip: $h (τ) \to h (- τ)$ .
Slide: $h (t - τ)$ is the flipped copy moved so that its original origin sits at $τ = t$ . Increasing $t$ moves it to the right.
For each range of $t$ , multiply $x (τ) h (t - τ)$ and integrate (or sum) over the overlap only.
The ranges change only where an edge of one signal crosses an edge of the other. If $x$ occupies $[a_{1}, b_{1}]$ and $h$ occupies $[a_{2}, b_{2}]$ , the breakpoints are $t = a_{1} + a_{2}$ , $a_{1} + b_{2}$ , $b_{1} + a_{2}$ and $b_{1} + b_{2}$ , plus any interior corners of piecewise signals.
Check the result. $y$ is zero outside $[a_{1} + a_{2}, b_{1} + b_{2}]$ . If neither signal contains an impulse (bounded, piecewise-continuous signals), $y$ has no jumps, so neighbouring pieces must agree at every breakpoint. The area of $y$ is $(\int x d t) (\int h d t)$ when both areas are finite.

COMMUTATIVE, ASSOCIATIVE AND DISTRIBUTIVE LAWS

x * h = h * x, (x * h_{1}) * h_{2} = x * (h_{1} * h_{2}), x * (h_{1} + h_{2}) = x * h_{1} + x * h_{2}

Hold for continuous and discrete convolution. Associativity gives the cascade rule and distributivity gives the parallel rule.

CONVOLUTION WITH AN IMPULSE

x (t) * δ (t) = x (t), x (t) * δ (t - t_{0}) = x (t - t_{0}), δ (t - t_{1}) * δ (t - t_{2}) = δ (t - t_{1} - t_{2})

Discrete time: $x [n] * δ [n - n_{0}] = x [n - n_{0}]$ . Convolving with an impulse train repeats the signal: $x (t) * \sum_{k} δ (t - k T_{0}) = \sum_{k} x (t - k T_{0})$ . Do not confuse convolution with multiplication by an impulse, which samples: $x (t) δ (t - t_{0}) = x (t_{0}) δ (t - t_{0})$ , for $x$ continuous at $t_{0}$ .

SHIFT PROPERTY

x (t) * h (t) = y (t) \Rightarrow x (t - t_{1}) * h (t - t_{2}) = y (t - t_{1} - t_{2})

Shifts add. The same holds in discrete time with integer shifts.

CONVOLUTION WITH THE UNIT STEP

x (t) * u (t) = \int_{- \infty}^{t} x (τ) d τ, x [n] * u [n] = k = - \infty \sum n x [k]

Convolving with $u$ is running integration (or running summation). Convolving with $δ^{'} (t)$ differentiates: $x (t) * δ^{'} (t) = x^{'} (t)$ .

DIFFERENTIATION PROPERTY

\frac{d}{d t} [x (t) * h (t)] = \frac{d x}{d t} * h = x * \frac{d h}{d t}

Useful for piecewise-constant signals, whose derivatives are impulses. A result found this way must be integrated back, using the fact that $y$ is zero before $t = a_{1} + a_{2}$ .

AREA (SUM) PROPERTY

\int_{- \infty}^{\infty} y (t) d t = (\int_{- \infty}^{\infty} x (t) d t) (\int_{- \infty}^{\infty} h (t) d t), n \sum y [n] = (n \sum x [n]) (n \sum h [n])

Holds whenever both areas (sums) are finite, for example when $x$ and $h$ are absolutely integrable (summable). The quickest check on any such convolution. In transform terms it is $Y (0) = X (0) H (0)$ .

WIDTH AND LENGTH RULE

x on [a_{1}, b_{1}], h on [a_{2}, b_{2}] \Rightarrow y on [a_{1} + a_{2}, b_{1} + b_{2}] lengths N_{1}, N_{2} \Rightarrow N_{1} + N_{2} - 1 samples, from n_{1} + n_{2} to n_{1} + n_{2} + N_{1} + N_{2} - 2

Continuous time: the widths add, $T_{y} = T_{x} + T_{h}$ . Discrete time: $n_{1}$ and $n_{2}$ are the first indices of $x$ and $h$ , and each length is counted from the first nonzero sample to the last. The first and last output samples are then single products of nonzero end samples, so they are nonzero; interior samples may be zero.

TIME-SCALING PROPERTY

x (a t) * h (a t) = \frac{1}{∣ a ∣} y (a t), a \neq = 0

Continuous time only. Scaling both signals by $a$ compresses the output time axis by $a$ and scales its height by $1/∣ a ∣$ .

CONVOLUTION THEOREM

y = x * h \leftrightarrow Y (j ω) = X (j ω) H (j ω), Y (s) = X (s) H (s), Y (z) = X (z) H (z)

The ROC of $Y$ contains the intersection of the ROCs of $X$ and $H$ . Multiplication in time is convolution in frequency: $x (t) p (t) \leftrightarrow \frac{1}{2 π} X (j ω) * P (j ω)$ in the $ω$ convention, and $X (f) * P (f)$ , without the $1/2 π$ , in the $f$ convention.

Convolution	Result	Condition
$u (t) * u (t)$	$t u (t)$ , the unit ramp	none
$e^{- a t} u (t) * u (t)$	$\frac{1 - e ^{- a t}}{a} u (t)$	$a \neq = 0$
$e^{- a t} u (t) * e^{- b t} u (t)$	$\frac{e ^{- a t} - e ^{- b t}}{b - a} u (t)$	$a \neq = b$
$e^{- a t} u (t) * e^{- a t} u (t)$	$t e^{- a t} u (t)$	none
$rect (t / T) * rect (t / T)$	$T Λ (t / T)$ : a triangle on $[- T, T]$ with peak $T$ at $t = 0$	$rect (t / T) = 1$ for $∣ t ∣ < T /2$ ; $Λ (t / T) = 1 - ∣ t ∣/ T$ for $∣ t ∣ \leq T$
Unit-height pulses of widths $T_{1} \leq T_{2}$	A trapezoid of total width $T_{1} + T_{2}$ with a flat top of height $T_{1}$ and length $T_{2} - T_{1}$	Starts at the sum of the two start times
$u [n] * u [n]$	$(n + 1) u [n]$	none
$a^{n} u [n] * u [n]$	$\frac{1 - a ^{n + 1}}{1 - a} u [n]$	$a \neq = 1$
$a^{n} u [n] * b^{n} u [n]$	$\frac{a ^{n + 1} - b ^{n + 1}}{a - b} u [n]$	$a \neq = b$
$a^{n} u [n] * a^{n} u [n]$	$(n + 1) a^{n} u [n]$	none

WORKED EXAMPLE

An LTI system has $h (t) = e^{- t} u (t)$ . The input is the pulse $x (t) = u (t) - u (t - 2)$ . Find $y (t)$ for all $t$ and the value $y (3)$ .

Solution

Flip the exponential: $h (t - τ) = e^{- (t - τ)}$ for $τ < t$ and zero for $τ > t$ . The pulse $x (τ) = 1$ on $0 < τ < 2$ .
Breakpoints: $x$ starts at 0 and ends at 2, and $h$ starts at 0 and never ends. So the ranges are $t < 0$ , $0 \leq t \leq 2$ and $t > 2$ .
$t < 0$ : the flipped exponential lies entirely to the left of the pulse, so $y (t) = 0$ .
$0 \leq t \leq 2$ : the overlap is $0 < τ < t$ , so $y (t) = \int_{0}^{t} e^{- (t - τ)} d τ = e^{- t} (e^{t} - 1) = 1 - e^{- t}$ .
$t > 2$ : the overlap is the whole pulse, $0 < τ < 2$ , so $y (t) = e^{- t} \int_{0}^{2} e^{τ} d τ = e^{- t} (e^{2} - 1) = e^{- (t - 2)} - e^{- t}$ .
$y (3) = e^{- 1} - e^{- 3} = 0.3679 - 0.0498 = 0.3181$ .
Shortcut by linearity and time invariance: the step response is $s (t) = (1 - e^{- t}) u (t)$ and $x = u (t) - u (t - 2)$ , so $y (t) = s (t) - s (t - 2)$ . For $t > 2$ this gives $(1 - e^{- t}) - (1 - e^{- (t - 2)}) = e^{- (t - 2)} - e^{- t}$ , as above.
Checks: both pieces give $1 - e^{- 2}$ at $t = 2$ , so there is no jump. The area is $(1 + e^{- 2}) + (1 - e^{- 2}) = 2 = (area of x) (area of h) = 2 \times 1$ .

Answer: $y (t) = 0$ for $t < 0$ , $1 - e^{- t}$ for $0 \leq t \leq 2$ , and $e^{- (t - 2)} - e^{- t}$ for $t > 2$ . $y (3) = e^{- 1} - e^{- 3} \approx 0.318$ .

WORKED EXAMPLE

$x [n] = {1, 2, 3}$ for $n = 0, 1, 2$ and $h [n] = {1, - 1, 2}$ for $n = - 1, 0, 1$ . Both are zero elsewhere. Find $y [n] = x [n] * h [n]$ .

Solution

Length rule: $3 + 3 - 1 = 5$ samples. The first index is $0 + (- 1) = - 1$ and the last is $2 + 1 = 3$ .
Multiply as polynomial coefficients, $(1, 2, 3) \times (1, - 1, 2)$ , adding the products whose positions add to the same total.
$y [- 1] = 1 \cdot 1 = 1$ .
$y [0] = 1 \cdot (- 1) + 2 \cdot 1 = 1$ .
$y [1] = 1 \cdot 2 + 2 \cdot (- 1) + 3 \cdot 1 = 3$ .
$y [2] = 2 \cdot 2 + 3 \cdot (- 1) = 1$ .
$y [3] = 3 \cdot 2 = 6$ .
Sum check: $\sum x = 6$ and $\sum h = 2$ , so $\sum y$ must be 12. Indeed $1 + 1 + 3 + 1 + 6 = 12$ .

Answer: $y [n] = {1, 1, 3, 1, 6}$ for $n = - 1, 0, 1, 2, 3$ , and zero elsewhere.

EXAM TIP

Symmetry, position and period shortcuts. An even signal convolved with an even signal is even, odd with odd is even, and even with odd is odd. If $x$ is symmetric about $t = c_{1}$ and $h$ about $t = c_{2}$ , then $y$ is symmetric about $t = c_{1} + c_{2}$ . Two unit-height pulses convolve to a trapezoid whose peak value equals the width of the shorter pulse, or to a triangle if the widths are equal. If $x$ is periodic with period $T_{0}$ and $h$ is absolutely integrable, $y$ is periodic with the same period $T_{0}$ .

EXAM TIP

One sample only. If a question asks for $y [n_{0}]$ alone, do not compute the whole convolution. $y [n_{0}] = \sum_{k} x [k] h [n_{0} - k]$ needs only the pairs of samples whose indices add to $n_{0}$ . The same idea works in continuous time: $y (t_{0}) = \int x (τ) h (t_{0} - τ) d τ$ is a single integral over the overlap at $t = t_{0}$ .

5. Eigenfunctions and the response to a sinusoid

Complex exponentials are the eigenfunctions of every LTI system. They come out with the same shape, multiplied by a complex constant that depends only on the system and the frequency. That constant is the transfer function evaluated at the input's complex frequency. Convention used here: the Fourier transform is written in $ω$ (rad/s), $H (j ω) = \int h (t) e^{- j ω t} d t$ , with the $1/2 π$ in the inverse transform. In the $f$ (Hz) convention, $H (f) = H (j ω)_{ω = 2 π f}$ and the transform pair has no $1/2 π$ . In discrete time the frequency response is $H (e^{j Ω})$ , with $Ω$ in rad/sample; it is periodic in $Ω$ with period $2 π$ . The Laplace and z-transforms in this section are bilateral, and each result holds for $s$ or $z$ in the ROC of $H$ .

EIGENFUNCTION PROPERTY (CONTINUOUS TIME)

x (t) = e^{s_{0} t} \Rightarrow y (t) = H (s_{0}) e^{s_{0} t}, H (s) = \int_{- \infty}^{\infty} h (τ) e^{- s τ} d τ

Holds for all $t$ when $s_{0}$ lies in the ROC of $H (s)$ , the bilateral Laplace transform of $h$ . The input exists for all time, $- \infty < t < \infty$ .

EIGENFUNCTION PROPERTY (DISCRETE TIME)

x [n] = z_{0}^{n} \Rightarrow y [n] = H (z_{0}) z_{0}^{n}, H (z) = k = - \infty \sum \infty h [k] z^{- k}

$z_{0}$ must lie in the ROC of $H (z)$ . With $z_{0} = e^{j Ω_{0}}$ the eigenvalue is the frequency response $H (e^{j Ω_{0}})$ .

FREQUENCY RESPONSE

H (j ω) = \int_{- \infty}^{\infty} h (t) e^{- j ω t} d t = ∣ H (j ω) ∣ e^{j ∠ H (j ω)}, H (e^{j Ω}) = n = - \infty \sum \infty h [n] e^{- j Ω n}

Exists as an ordinary function when $h$ is absolutely integrable (summable), that is, when the system is BIBO stable. For a real $h$ , $∣ H ∣$ is even and $∠ H$ is odd in frequency.

LTI RESPONSE TO A SINUSOID

A cos (ω_{0} t + ϕ) \to A ∣ H (j ω_{0}) ∣ cos (ω_{0} t + ϕ + ∠ H (j ω_{0}))

Needs a real $h$ and a stable system. Discrete time: $A cos (Ω_{0} n + ϕ) \to A ∣ H (e^{j Ω_{0}}) ∣ cos (Ω_{0} n + ϕ + ∠ H (e^{j Ω_{0}}))$ . For a sum of sinusoids, apply the rule to each term and add (superposition).

RESPONSE TO A CONSTANT (DC GAIN)

x (t) = K \Rightarrow y (t) = K H (0) = K \int_{- \infty}^{\infty} h (t) d t, x [n] = K \Rightarrow y [n] = K k \sum h [k]

The $ω = 0$ (or $Ω = 0$ ) case of the sinusoid rule. For $x [n] = (- 1)^{n}$ use $Ω = π$ : $y [n] = H (e^{j π}) (- 1)^{n}$ , with $H (e^{j π}) = \sum_{k} (- 1)^{k} h [k]$ .

WORKED EXAMPLE

An LTI system has $h (t) = 2 e^{- 2 t} u (t)$ . Find its output for the input $x (t) = 3 + 4 cos 2 t$ , applied for all time.

Solution

The system is causal and stable ( $\int ∣ h ∣ d t = 1$ ), so $H (j ω)$ exists: $H (j ω) = \int_{0}^{\infty} 2 e^{- 2 t} e^{- j ω t} d t = \frac{2}{2 + j ω}$ .
DC term: $H (0) = 1$ , so the constant 3 passes as 3.
At $ω_{0} = 2$ rad/s: $H (j 2) = \frac{2}{2 + j 2} = \frac{1}{1 + j}$ , so $∣ H (j 2) ∣ = \frac{1}{2}$ and $∠ H (j 2) = - tan^{- 1} (1) = - \frac{π}{4}$ .
The cosine term becomes $4 \cdot \frac{1}{2} cos (2 t - \frac{π}{4}) = 22 cos (2 t - \frac{π}{4})$ .
Add the two responses (superposition).

Answer: $y (t) = 3 + 22 cos (2 t - π /4) \approx 3 + 2.828 cos (2 t - 4 5^{\circ})$ .

EXAM TIP

Sinusoids in, the same frequencies out. $e^{j ω_{0} t}$ is an eigenfunction of every LTI system, but for a real $h$ , $cos ω_{0} t$ is one only when $H (j ω_{0})$ is real; otherwise its phase shifts. An LTI system cannot create a frequency that is not in its input. If the output contains a new frequency, the system is nonlinear or time-varying: $x^{2}$ turns $cos ω_{0} t$ into a DC term plus a term at $2 ω_{0}$ , and $x (t) cos ω_{c} t$ moves $ω_{0}$ to $ω_{c} \pm ω_{0}$ .

EXAM TIP

Eternal versus switched-on sinusoid. The rule above gives the whole output only when the sinusoid has been applied for all time. For $cos (ω_{0} t) u (t)$ into a causal, stable system it gives only the steady-state output; a transient that decays to zero is added to it. A zero of $H$ on the $j ω$ -axis (or on the unit circle) blocks that frequency completely, so such a system cannot be invertible.

6. Autocorrelation and cross-correlation

Correlation measures how alike two signals are as one is slid past the other. Unlike convolution, neither signal is flipped. Convention used here: $R_{x y} (τ) = \int x (t + τ) y^{*} (t) d t$ . Some texts define the cross-correlation as $\int x (t) y^{*} (t + τ) d t$ , which equals $R_{x y} (- τ)$ (and $R_{y x}^{*} (τ)$ ) in this convention. When a question states its own definition, use that one. For real signals, drop the conjugates. The first formulas are for energy signals. Power signals use a time average, given further down.

CROSS-CORRELATION (ENERGY SIGNALS)

R_{x y} (τ) = \int_{- \infty}^{\infty} x (t + τ) y^{*} (t) d t = \int_{- \infty}^{\infty} x (t) y^{*} (t - τ) d t = x (τ) * y^{*} (- τ)

Correlation is convolution with the second signal reversed and conjugated. Because of the reversal it is not commutative: $R_{y x} (τ) = R_{x y}^{*} (- τ)$ .

AUTOCORRELATION (ENERGY SIGNALS)

R_{xx} (τ) = \int_{- \infty}^{\infty} x (t + τ) x^{*} (t) d t = x (τ) * x^{*} (- τ)

For a real, even signal, $x (- τ) = x (τ)$ , so $R_{xx} = x * x$ .

DISCRETE-TIME CORRELATION

R_{x y} [m] = n = - \infty \sum \infty x [n + m] y^{*} [n] = x [m] * y^{*} [- m], R_{xx} [m] = n = - \infty \sum \infty x [n + m] x^{*} [n]

$m$ is the integer lag. The length rule of convolution applies: sequences of lengths $N_{1}$ and $N_{2}$ give $N_{1} + N_{2} - 1$ lags, so the autocorrelation of an $N$ -sample sequence has $2 N - 1$ lags.

PROPERTIES OF AUTOCORRELATION

R_{xx} (0) R_{xx} (- τ) ∣ R_{xx} (τ) ∣ \int_{- \infty}^{\infty} R_{xx} (τ) d τ = \int_{- \infty}^{\infty} ∣ x (t) ∣^{2} d t = E_{x} = R_{xx}^{*} (τ) \leq R_{xx} (0) = \int_{- \infty}^{\infty} x (t) d t^{2}

For a real signal, $R_{xx}$ is real and even, and its largest value is at $τ = 0$ . The last line is the area property of convolution; it needs $x$ absolutely integrable.

PROPERTIES OF CROSS-CORRELATION

R_{x y} (τ) ∣ R_{x y} (τ) ∣^{2} = R_{y x}^{*} (- τ) \leq E_{x} E_{y}

The inequality is the Cauchy-Schwarz inequality. Unlike $R_{xx}$ , the cross-correlation need not peak at $τ = 0$ .

CORRELATION AND ENERGY SPECTRAL DENSITY

R_{xx} (τ) \leftrightarrow ∣ X (j ω) ∣^{2}, R_{x y} (τ) \leftrightarrow X (j ω) Y^{*} (j ω), E_{x} = \frac{1}{2 π} \int_{- \infty}^{\infty} ∣ X (j ω) ∣^{2} d ω

$∣ X (j ω) ∣^{2}$ is the energy spectral density. In the $f$ convention, $E_{x} = \int ∣ X (f) ∣^{2} df$ (Parseval). Discrete time: $R_{xx} [m] \leftrightarrow ∣ X (e^{j Ω}) ∣^{2}$ and $E_{x} = \sum_{n} ∣ x [n] ∣^{2} = \frac{1}{2 π} \int_{- π}^{π} ∣ X (e^{j Ω}) ∣^{2} d Ω$ . Because $∣ X ∣^{2}$ discards phase, $x (t)$ and every shifted copy $x (t - t_{0})$ have the same autocorrelation.

AUTOCORRELATION OF POWER SIGNALS

R_{xx} (τ) = T \to \infty lim \frac{1}{T} \int_{- T /2}^{T /2} x (t + τ) x^{*} (t) d t, R_{xx} (0) = P_{x}

For a periodic signal of period $T_{0}$ , average over one period; $R_{xx}$ is then periodic with period $T_{0}$ and returns to its peak $R_{xx} (0)$ at every multiple of $T_{0}$ . The cross-correlation of power signals uses the same average with $y^{*} (t)$ in place of $x^{*} (t)$ . The power spectral density is the Fourier transform of the autocorrelation, $S_{xx} (ω) = F {R_{xx} (τ)}$ , and $P_{x} = \frac{1}{2 π} \int_{- \infty}^{\infty} S_{xx} (ω) d ω$ . Example: $A cos (ω_{0} t + ϕ)$ has $R_{xx} (τ) = \frac{A ^{2}}{2} cos ω_{0} τ$ , whatever $ϕ$ is, so $R_{xx} (0) = A^{2} /2$ is its average power and $S_{xx} (ω) = \frac{π A ^{2}}{2} [δ (ω - ω_{0}) + δ (ω + ω_{0})]$ .

CORRELATION THROUGH AN LTI SYSTEM

R_{y x} (τ) = h (τ) * R_{xx} (τ), R_{y y} (τ) = R_{hh} (τ) * R_{xx} (τ), R_{hh} (τ) = h (τ) * h^{*} (- τ)

For $y = h * x$ , in the convention above. In frequency, $S_{y x} (ω) = H (j ω) S_{xx} (ω)$ and $S_{y y} (ω) = ∣ H (j ω) ∣^{2} S_{xx} (ω)$ , where $S$ is the energy (or power) spectral density.

Feature	Convolution $x * h$	Correlation $R_{x y}$
Definition	$\int x (τ) h (t - τ) d τ$	$\int x (t + τ) y^{*} (t) d t$
Reversal	One signal is flipped	No flip, only conjugation
Commutative	Yes	No: $R_{y x} (τ) = R_{x y}^{*} (- τ)$
Fourier transform	$X (j ω) H (j ω)$	$X (j ω) Y^{*} (j ω)$
Support, for $x$ on $[a_{1}, b_{1}]$ and the other signal on $[a_{2}, b_{2}]$	$[a_{1} + a_{2}, b_{1} + b_{2}]$	$[a_{1} - b_{2}, b_{1} - a_{2}]$
Typical use	Output of an LTI system	Similarity, delay estimation, energy and power

WORKED EXAMPLE

Find the autocorrelation of $x (t) = 2 [u (t) - u (t - 3)]$ and check it against the properties.

Solution

$x (t + τ)$ is the pulse moved to occupy $[- τ, 3 - τ]$ . For $0 \leq τ \leq 3$ it overlaps $x (t)$ on $[0, 3 - τ]$ .
On the overlap the product is $2 \times 2 = 4$ , so $R_{xx} (τ) = 4 (3 - τ)$ for $0 \leq τ \leq 3$ , and $0$ for $τ > 3$ .
$x$ is real, so $R_{xx}$ is even: $R_{xx} (τ) = 4 (3 - ∣ τ ∣)$ for $∣ τ ∣ \leq 3$ , and zero otherwise. It is a triangle on $[- 3, 3]$ .
Checks: $R_{xx} (0) = 12 = 2^{2} \times 3 = E_{x}$ . The width is $3 + 3 = 6$ , as the convolution width rule requires. The area is $\frac{1}{2} \times 6 \times 12 = 36 = (\int x d t)^{2} = 6^{2}$ .

Answer: $R_{xx} (τ) = 4 (3 - ∣ τ ∣)$ for $∣ τ ∣ \leq 3$ and zero otherwise, with its peak $E_{x} = 12$ at $τ = 0$ .

EXAM TIP

Delay estimation depends on the convention. If $y (t) = x (t - t_{d})$ , then $R_{y x} (τ) = R_{xx} (τ - t_{d})$ peaks at $τ = t_{d}$ , while $R_{x y} (τ) = R_{xx} (τ + t_{d})$ peaks at $τ = - t_{d}$ . Check the question's definition before reading a delay off a peak. Autocorrelation alone cannot locate a signal in time, because $x (t)$ and $x (t - t_{0})$ have the same $R_{xx}$ .

PREVIOUS-YEAR QUESTIONS

31 questions from the EC, EE and IN papers

Try each question first, then open the answer. 3 show the official answer only. 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 2023 · GATE ECE 2022 · GATE ECE 2020 · GATE EE 2026 · GATE EE 2025 · GATE EE 2024 · GATE EE 2023 · GATE EE 2022 · GATE EE 2020 · GATE EE 2018 · GATE IN 2026 · GATE IN 2025 · GATE IN 2024 · GATE IN 2023 · GATE IN 2022 · GATE IN 2020

GATE 2026 EC Q.47

GATE 2026 · ECQ.47 · 2 marks · MCQ

Consider the discrete time system (S) with input $x [n]$ and output $y [n]$ as shown in the Figure. The two sub-systems represented by their impulse responses $h_{1} [n]$ and $h_{2} [n]$ are linear and time invariant.

Which of the following statements is necessarily TRUE?

GATE 2026 EC question 47: Block diagram. — Figure from the GATE 2026 EC paper. It is part of the question.

(A)
$S$ is causal.
(B)
$S$ is linear and time invariant.
(C)
$S$ is linear and time varying.
(D)
$S$ is non-linear.

Answer and solution

GATE 2026 EC Q.47: answer and solution

GATE 2026 EE Q.14

GATE 2026 · EEQ.14 · 1 mark · MCQ

Consider the following differential equation:

t^{2} \frac{d ^{2} y}{d t ^{2}} + 7 t \frac{d y}{d t} + 8 t y = 10 sin (t)

Which one of the following options is correct?

(A)
It is a linear differential equation
(B)
It is a nonlinear differential equation
(C)
It is a time-invariant differential equation
(D)
It is a second-order partial differential equation

Answer and solution

GATE 2026 EE Q.14: answer and solution

See the question in the paper

LTI systems

1. System properties and the test for each

2. Property tests on typical system equations

3. The impulse response describes an LTI system completely

4. Convolution: graphical method, properties and standard results

5. Eigenfunctions and the response to a sinusoid

6. Autocorrelation and cross-correlation

31 questions from the EC, EE and IN papers

5 questions to test yourself