Convolution and LTI Systems for GATE: Formulas, Properties and Solved Examples

Convolution for GATE made simple: LTI systems, flip-and-slide, key formulas and properties, causality and stability, common traps, and practice questions with answers.

Every year I see students who can recite the convolution integral in their sleep, and then freeze when a GATE question hands them two rectangles and asks for the output at $t = 3$. The formula was never the hard part. Picturing what it does is.

So in this post we'll do convolution the way I do it on the board: draw, flip, slide, add. By the end you should be able to solve most GATE convolution questions in a couple of minutes, and many of them without writing a single integral.

This is a core topic of Signals and Systems in GATE EC, and it also comes up in EE and IN. Everything here builds on it: Laplace, Fourier and Z-transforms all make more sense once convolution does.

First, what is an LTI system?

LTI stands for linear time-invariant. Two tests, and a system must pass both.

Linear means superposition works. If input $x1$ gives $y1$ and $x2$ gives $y2$, then

Time-invariant means the system behaves the same today as tomorrow. Delay the input by $t0$ and the output is simply delayed by $t0$, with nothing else changing:

A quick way to spot trouble: if you see $t$ multiplying something, as in $y(t) = t\,x(t)$, or a squeezed time axis, as in $y(t) = x(2t)$, the system is usually not time-invariant. If you see a constant added, as in $y(t) = x(t) + 3$, it is not linear, because a zero input should give a zero output.

Why the impulse response is all you need

Here's the beautiful part. Feed an LTI system a unit impulse $\delta(t)$ and call what comes out the impulse response, $h(t)$. That one signal tells you everything about the system.

Why? Any input can be thought of as a long row of tiny, scaled, shifted impulses. Time invariance says each shifted impulse produces a shifted copy of $h(t)$. Linearity says we can add all those copies up. That sum is convolution:

For discrete-time signals the integral becomes a sum:

Convolution in four steps: draw, flip, slide, add

Read the integral slowly and it tells you exactly what to do.

Draw $x(\tau)$ and $h(\tau)$ on a $\tau$ axis. Think of $t$ as a fixed number for now. Flip $h$ to get $h(-\tau)$. Slide it right by $t$ to get $h(t - \tau)$. Multiply the two curves point by point and find the area of the product. That area is $y(t)$.

Then let $t$ move from $-\infty$ to $\infty$ and watch how the overlap changes. Most of the work is spotting where the overlap starts and stops. Those points become the limits of your integral.

A worked example you should know by heart

Take $x(t) = u(t)$ and $h(t) = e^{-2t}\,u(t)$, the two signals in Fig. 1.

For $t < 0$ the flipped $h$ hasn't reached the step yet, so there's no overlap and $y(t) = 0$. For $t \ge 0$ the overlap runs from $\tau = 0$ to $\tau = t$:

So $y(t) = \tfrac{1}{2}\left(1 - e^{-2t}\right)u(t)$. Notice that the output rises smoothly and settles at $\tfrac{1}{2}$, which is exactly the area under $h(t)$. That's no accident, and it's a shortcut we'll use in a moment.

The discrete version, done with a table

Discrete convolution is even friendlier, because you can do it without flipping anything. Take

with both starting at $n = 0$. Write $x$ along the top, $h$ down the side, fill in the products, and add along the diagonals:

$x[0] = 1$ $x[1] = 2$ $x[2] = 1$ ------------ $h[0] = 1$ 1 2 1 $h[1] = 1$ 1 2 1

The diagonals give $y[n] = \{1,\ 3,\ 3,\ 1\}$ for $n = 0$ to $3$.

Two quick checks catch most mistakes here. The output has $3 + 2 - 1 = 4$ samples, and the sum of $y$ is $8$, which equals the sum of $x$ times the sum of $h$, $4 \times 2$.

Properties that save you time in the exam

These are the ones I'd actually memorise.

Commutative: $x h = h x$. Flip whichever signal is simpler. Associative: $(x h1) h2 = x (h1 h2)$. Two systems in series act like one system with impulse response $h1 h2$. Distributive: $x (h1 + h2) = x h1 + x h2$. Two systems in parallel act like one system with impulse response $h1 + h2$. Shift: $x(t) \delta(t - t0) = x(t - t0)$. Convolving with a shifted impulse just shifts the signal. Area: the area under $y$ equals the area under $x$ times the area under $h$. For sequences, the sum of $y$ equals the sum of $x$ times the sum of $h$. Width: if $x$ lasts $T1$ seconds and $h$ lasts $T2$, then $y$ lasts $T1 + T2$. In discrete time, lengths $N$ and $M$ give $N + M - 1$ samples. Start and end: $y$ starts where $x$ and $h$ start added together, and ends where their ends add together.

The series and parallel rules are easiest to remember as a picture. The chain below behaves exactly like one system with impulse response $h1 h2$:

Two systems side by side, with their outputs added, behave like one system with impulse response $h1 + h2$:

Causality and stability from h(t)

GATE loves asking these, and both come straight from the impulse response.

An LTI system is causal if it never reacts before the input arrives, which means

It is BIBO stable (every bounded input gives a bounded output) if the impulse response is absolutely integrable:

So $h(t) = e^{-2t}u(t)$ is causal and stable, since its area is $\tfrac{1}{2}$. But $h(t) = u(t)$, a pure integrator, is causal and unstable: feed it a constant and the output grows forever.

Common traps in GATE questions

I've watched good students lose marks to every one of these.

Forgetting the $u(t)$ in the answer. $\tfrac{1}{2}(1 - e^{-2t})$ without the $u(t)$ is wrong for $t < 0$. Wrong limits. The limits come from where the flipped signal overlaps the other one, not from $-\infty$ to $\infty$. Sketch it first, every time. Two equal rectangles give a triangle, unequal ones give a trapezoid. The flat top of the trapezoid is as wide as the difference in widths. $u(t) u(t) = t\,u(t)$, a ramp, not another step. The length of a discrete result is $N + M - 1$, not $N + M$. "Linear" doesn't mean "straight line". $y(t) = x(t) + 3$ looks simple but fails the linearity test. $y(t) = t\,x(t)$ looks odd but is linear, just not time-invariant. Stable is about $h$, not $h$. Always check the absolute value, especially when $h$ alternates in sign.

Practice questions

Try these before looking at the answers. Give yourself about two minutes each, which is roughly what you'll have in the exam.

Find $y[n] = x[n] h[n]$ for $x[n] = \{1,\ 2,\ 3\}$ and $h[n] = \{1,\ -1\}$, both starting at $n = 0$. $x(t) = 1$ for $t < 1$ and $0$ elsewhere. Find $x(t) x(t)$. What is its peak value, and over what range is it non-zero? Find $e^{-t}u(t) e^{-2t}u(t)$. Is the system with $h[n] = (0.5)^n\,u[n]$ stable? What about $h[n] = u[n]$? Is $y(t) = x(t)\cos(\omega0 t)$ linear? Is it time-invariant?

Answers

$y[n] = \{1,\ 1,\ 1,\ -3\}$. Check: the sum of $y$ is $0$, and the sum of $x$ times the sum of $h$ is $6 \times 0 = 0$. A triangle, zero outside $-2 < t < 2$, with a peak of $2$ at $t = 0$. The width is $2 + 2 = 4$ and the area is $2 \times 2 = 4$, which match the width and area rules. $\left(e^{-t} - e^{-2t}\right)u(t)$. For $t \ge 0$, $\int0^t e^{-\tau} e^{-2(t - \tau)}\,d\tau = e^{-2t}\left(e^{t} - 1\right)$. The first is stable, because $\sum (0.5)^n = 2$. The second is unstable, because $\sum u[n]$ never stops growing. Linear, yes. Time-invariant, no: delaying the input doesn't delay the $\cos(\omega0 t)$ along with it.

If you got four or five right, you're in good shape for this topic. If question 2 caught you out, go back to Fig. 1 and do the flip and slide by hand with two rectangles. It clicks after one or two tries. Leave a comment with the question that gave you the most trouble, and I'll pick it up in a later post.