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Unit 2 · Time-Domain Analysis

Convolution, without the dread

Convolution looks scary written down, but the idea is simple: it is how a system echoes its input.

One response to rule them all

An LTI system (linear, time-invariant) is fully described by a single thing: its impulse response h(t)h(t), the output you get when you poke it with one sharp impulse δ(t)\delta(t).

Here is the trick. Any input signal can be seen as a dense train of impulses, one at each instant, each scaled by the signal’s value there. Because the system is linear, each of those input impulses produces its own scaled copy of hh, and because it is time-invariant, that copy is just hh shifted to where the impulse was. The output is all those echoes added up:

y(t)=(xh)(t)=x(τ)h(tτ)dτy(t) = (x * h)(t) = \int_{-\infty}^{\infty} x(\tau)\, h(t - \tau)\, d\tau

Read it literally: for every past input value x(τ)x(\tau), add its echo h(tτ)h(t - \tau). That is all the integral is doing.

The flip-slide recipe

To compute yy at one time tt by hand:

  1. Flip one signal, usually h(τ)h(τ)h(\tau) \to h(-\tau).
  2. Slide it to position tt, giving h(tτ)h(t - \tau).
  3. Multiply it point by point with x(τ)x(\tau).
  4. Sum (integrate) the overlap. That number is y(t)y(t).

Slide all the way through and you trace out the whole output.

The example to remember

A rectangular pulse convolved with a rectangular pulse is a triangle. As you slide one box across the other, the overlap grows, peaks when they line up, then shrinks:

x(t)*h(t)=y = x ∗ h

If you remember just one convolution picture, make it this one. It shows the whole story: the output is widest and largest where the two signals overlap the most.

Why bother

Two reasons. First, it means you never need the system’s internals, only h(t)h(t). Second, the moment you move to the Laplace or Fourier domain, this whole integral collapses into a plain multiplication, Y=XHY = X \cdot H. That is the real reason those transforms exist.