MIT 6.300 — Lecture 9

Unit-Sample Response and Convolution

§ Lecture and Recitation: Unit-Sample Response and Convolution

It turns out that there's a simple classification of all LTI systems on DT signals.

Definition. Given two signals $a[n]$ and $b[n]$ , their convolution is $(a * b)[n] := \sum_{m = -\infty}^{\infty} a[m] \cdot b[n - m]$ .

Theorem. (LTI = Convolution) Suppose $\delta[n] \mapsto h[n]$ under an LTI system on DT signals. Then:

x[n] \mapsto \sum_{m = -\infty}^{\infty} x[m] \cdot h[n - m] = (x * h)[n].

Proof: Immediate from linearity and time-invariance after writing $x[n] = \sum_{m = -\infty}^{\infty} x[m] \delta[n - m]$ . $\blacksquare$

Definition. The unit-sample response of a system on DT signals is the image $h[n]$ of the Kronecker delta signal $\delta[n]$ .

It's not hard to generalize this to CT signals, too.

Definition. Given two signals $f(t)$ and $g(t)$ , their convolution is $(f * g)(t) := \int_{-\infty}^{\infty} f(\tau) \cdot g(t - \tau) \, \mathrm{d}\tau$ .

Definition. The impulse response of a system on CT signals is the image $h(t)$ of the Dirac delta signal $\delta(t)$ .

Theorem. (LTI = Convolution) Suppose $\delta(t) \mapsto h(t)$ under an LTI system on CT signals. Then:

x(t) \mapsto \int_{-\infty}^{\infty} x(\tau) \cdot h(t - \tau) \, \mathrm{d}\tau = (x * h)(t).

The point is that every LTI system is a convolution, defined solely by a single unit-sample / impulse response.