Discrete-time systems review

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An LTI system is fully described by its impulse response $h[n]$: the output is convolution $y[n] = \sum_k h[k]\,x[n-k]$. Causality: $h[n] = 0$ for $n < 0$; BIBO stability: $\sum |h[n]|$ converges.

Example

$h = [1,2,1]$ on a unit step gives $y = 1, 3, 4, 4, \ldots$ — it settles at $\sum h = 4$, the DC gain $H(1)$ ✓.

{exercise:uni-ee1201-settle}

Luyện tập

The filter h = [1, 2, 1] is fed a unit step. Where does the output settle?

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English

The averager $y[n] = (y[n-1] + x[n])/2$ has $h[n] = 0.5^{n+1}$: its step response is $1 - 2^{-n}$ (0.5, 0.75, 0.875, …), because its one pole sits at $z = 0.5$.
Exponentials are eigenfunctions: $H(e^{j\omega}) = \sum h[n]e^{-j\omega n}$ is the frequency response — the DTFT of the taps.

{exercise:uni-ee1201-pole} {exercise:uni-ee1201-bibo}

Luyện tập

The averager y[n] = (y[n−1] + x[n])/2 has a pole at z = 0.5, so its step response is —

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