LTI properties; BIBO stability condition

Properties of LTI systems

For an LTI system with impulse response $h[n]$ (output $y=x*h$):

1. Commutative: $x*h=h*x$.
2. Associative: $(x*h_1)*h_2=x*(h_1*h_2)$ (cascade connection).
3. Distributive: $x*(h_1+h_2)=x*h_1+x*h_2$ (parallel connection).
4. Memoryless if and only if $h[n]=K\delta[n]$.
5. Causal if and only if $h[n]=0$ for $n<0$.
6. Stable (BIBO) if and only if $\sum_n|h[n]|<\infty$.
7. Invertible if an $h_i[n]$ exists with $h*h_i=\delta[n]$.
8. Step response: $s[n]=\sum_{k=-\infty}^{n}h[k]$.

Proof: absolutely summable $h[n]$ gives a stable system

A system is BIBO stable if every bounded input produces a bounded output.

Let the input be bounded: $|x[n]|\le B_x<\infty$ for all $n$. The output is

$$y[n]=\sum_{k=-\infty}^{\infty}h[k]\,x[n-k]$$

Take the magnitude and use the triangle inequality ($|\sum a_k|\le\sum|a_k|$):

$$\begin{aligned} |y[n]| &= \left|\sum_{k}h[k]\,x[n-k]\right| \le \sum_{k}|h[k]|\,|x[n-k]| \\ &\le B_x\sum_{k=-\infty}^{\infty}|h[k]| \end{aligned}$$

If $h[n]$ is absolutely summable, $\sum_k|h[k]|=S<\infty$, then

$$|y[n]|\le B_xS<\infty \quad\text{for all } n$$

so the output is bounded and the system is stable.

Converse (necessity): if $S=\infty$, choose the bounded input $x[n]=\mathrm{sgn}(h[-n])$ (values $\pm1$). Then $y[0]=\sum_k h[k]\,\mathrm{sgn}(h[k])=\sum_k|h[k]|=\infty$, so a bounded input gives an unbounded output. Hence absolute summability is necessary and sufficient for BIBO stability of an LTI system.

Examples

• $h[n]=(0.5)^n u[n]$: $S=\frac{1}{1-0.5}=2<\infty$, stable.
• $h[n]=u[n]$ (accumulator): $S=\sum_{n\ge0}1=\infty$, unstable (input $u[n]$ gives $y[n]=(n+1)u[n]$, which grows without bound).