Stability

### Definition

Impulse response - $h[n]$ or $h(t)$

(1)
\begin{align} y[n]=\sum_{k=-\infty}^{\infty}h[k]x[n-k] \nonumber \end{align}

We want $y[n]$ to be bounded if $x[n]$ is bounded.

(2)
\begin{align} |y[n]|=|\sum h[k]x[n-k]| \leq \sum |h[k]x[n-k]| \nonumber \end{align}

Therefore,

(3)
\begin{align} |y[n]|=\sum_{k=-\infty}^{\infty} |h[k]x[n-k]| \nonumber \end{align}

Since $x[n]$ is bounded , $|x[n-k]|< M_x$.So,

(4)
\begin{align} |y[n]| \leq M_x \big(\sum_{k=-\infty}^{\infty}|h[k]|\big) \end{align}

If $\sum_{n=-\infty}^{\infty}|h[n]| < M_h$ (is bounded) $\int_{n=-\infty}^{\infty}|h(t)| < M_h$ or , then the LTI System is stable.

### Examples

1. $h(t)=u(t)$

$\int_{n=-\infty}^{\infty}|h(t)|dt=\int_{0}^{\infty}1.dt$ is not bounded $\Rightarrow$ Unstable.
2. $h[n]=\big(\frac{1}{2}\big)^n u[n]$
$\sum_{n=-\infty}^{\infty}|h[n]|=\sum_{n=-\infty}^{\infty}\big(\frac{1}{2}\big)^n=\frac{1}{1-\frac{1}{2}}=2 \Rightarrow$ bounded $\Rightarrow$ The system is stable._

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page revision: 1, last edited: 26 Aug 2016 15:36