BIBO stability

For a continuous time linear time-invariant (LTI) system, the condition for BIBO stability is that the impulse response, ${\displaystyle h(t)}$ , be absolutely integrable, i.e., its L¹ norm exists.

${\displaystyle \int _{-\infty }^{\infty }\left|h(t)\right|\,{\mathord {\operatorname {d} }}t=\|h\|_{1}\in \mathbb {R} }$

For a discrete time LTI system, the condition for BIBO stability is that the impulse response be absolutely summable, i.e., its ${\displaystyle \ell ^{1}}$ norm exists.

${\displaystyle \ \sum _{n=-\infty }^{\infty }|h[n]|=\|h\|_{1}\in \mathbb {R} }$

Given a discrete time LTI system with impulse response ${\displaystyle \ h[n]}$ the relationship between the input ${\displaystyle \ x[n]}$ and the output ${\displaystyle \ y[n]}$ is

${\displaystyle \ y[n]=h[n]*x[n]}$

where ${\displaystyle *}$ denotes convolution. Then it follows by the definition of convolution

${\displaystyle \ y[n]=\sum _{k=-\infty }^{\infty }h[k]x[n-k]}$

Let ${\displaystyle \|x\|_{\infty }}$ be the maximum value of ${\displaystyle \ |x[n]|}$, i.e., the ${\displaystyle L_{\infty }}$-norm.

${\displaystyle \left|y[n]\right|=\left|\sum _{k=-\infty }^{\infty }h[n-k]x[k]\right|}$

${\displaystyle \leq \sum _{k=-\infty }^{\infty }\left|h[n-k]\right|\left|x[k]\right|}$ (by the triangle inequality)

${\displaystyle {\begin{aligned}&\leq \sum _{k=-\infty }^{\infty }\left|h[n-k]\right|\|x\|_{\infty }\\&=\|x\|_{\infty }\sum _{k=-\infty }^{\infty }\left|h[n-k]\right|\\&=\|x\|_{\infty }\sum _{k=-\infty }^{\infty }\left|h[k]\right|\end{aligned}}}$

If ${\displaystyle h[n]}$ is absolutely summable, then ${\displaystyle \sum _{k=-\infty }^{\infty }{\left|h[k]\right|}=\|h\|_{1}\in \mathbb {R} }$ and

${\displaystyle \|x\|_{\infty }\sum _{k=-\infty }^{\infty }\left|h[k]\right|=\|x\|_{\infty }\|h\|_{1}}$

So if ${\displaystyle h[n]}$ is absolutely summable and ${\displaystyle \left|x[n]\right|}$ is bounded, then ${\displaystyle \left|y[n]\right|}$ is bounded as well because ${\displaystyle \|x\|_{\infty }\|h\|_{1}\in \mathbb {R} }$.

The proof for continuous-time follows the same arguments.

For a rational and continuous-time system, the condition for stability is that the region of convergence (ROC) of the Laplace transform includes the imaginary axis. When the system is causal, the ROC is the open region to the right of a vertical line whose abscissa is the real part of the "largest pole", or the pole that has the greatest real part of any pole in the system. The real part of the largest pole defining the ROC is called the abscissa of convergence. Therefore, all poles of the system must be in the strict left half of the s-plane for BIBO stability.

This stability condition can be derived from the above time-domain condition as follows:

${\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }\left|h(t)\right|\,dt&=\int _{-\infty }^{\infty }\left|h(t)\right|\left|e^{-j\omega t}\right|\,dt\\&=\int _{-\infty }^{\infty }\left|h(t)(1\cdot e)^{-j\omega t}\right|\,dt\\&=\int _{-\infty }^{\infty }\left|h(t)(e^{\sigma +j\omega })^{-t}\right|\,dt\\&=\int _{-\infty }^{\infty }\left|h(t)e^{-st}\right|\,dt\end{aligned}}}$

where ${\displaystyle s=\sigma +j\omega }$ and ${\displaystyle \operatorname {Re} (s)=\sigma =0.}$

The region of convergence must therefore include the imaginary axis.

For a rational and discrete time system, the condition for stability is that the region of convergence (ROC) of the z-transform includes the unit circle. When the system is causal, the ROC is the open region outside a circle whose radius is the magnitude of the pole with largest magnitude. Therefore, all poles of the system must be inside the unit circle in the z-plane for BIBO stability.

This stability condition can be derived in a similar fashion to the continuous-time derivation:

${\displaystyle {\begin{aligned}\sum _{n=-\infty }^{\infty }\left|h[n]\right|&=\sum _{n=-\infty }^{\infty }\left|h[n]\right|\left|e^{-j\omega n}\right|\\&=\sum _{n=-\infty }^{\infty }\left|h[n](1\cdot e)^{-j\omega n}\right|\\&=\sum _{n=-\infty }^{\infty }\left|h[n](re^{j\omega })^{-n}\right|\\&=\sum _{n=-\infty }^{\infty }\left|h[n]z^{-n}\right|\end{aligned}}}$

where ${\displaystyle z=re^{j\omega }}$ and ${\displaystyle r=|z|=1}$.

The region of convergence must therefore include the unit circle.