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BIBO stability

BIBO Stability is a form of stability for signals and systems. BIBO stands for Bounded Input/Bounded Output. If a system is BIBO stable (e.g., a low-pass filter) then the output will be bounded provided that the input to the system is bounded.

Contents

Time domain condition

Continuous-time sufficient condition

In continuous-time, the condition for BIBO stability is that the impulse response must be absolutely integrable, i.e., its L1 norm must exist.

\int_{-\infty}^{\infty}{\left|\mathbf{h}(t)\right|dt} = \| h \|_{1} < \infty

Discrete-time sufficient condition

In discrete-time, the condition for BIBO stability is that the impulse response must be absolutely summable, i.e., its \ell^1 norm must exist.

\sum_{n=-\infty}^{\infty}{\left|\mathbf{h}(n)\right|} = \| h \|_{1} < \infty

Proof

Given a discrete, linear, time-invariant system with impulse response \mathbf{h}(n) the relationship between the input \mathbf{x}(n) and the output \mathbf{y}(n) is

\mathbf{y}(n) = \mathbf{h}(n) * \mathbf{x}(n)

where * denotes convolution. Then it follows by the definition of convolution

\mathbf{y}(n) = \sum_{k=-\infty}^{\infty}{\mathbf{h}(k) \mathbf{x}(n-k)}

Let \| x \|_{\infty} be the maximum value of \mathbf{x}(n), i.e., the infinity norm.

\left|\mathbf{y}(n)\right| = \left|\sum_{k=-\infty}^{\infty}{\mathbf{h}(n-k) \mathbf{x}(k)}\right| \le \sum_{k=-\infty}^{\infty}{\left|\mathbf{h}(n-k)\right| \left|\mathbf{x}(k)\right|} (by the triangle inequality)
\le \sum_{k=-\infty}^{\infty}{\left|\mathbf{h}(n-k)\right| \| x \|_{\infty}}= \| x \|_{\infty} \sum_{k=-\infty}^{\infty}{\left|\mathbf{h}(n-k)\right|}
= \| x \|_{\infty} \sum_{k=-\infty}^{\infty}{\left|\mathbf{h}(k)\right|}

If \mathbf{h}(n) is BIBO stable, then \sum_{k=-\infty}^{\infty}{\left|\mathbf{h}(k)\right|} = \| h \|_1  < \infty and

\| x \|_{\infty} \sum_{k=-\infty}^{\infty}{\left|\mathbf{h}(k)\right|} = \| x \|_{\infty} \| h \|_1

So if \| x \|_{\infty} < \infty (i.e., it is bounded) then \left|\mathbf{y}(n)\right| is bounded as well because \| x \|_{\infty} \| h \|_1 < \infty.


The proof for continuous-time follows the same arguments.

Frequency domain condition

Continuous signals

For a causal, rational, 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. (Largest here is defined so that the real part of the largest pole is greater than the real part of any other 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 :

\int_{-\infty}^{\infty}{\left|\mathbf{h}(t)\right| dt}  = \int_{-\infty}^{\infty}{\left|\mathbf{h}(t)\right| \left| e^{-j \omega t} \right| dt}  = \int_{-\infty}^{\infty}{\left|\mathbf{h}(t) (1 \cdot e)^{-j \omega t} \right| dt} =


\int_{-\infty}^{\infty}{\left|\mathbf{h}(t) (e^{\sigma + j \omega})^{- t} \right| dt}  = \int_{-\infty}^{\infty}{\left|\mathbf{h}(t) e^{-s t} \right| dt} \mbox{ where } s = \sigma + j \omega \mbox{ and Re}(s) = \sigma = 0

The region of convergence must therefore include the imaginary axis.

Discrete signals

For a causal, rational, 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 derivation:

\sum_{n = -\infty}^{\infty}{\left|\mathbf{h}(n)\right|}  = \sum_{n = -\infty}^{\infty}{\left|\mathbf{h}(n)\right| \left| e^{-j \omega n} \right|}  = \sum_{n = -\infty}^{\infty}{\left|\mathbf{h}(n) (1 \cdot e)^{-j \omega n} \right|} =


\sum_{n = -\infty}^{\infty}{\left|\mathbf{h}(n) (r e^{j \omega})^{-n} \right|}  = \sum_{n = -\infty}^{\infty}{\left|\mathbf{h}(n) z^{- n} \right|} \mbox{ where } z = r e^{j \omega} \mbox{ and } r = |z| = 1

The region of convergence must therefore include the unit circle.

See also

LTI system theory

10-26-2009 08:16:03
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