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# BIBO Stability

Module by: Richard Baraniuk. E-mail the author

Summary: An overview of BIBO stability.

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With finite length signals N , it is impossoble to make the system output blow up.

But this can happen with -length system.

## Example 1

` l p z l p z spaces are useful to understand what's going on...

Definition 1: Bounded-Input Bounded-Output (BIBO) Stable
When, in an LSI system, for any bounded input the output is always bounded.
yn= k =hnkxk y n k h n k x k <n< n bounded input/output
xy l x x y l x
(1)
So, the question is condition on h h such that (y=h*x) l x y h x l x with x l x x l x ?

## Note:

An LSI system is BIBO stable if and only if the impulse response h l 1 z h l 1 z .
ie: If x l z x l z and h l 1 z h l 1 z then (y=h*x) l x y h x l x

## Proof

Not that the fact is true "if and only if" there are two sides to the proof.

### Part 1

Sufficiency ie: if h l 1 z h l 1 z then system is BIBO. y=max nz nz |yn| y n z y n So look at |yn| y n and show it is < .

Use convolution formula |yn|=| k =xnkhk| y n k x n k h k k =|xnk||hk| k x n k h k Where x l z(|hk|x) x l z h k x k =|hk|x k h k x Since x x is constant, =x k =|hk| x k h k Where k =|hk|=h1 k h k 1 h Therefore, |yn|xh1 y n x 1 h But x l z(x<) x l z x h l 1 z(h1<) h l 1 z 1 h Therefore |yn|< y n Therefore max nz nz |yn|< n z y n Therefore y< y and the bounded input x x created a bounded output y y.

### Part 2:

necessity ie: if h l 1 z h l 1 z then the system is not BIBO.

#### Note:

This part is tricky!!!
So, assume that h1= 1 h and show that there exists a signal x l z x l z such that y l z y l z .

#### Note:

We only have 1 x x to show this.
Some creativity suggests the following signal: xk={hk¯|hk|  if  hk0 0   if  h=0 x k h k h k h k 0 0 h 0 For real h h, x x is the sign of time-reversed h h.

#### Note:

x x takes on values -1, 1, 0, and so x=1< x 1 ie: bounded input.
Compute yn y n at n=0 n 0 : yn= k =xnkhk y n k x n k h k y0= k =xkhku y 0 k x k h ku xk=|hk¯hk| x k h k h k = k =hk¯hk|hk| k h k h k h k = k |hk|2|hk| k h k 2 h k = k =|hk| k h k y0=h= y 0 h by assumption.

y l z y l z ie: not bounded system is not BIBO.

## BIBO Stablility Bottom Line

x l z x l z bounded and y l z y l z bounded.

Need: h l 1 z h l 1 z ie: h1< 1 h and h h must decay fast enough as |n| n .

BIBO stable?

BIBO stable?

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