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BIBO Stability

Module by: Richard Baraniuk. E-mail the author

Summary: An overview of BIBO stability.

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

But this can happen with -length system.

Example 1

Figure 1
Figure 1 (fig1.png)
yn= k =hnkxk= k =1×1= y n k h n k x k k 1 1 for all n n

l p Z l p 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.
Figure 2
Figure 2 (fig2.png)
yn= k =hnkxk y n k h n k x k <n< n
Figure 3
Figure 3 (fig3.png)
bounded input/output
x, y l Z x, y l
So, the question is what is the condition on h h such that (y=h*x) l Z y h x l with x l Z x l ?


An LSI system is BIBO stable if and only if the impulse response h l 1 Z h l 1 .
That is, if x l x l and h l 1 Z h l 1 then (y=h*x) l Z y h x l


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

Part 1

Sufficiency: If h l 1 Z h l 1 then system is BIBO. y=max nZ nZ |yn| y n 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(|xnk|x) x l x n 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 x h l 1 Z(h1<) h l 1 1 h Therefore |yn|< y n Therefore max nZ nZ |yn|< n y n Therefore y< y and the bounded input x x created a bounded output y y.

Part 2:

Necessity: If h l 1 Z h l 1 then the system is not BIBO.


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


We only have 1 x x to show this.
Some creativity suggests the following signal: xn={hn¯|hn|  if  hn0 0   if  h=0 x n h n h n h n 0 0 h 0 For real h h, x x is the sign of time-reversed h h.
Figure 4
Figure 4 (fig4.png)


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 =xkhk y 0 k x k h k 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=h1= y 0 1 h by assumption.

y l Z y l This means that it is not bounded system is not BIBO.

BIBO Stablility Bottom Line

Figure 5
Figure 5 (fig5.png)
x l Z x l bounded and y l Z y l bounded.

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

Example 2: Moving Window Filter (FIR)

Is it BIBO stable?

Example 3: Recursive Filter (IIR)

Is it BIBO stable?

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