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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

Figure 1
Figure 1 (fig1.png)

` 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.
Figure 2
Figure 2 (fig2.png)
y⁢n=∑ k =−∞∞h⁢n−k⁢x⁢k y n k h n k x k −∞<n<∞ n
Figure 3
Figure 3 (fig3.png)
bounded input/output →
x⁢y∈ 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 n∈z n∈z |y⁢n| y n z y n So look at |y⁢n| y n and show it is <∞ .

Use convolution formula |y⁢n|=|∑ k =−∞∞x⁢n−k⁢h⁢k| y n k x n k h k ≤∑ k =−∞∞|x⁢n−k|⁢|h⁢k| k x n k h k Where x∈ l ∞ ⁢z→(|h⁢k|≤∥x∥∞) x l z h k x ≤∑ k =−∞∞|h⁢k|⁢∥x∥∞ k h k x Since ∥x∥∞ x is constant, =∥x∥∞⁢∑ k =−∞∞|h⁢k| x k h k Where ∑ k =−∞∞|h⁢k|=∥h∥1 k h k 1 h Therefore, |y⁢n|≤∥x∥∞⁢∥h∥1 y n x 1 h But x∈ l ∞ ⁢z→(∥x∥∞<∞) x l z x h∈ l 1 ⁢z→(∥h∥1<∞) h l 1 z 1 h Therefore |y⁢n|<∞ y n Therefore max n∈z n∈z |y⁢n|<∞ 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 ∥h∥1=∞ 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: x⁢k={h⁢−k¯|h⁢−k|  if  h⁢k≠0 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.
Figure 4
Figure 4 (fig4.png)

Note:

x x takes on values -1, 1, 0, and so ∥x∥∞=1<∞ x 1 ie: bounded input.
Compute y⁢n y n at n=0 n 0 : y⁢n=∑ k =−∞∞x⁢n−k⁢h⁢k y n k x n k h k y⁢0=∑ k =−∞∞x⁢−k⁢h⁢ku y 0 k x k h ku x⁢−k=|h⁢−k¯h⁢−k| x k h k h k =∑ k =−∞∞h⁢−k¯⁢h⁢k|h⁢−k| k h k h k h k =∑ k |h⁢k|2|h⁢−k| k h k 2 h k =∑ k =−∞∞|h⁢−k| k h k y⁢0=∥h∥∞=∞ y 0 h by assumption.

→y∉ l ∞ ⁢z y l z ie: not bounded → system is not BIBO.

BIBO Stablility Bottom Line

Figure 5
Figure 5 (fig5.png)
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: ∥h∥1<∞ 1 h and h h must decay fast enough as |n|→∞ n .

Example 2: Mocing Window Filter (FIR)

BIBO stable?

Example 3: Recursive Filter (IIR)

BIBO stable?

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