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BIBO Stability

Module by: Richard Baraniuk

Summary: An overview of BIBO stability.

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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=-∞∞hn−kxk=∑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=-∞∞hn−kxk 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 ∞ (1)
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 ∞ ?

Fact:

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 ∞

Proof

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∥∞=maxn∈ℤ{|yn|} y n y n So look at |yn| y n and show it is <∞ .

Use convolution formula |yn|=|∑k=-∞∞xn−khk| y n k x n k h k ≤∑k=-∞∞|xn−k||hk| k x n k h k Where x∈ l ∞ ℤ→|xn−k|≤∥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|=∥h∥1 k h k 1 h Therefore, |yn|≤∥x∥∞∥h∥1 y n x 1 h But x∈ l ∞ ℤ→∥x∥∞<∞ x l ∞ x h∈ l 1 ℤ→∥h∥1<∞ h l 1 1 h Therefore |yn|<∞ y n Therefore maxn∈ℤ{|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.

Note:

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

Note:

We only have 1 x x to show this.
Some creativity suggests the following signal: xn= h-n¯|h-n|ifhn≠0 0 ifh=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)

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=-∞∞xn−khk y n k x n k h k y0=∑k=-∞∞x-khk y 0 k x k h k x-k=h-k¯|h-k| x k h k h k =∑k=-∞∞hk¯hk|hk| k h k h k h k =∑|hk|2|hk| k h k 2 h k =∑k=-∞∞|hk| k h k y0=∥h∥1=∞ 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: ∥h∥1<∞ 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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