2.5c BIBO Stability

Bounded-Input, Bounded-Output (BIBO) Stability

A system $H$ is BIBO stable if for any input $x\in \mathbb{X}$ bounded by a number $B_x\in {\mathbb{R}}^{+}$, such that $|x(t)|\le B_x$ $\forall t\in \mathbb{R}$, there exists a number $B_y\in {\mathbb{R}}^{+}$ for which the output $|y(t)|\le B_y$ $\forall t\in \mathbb{R}$.

In addition to this, any discrete-time LTI system is BIBO stable if and only if $h$ is absolutely summable ($h\in {\ell }^1$), such that for some $B_y\in \mathbb{R}$,
${\sum }_{k=-\infty }^{\infty }|h(k)|\le B_y.$

Equivalently, any continuous-time LTI system is BIBO stable if and only if $h$ is absolutely integrable ($h\in {\alpha }^1$), such that
${\int }_{-\infty }^{\infty }|h(\tau )|d\tau \le B_y.$

Intuition

The proof for the BIBO stability of discrete-time LTI systems can be easily seen by showing that the triangle inequality holds, such that

|y(n)|=\sum^{\infty}_{k=-\infty}|x(n-k)h(k)| &\leq \sum^{\infty}_{k=-\infty}|x(n-k)||h(k)| \\ &\leq B_{x}\sum^{\infty}_{k=-\infty}|h(k)| \\ &\leq B_{y}. \end{align}$$ However, this is only half of proving the dual relationship. Therefore, we need something else to prove that $\text{H is BIBO stable} \implies \sum^{\infty}_{k=-\infty}|\delta(k)|\leq B_{y}$. We cam proceed by taking the contrapositive of this statement, such that $\sum^{\infty}_{k=-\infty}|\delta(k)| = \infty \implies \text{H is not BIBO stable.}$ We need to prove that a bounded input can lead to an unbounded output. To proceed, we must make $y(0)$ to blow up using a bounded input $x(n)$: $$y(0)=\sum^{\infty}_{k=-\infty}x(-k)h(k)$$ We choose $x(-n)$ carefully, such that $x(-n)=\text{sgn}(h(n))$, where $\text{sgn}(\cdot)$ is the sign of each value of $h(n)$. This turns the product $x(-k)h(k)$ into the absolute value of the filter, such that $$\begin{align} y(0) &= \sum^{\infty}_{k=-\infty}\text{sgn}(h(n))h(k) \\ \\ &= \sum^{\infty}_{k=-\infty}|h(k)| \\ &= \infty. \end{align}$$ But what if $h(n)$ is not a real-valued filter? Then, we can set $x(-n) = \frac{h\textasteriskcentered(n)}{|h(n)|}$, which yields the equation $$\begin{align} y(0) &= \sum^{\infty}_{k=-\infty}h(k)\frac{h\textasteriskcentered(k)}{|h(k)|} \\ &= \sum^{\infty}_{k=-\infty}\frac{|h(k)|^2}{|h(k)|} \\ &= \sum^{\infty}_{k=-\infty}h(k) \\ &= \infty. \end{align}$$ # Properties #### (*Frequency Response*) If $h \in \ell^1$ is BIBO stable, then $|H(\omega)| < \infty$ $\forall \omega \in \mathbb{R}$ and $H(\omega)$ is continuous in $\omega$. # Examples >[!example] Is $x(n) = \frac{x(n) + x(n-1)}{2}$ BIBO stable? > >We prove by the definition of BIBO stability: >$$|y(n)|=\frac{1}{2}|x(n) + x(n-1)|.$$ >We see by the triangle theorem that >$$\begin{align} \frac{1}{2}|x(n) + x(n-1) &\leq \frac{1}{2}|x(n)| + \frac{1}{2}|x(n-1)| \\ &\leq \frac{1}{2}|B_{x}|+\frac{1}{2}|B_{x}| \\ &\leq B_{x}. \end{align}$$ >Therefore, in this case, $B_{y}=B_{x}$.