BIBO stability

A system is BIBO stable (bounded-input, bounded-output) if every bounded input produces a bounded output. “Bounded” means “stays between some finite bounds” — the input never goes off to infinity, and neither does the output.

Formally: a system is BIBO stable if, for every input $x(t)$ satisfying $|x(t)|\le M_x$ for some finite $M_x$ and all $t$, the output $y(t)$ satisfies $|y(t)|\le M_y$ for some finite $M_y$ and all $t$.

Why it matters

Real engineering systems have to behave reasonably for reasonable inputs. An audio amplifier whose output goes to infinity for a finite-amplitude input is unusable. A control system whose output blows up for a bounded reference signal is dangerous. BIBO stability is a practical requirement, not a mathematical curiosity.

Test from the impulse response

For an LTI system, BIBO stability has a clean criterion:

${\int }_{-\infty }^{\infty }|h(t)|dt<\infty ⟺\text{system is BIBO stable}.$

The impulse response must be absolutely integrable.

Proof sketch. If $|x(t)|\le B$ for all $t$, then from the convolution integral

$|y(t)|=\left|\int x(t-\tau )h(\tau )d\tau \right|\le \int |x(t-\tau )||h(\tau )|d\tau \le B\int |h(\tau )|d\tau .$

If the integral on the right is finite, $y$ is bounded for every $t$. The converse (BIBO stability ⇒ absolute integrability) takes a slightly more delicate argument.

Examples

• $h(t)=e^{-3t}u(t)$: ${\int }_0^{\infty }{e}^{-3t}dt=1/3$, finite. Stable.
• $h(t)=u(t)$ (pure integrator): ${\int }_0^{\infty }1dt=\infty$. Not stable — a constant (bounded) input gives a ramp output (unbounded), confirming this directly.
• $h(t)=e^{3t}u(t)$: integral diverges. Not stable — pole in the right half-plane.

Pole-location criterion

For an LTI system described by a linear constant-coefficient ODE, the impulse response is a sum of exponentials (and possibly polynomial-times-exponentials, for repeated poles) of the form $e^{s_{k}t}$, where $s_k$ are the roots of the characteristic polynomial. The integral $\int |h|dt$ is finite iff every $e^{s_{k}t}u(t)$ tail decays — iff every $s_k$ has strictly negative real part.

So BIBO stability ⟺ all poles of $H(s)$ in the open left half-plane. This is the standard stability criterion used throughout control theory and filter design.

Marginal cases — simple poles on the imaginary axis — are not BIBO stable (the impulse response is a sustained sinusoid, not absolutely integrable), even though they don’t grow. They’re called marginally stable.

RC lowpass: stable

The RC lowpass has characteristic equation $RCs+1=0$, giving $s=-1/(RC)$ on the negative real axis. For any positive $R$ and $C$, the pole is in the open left half-plane, so the system is BIBO stable. Every real RC lowpass is stable — there’s no way to build an unstable one out of passive components.

In the wider picture

BIBO stability tells us whether the system will behave sensibly in service. Causality tells us whether it can be implemented in real time. Memory tells us whether it has internal dynamics. The three combined describe most of what we need to know about an LTI system’s structural behavior.