Bounded input, bounded output: what is being promised?
BIBO stability is a statement about every bounded input and the observed output, not about one attractive plot.
Use an impulse-response bound to explain what BIBO stability guarantees—and what it does not say about hidden states.
One loop can settle in a step test. BIBO stability makes a broader promise about every bounded input and the selected output.
Review the previous lesson: Stability and steady-state errorFirst, picture it
A small bounded push should not make the observed response grow without bound. A motor that looks calm for twelve seconds under one load has passed only that test; another input, a longer time or an unobserved internal state may tell a different story.
Build the idea step by step
BIBO stands for bounded input, bounded output. Bounded means there is one finite limit the signal never exceeds in magnitude. The claim must hold for every bounded input, from zero initial state. For a causal LTI system, add up the absolute size of its impulse response over all future time. If that area is finite, it gives an output bound. This checks the chosen output, not every internal state that a sensor might fail to reveal.
Read the equations
For a causal zero-state LTI system with an integrable impulse response, this inequality bounds its output for any input bounded by M.
What the symbols mean
- input magnitude bound, in input units
- impulse response, in output per input per second
- input signal
- selected output signal
The bound can depend on the input's allowed size and the system, but not on elapsed time. Our ordinary impulse-response example has no direct feedthrough. A finite direct gain is also bounded and can be handled separately.
Bound a first-order output
Take a causal normalized system whose impulse response is the decaying exponential. Assume the input never exceeds 2 in magnitude.
The area under the absolute impulse response is one.
The chosen output stays within magnitude 2 from zero initial state. This bound says nothing by itself about hidden states in a nonminimal realization.
More detail and model assumptions
Two small counterexamples
An ideal integrator has y(t)=∫₀ᵗu(τ)dτ. Feed it the bounded constant input u=1 and its output becomes y=t, which is unbounded. 'The input is harmless-looking' is not enough.
Now imagine two internal states: ẋ₁=−x₁+u, ẋ₂=+x₂, while the sensor reports only y=x₁. The input-to-output relation can look stable, but an initial x₂≠0 grows invisibly. This is why a canceled or unobservable unstable mode matters even when a transfer-function plot looks safe.
What the Lab can and cannot show
You can pin runs, compare bounded load pulses, inspect finite-duration output and watch for rail exit. These are diagnostic experiments, not a proof over every bounded input and infinite time. Use an analytic model and its assumptions for a mathematical claim, then use experiments to challenge those assumptions.
Connect the example to the lab
Apply one small and one larger load pulse to the motor, keeping controller settings fixed. Report finite-run behavior without labeling either result a BIBO proof.
Compare finite load testsMake a prediction before moving a control.
- PREDICTWill this load pulse make the response diverge in the displayed window?
- CHANGE ONE THINGChange only the pulse magnitude and pin both runs.
- OBSERVERecord the window length, output, input limit and any fault.
- EXPLAINSeparate your observation from the unproved all-input BIBO claim.
A stable transfer function can conceal unstable internal dynamics when the realization is not minimal. Conversely, a single diverging nonlinear saturated run is not automatically a proof about an LTI transfer function.
Check your understanding+
Is a bounded 12-second motor trace sufficient to establish BIBO stability?
No. BIBO quantifies over all bounded inputs and all time in a stated input-output model.Why the next lesson follows
For polynomial closed-loop models, the Routh test checks pole location without calculating each root.