Signals and Systems · 144 h · Topic 19 of 34

Stability

Every bounded input has to produce a bounded output, and a single exception is enough to rule out stability. An ideal integrator is the closest example: it does not satisfy it.

BIBO Integrability Worst case Poles Integrator

01Bounded input, bounded output

The stability criterion used in signals has a name of its own: BIBO, for bounded input, bounded output. A system is stable if every bounded input produces a bounded output:

\[ |x(t)| \le M \lt \infty \;\;\text{for all } t \;\Longrightarrow\; |y(t)| \le K \lt \infty \;\;\text{for all } t \] The key word is every. Finding a single bounded input that makes the output grow is enough for the system to be unstable.

That last point is what makes verification delicate: a system behaving well with the usual test signals proves nothing. The condition has to hold for any input, and there is always one that is worse than the others.

02The condition on the impulse response

For an LTI system, everything boils down to an integral. The output is \( y(t) = \int h(\tau)x(t-\tau)d\tau \), and bounding in absolute value:

\[ |y(t)| \le \int_{-\infty}^{\infty} |h(\tau)|\,|x(t-\tau)|\,d\tau \le M \int_{-\infty}^{\infty} |h(\tau)|\,d\tau \]

So if that integral is finite, the output is bounded. And the converse also holds, because the input that forces equality can always be constructed. The conclusion is clean:

\[ \text{BIBO stable} \iff \int_{-\infty}^{\infty} |h(t)|\,dt \lt \infty \qquad\text{or, in discrete time,}\qquad \sum_{n=-\infty}^{\infty} |h[n]| \lt \infty \] It is said that \( h \) has to be absolutely integrable, or absolutely summable.

The worst input for it

What is the bounded signal that maximizes the output at an instant \( T \)? The one that at every moment has the same sign as the \( h \) that will multiply it: \( x(t) = \operatorname{sgn} \big(h(T-t)\big) \). With that input, all the contributions to the integral add up instead of canceling, and the result is exactly \( y(T) = \int |h| \). If that integral diverges, the output goes to infinity with an input that never exceeded 1 in absolute value. The lab constructs that input and applies it.

03Watching the integral converge, or not

Lab · the integral of |h| decides

Above, the impulse response and the worst-case input that corresponds to it. Below, the cumulative integral of \( |h| \), which is exactly the output that input produces. If the curve flattens out, the system is stable; if it keeps climbing, it is not.

Up to T = 10.0 s

04Read in the poles

For a rational system, the integral condition translates into the location of the poles, which is how it is used in practice:

Poles	\( \int \|h\| \)	Stability	Example
All with \( \operatorname{Re} \lt 0 \)	Converges	Stable	RC, damped RLC
Any with \( \operatorname{Re} \gt 0 \)	Diverges	Unstable	Positive feedback
Simple, on the \( j\omega \) axis	Diverges	Not BIBO stable	Integrator, ideal oscillator
Double, on the axis	Diverges	Unstable	Double integrator

The third row deserves attention because it contradicts a habit. A system with simple poles on the imaginary axis is usually called marginally stable, and that is not wrong as a description —its impulse response does not grow— but it is not BIBO stable. An integrator with a constant input, which is a perfectly bounded signal, gives a ramp that grows without limit. And an ideal oscillator driven exactly at its frequency goes into resonance and also runs away.

Why a real integrator does not blow up

Because it is not ideal. An op amp integrator has a resistor in parallel with the capacitor —sometimes just the leakage resistance— that moves the pole slightly to the left of the axis. That makes it BIBO stable and bounds the output to the DC value set by that resistor. When that path does not exist, the offset voltage is integrated and the output ends up against the power supply: the classic integrator that goes into saturation.

05In discrete time

The condition is \( \sum |h[n]| \lt \infty \), and in terms of poles, all inside the unit circle. An FIR filter is always stable, regardless of its coefficients: since its impulse response has a finite number of terms, the sum cannot diverge. An IIR, on the other hand, can be unstable, and that is why the position of its poles has to be checked.

\[ y[n] = a\,y[n-1] + x[n] \;\Longrightarrow\; h[n] = a^{n}u[n], \qquad \sum_{n=0}^{\infty}|a|^{n} = \frac{1}{1 - |a|} \;\text{ if } |a| \lt 1 \] The simplest first-order filter: stable if and only if \( |a| \lt 1 \), which is the pole inside the circle.

With \( |a| = 1 \) the sum diverges and with \( |a| \gt 1 \) the system grows exponentially. In a fixed-point implementation this shows up right away: the accumulator overflows within a few samples.

06In the lab

Exercise 1 · The integral, by hand

Compute \( \int_0^{\infty} |h| \) for \( h(t) = e^{-2t}u(t) \) and for \( h(t) = e^{-0.2t}\cos(10t)u(t) \). Check that both converge and compare the values with those given by the lab.

Exercise 2 · The integrator that runs away

Build an op amp integrator with no DC feedback resistor and observe how long the output takes to reach saturation with the input grounded. Then add a 1 MΩ resistor in parallel with the capacitor and measure the new quiescent value.

Exercise 3 · The limit of an IIR

Program \( y[n] = a\,y[n-1] + x[n] \) with \( a = 0.9 \), \( 0.99 \), \( 1.0 \) and \( 1.01 \). Measure how many samples each one takes to overflow a 16-bit accumulator with a constant input of value 1. The first two never overflow: they settle at \( 1/(1-a) \). Write down that value, and for the other two, at which sample they overflow.

07Common mistakes

Testing with a single input and concluding it is stable. The condition holds for every bounded input.
Calling a system with poles on the axis stable. It is marginally stable, and it is not BIBO stable.
Believing that a bounded \( h \) is enough. The step is bounded and its integral diverges.
Forgetting that an FIR is always stable. Its sum is finite by construction.
Confusing stability with causality. They are independent.

08Self-assessment

What does the BIBO criterion say?

That every bounded input must produce a bounded output. A single exception is enough to rule out stability.

What is the condition on \( h \)?

That it be absolutely integrable: \( \int|h(t)|dt \lt \infty \). In discrete time, absolutely summable.

Is an integrator BIBO stable?

No. With a constant —bounded— input the output is a ramp that grows without limit.

What is the worst input for a system?

\( x(t) = \operatorname{sgn}(h(T-t)) \): it makes all the contributions add up instead of canceling.

For which values of \( a \) is \( y[n] = a\,y[n-1] + x[n] \) stable?

For \( |a| \lt 1 \). There the geometric sum converges to \( 1/(1-|a|) \).

Why is an FIR always stable?

Because its impulse response has a finite number of terms, and a finite sum never diverges.

09Further reading

Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1998. The stability condition proved in both directions, with the worst-case input.
Katsuhiko Ogata. Modern Control Engineering. 5th ed. (Spanish edition, “Ingeniería de control moderna”), Pearson, 2010. Stability in terms of poles, with Routh-Hurwitz and the phase and gain margins.
John G. Proakis and Dimitris G. Manolakis. Digital Signal Processing. 4th ed. (Spanish edition, “Tratamiento digital de señales”), Pearson, 2007. Stability of IIR filters and the effects of finite-precision arithmetic.

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