BIBO Stability

BIBO stability is one of the core properties tested in both theoretical analysis and GATE examinations for LTI systems. A system is said to be BIBO stable if every bounded input produces a bounded output. This property is critical for practical system design because an unstable system can produce outputs that grow without bound, making it useless or even dangerous in real hardware implementations.

Core Concept: BIBO Stability Defined

A system is BIBO stable if and only if, for every input x(t) satisfying |x(t)| is less than or equal to Mx for some finite Mx, the output y(t) satisfies |y(t)| is less than or equal to My for some finite My. The critical insight is that this definition does not say the output is small. It only guarantees the output does not grow unboundedly. A bounded input can produce a large but still finite output in a stable system.

For LTI systems specifically, the BIBO condition translates entirely into a condition on the impulse response h(t). Since y(t) = x(t) * h(t) (convolution), we can bound the output as |y(t)| is less than or equal to Mx times the integral of |h(tau)| from minus infinity to plus infinity. For the output to remain bounded for all bounded inputs, this integral must itself be finite. This is the central mathematical condition for BIBO stability.

Mathematical Condition for BIBO Stability

The necessary and sufficient condition for BIBO stability of a continuous-time LTI system is that the impulse response h(t) must be absolutely integrable. This is written as:

Integral from minus infinity to plus infinity of |h(t)| dt < infinity

For discrete-time LTI systems, the equivalent condition is that the sum of |h[n]| over all integers n must be finite. This is the l1-norm of the impulse response sequence. If this sum diverges, the system is BIBO unstable. The connection to pole locations is direct: for a rational transfer function, a continuous-time system is BIBO stable if and only if all poles of H(s) lie strictly in the left-half s-plane. For discrete-time, all poles of H(z) must lie strictly inside the unit circle.

A subtle but important distinction exists between BIBO stability and Lyapunov stability. Marginally stable systems (poles on the imaginary axis) are Lyapunov stable but not BIBO stable. For example, an integrator with H(s) = 1/s has a pole at s=0. A bounded input like a unit step produces an output that grows linearly without bound, making it BIBO unstable even though the system is not unstable in the Lyapunov sense.

Practical Understanding: Pole Locations and Stability

In practice, BIBO stability is checked by examining the poles of the transfer function H(s) or H(z). For a causal continuous-time system described by a rational H(s), the system is BIBO stable if all poles have strictly negative real parts, meaning they lie in the open left-half plane. For a causal discrete-time system with rational H(z), all poles must have magnitude strictly less than one, placing them inside the unit circle.

A common example: h(t) = e^(-a t) u(t) with a greater than zero. The absolute integral from 0 to infinity of e^(-at) dt equals 1/a, which is finite. So this system is BIBO stable. In contrast, h(t) = e^(at) u(t) with a greater than zero produces an integral that diverges to infinity, confirming BIBO instability.

Given:
h(t) = e^(-4t) u(t)   (impulse response of a causal LTI system)
Check BIBO stability.

Why this formula applies:
BIBO stability requires integral from -inf to +inf of |h(t)| dt < infinity.
Since h(t) = 0 for t < 0 (causal), lower limit becomes 0.

Formula:
I = ∫₀^∞ |h(t)| dt = ∫₀^∞ e^(-4t) dt

Substitution:
I = [ -1/4 * e^(-4t) ] from 0 to ∞

Calculation:
At t → ∞: e^(-4t) → 0
At t = 0:  e^(0) = 1
I = -1/4 * (0 - 1) = 1/4

Final Answer:
I = 0.25 (finite)
The system IS BIBO stable.
Pole of H(s) = 1/(s+4) is at s = -4 (left-half plane, confirms stability).

Exam Tip: For GATE, h(t) = u(t) (unit step) is the classic BIBO unstable example. Its absolute integral diverges because the integrand is 1 for all t greater than 0. An integrator H(s) = 1/s has a pole at origin and is BIBO unstable. Always check poles for real part strictly less than zero, not less than or equal to zero.

Quick Revision

• BIBO stability condition: integral of |h(t)| dt from -inf to +inf must be finite (absolute integrability).
• For discrete systems: sum of |h[n]| over all n must be finite.
• Stable pole condition (continuous): all poles of H(s) in strict left-half plane (Re(s) < 0).
• Stable pole condition (discrete): all poles of H(z) strictly inside unit circle (|pole| < 1).
• Marginally stable systems (poles on imaginary axis) are NOT BIBO stable, e.g., integrator H(s) = 1/s.
• Key formula: |y(t)| <= Mx * integral|h(tau)|dtau, so finite integral guarantees bounded output.
• Exam trap: h(t) = u(t) or h(t) = cos(t)u(t) are BIBO unstable despite appearing bounded, because their integrals diverge.