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Stability and causality of LTI systems

Short answer

An LTI system is causal iff h[n]=0 for n<0, equivalently ROC of H(z) is outside a circle; it is BIBO stable iff sum |h[n]| is finite, equivalently ROC includes the unit circle. For a causal rational system, stability requires all poles inside the unit circle; e.g. (0.5)^n u[n] is causal and stable, 2^n u[n] is causal but unstable.

Describe stability and causality characteristics of LTI system in terms of impulse response and ROC of its transfer function with suitable examples.

Causality

An LTI system is causal if its output at time n depends only on present and past inputs.

  • Time domain: h[n] = 0 for n < 0.
  • z-domain: H(z) is the z-transform of a right-sided sequence, so the ROC is the exterior of a circle, |z| > r (extending to ∞). (For a rational H(z), also the numerator degree must not exceed the denominator degree.)
  • Example: h[n] = (0.5)ⁿ u[n] gives H(z) = 1/(1 − 0.5z⁻¹), ROC |z| > 0.5: causal. h[n] = (0.5)ⁿ u[−n−1] has ROC |z| < 0.5: not causal.

Stability (BIBO)

An LTI system is stable if every bounded input gives a bounded output.

  • Time domain: Σ |h[n]| < ∞ (absolutely summable).
  • z-domain: the ROC of H(z) includes the unit circle |z| = 1 (so the frequency response exists).
  • Example: h[n] = (0.5)ⁿ u[n], Σ|h| = 2 < ∞, ROC |z| > 0.5 includes |z| = 1: stable. h[n] = 2ⁿ u[n] has ROC |z| > 2, which excludes the unit circle: unstable.

Both together

A causal LTI system with a rational H(z) is stable if and only if all poles lie inside the unit circle, because the ROC |z| > (largest pole magnitude) must include |z| = 1.

h[n]H(z)ROCCausalStable
(0.5)ⁿ u[n]1/(1−0.5z⁻¹)|z| > 0.5YesYes
2ⁿ u[n]1/(1−2z⁻¹)|z| > 2YesNo
−2ⁿ u[−n−1]1/(1−2z⁻¹)|z| < 2NoYes

The same H(z) can be stable or causal depending on the ROC chosen.

✓ Answer checked by Claude Opus 5.5 (high effort)

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