- 2072 Chaitra · 2+3+3 marks
Stable/unstable systems; test stability
Short answer
A BIBO stable LTI system satisfies ∫|h| dt < ∞; otherwise unstable. For ψ>0: (i) h(t)=e^(ψt)u(t) has infinite integral → unstable; (ii) h(t)=e^(-ψt)u(t) has ∫=1/ψ < ∞ → stable.
Define stable and unstable systems. Test the stability of the LTI systems whose impulse responses are given as (i) h(t) = e^(ψt)u(t) (ii) h(t) = e^(-ψt)u(t) [exponent as printed, unclear in scan]
Definitions
- Stable system (BIBO): every bounded input |x(t)| ≤ M₁ produces a bounded output |y(t)| ≤ M₂. For an LTI system, this holds if and only if ∫|h(t)| dt < ∞.
- Unstable system: some bounded input produces an unbounded output (∫|h| dt = ∞). Example: an ideal integrator, h(t) = u(t).
Tests
The exponent printed as ψ is unclear in the scan; I treat it as a real constant ψ > 0 (for example 2) and give the general condition.
(i) h(t) = e^(ψt) u(t)
∫₀^∞ e^(ψt) dt = [e^(ψt)/ψ]₀^∞ = ∞ for ψ > 0
The integral diverges, so the system is unstable. (It would be stable only if ψ were negative, ψ < 0, giving 1/|ψ|.)
(ii) h(t) = e^(-ψt) u(t)
∫₀^∞ e^(-ψt) dt = [-e^(-ψt)/ψ]₀^∞ = 1/ψ < ∞ for ψ > 0
The system is stable. Its pole s = -ψ lies in the left half-plane.
Answer: for ψ > 0, (i) is unstable and (ii) is stable with ∫|h| dt = 1/ψ.
✓ Answer checked by Claude Opus 5.5 (high effort)
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