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Recursive/nonrecursive systems; stability

Short answer

Nonrecursive systems depend only on present/past inputs; recursive systems also depend on past outputs. For h(t)=e^(-t) sin(t) u(t), ∫|h| dt ≤ ∫₀^∞ e^(-t) dt = 1 < ∞, so the system is stable.

What are Recursive and Nonrecursive systems? Test the stability of the CT system whose impulse response is given as h(t) = e^(-t)·sin(t)·u(t).

Recursive and non-recursive systems

  • Non-recursive system: the present output depends only on present and past inputs, not on past outputs. There is no feedback. Example: y[n] = x[n] + 2x[n - 1] (FIR filter, finite impulse response).
  • Recursive system: the present output depends also on past outputs, so there is feedback. Example: y[n] = 0.5y[n - 1] + x[n] (IIR filter, impulse response of infinite length).

Stability test of h(t) = e^(-t) sin(t) u(t)

BIBO stability needs ∫|h(t)| dt < ∞.

∫|h(t)| dt = ∫₀^∞ e^(-t) |sin t| dt

Since |sin t| ≤ 1:

∫₀^∞ e^(-t) |sin t| dt ≤ ∫₀^∞ e^(-t) dt = 1 < ∞

(Evaluated exactly, the integral equals (1 + e^(-π)) / (2(1 - e^(-π))) ≈ 0.545.)

Also, H(s) = 1/((s + 1)² + 1), whose poles s = -1 ± j lie in the left half-plane.

Answer: the system is stable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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