Stability of h(t) = t·e^(-3t)u(t)

Condition for stability

An LTI system is BIBO stable if and only if its impulse response is absolutely integrable:

∫ |h(t)| dt < ∞, integrated over -∞ to ∞.

(Equivalently, all poles of H(s) lie in the left half of the s-plane and the ROC includes the jω axis.)

Test for h(t) = t·e^(-3t)·u(t)

Since h(t) ≥ 0 for t ≥ 0,

∫|h(t)| dt = ∫₀^∞ t e^(-3t) dt

Integrate by parts (u = t, dv = e^(-3t) dt):

= [-t e^(-3t)/3]₀^∞ + (1/3) ∫₀^∞ e^(-3t) dt = 0 + (1/3)(1/3) = 1/9

Standard result: ∫₀^∞ tⁿ e^(-at) dt = n!/a^(n+1), so with n = 1, a = 3 it gives 1/9.

The value is finite. (H(s) = 1/(s + 3)², a double pole at s = -3 in the left half-plane.)

Answer: the system is BIBO stable (∫|h| dt = 1/9).