Why does e^(-a|t|)u(t) reduce to e^(-at)u(t)?

u(t) removes t < 0 where |t| = t

Energy or power: e^(-a|t|)u(t)

Short answer

For t≥0, x(t)=e^{-at}u(t). Energy E = ∫₀^∞ e^{-2at} dt = 1/(2a) J; average power P = 0. Finite energy, zero power → energy signal.

Justify whether a signal x(t) = e^(-a|t|)·u(t) (a > 0) is energy or power signal.

x(t) = e^(-a|t|) u(t), a > 0

Since u(t) is zero for t < 0, only t ≥ 0 counts, where |t| = t. So x(t) = e^(-at) u(t).

E = ∫₀^∞ e^(-2at) dt = [-e^(-2at)/(2a)]₀^∞ = 1/(2a) (finite and non-zero)

P = lim (T→∞) (1/2T) ∫₀^T e^(-2at) dt = lim (1/2T) · (1 - e^(-2aT))/(2a) = 0

Answer: E = 1/(2a) < ∞ and P = 0, so x(t) is an energy signal.

✓ Answer checked by Claude Opus 5.5 (high effort)

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