Which standard result is the starting point?

Z[sin wt] = z sin wT/(z^2 - 2z cos wT + 1)

Z[sin wt] = z cos wT/(z^2 - 2z sin wT + 1)

Z[sin wt] = z sin wT/(z^2 + 2z cos wT + 1)

After simplifying, Z[e^(-at) sin wt] is

z sin wT/(z^2 - 2z e^(-aT) cos wT + 1)

z e^(-aT) cos wT/(z^2 - 2z e^(-aT) sin wT + e^(-2aT))

Short answer

By the damping rule, Z[e^{-at} sin wt] = z e^{-aT} sin wT / (z^2 - 2z e^{-aT} cos wT + e^{-2aT}), with region of convergence |z| > e^{-aT}.

Find the z-transform of e^{-at} sin wt.

Take sampling period T, so x(kT) = e^(−akT) sin(wkT).

Step 1. Z[sin wt] = z sin wT/(z² − 2z cos wT + 1).

Step 2. Damping rule: Z[e^(−at) x(t)] = X(z e^(aT)). Replace z by z e^(aT):

Z[e^(−at) sin wt] = (z e^(aT)) sin wT / ( z² e^(2aT) − 2 z e^(aT) cos wT + 1 )

Multiply numerator and denominator by e^(−2aT):

Answer:

Z[e^(−at) sin wt] = z e^(−aT) sin wT / ( z² − 2z e^(−aT) cos wT + e^(−2aT) ), |z| > e^(−aT)

(Direct check: write sin wt = (e^(iwt) − e^(−iwt))/(2i) and sum the two geometric series with ratio e^(−aT ± iwT)/z.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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