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z-transform of e^{at} cos wt

Short answer

Sample at t=kT: x(k)=e^{akT} cos(wkT). Apply Euler's formula and Z[b^k]=z/(z-b) to get X(z)=z(z - e^{aT} cos wT)/(z^2 - 2z e^{aT} cos wT + e^{2aT}), |z| > e^{aT}.

Find the z-transform of e^{at} cos wt for t ≥ 0.

Setup. Sample at t = kT. Then x(k) = e^{akT} cos(wkT), k ≥ 0.

Method. Write cos(wkT) = (e^{jwkT} + e^{−jwkT})/2 and use Z[bᵏ] = z/(z − b):

Z[e^{akT} cos wkT]
 = ½ [ z/(z − e^{(a+jw)T}) + z/(z − e^{(a−jw)T}) ]
 = ½ · z [ 2z − e^{aT}(e^{jwT} + e^{−jwT}) ]
       / [ z² − z e^{aT}(e^{jwT} + e^{−jwT}) + e^{2aT} ]

Since e^{jwT} + e^{−jwT} = 2 cos wT:

Answer: X(z) = z (z − e^{aT} cos wT) / (z² − 2z e^{aT} cos wT + e^{2aT}), for |z| > e^{aT}.

With T = 1 this reads z(z − eᵃ cos w)/(z² − 2z eᵃ cos w + e²ᵃ). (Equivalent to the shifting rule: replace z by z e^{−aT} in Z[cos wkT].)

✓ Answer checked by Claude Opus 5.5 (high effort)

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