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Z-transform property proof

Short answer

Using initial and final value theorems on X(z)=z/(z−1) − z/(z−e^(−T)), x(0)=lim_{z→∞} X(z)=0, x(∞)=lim_{z→1} (z−1)X(z)=1. Directly, 1−e^0=0 and 1−e^(−∞)=1.

Find x(0) and x(∞) of the Z-transform x(t) = 1 − e^(−t).

Take x(t) sampled at t = kT, so x(k) = 1 − e^(−kT). (Here "x(0)" and "x(∞)" are found by the initial and final value theorems.)

Z-transform:

X(z) = z/(z − 1) − z/(z − e^(−T))

Initial value: x(0) = lim_{z→∞} X(z) = 1 − 1 = 0.

Final value: x(∞) = lim_{z→1} (z − 1) X(z).

(z − 1) X(z) = z − z(z − 1)/(z − e^(−T))  →  1 − 0 = 1   as z → 1

Answer: x(0) = 0 and x(∞) = 1. (Directly: 1 − e^0 = 0 and 1 − e^(−∞) = 1.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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