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Z-transform property proof

Short answer

Expand X(z)=x(0)+x(1)z^{-1}+x(2)z^{-2}+... . Then z[X(z)-x(0)]= x(1)+x(2)z^{-1}+... . As z→∞, all terms with negative powers vanish, leaving x(1).

If Z[x(k)] = X(z), show that x(1) = lim_{z→∞} z[X(z) − x(0)], where the symbols have their usual meanings.

Proof. By definition,

X(z) = Σ_{k=0}^∞ x(k) z^(−k) = x(0) + x(1) z⁻¹ + x(2) z⁻² + x(3) z⁻³ + ...

Subtract x(0) and multiply by z:

z [X(z) − x(0)] = x(1) + x(2) z⁻¹ + x(3) z⁻² + ...

Now let z → ∞. Every term containing a negative power of z tends to 0, so

lim_{z→∞} z [X(z) − x(0)] = x(1)

Hence x(1) = lim_{z→∞} z[X(z) − x(0)]. ∎

(Here x(0) = lim_{z→∞} X(z) is the initial value theorem; the result is its next-order extension.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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