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Derive time shifting property of z-transform

Short answer

x[n-n0]↔z^(-n0)X(z), with the same ROC except possibly at z=0 or ∞. Derivation: Σ_n x[n-n0] z^(-n) = Σ_m x[m] z^(-(m+n0)) = z^(-n0)X(z) by substitution m=n-n0.

Derive the time shifting property of z-transform.

Statement: If x[n] ↔ X(z) with ROC R, then

x[n − n₀] ↔ z⁻ⁿ⁰ X(z), with the same ROC (except for the possible addition or removal of z = 0 or z = ∞).

Derivation. By definition, the z-transform of y[n] = x[n − n₀] is

Y(z) = Σ(n = −∞ to ∞) x[n − n₀] z⁻ⁿ

Let m = n − n₀, so n = m + n₀. As n runs over all integers, so does m:

Y(z) = Σ(m) x[m] z⁻⁽ᵐ⁺ⁿ⁰⁾ = z⁻ⁿ⁰ Σ(m) x[m] z⁻ᵐ = z⁻ⁿ⁰ X(z)

Remarks:

  • A delay of one sample multiplies X(z) by z⁻¹; an advance of one sample multiplies it by z.
  • Multiplying by z⁻ⁿ⁰ can add a pole (or zero) at z = 0 or ∞, so the ROC can change only at these two points.
  • For the one-sided (unilateral) z-transform, the delay rule has extra initial-condition terms, e.g. x[n−1] ↔ z⁻¹X(z) + x[−1].

Example: δ[n − 3] ↔ z⁻³ (since δ[n] ↔ 1).

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