- 2069 Chaitra · 3 marks
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).
✓ Answer checked by Claude Opus 5.5 (high effort)
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