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Derive convolution property of z-transform

Short answer

If x1[n]↔X1(z) with ROC R1 and x2[n]↔X2(z) with ROC R2, then y[n]=x1[n]*x2[n] has transform Y(z)=X1(z)X2(z) with ROC at least R1∩R2. Proof: Σ_n (Σ_k x1[k]x2[n-k]) z^(-n) = Σ_k x1[k]z^(-k) Σ_m x2[m]z^(-m).

Derive the convolution property of z-transform.

Statement: If x₁[n] ↔ X₁(z) with ROC R₁ and x₂[n] ↔ X₂(z) with ROC R₂, then

x₁[n] * x₂[n] ↔ X₁(z) X₂(z), with ROC containing R₁ ∩ R₂.

Derivation. Let y[n] = x₁[n] * x₂[n] = Σ(k = −∞ to ∞) x₁[k] x₂[n − k]. Then

Y(z) = Σ(n) y[n] z⁻ⁿ = Σ(n) [ Σ(k) x₁[k] x₂[n − k] ] z⁻ⁿ

Interchange the order of summation (valid inside the common ROC, where both converge absolutely):

Y(z) = Σ(k) x₁[k] [ Σ(n) x₂[n − k] z⁻ⁿ ]

Put m = n − k, so n = m + k and z⁻ⁿ = z⁻ᵐ z⁻ᵏ:

Y(z) = Σ(k) x₁[k] z⁻ᵏ · Σ(m) x₂[m] z⁻ᵐ = X₁(z) X₂(z)

The ROC is at least the overlap R₁ ∩ R₂; it can be larger if a pole of one is cancelled by a zero of the other.

Use: the output of an LTI system is Y(z) = X(z)H(z), so convolution in time becomes simple multiplication in the z-domain.

✓ Answer checked by Claude Opus 5.5 (high effort)

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