Convolution theorem for z-transform

Statement

If x(k) and y(k) are causal sequences with Z[x(k)] = X(z) and Z[y(k)] = Y(z), then the z-transform of their convolution is the product of the transforms:

Z[x(k) * y(k)] = X(z) · Y(z)

where the convolution is (x * y)(k) = Σ_{m=0}^{k} x(m) y(k − m).

Equivalently, Z⁻¹[X(z)Y(z)] = x(k) * y(k).

Proof

By definition,

Z[x * y] = Σ_{k=0}^{∞} (x * y)(k) z⁻ᵏ = Σ_{k=0}^{∞} Σ_{m=0}^{k} x(m) y(k − m) z⁻ᵏ

Since x and y are causal (y(n) = 0 for n < 0), the inner sum can be taken over all m from 0 to ∞. Interchange the order of summation (valid inside the common region of convergence):

Z[x * y] = Σ_{m=0}^{∞} x(m) Σ_{k=m}^{∞} y(k − m) z⁻ᵏ

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

Z[x * y] = Σ_{m=0}^{∞} x(m) z⁻ᵐ  ·  Σ_{n=0}^{∞} y(n) z⁻ⁿ
         = X(z) · Y(z)

The region of convergence contains the intersection of the ROCs of X(z) and Y(z). ∎

Use

It turns the convolution of two sequences into a simple product, so inverse transforms of products can be found. Example: Z⁻¹[z/(z−1) · z/(z−2)] = 1ᵏ * 2ᵏ = Σ_{m=0}^{k} 2^{k−m} = 2^{k+1} − 1.