Q9Advanced Engineering Mathematics I

Convolution Theorem for Z-Transform:

If $Z\{x_n\}=X(z)$ and $Z\{y_n\}=Y(z)$, then the Z-transform of their convolution sequence $(x\ast y{)}_n$ is the product of their individual Z-transforms.

$$Z\{x_n\ast y_n\}=X(z)\cdot Y(z)$$

Where the discrete convolution is defined as $(x\ast y{)}_n={\sum }_{k=0}^n{x}_k{y}_{n-k}$ (for causal sequences).