Z-transform

The Laplace transform is for continuous causal signals but in DSP we have sampled signals. So, we need to investigate what happens if we feed a sampled signal:

$$x(t)=\sum _{n=0}^{\infty }x(n)\delta (t-nT)$$

Into the laplace transform:

$$\begin{array}{rl}X(s)& =\sum _{n=0}^{\infty }x(n)e^{-snT}={\int }_0^{\infty }x(n)e^{-st}dt\\ & =\sum _{n=0}^{\infty }x(n)\underset{z^{-1=e^{-sT}}}{\underbrace{(e^{-sT}{)}^n}}\\ & =\sum _{n=0}^{\infty }x(n)(z^{-1}{)}^n\end{array}$$

Where $z^{-1}$ represents a unit delay