The 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:

  $$x(t) = \sum_{n = 0}^{\infty} x(n) \delta(t - nT) \qquad \mbox{Sampled signal}$$ (93)

into the Laplace transform:

$$X(s) = \sum_{n = 0}^{\infty} x(n) e^{-snT}$$ (94)

$$= \int_{0}^{\infty} x(n) e^{-st} dt$$ (95)

$$= \sum_{n = 0}^{\infty} x(n) \underbrace{(e^{-sT})^{n}}_{z^{-1} = e^{-sT}}$$ (96)

$$= \sum_{n = 0}^{\infty} x(n) (z^{-1})^{n} \qquad \mbox{z-transform}$$ (97)

What is $e^{-sT} = z^{-1}$? It's a unit delay (see Eq. 80).

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