Relation to the z Transform

The Laplace transform is used to analyze continuous-time systems. Its discrete-time counterpart is the $z$ transform:

$$X_d(z) \mathrel{\stackrel{\mathrm{\Delta}}{=}}\sum_{n=0}^\infty x_d(nT) z^{-n}$$

If we define $z=e^{sT}$, the $z$ transform becomes proportional to the Laplace transform of a sampled continuous-time signal:

$$X_d(e^{sT}) = \sum_{n=0}^\infty x_d(nT) e^{-snT}$$

As the sampling interval $T$ goes to zero, we have

$$\begin{eqnarray*} \lim_{T\to 0} X_d(e^{sT})T &=& \lim_{\Delta t\to 0} \sum_{n=0... ...ty x_d(t) e^{-st} dt \mathrel{\stackrel{\mathrm{\Delta}}{=}}X(s) \end{eqnarray*}$$

where $t_n\mathrel{\stackrel{\mathrm{\Delta}}{=}}nT$ and $\Delta t \mathrel{\stackrel{\mathrm{\Delta}}{=}}t_{n+1} - t_n = T$.

In summary,

$\parbox{0.8\textwidth}{the {\it z} transform\ (times the sampling in... ...o0$, the Laplace Transform\ of the underlying continuous-time signal $x_d(t)$.}$

Note that the $z$ plane and $s$ plane are related by

$$\zbox{z = e^{sT}}$$

In particular, the discrete-time frequency axis $\omega_d \in(-\pi/T,\pi/T)$ and continuous-time frequency axis $\omega_a \in(-\infty,\infty)$ are related by

$$\zbox{e^{j\omega_d T} = e^{j\omega_a T}}$$

For the mapping $z=e^{sT}$ from the $s$ plane to the $z$ plane to be invertible, it is necessary that $X(j\omega_a )$ be zero for all $\vert\omega_a \vert\geq \pi/T$. If this is true, we say $x(t)$ is bandlimited below half the sampling rate. As is well known, this condition is necessary to prevent aliasing when sampling the continuous-time signal $x(t)$ at the rate $f_s=1/T$ to produce $x(nT)$, $n=0,1,2,\ldots\,$