# What are the advantages and disadvantages of Laplace transform over Z transform?

Laplace transform for continuous signal $x(t)$ is given by

$$X(s) = \int\limits_{-\infty}^{+\infty} x(t) e^{-s t} dt. \quad (1)$$

Z-transform for discrete signal $x(n)$ is given by

$$X(z) = \sum\limits_{n=-\infty}^{+\infty} x[n] z^{-n}. \quad (2)$$

I can say that only difference between the two transform is Laplace used for continuous signal and Z transform for discrete signal .

But what are the advantages and disadvantages of one transform over the other? Where are they used?

• That's a bit like asking what the advantage and disadvantages of a cat over a bicycle are; they are different things, and are used for different things. Jun 8, 2016 at 10:40
• also, canonically, laplace is $\int\limits_{t=0}^{\infty}$, not integration over the complete $\mathbb R$; that would usually yield unbounded results (for practically all functions but those that are Fourier transformable). Jun 8, 2016 at 10:48
• @MarcusMüller: I kinda agree with your first comment, but not so much with the second one. You have the bilateral and the unilateral Laplace transform. For causal function they're of course identical, but there are certain two-sided functions for which the bilateral version is useful. Jun 8, 2016 at 10:58
• @MarcusMüller: And concerning the comparison with the Fourier transform, there are functions for which the bilateral Laplace transform exists but the Fourier transform doesn't, and there are also functions for which the Fourier transform exists but not the bilateral Laplace transform. Jun 8, 2016 at 11:00
• @MarcusMüller: It's often just those 'overly hasty' comments that keep this place alive and fun ... :) Jun 8, 2016 at 11:33

Both transforms are equivalent tools, but the Laplace transform is used for continuous-time signals, whereas the $\mathcal{Z}$-transform is used for discrete-time signals (i.e, sequences).

You can see that they are equivalent by using the continuous-time representation of a discrete-time signal, and then applying the Laplace transform to that signal. The continuous-time representation of a discrete-time signal is a weighted Dirac comb:

$$x_d(t)=\sum_{n=-\infty}^{\infty}x[n]\delta(t-nT)\tag{1}$$

where $x[n]$ is the discrete-time signal, $\delta(t)$ is the Dirac delta impulse, and $T$ is the sampling period.

The (bilateral) Laplace transform of $(1)$ is

\begin{align}X_d(s)&=\int_{-\infty}^{\infty}x_d(t)e^{-st}dt\\&=\int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}x[n]\delta(t-nT)e^{-st}dt\\&=\sum_{n=-\infty}^{\infty}x[n]\int_{-\infty}^{\infty}\delta(t-nT)e^{-st}dt\\&=\sum_{n=-\infty}^{\infty}x[n]e^{-snT}\tag{2}\end{align}

which equals the (bilateral) $\mathcal{Z}$-transform of $x[n]$

$$X(z)=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{3}$$

for $z=e^{sT}$.