Time Setting of $z$ and Laplace Transforms

Question

I'm aware that the z-transform and the Laplace Transform have an analogous relationship but I want to be doubly-sure that the z-transform only works in discrete-time and that the Laplace transform only works in the continuous time settings?

Thanks.

Answer 1

In typical (or all) applications, the Laplace Transform is used for continuous time systems and the z-Transform for discrete systems. However, the Laplace Transform for Discrete Time Systems certainly exists but would be more complicated than necessary to solve. The z transform exists as a mathematical simplification of the Laplace Transform that can be applied if a consistent sampling rate is used. This may be clearer with the formulas given below:

The (one-sided, for causal systems) Laplace Transform is given as:

$$X(s) = \int_{t=0}^\infty x(t) e^{-st}dt$$

Thus the Laplace Transform for a discrete time system is given simply by setting t= nT for integer n (where the integral then becomes a summation):

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

At this point we can make a substitution to get rid of the more complicated exponential (which is one possible mapping from s to z):

$$z = e^{sT}$$

resulting in the z-Transform for a discrete-time sequence:

$$X(z) = \sum_{n=0}^\infty x(nT)z^{-n}$$

Equivalently the z-Transform for a discrete sequence (no units of time, simply sample number) is:

$$X(z) = \sum_{n=0}^\infty x(n)z^{-n}$$

Thus we see that the Laplace Transform of a discrete time system exists (but we have no need to solve that typically as working with the z transform once discrete is much simpler). I am not aware of a reverse case although we may be able to create one by taking the limit $T\rightarrow 0$ of the discrete time z-Transform. I haven't worked through this but we may find the same situation (we can force it to be mathematically equivalent but the math involved would be more complicated than necessary vs taking the Laplace Transform directly).