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?


  • $\begingroup$ Per definition, ZT applies to a sequence and LT applies to a function of a real valued variable. What else are you looking for? $\endgroup$ – Juancho Apr 20 '19 at 14:25
  • $\begingroup$ I just wanted to make sure that I'm inferring the time domains correctly seeing as this is not my area of expertise. $\endgroup$ – Janitt Apr 21 '19 at 11:44

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).

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    $\begingroup$ Related: this answer on the discrete Laplace transform. $\endgroup$ – Matt L. Apr 20 '19 at 16:32
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    $\begingroup$ @MattL. Ha! Our answers repeat like the frequency response using the z-transform $\endgroup$ – Dan Boschen Apr 20 '19 at 16:47

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