A continuous time domain system is well described by the Laplace transform. It allows to express any continuous signal x(t) as the integral sum of weighted complex and exponentially growing/decaying sine waves $e^{st} = e^{\sigma t} \cdot e^{j\omega t}$:

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Where X(s) is the Laplace Transform and may be evaluated as:

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The variable of interest (on which the Laplace Transform depends) is the complex angular frequency $s = \sigma + j\omega$. If $\sigma= 0$, the Laplace Analysis coincides with the Fourier analysis since $s = j\omega$

In a discrete time Frequency, the Z transform is usually used:

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  1. What does the synthesis equation of Z Transform mean? Does it mean that the discrete time domain sequence is expressed as the sum of weighted complex power functions $z^{n-1}$?

  2. Is $z$ the variable of interest? Often I read that if we replace z with $e^{j\omega t}$, we get the DFT. Correct, but the DFT is an exponential function of $\omega$, and not a power function of $z$. Which is (between $z$ and $\omega$) the variable of interest for a discrete time sequence and why?

  3. I think understanding which is the significant variable (between $z$ and $\omega$) is crucial to understand Frequency warping, that is the frequency distorsion due to the fact that the real angular frequency axis $[-\infty;+\infty]$ becames the unit circumference $z = e^{j\omega t}$ . Well, but this is due only to the fact that we decide that the variable of interest for a discrete-time sequence is z instead of $\omega$. Also Fourier (and Laplace) Transform of continuous signals contains $e^{j\omega t}$, but we don't say "We put $z = e^{j\omega t}$ hence there is distorsion", and the variable of interest is assumed to be $\omega$. I've never seen people complaining about this for continuous signals.

It seems that: "Until it's continuous, $\omega$ is important. When your signal becomes discrete, $z$ is important". But I do not understand why.


1 Answer 1


The z-transform is the discrete version of the Laplace transform and in both cases z and s are the set of all complex numbers, and as such we map with the transform the time domain function into the domain of complex frequencies; signals that change in rotation only which is the Fourier Transform and in addition to that such signals that can grow and decay in time. This leads to great mathematical simplifications, for example doing this will translate integral-differential equations into simple algebra.

The question is better phrased "why use z instead of s", as what we would refer to as the frequency “axis” is a one dimensional subset of the complex surface for both cases: In the s plane, the frequency axis is the $j\omega$ axis (the imaginary axis of the s-plane) and in the z-plane, the frequency axis is the unit circle.

The significant convenience of the z-plane is clear when you compare the equations for the Laplace Transform, to the Laplace Transform for a discrete time sequence, and finally with a simple substitution of $z= e^{sT}$ in the Laplace Transform for the discrete time sequence we arrive at the z-transform. Notably, this translates a transcendental equation to a simple polynomial that we can resolve to finite poles and zeros and takes advantage of the repetition in discrete time resulting in a much simpler equation for further manipulation. One can continue to process everything using Laplace, both continuous time and discrete time waveforms, but why take that punishment when the z-transform can be used instead for the discrete time cases?

Consider the Laplace Transform for a causal continuous time sequence:

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

The Laplace Transform when applied to a discrete time sequence becomes:

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

Note this simple case of applying this formula to solve for the Laplace Transform of a discrete-time two sample moving average, and then solving for the poles and zeros:


If we substitute $z = e^{sT}$ the transform then becomes the "z Transform" as:

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

And the example above becomes the much simpler polynomial ratio as:

$$H(z) = \frac{1}{2}+\frac{1}{2}z^{-1} = \frac{z+1}{2z}$$

This simplification is not limited to discrete-time systems as we can have similar complexities in continuous time of transfer functions involving fixed delays (with the same challenge of having a transcendental equation versus simpler polynomials of fixed order), in which case we could make the same substitution to proceed with further processing and mathematical manipulation. The point is that in discrete time this always occurs that we are working with systems of consistent unit delays, and the z-transform abstracts that exponential that is always there and avoids us from carrying it through every computation unnecessarily. It is much simpler. This is no different from other mathematical mappings and change of basis with the main objective that in the new space it is much easier to manipulate the equations and find solutions, and if the original space was important, we can transform back to that after all the heavy lifting is done.

Frequency warping only occurs if we choose to map transfer functions from continuous time to discrete time using the Bilinear Transform method and this is only because this method will provide a one to one mapping from the unique continuous frequency domain which extends to $\pm \infty$ to the unique frequency domain for sampled systems which extends from $-f_s/2$ to $+f_s/2$. Other mappings exist with no warping but will instead have aliasing.

  • $\begingroup$ Thank you very much. The last question is: why shouldn't we put z = exp(st) also for continuous signals? Also in this case it would have converted exponentials to polynomials $\endgroup$
    – Kinka-Byo
    Jun 16, 2021 at 19:03
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    $\begingroup$ @Kinka-Byo that will be very clear when I put in the equations and ask you to solve each for the poles and zeros. I am on my phone so will update soon when I can type the equations easily- it will be very clear to you from that. $\endgroup$ Jun 16, 2021 at 19:13
  • $\begingroup$ Perfect, thank you very much! $\endgroup$
    – Kinka-Byo
    Jun 16, 2021 at 19:24
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    $\begingroup$ @Kinka-Byo Please also see this detailed explanation on z and the various transforms and how they are all related: dsp.stackexchange.com/questions/31830/… $\endgroup$ Jun 17, 2021 at 3:45

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