I would like to know if

$$ \text {Z-Transform ( } G(s)H(s) \text{ )} = \text {Z-Transform (}G(s) \text{)} \text { Z-Transform (} H(s) \text{) } = G(z)H(z) $$

where G(s), H(s) are the Laplace transform representations of g and h, and G(z) and H(z) are the Z-transform representation of g and h.

Is this relation true?

  • 1
    $\begingroup$ The Laplace transform encodes a continuous signal, the Z transform encodes a discrete signal. How do you propose the two are related? $\endgroup$ – Lutz Lehmann Feb 18 '14 at 11:19
  • $\begingroup$ The relation between s and z is $s = \frac{1}{T}ln(z)$ is it not? $\endgroup$ – KillaKem Feb 18 '14 at 15:01
  • $\begingroup$ This assumes a change in the quality of the underlying signal, i.e., sampling with a sampling step of $T$. Then the variable replacement will not change the functional relationship. But use different symbols for the functions, $G(s)$ is a different function than what you named $G(z)$, which then really is $G(\tfrac 1T\ln(z))$. $\endgroup$ – Lutz Lehmann Feb 18 '14 at 15:11

Your question makes no sense. Z transform is performed on a discrete signal/series.

Since $H(s)$ is a continuous function, you can't just calculate a Z-transform of $H(s)$ without first sampling it, to make it discrete. Also, it doesn't make much sense to do a time->spectrum transform (such as a Z-transform) on a spectral representation ($H(s)$)

I'm assuming that when you write "$Z-Transform(H(s))$", what you really want to do is to convert $H(s)\to H(z)$, meaning to calculate the Z-transform of $h[nT]$, where $h[nT]$ is $h(t)$, sampled at intervals of $T$, and $h(t)$ is the inverse Laplace transform of $H(s)$.

If I'm correct in my assumption, the transformation you are seeking is known as "star transform", which would provide a transform function $H^{*}(s)$, in terms of $e^{sT}$, which may be easily converted to $H(z)$ by way of the substitution $z=e^{sT}$.

Edit - some elaboration on the conversion process: what you need to do is calculate $H^{*}(s)$ from $H(s)$ using one of the two relations described in "Relation to Laplace transform" in the Wikipedia article, then do the substitution $z=e^{sT}$, to get $H(z)$.

  • $\begingroup$ Thanks for clarifying that z transform is for discrete signals while Laplace transform is for continuous. $\endgroup$ – gdaras Jun 19 '18 at 19:18
  • $\begingroup$ Are you sure? This video explains just "one of the methods" to convert from Laplace Domain to Z-domain youtube.com/watch?v=88tWmyBaKIQ . I'm not an expert, but also there are other lectures about converting continuous controllers into discrete controllers, and so on. Furthermore, even MATLAB has a function to do such conversions: mathworks.com/matlabcentral/answers/… . So I don't think this question "makes no sense" $\endgroup$ – Bersan Jan 25 at 1:22
  • $\begingroup$ @Bersan, you keep referring to a process where a function in the Laplace domain is converted to the Z-domain, which is exactly what my answer describes. The OP's question is about Z-transform operating on the Laplace domain, which makes no sense because Z-transform operates on a discrete signal, as I explained. $\endgroup$ – Sagie Jan 26 at 12:40
  • $\begingroup$ Also, by calling these processes simply "conversions" we lose details of what actually is happening. The bilinear transformation in the video you referred to is a mapping of a function from one domain to another. The MATLAB function you referred to calculates what the Z transform would've been for a system with the provided transfer function, if a ZOH was connected to its input. The starred transform in my answer calculates what the Z transform would've been for a system with the provided transfer function, if an ideal sampler was connected to its input. $\endgroup$ – Sagie Jan 26 at 13:09
  • $\begingroup$ @Sagie I don't think your answer is wrong, but I think it's confusing, I wasn't sure what you meant until after reading it multiple times. $Z\{G(s)\}$ is just the Z-transform of the inverse laplace transform of $G(s)$, sampled at time $T$. It's what this guy does here youtu.be/rL_1oWrOplk?t=228 So the answer to OP's question would be no. What you could say instead is $Z\{G(s) + H(s)\} = Z\{G(s)\} + Z\{H(s)\}$ $\endgroup$ – Bersan Feb 12 at 0:53

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