In Levine book "The control handbook" it is shown that, for discretizing a transfer function $\frac{1}{s}$ using Forward Euler i simply have to replace s with $\frac{z-1}{T}$. How can extend the proof and show that the substitution works for every possible transfer function $G(s)$ ?
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What do you mean by "works"? Of course you can replace $s$ by that function and see what you get. It is just one out of several ways to discretize a continuous system. There are other transformations (such as backward Euler) that will guarantee that a stable system transforms to a stable system. That is not the case with forward Euler. – Matt L. Jan 12 '20 at 17:52
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I don't understand what's the proof for the subtitution, to me laplace and z-transform are completely different things, so i don't understand why is the correct to apply that substitution. In my textbook, the proof is derived from a particular case, and i would like to generalize it. (Levine - The Control Handbook. Volume 1 pag 283) – themagiciant95 Jan 12 '20 at 18:22
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Does my answer clarify things a bit? – Matt L. Jan 12 '20 at 19:05
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Read this answer to see that the Laplace and Z-transforms are very much related. The Z-transform is in the discrete domain what the Laplace transform is in the continuous domain. – Matt L. Jan 12 '20 at 19:08
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@MattL., maybe here my doubt is more clear .... https://dsp.stackexchange.com/questions/63229/derive-the-forward-euler-substitution-for-transfer-function – themagiciant95 Jan 14 '20 at 10:22
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Another way to see how the forward Euler method approximates a continuous-time system is by considering the "ideal" mapping of the $s$-plane to the $z$-plane (why?):
$$z=e^{sT}\tag{1}$$
For frequencies that are much smaller than the sampling frequency (i.e., $|s|T\ll 1$) we can approximate $e^{sT}$ by its first order Taylor series:
$$z\approx 1+sT\tag{2}$$
which is exactly the forward Euler mapping, resulting in
$$s=\frac{z-1}{T}\tag{3}$$
Matt L.
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