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Jan 14, 2020 at 10:22 comment added themagiciant95 @MattL., maybe here my doubt is more clear .... dsp.stackexchange.com/questions/63229/…
Jan 12, 2020 at 19:08 comment added Matt L. 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.
Jan 12, 2020 at 19:05 comment added Matt L. Does my answer clarify things a bit?
Jan 12, 2020 at 18:22 comment added themagiciant95 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)
Jan 12, 2020 at 18:00 answer added Matt L. timeline score: 2
Jan 12, 2020 at 17:52 comment added Matt L. 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.
Jan 12, 2020 at 17:28 history asked themagiciant95 CC BY-SA 4.0