[Z transform] problem with MATLAB

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Franklin Ngo
Franklin Ngo 2017-9-11
编辑： Jeovane Sousa 2021-4-21
Hi experts, I have a question about Z-transform on MALTAB. When I convert a Laplace function F(s)=1/s to Z function, MATLAB says it is T/(z-1), but the Laplace-Z conversion table show that is z/(z-1). I know MATLAB cannot wrong because I drew a step graph of all these three functions. But all the books I found about Laplace and Z-transform also say the conversion table is right.
It's confusing me and I need an answer. Could anyone give me an explain for that? Thank in advanced.

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Star Strider
Star Strider 2017-9-11
The Symbolic Math Toolbox doesn’t know you’re trying to convert it from the Laplace domain to the ‘z’ domain. You have to play by its rules.
This works:
syms s t z
T1(s) = 1/s
T2(t) = ilaplace(T1)
T3(z) = ztrans(T2)
T1(s) =
1/s
T2(t) =
1
T3(z) =
z/(z - 1)
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Star Strider
Star Strider 2017-9-12
As always, my pleasure.
If my Answer hslped you solve your problem, please Accept it!
Jeovane Sousa
Jeovane Sousa 2021-4-21
编辑：Jeovane Sousa 2021-4-21
As I understand from here https://www.mathworks.com/help/control/ug/continuous-discrete-conversion-methods.html#bs78nig-12
Matlab uses the following method to calculate c2d by ZOH
The ZOH block generates the continuous-time input signal u(t) by holding each sample value u(k) constant over one sample period:
The signal u(t) is the input to the continuous system H(s). The output y[k] results from sampling y(t) every Ts seconds.
Resolving to :
each term will look something like that
thus,
And
Or remember the z-transform of a signal tha goes through a sample and holder, tha is:
This is what I think how c2d was implemented. No garantees it is correct or valid.
I hope I have made the correct considerations. Feel free to complement or point any mistake. Sorry about the poor English, it is not my primary language.

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