[Z transform] problem with MATLAB
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Franklin Ngo
el 11 de Sept. de 2017
Editada: Jeovane Sousa
el 21 de Abr. de 2021
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
el 11 de Sept. de 2017
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
el 12 de Sept. de 2017
As always, my pleasure.
If my Answer hslped you solve your problem, please Accept it!
Jeovane Sousa
el 21 de Abr. de 2021
Editada: Jeovane Sousa
el 21 de Abr. de 2021
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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