[Z transform] problem with MATLAB
96 ビュー (過去 30 日間)
古いコメントを表示
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.
0 件のコメント
採用された回答
Star Strider
2017 年 9 月 11 日
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)
5 件のコメント
Star Strider
2017 年 9 月 12 日
As always, my pleasure.
If my Answer hslped you solve your problem, please Accept it!
Jeovane Sousa
2021 年 4 月 21 日
編集済み: Jeovane Sousa
2021 年 4 月 21 日
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.
その他の回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Number Theory についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!