Lsim inputs inversion provide the same result

2022 4 13
1 个回答
回答已采纳
更新时间：2022 4 14
7 次查看（30 天）

0 个投票

Hello to everybody,
I am using Matlab 2019b.
Provided that sys is a transfer function written using the command tf, u a time signal and t the corresponding time vector of the same length of u.
If I write: y=lsim(sys,u,t) I get the very same result of writing y=lsim(u,sys,t), with no warnings at all. Why is this happening?
Thank you very much in advance.

请先登录，再回答此问题。

 采纳的回答

Paul
Paul 2022-4-13
Ran in:

1 个投票

Ran in:
Interesting. A quick look at the code, at least for the case where an output argument is specified, shows that the function that parses the inputs to lsim() doesn't enforce that the first argument(s) in the list must be a dynamic system object. In fact, I don't think it enforces any order other than that u comes before t, and t comes before x0.
For example:
sys = tf(1,[1 1]);
t = 0:10;
u = rand(size(t));
method = 'foh';
y1 = lsim(sys,u,t,method);
y2 = lsim(method,u,sys,t);
isequal(y1,y2)
ans = logical
1
Not sure TMW would consider this a bug. Certainly undocumented behavior.
Did it cause a problem or just asking out of curiosity?

1 个评论
显示 -1更早的评论 隐藏 -1更早的评论

Lorenzo Bernardini
Lorenzo Bernardini 2022-4-14
Thank you very much for your answer Paul!
Actually, it was a curiosity. I found this behaviour very strange and I was wondering if I was the only one to experience it. The point is that sys and u are very different inputs, as you know, and it sounded remarkably weird that the inversion gave me equal outcomes to the case in which the inputs were inserted in the suggested way. With the inversion, I was expecting at least to get different results or a warning/error.
Again, thank you for your time and clear answer.

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Programming 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by