Different outputs with same inputs

5 次查看（过去 30 天）
Paolo De Ciccio
Paolo De Ciccio 2020-12-19
评论： Paolo De Ciccio 2020-12-19
Hello everyone,
, in
As you can see in image I'm trying to write in different forms a transfer function wich has {10+j, 10-j} poles. I obtained two different expressions, in facts in the 1st case I have {10+j, 10-j} poles, in the 2nd case I have {10, 10} poles, without no imag part.
I also tried an other case: a transfer function wich has {-10+j, -10-j} poles, but in this attempt I multiplyied the imag part with a number very close to 1, and I obtained the desiderd result: {-10+j, -10-j}. I believe that is so strange, so the question is: why I have different outputs with same inputs? It's a bug? Memory management issue? A typo?
My setup is Matlab R2020b, Windows 10 - 64 bit, Ryzen 7 4700U - 8 GB RAM

请先登录，再回答此问题。

采纳的回答

Cris LaPierre
Cris LaPierre 2020-12-19
Ran in:
Not sure what the question is, but something is not adding up in your examples. Try clearing your workspace and running your examples again.
s=tf('s');
1/((s+10)^2 +1)
ans = 1 ---------------- s^2 + 20 s + 101 Continuous-time transfer function.
% Your screenshot shows "+100" in the denominator
1/((s+i+10)*(s-i+10))
ans = 1 ---------------- s^2 + 20 s + 101 Continuous-time transfer function.
% Your screenshot shows "+100" in the denominator
1/((s+i-10)*(s-i-10))
ans = 1 ---------------- s^2 - 20 s + 101 Continuous-time transfer function.
If you know your zero, pole, and gain, you can also use that to create your transfer function?
Z = [];
P = [-10-1i -10+1i];
K = 1;
G = zpk(Z,P,K)
G = 1 ----------------- (s^2 + 20s + 101) Continuous-time zero/pole/gain model.
  12 个评论
显示 10更早的评论
Cris LaPierre
Cris LaPierre 2020-12-19
Ran in:
Technically yes. You can create custom functions that can be called by your code, but are not variables that appear in the workspace. That is likely not the case with i.
This is a computer program, though, so sometimes clearig the memory or just restarting the program can fix unexpeted behavior.
Recreating your eamples again:
s=tf('s');
i=0;
1/((s+i+10)*(s-i+10))
ans = 1 ---------------- s^2 + 20 s + 100 Continuous-time transfer function.
1/(s+1.00000000001i-10)/(s-1.00000000001i-10)
ans = 1 ---------------- s^2 - 20 s + 101 Continuous-time transfer function.
% Now fix the issue by clearing i
clear i
1/((s+i+10)*(s-i+10))
ans = 1 ---------------- s^2 + 20 s + 101 Continuous-time transfer function.
Paolo De Ciccio
Paolo De Ciccio 2020-12-19
I can accept only one answer, but I appreciated a lot all of yours

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

产品


版本

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by