How do I plot the bode plot of the transfer function with exp(sT)?

13 ビュー (過去 30 日間)
yklu008
yklu008 2019 年 5 月 6 日
回答済み: Raj 2019 年 5 月 6 日
The transfer function is show as following, where T=10e-9.
How could I get the bodeplot from this function?
59538682_243684739829449_8500004122919960576_n.jpg
  1 件のコメント
David Wilson
David Wilson 2019 年 5 月 6 日
Since you have the time delay element in the denominator, your don't have a polynomial transfer function anymore, so you will need to approximate it using pade.
You also might want to check that it is exp(-Ts), not the acausal version you have written above.

サインインしてコメントする。

採用された回答

Raj
Raj 2019 年 5 月 6 日
You can rearrange the equation as (e^-sT)(s+10)/s(1-(e^-sT))
Then your code should look something like this:
s = tf('s');
T=10e-9;
f = exp(-T*s);
sys=(f*(s+10))/(s*(1-f))
bode(sys)

その他の回答 (0 件)

製品


リリース

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by