Didviding and multiplying transfer functions

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Marko
Marko 2016 年 4 月 6 日
コメント済み: Syndi Katya 2024 年 2 月 21 日
Hi all,
I have a question on transfer function division and multiplication. If I calculate the closed loop function using:
G/(1+G*H)
I obtain different results than if I try calculating it by hand or using command.
feedback(G,H)
For example
4.244e-12 s^3 + 8.004e-08 s^2 + 8e-07 s
------------------------------------------------------------------------
1.801e-19 s^5 + 6.792e-15 s^4 + 6.407e-11 s^3 + 4.308e-08 s^2 + 0.0008 s
This is the closed loop using first command. This is with feedback command:
0.0001 s + 0.001
-------------------------------------------
4.244e-12 s^3 + 8.004e-08 s^2 + 8e-07 s + 1
While calculating it by hand gives:
0.0001 s + 0.001
-------------------------------------------
4.244e-12 s^3 + 1.224e-07 s^2 + 8e-07 s + 1
Do you know how is this possible?
Thanks!
  2 件のコメント
J. Carlos Aguado
J. Carlos Aguado 2020 年 10 月 7 日
You just have to apply "minreal" to your result, and you will see that minreal(G/(1+G*H)) = feedback(G, H). MatLab should you that automatically.
Syndi Katya
Syndi Katya 2024 年 2 月 21 日
Nice trick

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回答 (2 件)

Debarati Banerjee
Debarati Banerjee 2016 年 4 月 12 日
編集済み: Debarati Banerjee 2016 年 4 月 12 日
Can you first apply the function ' minreal ' on your 'G' and 'H' functions before applying the closed loop formula? This 'minreal' function will reduce the systems into minimal order.
Also, can you please provide the G and H functions with which you arrived at these results?
Cheers!
Debarati

yu-hsien chen
yu-hsien chen 2018 年 10 月 14 日
編集済み: Walter Roberson 2020 年 10 月 7 日
This is a really really late reply but the above link explained it.
  2 件のコメント
J. Carlos Aguado
J. Carlos Aguado 2020 年 10 月 7 日
That link does not work any more
Walter Roberson
Walter Roberson 2020 年 10 月 7 日

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