integral square error calculation
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i want to calculate ise of a transfer function for which i have to claculate inverse laplace
this is my code...
num=[3 2]
den=[155.0388 33.5652 2]
R2=tf(num,den)
num1=[1 8 20 16 3 2]
den1=[1 18.3 102.4 209.5 155.9 33.6 2]
G6=tf(num1,den1)
E=G6-R2
et=ilaplace(E)
gives an error
please help me out in calculation of ise for E
回答 (1 件)
Walter Roberson
2021 年 1 月 7 日
編集済み: Walter Roberson
2021 年 1 月 7 日
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3 件のコメント
apoorv srivastava
2021 年 1 月 7 日
Walter Roberson
2021 年 1 月 7 日
FT_time_domain is going to be symbolic -- in particular a symbolic sum of 8 roots of a polynomial is used multiple times in the expression.
collect() on it is not going to have any useful effect as there are no exp(-t) terms -- the terms are exp(something*t) and collect() is not smart enough to collect those. But doing the collect() does not hurt.
Either way the result of the collect() is going to be symbolic in the variable t .
You cannot trapz() a symbolic expression.
You could, however,
endtime = 1; %as appropriate
ise = trapz(0.1, subs(FT_time_domain, t, 0:0.1:endtime).^2)
apoorv srivastava
2021 年 1 月 7 日
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