Estimate poles/zeros given a transfer function
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Hi,
Giving a transfer function (vector of frequencies and values) - is it possible to obtain poles and zeros?
From a circuit simulator I get exported data as CSV with first column frequency, second real part and third imaginary.
I tried using the system identification toolbox:
data = csvread('export.matlab',1);
f = data(:,1);
H = data(:,2) + 1i*data(:,3);
fdata = iddata(H, ones(size(H)), 0, 'Frequency', 2*pi*f);
sys = tfest(fdata, 2, 1);
He = bode(sys, 2*pi*f);
semilogx(f, 20*log10(abs([ H , squeeze(He) ])));
legend('act','estimate');
However, this is the result:
Changing nr. of poles/zeros changes results significantly but none comes to the actual curve even close.
Am I doing something wrong?
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回答 (1 件)
Star Strider
2015 年 9 月 15 日
編集済み: Star Strider
2015 年 9 月 15 日
I may not completely understand your Question.
It’s been a while since I used the System Identification Toolbox, but you probably need to ask tfest to estimate a higher-order system, unless you know that it only has two poles and one zero.
The poles of the system are the roots of the denominator (‘sys.den’) of the transfer function, and the zeros are the roots of the numerator (‘sys.num’).
2 件のコメント
Star Strider
2015 年 9 月 15 日
You’re telling iddata that the sample time (Ts) is zero, meaning that it is continuously sampled. (I doubt any ADC is that fast!) That could be confusing it.
In my experience (and in the documentation), the data supplied to iddata is in the time domain. What is ‘data(:,1)’? Is it time or frequency?
What do you know about the circuit you’re simulating?
Are you certain that the output is complex? All the real-world output I’ve seen is real. What exactly are the outputs?
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