Bode form of a transfer function
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Suppose I have the transfer function k*(s-z1)/((s - p1)*(s - p2)), i.e., in the form that comes from the zpk() function. This form is convenient in that I can see immediately the location of the poles and zeros, but in order to obtain the gain at s=0, I need to compute k*(-z1)*(-p1)*(-p2). Is there a way to instead have the transfer function displayed in Bode form: k2*(-s/z1 + 1)/((-s/p1 + 1)*(-s/p2 + 1)), so that k2 = (-z1)*(-p1)*(-p2), and this represents the gain at s=0? I find this much nicer because I can see the poles, zeros, and the gain at s=0 without having to do any calculations in my head.
Thanks, Luke
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採用された回答
Arkadiy Turevskiy
2013 年 6 月 27 日
>> sys=zpk(1,[2 3],4)
sys =
4 (s-1)
-----------
(s-2) (s-3)
Continuous-time zero/pole/gain model.
>> sys.DisplayFormat='frequency';
>> sys
sys =
-0.66667 (1-s)
---------------
(1-s/2) (1-s/3)
Continuous-time zero/pole/gain model.
その他の回答 (1 件)
David Sanchez
2013 年 6 月 27 日
s=tf('s');
% dummy values for the parameters
k=1;
z1=1;
p1=1;
p2=1;
H=k*(s-z1)/((s - p1)*(s - p2));
k2= (-z1)*(-p1)*(-p2);
H2=k2*(-s/z1 + 1)/((-s/p1 + 1)*(-s/p2 + 1));
bode(H);
figure;
bode(H2);
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