Discrepancy between using function 'feedback()' and calculating manually
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Hello,
I am working with MIMO systems and have made a comparison between the results obtained using the Matlab function 'feedback()' and computing the closed loop transfer function 'by hand'. There are differences between them I don't understand.
The code below illustrates the issue. I define a forward path array of transfer functions G and close the loop with K in the feedback path.
To compute the feedback manually I calculate CL = G/(I+K*G).
If you run this code fragment you will see that 'cl1' and 'cl2' do not agree. I don't understand why not. Can anyone enlighten me please?
Thanks,
Chris
g11 = tf(1,[1 0]);
g12 = tf(2,[1 0]);
g21 = tf(3,[1 0]);
g22 = tf(4,[1 0]);
G = [g11 g12;g21 g22];
k = 2;
K = [k 0;0 0];
cl1 = feedback(G,K,-1)
I = eye(2);
cl2 = G / (I + K*G)
采纳的回答
Paul
2011-4-5
0 个投票
cl2 can be simplfied by canceling s out of the numerator and denominator of each entry. After that you'll see that cl2 and cl1 are the same to within some roundoff. Or are you concerned about the roundoff?
3 个评论
Christopher
2011-4-6
Paul
2011-4-6
For most intents and purposes, the Bode plots are identical except at very, very low frequencies. The biggest discrepancy is in the (1,2) term and arises because of that phantom zero at 4.4e-16. If the difference at those freuqencies is an issue, then do as Arnaud suggested and take the minreal of both. But note that
isequal(minreal(cl1),minreal(cl2)) will still evaluate to false because of the roundoff errors.
Christopher
2011-4-7
更多回答（2 个）
Arnaud Miege
2011-4-6
Use minreal on both and you'll get the same answer:
g11 = tf(1,[1 0]);
g12 = tf(2,[1 0]);
g21 = tf(3,[1 0]);
g22 = tf(4,[1 0]);
G = [g11 g12;g21 g22];
k = 2;
K = [k 0;0 0];
I = eye(2);
cl1 = minreal(feedback(G,K,-1));
cl2 = minreal(G / (I + K*G));
bode(cl1,cl2)
HTH,
Arnaud
Walter Roberson
2011-4-5
0 个投票
I don't have whichever toolbox you are using, but if you were to indicate the outputs possibly I might have some ideas.
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2011-4-5
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