Discrepancy between using function 'feedback()' and calculating manually

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)

Answer 1

[Paul]

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?

[Christopher]

I am concerned because the Bode plots that these transfer functions represent are therefore not identical (that is the purpose of these calculations).

[Paul]

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]

Thank you, your explanation is clear and helpful. Much appreciated.

Chris

Answer 2

[Arnaud Miege]

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

[Christopher]

Thank you, Arnaud - very helpful.

Chris

Answer 3

[Walter Roberson]

I don't have whichever toolbox you are using, but if you were to indicate the outputs possibly I might have some ideas.

Have you read the numeric FAQ

[Christopher]

Hello, I am using the Control toolbox (which contains 'feedback()'.

Here is the output of the script:

Transfer function from input 1 to output...

1

#1: -----

s + 2

3

#2: -----

s + 2

Transfer function from input 2 to output...

2 s + 4.441e-016

#1: ----------------

s^2 + 2 s

4 s - 4

#2: ---------

s^2 + 2 s

Transfer function from input 1 to output...

s

#1: ---------

s^2 + 2 s

3 s

#2: ---------

s^2 + 2 s

Transfer function from input 2 to output...

2 s^2

#1: -----------

s^3 + 2 s^2

4 s^2 - 4 s

#2: -----------

s^3 + 2 s^2

The transfer functions should match, and they do not.

I'm afraid I have not read the FAQ, but I will.

Thank you,

Chris