Re: Feedback for octave

[Doug Stewart]

Andrei Gutierrez wrote:
> I have a problem to define the arguments for the feedback for control.
> for example:
> num=1;
> den=[1 10 1];
> feedback(tf(num,den),tf(1,1))
>
> It is OK, but if My H(s) is not 1, for example H(s) = s + 1 ...
>
> numh=[1 1];
> denh=1;
> feedback(tf(num,den),tf(numh,denh))
>
> Gives me the error that the zeros are more than the poles, but in the
> feedback its common to have a numerator like this.
> Any idea?-.
> Thank you.
>
>
> _______________________________________________
> Help-octave mailing list
> address@hidden
> https://www.cae.wisc.edu/mailman/listinfo/help-octave
>
>   
You can get around this problem by adding a pole out at a "high" 
frequency so that its influence on the system will be minimal.
So try this



num = 1
den=[1 10 1]

numh=[1 1]
denh=1

%%now make a pole at 100
%%with a gain of 1
np=100
dp=[1 100]
%% now make H  as:
nh=conv(numh,np)
dh=conv(1,dp)

sys1=feedback(tf2sys(num,den),tf2sys(nh,dh));

this makes a pole and a zero at about   -100 and they cancel each other out.
and your step  response and  and bode will be OK in the region that you 
are concerned about.


the zero is at -100
the pole of the closed loop using this method are:

 -98.8865
 -10.9285
 -0.1851

the theoretical poles are
 -10.8151
 -0.1849

If you put the added pole out farther then it even has less effect.

Doug Stewart