Consider acontrol system with the following transfer matrix:
\$G(s)=\begin{pmatrix} G_{11}(s) & G_{12}(s)\\ 0&G_{22}(s) \end{pmatrix}\$
and suppose I want to get the transfer functions for the two channels in order to do some analysis in Matlab, for example look if there are pole at the origins or RHP zeros, or look at the effect of coupling, I am not sure if I should do as follows:
\$y_{2}(s)=\frac{G_{22}(s)}{1+G_{22}(s)}u_2(s)\$
and for the first channel is:
\$y_{1}(s)=\frac{G_{11}(s)+G_{12}(s)}{1+G_{11}(s)+G_{12}(s)}\$
Or I have to do something like:
\$y_{1}(s)=\frac{G_{11}(s)}{1+G_{11}(s)}u_1(s)+\frac{G_{12}(s)}{1+G_{12}(s)}u_2(s)\$
I ask because I have not clear how to deal with the effect of coupling when I want to define the transfer functions for each channel.
Using Matlab, I have done:
F4_cross = (F4_min(1,1)/(1+F4_min(1,1)))+(F4_min(1,2)/(1+F4_min(1,2)));
where I considered \$F4_{min}(1,1)\$ as \$G(1,1)\$ and the same for \$G(1,2)\$ for reasons due to previous code. The problem is that \$F4_{min}(1,1)\$ has a pole at the origin, while \$F4_{min}(1,2)\$ doesn't. The controller then should have a pole at the origin for \$F4_{min}(1,2)\$ and for the other one not, if the steady state error has to be zero. Moreover, the step response for the coupled system F4_cross
is :
so it is present a steady state error.
If I consider for example:
\$F4_{min}(1,1)+F4_{min}(1,2)\$
as open loop, I think this is wrong since it is a system with two inputs which are differents.
If I plot the step response I have two different step responses, one with infinite steady sate error and one with zero steady state error.
The way of operating should be to apply the decoupling, bit if I don't want to apply it I donìt see a way out.
I have tried doing:
loopsens(F4_cross_min,k1_1/s)
and this gives me zero steady state error, since F4_cross_min
does not have a pole at the origin, but the syntax for loopsens
is loopsens(plant,controller)
, and it closes the loop. But in this case F4_cross_min
it is a closed loop, but this is the only solution I have found that works at least on the calculations, but I think it is conceptually wrong.
Can somebody please help me?