Hi, I was attempting to calculate the gain matrix and transfer function for full

Question

Hi, I was attempting to calculate the gain matrix and transfer function for full-state feedback control with a system with design specifications indicated below. I have provided the code I have written in order to obtain the gain matrix and the transfer function. It us reprsented in italics below.

% assuming 25% O.S
po= 0.25;
% assuming Ts =< 1.5 sec (aka 1.3)settling time
Ts= 1.3
% find your sigma/damping ration using formula found online everywhere
Sigma=(-log(po))/(sqrt(pi^2 + log(po)^2))
%now we can find our Wn (natural angular frequency) using this formula Wn =
%4/(sigma*Ts)
Wn = 4/(Sigma *Ts)
%we can start by writting our characteristic polynomial
%you'll have a 2nd order polynomuial like this (s^2 + 2*Sigma*Wn + Wn^2)
%in general second order C.P's roots can be found like this
%(-Sigma*Wn) +- Wn*j*sqrt(1-Sigma^2) with Wd = Wn*sqrt(1-Sigma^2)

root1sec = (-Sigma*Wn) + Wn*j*sqrt(1-Sigma^2)
root2sec = (-Sigma*Wn) - Wn*j*sqrt(1-Sigma^2)

%for a third order system, we split it into a 2nd order system
%( aka s^2 +2*Sigma*Wn + Wn^2) times a third root
%reference:
%http://www.cds.caltech.edu/~murray/courses/cds101/fa02/caltech/astrom-ch4.pdf
%page 166
%the third root is simply (s + Sigma*Wn)
root3 = -Sigma*Wn

%finally for the fourth root, we'll set the value to be -25 since it has to
%be negative and hope for the best
root4= -29

%now that we got our roots we can use the original A and B matrices from 1)
%we can now find our gain matrix K
gr = 1.5;
gr_prime= 1.5;
Jd = 0.0004;
Jl = 0.0065;
Jp = 0.00055;
k=5.8;
Jd_star = 0.0016;
s = tf('s');

A = [0 1 0 0; -k*gr^-2/Jd_star -c1/Jd_star k*gr^-1/Jd_star 0; 0 0 0 1; k*gr^-1/Jl 0 -k/Jl -c2/Jl]
B = [0 ; 1/Jd_star; 0 ; 0]
C = [1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1];
D = [0; 0; 0; 0];

%finding gain

p = [root1sec root2sec root3 root4]
K = place (A,B,p)

% finding the transfer function

Af = A-B*K
I = eye(4)
T=(C*((inv((s*I)-Af))*B))+D

I am confused about how to interpret the result that it gives for the transfer function. It says that the transfer function is:

" From input to output...

              625 s^2 + 480.8 s + 5.577e05
   1: -------------------------------------------
       s^4 + 38.23 s^3 + 344.7 s^2 + 2412 s + 5183

       625 s^3 + 480.8 s^2 + 5.577e05 s - 1.315e-08
   2: --------------------------------------------
       s^4 + 38.23 s^3 + 344.7 s^2 + 2412 s + 5183

                        3.718e05
   3: -------------------------------------------
       s^4 + 38.23 s^3 + 344.7 s^2 + 2412 s + 5183

                 3.718e05 s - 1.166e-09
   4: -------------------------------------------
       s^4 + 38.23 s^3 + 344.7 s^2 + 2412 s + 5183 "

What does this mean? What would the transfer function end up being?

Explanation / Answer

Hello,
          Please find the answer attached as under. Please give a thumbs up rating if you find the answer useful! Have a rocking day ahead!
The system that you have provided in the question is a 1x4 system i.e. there is one input and 4 outputs. This is evident from the dimensions of the A, B and C matrix. Thus, there is nothing like " a single transfer function" of the system. There will be four transfer functions, and the transfer function describing this whole system will be a 4X1 matrix of the individual transfer functions.

Coming to your question of pole placement, take the dominant dynamics within these 4 systems (take those transfer functions which come within the feedback loop), and assess the position of the poles in each transfer function. They most probably will be the same as those you had placed. Thus, the trick is to individually evaluate each transfer function and see the location of the poles (desired v/s actual).

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