Codigo RLC ingenieria control

%Determine la matriz de ganancia k, para el sistema RLC, considerando que
%se desea controlar el sistema con un ts=3s y un Mp=0.1

clear all;
clc;
format short g

%  MODELO DEL SISTMA


R1=1E3;
R2=10E3;
L=10E-3;
C=10E-6;

a=(R1+R2)*L*C;
b=(R1*R2*C+L);
c=R1;
d=L;

num=[d/a 0];
den=[a/a b/a c/a];
Gt=tf(num,den)

A=[-b/a -c/a;1 0];
B=[1;0];
C=[d*a 0];
D=[0];
Sistema=ss(A,B,C,D)
step(Sistema)


%#  POLOS DESEADOS


Mp=0.1;
ts=3;
E=sqrt(((log(Mp))^2)/((pi^2)+(log(Mp))^2))
wn=4/(ts*E)
Gd=tf(wn^2,[1,2*E*wn,wn^2])

hold on
step(Gd)


%#  Ackerman(K)

I=eye(size(A));
Co=[B A*B];
Coi=inv(Co);
Pd=[A^2+(2*E*wn)*A+(wn^2)*I]
K=[0 1]*Coi*Pd

Gt =

           9.091 s
 ---------------------------
 s^2 + 9.092e04 s + 9.091e05

Continuous-time transfer function.


Sistema =

 A =
              x1          x2
  x1  -9.092e+04  -9.091e+05
  x2           1           0

 B =
      u1
  x1   1
  x2   0

 C =
           x1       x2
  y1  1.1e-05        0

 D =
      u1
  y1   0

Continuous-time state-space model.


E =

     0.59116


wn =

      2.2555


Gd =

         5.087
 ---------------------
 s^2 + 2.667 s + 5.087

Continuous-time transfer function.


Pd =

   8.265e+09    8.265e+10
      -90916  -9.0909e+05


K =

      -90916  -9.0909e+05

[pic 1]

Published with MATLAB® R2020b

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