782 Chapter 10 / Control Systems Design in State SpaceMATLAB Program 10–14
% Determination of transfer function of observer controller.
A = [0 1 0;0 0 1;0 -24 -10];
B = [0;10;-80];
Aaa = 0; Aab = [1 0]; Aba = [0;0]; Abb = [0 1;-24 -10];
Ba = 0; Bb = [10;-80];
Ka = 1.25; Kb = [1.25 0.19375];
Ke = [-1;6.25];
Ahat = Abb - Ke*Aab;
Bhat = AhatKe + Aba - KeAaa;
Fhat = Bb - Ke*Ba;
Atilde = Ahat - Fhat*Kb;
Btilde = Bhat - Fhat(Ka + KbKe);
Ctilde = -Kb;
Dtilde = -(Ka + Kb*Ke);
[num,den] = ss2tf(Atilde,Btilde,-Ctilde,-Dtilde)
num =
1.2109 11.2125 25.3125
den =
1.0000 6.0000 2.1406
Next, we shall obtain the transfer function of the observer controller. MATLAB
Program 10–14 produces this transfer function as follows:
=
1.2109(s+5.3582)(s+3.9012)
(s+5.619)(s+0.381)
Gc(s)=
1.2109s^2 +11.2125s+25.3125
s^2 +6s+2.1406
Notice that this is a stable controller. Define the system with this observer controller as
System 2. We shall proceed to obtain the response of System 2 to the given initial
condition:
By substituting into the state-space equation for the plant, we obtain
=Ax-BKbx- B (10–110)
0
e
Rr =Ax-BKx+BCKa KbDB
0
e
R
x
=Ax-BK x =Ax-BKB
xa
xb
R =Ax-BKB
xa
xb-e
R
u=-Kx
x(0)=C
1
0
0
S, e(0)= B
1
0
R
Openmirrors.com