Now draw simulation diagram:EE611
Solution to Homework #26
1a) Draw the simulation diagram for:
Now draw simulation diagram:
b) Determine the stability of 
if
i)
Therefore system is ASYMPTOTICALLY STABLE!!
ii)
Therefore system is UNSTABLE!!
iii)
Therefore system is UNSTABLE!!
c)
i) 
:
ii) No overshoot, or z real between 0 and 1:
iii) both i) and ii)
d) Find eigenvalues and eigenvectors:
e) Draw a simulation diagram for your system in decoupled form
f)
i) stable eigenvalues are {1/2, -1/2}
ii) controllable eigenvalue is {2}
iii) observable eigenvalues are {1/2, -1/2}
g) The system IS stabilizable, since all uncontrollable eigenvalues are stable!!
The system is NOT detectable, since the unobservable egeinvalue is unstable!!
2.
» A=(1/20)*[-4 -9 0 -9 -9;-18 -13 0 -18 -18;18 18 5 18 18;-9 -9 0 -4 -9;-9 -9 0 -9 -4]
A =
|
-0.2000 |
-0.4500 |
0 |
-0.4500 |
-0.4500 |
|
-0.9000 |
-0.6500 |
0 |
-0.9000 |
-0.9000 |
|
0.9000 |
0.9000 |
0.2500 |
0.9000 |
0.9000 |
|
-0.4500 |
-0.4500 |
0 |
-0.2000 |
-0.4500 |
|
-0.4500 |
-0.4500 |
0 |
-0.4500 |
-0.2000 |
» B=(1/5)*[0 -1;5 3;0 2;0 -1; 0 -1]
B =
|
0 |
-0.2000 |
|
1.0000 |
0.6000 |
|
0 |
0.4000 |
|
0 |
-0.2000 |
|
0 |
-0.2000 |
» C=[1 1 0 1 1;1 2 1 1 1;1 0 1 0 1]
C =
|
1 |
1 |
0 |
1 |
1 |
|
1 |
2 |
1 |
1 |
1 |
|
1 |
0 |
1 |
0 |
1 |
» M = [B A*B A*A*B A*A*A*B A*A*A*A*B]
M =
|
0 |
-0.2000 |
-0.4500 |
-0.0500 |
0.7875 |
-0.0125 |
-1.6031 |
-0.0031 |
3.1992 |
-0.0008 |
|
1.0000 |
0.6000 |
-0.6500 |
0.1500 |
1.6375 |
0.0375 |
-3.1906 |
0.0094 |
6.4023 |
0.0023 |
|
0 |
0.4000 |
0.9000 |
0.1000 |
-1.5750 |
0.0250 |
3.2063 |
0.0063 |
-6.3984 |
0.0016 |
|
0 |
-0.2000 |
-0.4500 |
-0.0500 |
0.7875 |
-0.0125 |
-1.6031 |
-0.0031 |
3.1992 |
-0.0008 |
|
0 |
-0.2000 |
-0.4500 |
-0.0500 |
0.7875 |
-0.0125 |
-1.6031 |
-0.0031 |
3.1992 |
-0.0008 |
» rank(M)
ans =
2
» Tccf = [M(:,1:2) [0;0;1;0;0] [0;0;0;1;0] [0;0;0;0;1]]
Tccf =
|
0 |
-0.2000 |
0 |
0 |
0 |
|
1.0000 |
0.6000 |
0 |
0 |
0 |
|
0 |
0.4000 |
1.0000 |
0 |
0 |
|
0 |
-0.2000 |
0 |
1.0000 |
0 |
|
0 |
-0.2000 |
0 |
0 |
1.0000 |
» Accf = inv(Tccf)*A*Tccf
Accf =
|
-2.0000 |
0.0000 |
0 |
-2.2500 |
-2.2500 |
|
2.2500 |
0.2500 |
0 |
2.2500 |
2.2500 |
|
0 |
0 |
0.2500 |
0 |
0 |
|
0 |
0 |
0 |
0.2500 |
0 |
|
0 |
0 |
0 |
0 |
0.2500 |
» Ac = Accf(1:2,1:2)
Ac =
|
-2.0000 |
0.0000 |
|||
|
2.2500 |
0.2500 |
» Ac_bar = Accf(3:5,3:5)
Ac_bar =
|
0.2500 |
0 |
0 |
||
|
0 |
0.2500 |
0 |
||
|
0 |
0 |
0.2500 |
» Bccf = inv(Tccf)*B
Bccf =
|
1 |
0 |
|||
|
0 |
1 |
|||
|
0 |
0 |
|||
|
0 |
0 |
|||
|
0 |
0 |
» Bc = Bccf(1:2,:)
Bc =
|
1 |
0 |
|
0 |
1 |
» eig(Ac)
ans =
-2.0000
0.2500
» %a) Controllable eigenvalues are {-2, 0.25}!!
» eig(Ac_bar)
ans =
0.2500
0.2500
0.2500
» %b) System is stabilizable!!
c) Design a feedback regulator
» %A is cyclic, so pick Ks:
» Ks = [1;1]
Ks =
1
1
» bs = Bc*Ks
bs =
1
1
» %check controllability of (Ac,bs):
» Ms = [bs Ac*bs]
Ms =
|
1.0000 |
-2.0000 |
|
1.0000 |
2.5000 |
» rank(Ms)
ans =
2
» %put in PV form:
» poly(Ac)
ans =
|
1.0000 |
1.7500 |
-0.5000 |
» Apv = [0 1;-ans(3) -ans(2)]
Apv =
|
0 |
1.0000 |
|
|
0.5000 |
-1.7500 |
» eig(Apv)
ans =
0.2500
-2.0000
» bpv =[0;1]
bpv =
0
1
» Mpv = [bpv Apv*bpv];
» poly_desired = poly([.1,.1])
poly_desired =
|
1.0000 |
-0.2000 |
0.0100 |
» Kpv = [poly_desired(3)+Apv(2,1) poly_desired(2)+Apv(2,2)]
Kpv =
|
0.5100 |
-1.9500 |
» Tpv_inverse = Mpv*inv(Ms)
Tpv_inverse =
|
-0.2222 |
0.2222 |
|
|
0.9444 |
0.0556 |
» Kc = Ks*Kpv*Tpv_inverse
Kc =
|
-1.9550 |
0.0050 |
|
|
-1.9550 |
0.0050 |
» eig(Ac-Bc*Kc)
ans =
0.1000
0.1000
» K = [Kc [0 0 0;0 0 0]]*inv(Tccf)
K =
|
-5.8900 |
-1.9550 |
0 |
0 |
0 |
|
-5.8900 |
-1.9550 |
0 |
0 |
0 |
» eig(A-B*K)
ans =
0.2500
0.1000
0.1000
0.2500
0.2500
» %settling time depends on eigenvalue closest to the unit circle:
» T=1/100;
d)
» ts = -4/((1/T)*log(.25))
ts =
0.0289
e)
» y_ref = [6;-4]
y_ref =
6
-4
» inv([[C(1:2,:) [0 0;0 0]];[A-eye(5,5) B]])*[1 0;0 1;0 0;0 0;0 0;0 0;0 0]
ans =
|
0.4000 |
-0.2000 |
|
-0.2000 |
0.6000 |
|
-0.8000 |
0.4000 |
|
0.4000 |
-0.2000 |
|
0.4000 |
-0.2000 |
|
3.0000 |
0.0000 |
|
-3.7500 |
0.7500 |
» Nx = ans(1:5,:)
Nx =
|
0.4000 |
-0.2000 |
|
-0.2000 |
0.6000 |
|
-0.8000 |
0.4000 |
|
0.4000 |
-0.2000 |
|
0.4000 |
-0.2000 |
» Nu = ans(6:7,:)
Nu =
|
3.0000 |
0.0000 |
|
-3.7500 |
0.7500 |
f)
g)
» t = [0:.01:2];
» a=A-B*K;
» u=[(Nu+K*Nx)*y_ref*ones(1,length(t))]';
» [y,x]=dlsim(a,B,C,0,u);
» u0=[(0+K*Nx)*y_ref*ones(1,length(t))]';
» [y0,x0]=dlsim(a,B,C,0,u0);
» figure(1)
» plot(t,y(:,1:2))
» grid
» title('1h) Plot of y1 and y2.')
» xlabel('Time (sec.)')
» ylabel('Output')
» figure(2)
» plot(t,y0(:,1:2))
» grid
» title('1i) Plot of y1 and y2, Nu=0')
» xlabel('Time (sec.)')
» ylabel('Output')
h) zero steady state error, ts is about .05 seconds
i) For Nu = 0:
