EE611

Solution to Homework #21

1a)

b)

2.a)

» A=[-2 -1 -1;-1 -2 -1;1 1 0]
A =

-2	-1	-1
-1	-2	-1
1	1	0

» B=[0 -1;-1 -1;1 2]

B =

0	-1
-1	-1
1	2

» C=[1 -1 0;0 1 1]

C =

1	-1	0
0	1	1

» %put in CCF:

» M=[B A*B A*A*B]

M =

0	-1	0	1	0	-1
-1	-1	1	1	-1	-1
1	2	-1	-2	1	2

» rank(M)

ans =

2

» Tccf = [M(:,1:2) [0;0;1]]

Tccf =

0	-1	0
-1	-1	0
1	2	1

» rank(Tccf)

ans =

3

» Accf = inv(Tccf)*A*Tccf

Accf =

-1	0	0
0	-1	1
0	0	-2

» Ac = Accf(1:2,1:2)

Ac =

-1	0
0	-1

» %since uncontrolable eigenvalue is stable (-2), system is stabilizable!!

b)
» Bccf = inv(Tccf)*B
Bccf =

1	0
0	1
0	0

» Bc = Bccf(1:2,:)

Bc =

1	0
0	1

» %set eigenvalues to -2, -2:

» Kc=[1 0; 0 1]

Kc =

1	0
0	1

» eig(Ac-Bc*Kc)

ans =

-2

-2

» K = [Kc [0;0]] * inv(Tccf)

K =

1	-1	0
-1	0	0

» eig(A-B*K)

ans =

-2.0000 + 0.0000i

-2.0000 - 0.0000i

-2.0000

c)

» %all eigenvalues are stable, therefore system is detectable!!

d)
» C
C =

1	-1	0
0	1	1

» R = [0 0 1]

R =

0

0

1

» Tro = inv([C;R])

Tro =

1	1	-1
0	1	-1
0	0	1

» Abar = inv(Tro)*A*Tro

Abar =

-1	0	0
0	-1	0
1	2	-2

» A11 = Abar(1:2,1:2)

A11 =

-1	0
0	-1

» A12 = Abar(1:2,3)

A12 =

0

0

» A21 = Abar(3,1:2)

A21 =

1

2

» A22 = Abar(3,3)

A22 =

-2

» %Find Kro to set the eigenvalues of A22-KroA12

» %But A12 = [0;0]!! Because only eigenvalue

» %of the reduced-order observer is unobservable!!

» %Doesn't matter what we pick for Kro, we cannot set eigenvalue

» %so pick an easy Kro:

» Kro = [0 0]

Kro =

0

0

» Bbar = inv(Tro)*B

Bbar =

1	0
0	1
1	2

» B1 = Bbar(1:2,:)

B1 =

1	0
0	1

» B2 = Bbar(3,:)

B2 =

1

2

» Aro = A22-Kro*A12

Aro =

-2

» Bro = B2-Kro*B1

Bro =

1

2

» Cro = (A22-Kro*A12)*Kro-Kro*A11+A21

Cro =

1

2

» Q1 = Tro(:,1:2)

Q1 =

1	1
0	1
0	0

» Q2 = Tro(:,3)

Q2 =

-1

-1

1