Sol'n to HW#23

1. Recall the continuous-time, state variable model from HW#16: (Please, use Matlab liberally on this first problem!)

a) Is the continuous model stable, marginally stable, or unstable?

» a=[3/2 -1/2 0;1/2 5/2 1;1 1 2]

a =

1.5000 -0.5000 0

0.5000 2.5000 1.0000

1.0000 1.0000 2.0000

» b=[1/2;1/2;0]

b =

0.5000

0.5000

0

» c=[2 2 0]

c =

2 2 0

» eig(a)

ans =

2.0000

3.0000

1.0000

» %1a) The continuous time system is UNSTABLE because eigenvalues are in the RHP.

b) Discretize the model using with Ts=ln(2) sec. Remember, if A^-1 exists then you can use the trick that

» T=log(2)

T =

0.6931

» Ahat=expm(a*T)

Ahat =

2.5000 -1.5000 -0.5000

2.5000 6.5000 3.5000

3.0000 3.0000 5.0000

» Bhat=inv(a)*[expm(a*T)-eye(3)]*b

Bhat =

0.4167

1.2500

0.6667

c) Is the discrete model stable, marginally stable, or unstable? Is it controllable (find )?

» eig(Ahat)

ans =

4.0000

8.0000

2.0000

» %1c) The discrete time system is UNSTABLE because eigenvalues are outside U.C.

» M=[Bhat Ahat*Bhat Ahat^2*Bhat]

M =

0.4167 -1.1667 -24.3333

1.2500 11.5000 101.0000

0.6667 8.3333 72.6667

» rank(M)

ans =

3

» %1c) The discrete system is completely controllable

» s_desired=[-2 -4 -5]

s_desired =

-2 -4 -5

d) Pick closed-loop eigenvalues in the S-plane of {-2 -4 -5} and map these values into the Z-plane.

» z_desired=exp(s_desired*T)

z_desired =

0.2500 0.0625 0.0312

e) Using these eigenvalues in the Z-plane, design a feedback regulator, Vin=-Kxk, such that the closed-loop discrete system has these eigenvalues (i.e., the eigenvalues of ) (If you are hazy, follow the procedures given in HW#16 to do this)

» %1d) First, find Apv and Bpv

» Bpv=[0;0;1]

Bpv =

0

0

1

» poly(Ahat)

ans =

1.0000 -14.0000 56.0000 -64.0000

» Apv=[0 1 0;0 0 1;-ans(4) -ans(3) -ans(2)]

Apv =

0 1.0000 0

0 0 1.0000

64.0000 -56.0000 14.0000

» %Now we can find Mpv and Tpv

» Mpv=[Bpv Apv*Bpv Apv^2*Bpv]

Mpv =

0 0 1.0000

0 1.0000 14.0000

1.0000 14.0000 140.0000

» Tpv=M*inv(Mpv)

Tpv =

15.3333 -7.0000 0.4167

10.0000 -6.0000 1.2500

-6.6667 -1.0000 0.6667

» %Check the value of Tpv by finding Apv and Bpv:

» inv(Tpv)*Ahat*Tpv

ans =

0.0000 1.0000 0.0000

0.0000 0.0000 1.0000

64.0000 -56.0000 14.0000

» %This is correct for Apv

» inv(Tpv)*Bhat

ans =

0.0000

0

1.0000

» %This is correct for Bpv

» %Next, find desired characteristic polynomial

» poly(z_desired)

ans =

1.0000 -0.3438 0.0254 -0.0005

» %Now find Kpv

» Kpv=[ans(4)+Apv(3,1) ans(3)+Apv(3,2) ans(2)+Apv(3,3)]

Kpv =

63.9995 -55.9746 13.6562

» %Check by finding the eigenvalues of Apv-Bpv*Kpv

» eig(Apv-Bpv*Kpv)

ans =

0.0313

0.0625

0.2500

» %They check!

» %As a last step, find w_k=-Kpv*inv(Tpv)x_k

» Kpv*inv(Tpv)

ans =

-0.4570 9.3099 3.3139

» %Check by finding the eigenvalues of Ahat-Bhat*K

» K=Kpv*inv(Tpv)

K =

-0.4570 9.3099 3.3139

f) How does your gain K compare to the analog gain K given in the solution to HW#16?

» %1f) The value of K in HW#16 was K = [-151.0000 185.0000 -73.0000]

» % This is not very close, but obviously works!

h) Please check your answer by calculating the eigenvalues . What is the settling time of the closed-loop discrete system.

» eig(Ahat-Bhat*K)

ans =

0.0313

0.0625

0.2500

» %They check, too!

Let's find a useful result before we solve this problem:

Let's use long division on this result:

As we can see from the long division,

d)

let