as below

November 23, 2015

EEL 5613 – Modern Control

Homework 4

1.1)

∴

Desired eigenvalues are {-1, -2, -2}; thus the desired characteristic polynomial is:

∴

The gain matrix for the controllable canonical form, then, is:

We next find Q, where .

The gain matrix, k, is found by multiplying:

Matlab verification can be achieved numerous ways. First, simply solving for the eigenvalues of the new system, we have the following:

>> Z=A+B*K

Z =

-2.6000 -1.6000 -2.8000

0 -1.0000 1.0000

0.2000 0.2000 -1.4000

>> eig(Z)

ans =

-2.0000

-2.0000

-1.0000

Going further, we can actually plot the new system with ltiview, and then choose the zero-pole graph to represent the zero and pole locations of that system:

We see the poles at -1 and -2, and we also note zeros at -0.5 and -1.

1.2) The system is observable iff the matrix is full rank. Indeed,

, so the system is observable.

For the observer to have poles at {-2, -2, -3}, we require that:

∴

14

∴

Confirming in Matlab, we have:

>> L=[6;-2;1];

>> Z=A-L*C

Z =

-7 -8 -2

2 1 1

0 -1 -1

>> eig(Z)

ans =

-3.0000 + 0.0000i

-2.0000 + 0.0000i

-2.0000 - 0.0000i

Thus, the entire system equation becomes:

Using Simulink to simulate the system and the observer, we create the following schematic, and the respective comparisons to xi and , i=1,2,3 follow:

1.3) Comparing the system in 1.1 with that in 1.2, we can see how the outputs behave:

There is a transient period in which the two signals differ but they eventually approach their respective steady states. It should be noted that those steady states are not equal. The steady state of the system in 1.1 differs from that of the original system (0.25 compared to 0.20), for a steady state tracking error of 25%, but the steady state of the system in 1.2 has zero steady state tracking error, perfectly matching the steady state of the original system. These steady states were expected given the transfer functions of the original and calculated systems.

2.1)

A) The system is unstable because all of the eigenvalues are positive (λ=1,2,3).

B) The system can be stabilized if the system is fully controllable. To determine this, we find:

>> A=[1 0 0; 0 2 -2; 0 0 3];

>> B=[1 0; 0 1; 0 1];

>> rank([B A*B (A^2)*B])

ans =

3

Given that the system is full rank, it is fully controllable, and thus stabilizable.

C) It is only possible to carry out a SISO state feedback design if, using either B=u1 or B=u2 produces a full rank system. Trying B=u1 first, we have:

>> A=[1 0 0; 0 2 -2; 0 0 3];

>> B=[1;0;0];

>> rank([B A*B (A^2)*B])

ans =

1

Clearly, the system using just B=u1 is not full rank.

Trying B=u2 now, we have:

>> A=[1 0 0; 0 2 -2; 0 0 3];

>> B=[0;1;1];

>> rank([B A*B (A^2)*B])

ans =

2

Again, the system using just B=u2 is not full rank. Thus, we must use a MIMO state feedback design for this problem.

Choose . Then, finding one more linearly independent column, we choose (because , which is not linearly independent). Thus, so that

.

P=2 because there are two l’s, namely l1 and l2.

S is a 2x3 matrix because m (inputs)=2, n (state variables)=3

We know that the vectors in columns of the S matrix form a (p-1)x(p-1), that is 2x2, identity matrix if the first row and the last n-p (or 3-2=1) rows are deleted. Thus, s equals:

Solving for (using matlab), we have:

Thus, the new system is represented by

Choosing , we continue with pole placement on the modified system.

∴

Desired eigenvalues are {-1, -2, -3}; thus the desired characteristic polynomial is:

∴

The gain matrix for the controllable canonical form, then, is:

We next find Q, where .

The gain matrix, k, is found by multiplying:

Verifying in Matlab, we have:

Pnew =

0 -0.3333 0.3333

-0.0000 -0.6667 1.6667

1.0000 -1.3333 6.3333

>> Knew=kbar*Pnew

Knew =

-12.0000 20.0000 -80.0000

>> zz=Anew+b*Knew

zz =

-11.0000 20.0000 -80.0000

1.0000 2.0000 -2.0000

1.0000 0 3.0000

>> eig(zz)

ans =

-3.0000

-2.0000

-1.0000

2.2)

Let F=0, B1 =[1 0 0]’

, .

Let ,

Choose

Let (arbitrarily), now:

Now, (A+BF, b1) should be a controllable pair. Verifying:

From this point, we can proceed to complete pole placement exactly as above.