EE422

Solution to HW#13

 

0. Pick a lab partner to work with Your first lab is on its way!!!

 

1. Given the following state variable model:

 

a) Find the characteristic equation for A, , then verify the Cayley-Hamilton theorem by substituting A for s.

Substituting A for s we find:

Thus, the Cayley-Hamilton Theorem is satisfied.

b) Use your answer to part a) to help you find the eigenvalues and eigenvectors of A (you may check your answer on Matlab)

Check on Matlab:

 

» a=[4 6;-3 -5]

 

a =

 

4 6

-3 -5

 

» [P,S]=eig(a)

 

P =

 

0.8944 -0.7071

-0.4472 0.7071

 

S =

 

1 0

0 -2

 

» P(:,1)*sqrt(5)

 

ans =

 

2.0000

-1.0000

 

» P(:,2)*sqrt(2)

 

ans =

 

-1.0000

1.0000

 

 

c) Is the system asymptotically stable, marginally stable, or unstable?

 

System is unstable because we have an eigenvalue in the RHP (s=1)

 

  1. Using any of the three methods, find the state transition matrix,
    2.

e) Use the equation, to find x(t) when w(t)=u(t) and identify the zero-input and zero-state portion of your solution. Plug x(t) into the output equation to find y(t).

 

First, let's find the zero-input solution:

 

Next, let's solve for xzero-state(t) if we input w(t) = u(t) into our system:

The entire solution is given by x(t)=xzero-input(t)+xzero-state(t)=

and is valid for t > 0.

The output, y(t) is given by y = Cx = [1 1]x(t) = 2et

 

f) Check your answers using Laplace Transforms

 

Using the Laplace transform:

 

This checks with our previous answer for x(t).

 

g) Use the property of Linearity and Time Invariance to solve,

when the input is w(t)=3u(t-4).

 

Note: Although the entire system is not LTI, each component of x(t) (i.e., xzero-input(t) and xzero-state(t)) behaves like an LTI system if we consider the initial condition as an input. Thus, if we multiply the initial state by 2 and delay the initial time by 1 second and we also multiply the input by 3 and delay by 4 seconds, the new solution is related to the old zero-input and zero-state solutions by:

xnew(t)=2xzero-input(t-1)+3xzero-state(t-4)=

 

2. Consider the system,

 

a) Use the similarity transformation, x=Pz, (where P is the matrix of eigenvectors) to decouple the system as follows:

 

 

b) Solve the decoupled system in problem 2a) then find x(t) from x(t)=Pz(t). Does your answer agree with problem 1e)?

This is the same answer as we obtained before!

 

 

c) Assuming that the intial state is zero, draw a block diagram of the decoupled system in problem 2a) and identify which eigenvalues are: i) controllable ii) observable iii) stable

i)

controllable: s2 = 1; (or uncontrollable: s1 = -2)

ii)

 observable: s1 = -2; (or unobservable: s2 = 1)

iii)

 stable s1 = -2; (or unstable: s2 = 1)

 

d) From your block diagram, find the transfer function H(s)=Y(s)/W(s). Are the poles of H(s) the same as the eigenvalues? Why not?

 

Since the block diagram is in parallel form, it is easy to reduce:

 

 

Our transfer function is H(s)=0!!

Thus, it has no poles at all!! The poles of the transfer function would have been {-2,1} if the numerator hadn't been zero for each term. Thus, because the first eigenvalue s1 = -2, was uncontrollable the corresponding numerator became zero in the transfer function and it did not appear as a pole. Similarly, because the second eigenvalue, s2 = -1, was unobservable the corresponding numerator became zero in the transfer function and it too did not appear as a pole.

 

Conclusion: The poles of the transfer function will be only those eigenvalues which are both controllable and observable simultaneously!