# State space

• posted

Hi people

I am following a course on control theory, but I got stuck on the following problem

In my book, it is given that (s / [(s-a)(s-a*)]) can be realized in state-space as

sX(s) =AX(s)+BU(s) Y(s) =CX(s)

(provided that X is a vector of 2 states)

with

A = [ Real(a) Real(a)-abs(a) ; Real(a)+abs(a) Real(a) ]; B = [ 1 ; 1 ]; C = [ 1 1 ];

I know this representation is not unique, but I don't understand how to obtain it. I found a book how I can realize the transfer function in companion form (or e.g. with A diagonal matrix), but then A has a different structure.

Can someone enlighten me here, how they found the A,B,C matrix in this form?

Thanks Robert

• posted

It takes a very short amount of dinking around with a 2x2 system to arrive at that. I suspect no one bothered to make an algorithm. Just start with

( s - a -b ) det ( ) ( -c s - d )

(my a not equal to your a) and look at the ways you can match up a, b, c and d to your polynomial of choice.

• posted

Ok, so I work out the determinant, and if I choose the a,b,c,d in such way that eigenvalues of [a b ; c d] equal the poles a and a*, the entries as they are given in my A matrix seem to give a possible (although not unique) solution.

PS. I made a typo in my first post : the transfer function is (2s/[(s-a)(s-a*)])

Now I'm still stuck with the problem how to find the B and C vector of the state space realization?

Thanks aga> Robert wrote:

• posted

- top posting fixed -

Yes, you are. And like the A matrix, between the B and C vectors you have four different variables that you can diddle with and an answer with two degrees of freedom.

I use a construct like this to make notch filters in sampled time, and I set the A matrix up in the equivalent way, I set up the B vector so that the states don't overflow, and I set up the C vector to achieve the transfer function that I'm looking for. MathCad, Maple, Mathmatica, etc. is very handy here. Constraint on myself to keep the gain from input to state bounded sets some limits on B. Once B is chosen then the values of the C vector are dictated.

• posted

Robert,

Why not try to realize the transfer function into the companion form, and then find a transformation matrix to convert the A-matrix in companion form into the A-matrix as given. You can easily see the the two A-matrix has the same eigen-values. This will be a little linear algebra execise.

HTH,

-RT

Robert wrote:

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.