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Robert
science forum beginner
Joined: 11 Jun 2005
Posts: 14
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 Posted: Sat Jul 09, 2005 9:16 am Post subject:
State space
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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 |
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Tim Wescott
science forum Guru Wannabe
Joined: 03 May 2005
Posts: 292
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| Posted: Sat Jul 09, 2005 1:49 pm Post subject:
Re: State space
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Robert wrote:
| Quote: | 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
|
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.
--
-------------------------------------------
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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Robert
science forum beginner
Joined: 11 Jun 2005
Posts: 14
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| Posted: Sat Jul 09, 2005 6:23 pm Post subject:
Re: State space
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Thank you for your help.
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 again in advance
Robert
In article <11cvsfqp6emes67@corp.supernews.com>, "Tim Wescott"
<tim@seemywebsite.com> wrote:
| Quote: | Robert wrote:
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
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.
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Tim Wescott
science forum Guru Wannabe
Joined: 03 May 2005
Posts: 292
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| Posted: Sat Jul 09, 2005 7:13 pm Post subject:
Re: State space
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Robert wrote:
- top posting fixed -
| Quote: | In article <11cvsfqp6emes67@corp.supernews.com>, "Tim Wescott"
tim@seemywebsite.com> wrote:
Robert wrote:
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
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.
Thank you for your help.
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?
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.
--
-------------------------------------------
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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RT
science forum beginner
Joined: 12 May 2005
Posts: 3
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| Posted: Tue Jul 19, 2005 5:12 pm Post subject:
Re: State space
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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:
| Quote: | 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 |
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