# basic control problem

I have a basic question from ogata. i hope you know the answer.
consider the system defined by matrices: A=[0 1 0; 0 0 1;
-6 -11 -6] B=[0; 0; 1] C=[c1 c2 c3]
Except for the obvious choice of c1��, find an example set of c1, c2, c3 that will make the system unobservable??
To check the observability, i have formed observ.matrix Obsv=[C; CA; CA^2] the determinant of the "Obsv" must be nonzero. and it must be full rank. When i formed the "Obsv" matrix, such complicated matrix was become. Do you have any suggestions for c1, c2 and c3? Thank you
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temcon wrote:

Homework?
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Tim Wescott
Wescott Design Services
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I'm studying for the exam.

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temcon wrote:

Ah. See my response to your original post, in a minute.
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Tim Wescott
Wescott Design Services
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Can't you use software?
A = {{0, 1, 0}, {0, 0, 1}, {-6, -11, -6}}; cmat = {{c1, c2, c3}}; obsrv = {Flatten[cmat], Flatten[cmat . A], Flatten[cmat . MatrixPower[A, 2]]}; Det[obsrv]
c1^3 - 6*c1^2*c2 + 11*c1*c2^2 - 6*c2^3 + 14*c1^2*c3 - 48*c1*c2*c3 + 36*c2^2*c3 + 49*c1*c3^2 - 66*c2*c3^2 + 36*c3^3
Reduce[% == 0, {c1, c2, c3}]
c3 == -(c1/9) + c2/3 || c3 == -(c1/4) + c2/2 || c3 == -c1 + c2
--Nasser
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temcon wrote:

If the system matrix were in diagonal form you'd have no problem, right? Because if the C matrix has a zero in it, then the corresponding mode (which only maps to one state) will be unobservable.
So one way you could do this is to find a similarity transformation to a diagonal form, zero out one or more columns in the C matrix, then transform the system back to the one you have.
(it worked for me, just like magic).
--

Tim Wescott
Wescott Design Services
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Stretch my old ring out.
I Am Kirk Johnson. "Anal Stretching, Wrenching & Expanding Specialist" http://www.imagefap.com/image.php?id 88478267