Difference between revisions of "Hw 5 ex 2"

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Hint on how to solve ex 2: assume that the system is observable, and try an argument by contradiction. If the controller makes the system unstable, then the corresponding matrix <math> \tilde{A}=A-BK</math> must have an eigenvalue with positive real part, to which corresponds a certain eigenvector <math>v</math>. Writing down the algebraic Riccati equation with <math> \tilde{A}</math>, and pre-post multiplying by the unstable eigenvector (as if you were evaluating a quadratic form), you will see that the only case in which the corresponding form can be zero is only if P=0 and <math>v^* Q_v v</math> is zero. Which contradicts the initial assumption.
Hint on how to solve ex 2: assume that the system is observable, and try an argument by contradiction. If the controller makes the system unstable, then the corresponding matrix <math> \tilde{A}=A-BK</math> must have an eigenvalue with positive real part, to which corresponds a certain eigenvector <math>v</math>.
 
One can rewrite the Algebraic Riccati equation using  <math> \tilde{A}</math>, where you should note the changes of signs: 
 
<math>P\tilde{A} + \tilde{A}^T P + PBQ_u^{-1}B^T P + Qx=0 </math>
 
 
Pre-post multiplying by the unstable eigenvector (as if you were evaluating a quadratic form), you will see that the only case in which the corresponding form can be zero is only if P=0 and <math>v^* Q_v v</math> is zero. Which contradicts the initial assumption.




--[[User:Franco|Elisa]]
--[[User:Franco|Elisa]]
[[Category: CDS 110b FAQ - Homework 6 ]]
[[Category:CDS 110b FAQ - Homework 5]]

Latest revision as of 01:28, 14 February 2007

Hint on how to solve ex 2: assume that the system is observable, and try an argument by contradiction. If the controller makes the system unstable, then the corresponding matrix must have an eigenvalue with positive real part, to which corresponds a certain eigenvector .

One can rewrite the Algebraic Riccati equation using , where you should note the changes of signs:


Pre-post multiplying by the unstable eigenvector (as if you were evaluating a quadratic form), you will see that the only case in which the corresponding form can be zero is only if P=0 and is zero. Which contradicts the initial assumption.


--Elisa