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.
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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>.
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One can rewrite the Algebraic Riccati equation using  <math> \tilde{A}</math>, where you should note the changes of signs: 
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<math>P\tilde{A} + \tilde{A}^T P + PBQ_u^{-1}B^T P + Qx=0 </math>
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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 5]]
 
[[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 \( \tilde{A}=A-BK\) must have an eigenvalue with positive real part, to which corresponds a certain eigenvector \(v\).

One can rewrite the Algebraic Riccati equation using \( \tilde{A}\), where you should note the changes of signs\[P\tilde{A} + \tilde{A}^T P + PBQ_u^{-1}B^T P + Qx=0 \]


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 \(v^* Q_v v\) is zero. Which contradicts the initial assumption.


--Elisa