Difference between revisions of "FAQ: In Example 5.3, why is the A matrix in the given form?"

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{{subst: FAQ template}}
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{{FAQ page
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| chapter = Linear Systems
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| contributor = R. M. Murray
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| date = 5 Apr 08
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| answer =
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In Example 5.3, we consider the state space representation of the model
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<center><amsmath>
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  \ddot q + \omega_0^2 q = u.
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</amsmath></center>
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If we choose the state vector to be <amsmath>x = (q, \dot q)</amsmath> then the dynamics matrix for the system is given by
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<center><amsmath>
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  A = \begin{bmatrix} 0 & 1 \\ -\omega_0^2 & 0 \end{bmatrix},
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</amsmath></center>
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which is different than the form listed.  Instead, we can use a ''normalized'' set of coordinates
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<center><amsmath>
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  x = \begin{bmatrix} q \\ \dot q / \omega_0 \end{bmatrix}
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  \qquad\implies\qquad
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  A = \begin{bmatrix} 0 & \omega_0 \\ -\omega_0 & 0 \end{bmatrix}.
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</amsmath></center>
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A similiar form can also be obtained by rescaling time by <math>\omega_0</math>, as described in Example 2.6.
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}}

Revision as of 21:29, 5 April 2008

(Contributed by R. M. Murray, 5 Apr 08)

In Example 5.3, we consider the state space representation of the model

math

If we choose the state vector to be math then the dynamics matrix for the system is given by

math

which is different than the form listed. Instead, we can use a normalized set of coordinates

math

A similiar form can also be obtained by rescaling time by , as described in Example 2.6.