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Force-Driven Mass Reconsidered

Why not include position $ x(t)$ as well as velocity $ v(t)$ in the state-space model for the force-driven mass?

$\displaystyle \left[\begin{array}{c} \dot{x}(t) \\ [2pt] \dot{v}(t) \end{array}\right]
\eqsp \left[\begin{array}{cc} 0 & 1 \\ [2pt] 0 & 0 \end{array}\right]\left[\begin{array}{c} x(t) \\ [2pt] v(t) \end{array}\right]
+ \left[\begin{array}{c} 0 \\ [2pt] 1/m \end{array}\right] f(t)
$



We might expect this because we know from before that the complete physical state of a mass consists of its velocity $ v$ and position $ x$ !


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``Introduction to State Space Models'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2014-03-24 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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