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Impulse Response of State Space Models

As derived in Book II [449, Appendix G], the impulse response of the state-space model can be summarized as

$\displaystyle {\mathbf{h}}(n) \eqsp \left\{\begin{array}{ll} D, & n=0 \\ [5pt] CA^{n-1}B, & n>0 \\ \end{array} \right. \protect$ (2.10)

Thus, the $ n$th ``sample'' of the impulse response is given by $ C A^{n-1}
B$ for $ n\geq0$. Each such ``sample'' is a $ p\times q$ matrix, in general.

In our force-driven-mass example, we have $ p=q=1$, $ B=[0,T/m]^T$, and $ D=0$. For a position output we have $ C=[1,0]$ while for a velocity output we would set $ C=[0,1]$. Choosing $ C=\mathbf{I}$ simply feeds the whole state vector to the output, which allows us to look at both simultaneously:

\begin{eqnarray*}
{\mathbf{h}}(n+1) &=&\left[\begin{array}{cc} 1 & 0 \\ [2pt] 0 ...
...\left[\begin{array}{c} nT \\ [2pt] 1 \end{array}\right]
\protect
\end{eqnarray*}

Thus, when the input force is a unit pulse, which corresponds physically to imparting momentum $ T$ at time 0 (because the time-integral of force is momentum and the physical area under a unit sample is the sampling interval $ T$), we see that the velocity after time 0 is a constant $ v_n = T/m$, or $ m\,v_n=T$, as expected from conservation of momentum. If the velocity is constant, then the position must grow linearly, as we see that it does: $ x_{n+1} = n
(T^2/m)$. The finite difference approximation to the time-derivative of $ x(t)$ now gives $ (x_{n+1}-x_n)/T = T/m = v_n$, for $ n\ge0$, which is consistent.


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Next: Zero-Input Response of State Space Models

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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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