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Impulse Response of State Space Models

As derived in Book II [452, Appendix G], the impulse response of the state-space model can be summarized in the single-input, single-output (SISO) case 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 generalizes to a $ p\times q$ matrix in the multi-input, multi-output (MIMO) case ($ q$ inputs, $ p$ outputs); in such a case, the input signal in Eq.$ \,$ (1.8) is $ \underline{u}(n)=\mathbf{I}_p\delta(n)$ , which is a collection of $ p$ input vectors $ \uv_i(n)$ , for $ i=1,\ldots,p$ , each having dimension $ p\times 1$ , corresponding to an impulse signal $ \delta(n)$ being applied to the $ i$ th system input.

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 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & T \\ [2pt] 0 & 1 \end{array}\right]^n\left[\begin{array}{c} 0 \\ [2pt] T/m \end{array}\right]\\ [5pt]
&=&\left[\begin{array}{cc} 1 & nT \\ [2pt] 0 & 1 \end{array}\right]\left[\begin{array}{c} 0 \\ [2pt] T/m \end{array}\right]\\ [5pt]
&=&\frac{T}{m}\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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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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