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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:
{\mathbf{h}}(n+1) &=&\left[\begin{array}{cc} 1 & 0 \\ [2pt] 0 ...
...\left[\begin{array}{c} nT \\ [2pt] 1 \end{array}\right]
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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Zero-Input Response of State Space Models
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