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

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