Example State Space Filter Transfer Function

In this example, we consider a second-order filter ($N = 2$ ) with two inputs ($p=2$ ) and two outputs ($q=2$ ):

$$\begin{eqnarray*} A &=& g\left[\begin{array}{rr} c & -s \\ [2pt] s & c \end{array}\right]\quad \mbox{with $c^2+s^2=1$\ and $0<g<1$}\\ [10pt] B &=& \left[\begin{array}{cc} 1 & 0 \\ [2pt] 0 & 1 \end{array}\right]\qquad C = \left[\begin{array}{cc} 1 & 0 \\ [2pt] 0 & 1 \end{array}\right]\qquad D = \left[\begin{array}{cc} 0 & 0 \\ [2pt] 0 & 0 \end{array}\right] \end{eqnarray*}$$

so that

$$\begin{eqnarray*} \left[\begin{array}{c} x_1(n+1) \\ [2pt] x_2(n+1) \end{array}\right] &=& g\left[\begin{array}{rr} c & -s \\ [2pt] s & c \end{array}\right] \left[\begin{array}{c} x_1(n) \\ [2pt] x_2(n) \end{array}\right] + \left[\begin{array}{cc} 1 & 0 \\ [2pt] 0 & 1 \end{array}\right] \left[\begin{array}{c} u_1(n) \\ [2pt] u_2(n) \end{array}\right],\\ [10pt] \left[\begin{array}{c} y_1(n) \\ [2pt] y_2(n) \end{array}\right] &=& \left[\begin{array}{cc} 1 & 0 \\ [2pt] 0 & 1 \end{array}\right] \left[\begin{array}{c} x_1(n) \\ [2pt] x_2(n) \end{array}\right]. \end{eqnarray*}$$

From Eq.(G.5), the transfer function of this MIMO digital filter is then

$$\begin{eqnarray*} H(z) &=& C(zI-A)^{-1}B = (zI-A)^{-1} = \left[\begin{array}{cc} z-gc & gs \\ [2pt] -gs & z-gc \end{array}\right]^{-1}\\ [5pt] &=&\frac{1}{z^2-2gcz +g^2c^2+g^2s^2}\left[\begin{array}{cc} z-gc & -gs \\ [2pt] gs & z-gc \end{array}\right]\\ [5pt] &=&\left[\begin{array}{cc} \frac{\displaystyle z^{-1}-gcz^{-2}}{\displaystyle 1-2gcz^{-1}+g^2z^{-2}} & -\frac{\displaystyle sz^{-2}}{\displaystyle 1-2gcz^{-1}+g^2z^{-2}} \\ [5pt] \frac{\displaystyle gsz^{-2}}{\displaystyle 1-2gcz^{-1}+g^2z^{-2}} & \frac{\displaystyle z^{-1}-gcz^{-2}}{\displaystyle 1-2gcz^{-1}+g^2z^{-2}} \end{array}\right]. \end{eqnarray*}$$

Note that when $g=1$ , the state transition matrix $A$ is simply a 2D rotation matrix, rotating through the angle $\theta$ for which $c=\cos(\theta)$ and $s=\sin(\theta)$ . For $g<1$ , we have a type of normalized second-order resonator [51], and $g$ controls the ``damping'' of the resonator, while $\theta = 2\pi f_r/f_s$ controls the resonance frequency $f_r$ . The resonator is ``normalized'' in the sense that the filter's state has a constant $\ensuremath{L_2}$ norm (``preserves energy'') when $g=1$ and the input is zero:

$$\left\Vert\,{\underline{x}}(n+1)\,\right\Vert \isdef \sqrt{x_1^2(n+1) + x_2^2(n+1)} = \left\Vert\,A\,{\underline{x}}(n)\,\right\Vert \equiv \left\Vert\,{\underline{x}}(n)\,\right\Vert \protect$$ (G.6)

since a rotation does not change the $\ensuremath{L_2}$ norm, as can be readily checked.

In this two-input, two-output digital filter, the input $u_1(n)$ drives state $x_1(n)$ while input $u_2(n)$ drives state $x_2(n)$ . Similarly, output $y_1(n)$ is $x_1(n)$ , while $y_2(n)$ is $x_2(n)$ . The two-by-two transfer-function matrix $H(z)$ contains entries for each combination of input and output. Note that all component transfer functions have the same poles. This is a general property of physical linear systems driven and observed at arbitrary points: the resonant modes (poles) are always the same, but the zeros vary as the input or output location are changed. If a pole is not visible using a particular input/output pair, we say that the pole has been ``canceled'' by a zero associated with that input/output pair. In control-theory terms, the pole is ``uncontrollable'' from that input, or ``unobservable'' from that output, or both.