State Space Formulation

Rewriting the filter in state-space form,²we have

$$\left[\begin{array}{c} x(n+1) \\ [2pt] y(n+1) \end{array}\right] ... ...ray}\right] + \left[\begin{array}{c} b_1 \\ [2pt] b_2 \end{array}\right] u(n)$$

or

$$\underline{x}(n+1) = \mbox{\boldmath$A$} \underline{x}+ \mbox{\boldmath$B$} u(n)$$ (21)

$$y(n) = \mbox{\boldmath$C$} ^T \underline{x}(n) + D u(n)$$ (22)

where $C$$^T=[c_1,c_2]=[0, 1]$ and $D=0$. As is well known, the transfer function of a state-space model (A,B,C,D) is given by

$$H(z) = \mbox{\boldmath$C$} ^T (z \mbox{\boldmath$I$} - \mbox{\boldmath$A$} )^{-1} \mbox{\boldmath$B$} + D$$ (23)

$$= \frac{e_1z^{-1}+ e_2z^{-2}}% {1-2r_1\cos(\theta_1)z^{-1}+ r_1^2z^{-2}},$$ (24)

where

$$e_1 = c_1 b_1 + c_2 b_2$$ (25)

$$e_2 = -c_1 (b_1x_1 +b_2y_1) + c_2 (b_1y_1-b_2x_1).$$ (26)

Thus, under all choices of input or output, there is at most one real finite zero at $z=-e_2/e_1$.