Finding Zeros of System Function using Eigenvalues

mardi 25 février 2014



Hi all - working on this problem wanted to see if anyone had any advice - thanks!



As shown in section 4.4, the poles of the system [itex]H(z)[/itex] with state matrices [itex] \mathbf{A, b, c^t, } d [/itex] are given by the eigenvalues of [itex]\mathbf{A}[/itex].



Find: Show that, if [itex] d\neq0[/itex], the zeros of the system are given by the eigenvalues of the matrix [itex] \left (\mathbf{A},d^{-1},\mathbf{b},\mathbf{c^t} \right ) [/itex].



Hint: The poles of the inverse system [itex] H^{-1}(z) [/itex] equal the zeros of [itex]H(z)[/itex], and [itex]H^{-1}(z)[/itex] has the output [itex]x(n)[/itex] if its input is [itex] y(n) [/itex].








2. [itex] H(z)=\mathbf{c^t}(\mathbf{zI-A})^{-1}\mathbf{b}+d[/itex]







3. I understand why the poles of the system are eigenvalues of A. I have gone through this derivation in other work. I feel like there is something I am missing in the linear algebra that would simplify this problem. My attempt at a solution below stops short of solving for eigenvalues of the new matrix because i feel that proving this in generality must be cleaner than this brute force method.



[itex] H^{-1}(z)=\left [\mathbf{c^t}(\mathbf{zI-A})^{-1}\mathbf{b}+d \right ]^{-1}[/itex]



[itex] \left ( \mathbf{A}-d^{-1}\mathbf{bc^t} \right ) = \left (\begin{bmatrix}

a_{1 1} & \cdots & a_{1 N-1} \\

\vdots & \ddots & \vdots \\

a_{N-1 1} & \cdots & a_{N-1 N-1}

\end{bmatrix} - \begin{bmatrix}

\frac{b_1c_1}{d} & \cdots & \frac{b1c_{N-1}}{d} \\

\vdots & \ddots & \vdots \\

\frac{b_{N-1}c_1}{d} & \cdots & \frac{b_{N-1}c_{N-1}}{d}

\end{bmatrix}\right ) = \begin{bmatrix}

a_{1 1} - \frac{b_1c_1}{d} & \cdots & a_{1 N-1} - \frac{b1c_{N-1}}{d} \\

\vdots & \ddots & \vdots \\

a_{N-1 1} - \frac{b_{N-1}c_1}{d} & \cdots & a_{N-1 N-1} - \frac{b_{N-1}c_{N-1}}{d}

\end{bmatrix} = \mathbf{A'}[/itex]



And then some eigen decomposition leads towards....



[itex] |\mathbf{A'}-\lambda\mathbf{I}|=0 = det\begin{bmatrix}

a_{1 1} - \frac{b_1c_1}{d}-\lambda & \cdots & a_{1 N-1} - \frac{b1c_{N-1}}{d}-\lambda \\

\vdots & \ddots & \vdots \\

a_{N-1 1} - \frac{b_{N-1}c_1}{d} -\lambda & \cdots & a_{N-1 N-1} - \frac{b_{N-1}c_{N-1}}{d}-\lambda

\end{bmatrix}[/itex]





Is there something in the composition of [itex] \mathbf{A,b,c^t,} d [/itex] that I am missing?



Thanks all.









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