- #1
DSRadin
- 12
- 1
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 H(z) with state matrices \mathbf{A, b, c^t, } d are given by the eigenvalues of \mathbf{A}.
Find: Show that, if d\neq0, the zeros of the system are given by the eigenvalues of the matrix \left (\mathbf{A}-d^{-1}\mathbf{b}\mathbf{c^t} \right ).
Hint: The poles of the inverse system H^{-1}(z) equal the zeros of H(z), and H^{-1}(z) has the output x(n) if its input is y(n).
2. H(z)=\mathbf{c^t}(\mathbf{zI-A})^{-1}\mathbf{b}+d
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.
H^{-1}(z)=\left [\mathbf{c^t}(\mathbf{zI-A})^{-1}\mathbf{b}+d \right ]^{-1}
\left ( \mathbf{A}-d^{-1}\mathbf{bc^t} \right ) = \left (\begin{bmatrix}<br /> a_{1 1} & \cdots & a_{1 N-1} \\<br /> \vdots & \ddots & \vdots \\<br /> a_{N-1 1} & \cdots & a_{N-1 N-1}<br /> \end{bmatrix} - \begin{bmatrix}<br /> \frac{b_1c_1}{d} & \cdots & \frac{b1c_{N-1}}{d} \\<br /> \vdots & \ddots & \vdots \\<br /> \frac{b_{N-1}c_1}{d} & \cdots & \frac{b_{N-1}c_{N-1}}{d}<br /> \end{bmatrix}\right ) = \begin{bmatrix}<br /> a_{1 1} - \frac{b_1c_1}{d} & \cdots & a_{1 N-1} - \frac{b1c_{N-1}}{d} \\<br /> \vdots & \ddots & \vdots \\<br /> 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}<br /> \end{bmatrix} = \mathbf{A'}
And then some eigen decomposition leads towards...
|\mathbf{A'}-\lambda\mathbf{I}|=0 = det\begin{bmatrix}<br /> a_{1 1} - \frac{b_1c_1}{d}-\lambda & \cdots & a_{1 N-1} - \frac{b1c_{N-1}}{d}-\lambda \\<br /> \vdots & \ddots & \vdots \\<br /> 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<br /> \end{bmatrix}Is there something in the composition of \mathbf{A,b,c^t,} d that I am missing?
Thanks all.
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