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Finding Zeros of System Function using Eigenvalues

  1. Feb 25, 2014 #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 [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.


     
    Last edited: Feb 25, 2014
  2. jcsd
  3. Feb 25, 2014 #2

    donpacino

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    Gold Member

    Just using the equation is the cleanest method I know of.
    if you use matlab this method is extremely easy to implement.
    I just want to note that you made an error in your step #3
     
    Last edited: Feb 25, 2014
  4. Feb 25, 2014 #3
    Thank you for your response donpacino.

    Unfortunately I'm not seeing it. (both the way to implement the equation and the error in 3)
    I considered MATLAB but I was hesitant to take a deterministic approach and go with "if it works this once it will always work" kind of deal. Maybe i'l do that.
     
  5. Feb 26, 2014 #4

    donpacino

    User Avatar
    Gold Member

    I= the identity matrix, which means the diagonal is equal to 1, and zero everywhere else

    I=
    [1 0 0
    0 1 0
    0 0 1]

    so if A=
    [1 1 1
    1 1 1
    1 1 1]

    then |A-λI|=
    [1-λ, 1, 1;
    1, 1-λ , 1;
    1, 1, 1-λ]

    does that make sense?
     
    Last edited: Feb 26, 2014
  6. Feb 27, 2014 #5
    doh! Yes absolutely it makes sense.
    I got the answer now too - the key was to back it into state space equations and re-solve.

    Thank you for the response!
    -DR
     
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