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Math Help - Show that a system is controllable

  1. #1
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    Show that a system is controllable

    Hello, everyone. I got a problem from my coursework question which relevent to controllability.

    The question asks that use e^{At} to show that a system dx/dt=AX+BU for which A=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} and B=\begin{pmatrix}c \\ d\end{pmatrix} is controllable whatever values c and d, take except d = 0.

    I guess that the matrix A should be converted to a Jordan form, then e^{At}=Pe^{Jt}P^{-1} ? But how to achieve P, P^{-1} and jordan matix J in this question?

    Thanks for help.
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  2. #2
    Super Member Gamma's Avatar
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    It's already in its jordan form. So P=I_2 if you want .
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  3. #3
    Super Member Gamma's Avatar
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    Expounding

    As I am not entirely sure of where your understanding of why you use the Jordan form, I thought I would add a little more for ya.

    The important thing about the Jordan Form is that (provided your field contains all the eigenvalues, if not you can just pass to an algebraic closure and work there instead) you can write your Jordan Matrix J=D+N.

    D is a diagonal matrix and N is a nilpotent matrix. These are particularly important because it greatly reduces your computations when doing things like oh, I don't know, taking powers of a matrix! Which makes this method incredibly desirable for writing out the exponential of a matrix.

    So A=P(D+N)P^{-1}, but it gets even better because for powers of A you see all the interior P's cancel out and you get:
    A^k=P(D+N)^kP^{-1}

    This is really easy to calculate since D and N commute you can use binomial theorem like you would hope (this takes some work to show). D is diagonal so any power of D is just the powers of the diagonal and for N, it is nilpotent, so at some point it just terminates completely and is just 0 from then on.

    In your case
    A=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}<br />


    And
    \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}^2=\begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}
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  4. #4
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    I was thinking that my question may be similar to this example, shows below:

    Consider the system dX/dt=AX+Bu, which has matrix form:
    d\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}/dt=\begin{pmatrix}1 & -1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}+\begin{pmatrix}1 \\ 0 \end{pmatrix}u

    for which we can show that A=PJP^{-1}, where

    A=\begin{pmatrix}1 & -1 \\ 1 & -1 \end{pmatrix}, P=\begin{pmatrix}1 & 0 \\ 1 & -1 \end{pmatrix}=P^{-1}, and J=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}.

    Then e^{At}=Pe^{Jt}P^{-1}=\begin{pmatrix}1 & 0 \\ 1 & -1 \end{pmatrix}\begin{pmatrix}1 & t \\ 0 & 1 \end{pmatrix}\begin{pmatrix}1 & 0 \\ 1 & -1 \end{pmatrix}=\begin{pmatrix}1+t & -t \\ t & 1-t \end{pmatrix}

    Hence e^{At}B=\begin{pmatrix}1+t & -t \\ t & 1-t \end{pmatrix}\begin{pmatrix}1 \\ 0 \end{pmatrix}=\begin{pmatrix}1+t \\ t \end{pmatrix}.

    Suppose now that there is a vector X_1=\begin{pmatrix}a \\ b \end{pmatrix}

    such that X_1^{T}e^{At}B=\begin{pmatrix}a & b \end{pmatrix}\begin{pmatrix}1+t \\ t \end{pmatrix}=a(1+t)+bt=0.

    From the constant part we see that a = 0, while from the coefficient of t it follows that a+b=b=0.

    Since X_1 cannot be non-zero, we have shown that the above system is controllable.
    Then I thought A should be converted to a Jordan form first, just like the above example did. But as you said if the matrix A already being a Jordan form, then how can I show that the system is controllable?
    Last edited by jackw; May 4th 2009 at 05:50 PM.
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