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State Space: time dependent states but time-independent output

  1. Nov 2, 2014 #1
    Let:
    $$x_1=A\sin{\omega t}$$ $$x_2=\dot{x}_1=A\omega \cos{\omega t}$$ $$y=A\omega$$
    We want to represent this system in a state space model. The state transition matrix read:
    $$A=\begin{bmatrix} 0 & 1 &\\ -\omega^2 & 0 \\ \end{bmatrix}$$ I am not sure what the output matrix will be like. Can we say
    $$y=A\omega=\frac{-x_2}{\cos{\omega t}}$$
    So that:
    $$C=\begin{bmatrix} 0 & \frac{-1}{\cos{\omega t}} \end{bmatrix}$$
     
  2. jcsd
  3. Nov 3, 2014 #2

    donpacino

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    Looks good to me!
     
  4. Nov 3, 2014 #3
    But this blows up if cos(x)=0, although the original output would not!
     
  5. Nov 3, 2014 #4

    donpacino

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    So one point that I overlooked, state space representation is really only valid for linear systems.
    sinusoidal functions are not linear.
    the state space representation also should not contain time as a variable, as it demonstrates how the system responds over time.

    I was incorrect to say it looked fine :(
     
  6. Nov 3, 2014 #5

    donpacino

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    you need to modify your system, or representation of the system, such that it is linear and time independant
     
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