# State Space: time dependent states but time-independent output

1. Nov 2, 2014

### phys_student1

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. Nov 3, 2014

### donpacino

Looks good to me!

3. Nov 3, 2014

### phys_student1

But this blows up if cos(x)=0, although the original output would not!

4. Nov 3, 2014

### donpacino

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 :(

5. Nov 3, 2014

### donpacino

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