I'm trying to derive the state space model for the following system: $L\ddot{\theta}-g\sin(\theta)-\ddot{z}\sin(\theta) = 0$

I'm allowed to use that $\sin(\theta) \approx \theta$. And I must set $x_1 = \theta, x_2 = \dot{\theta}$, input $u = \ddot{z}$, output $y = \theta$.

This is what I've come up with: \begin{equation*} \dot{x} = \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = Ax + Bu = \begin{bmatrix} 0 & 1 \\ \frac{g}{L} & 0 \end{bmatrix} x + \begin{bmatrix} 0 \\ \frac{\theta}{L} \end{bmatrix} u \end{equation*} \begin{equation*} y = Cx + Du = \begin{bmatrix} 1 & 0 \end{bmatrix}x \end{equation*}

But I can't have the $\theta$ in $B$, since that is equal to $x_1 = y = \theta$, right? How should I solve it?

Thanks in advance!

  • $\begingroup$ Who is $g$? A constant? $\endgroup$ – Math1000 Nov 22 '16 at 21:16
  • $\begingroup$ Yes g is gravity! $\endgroup$ – user1415066 Nov 22 '16 at 21:18
  • $\begingroup$ As in $$g \approx 9.80665\ \mathrm m\ \mathrm s^{-2}\quad \large ? $$ $\endgroup$ – Math1000 Nov 22 '16 at 21:21
  • 1
    $\begingroup$ Yes, exactly! But $g\approx 9.82 ms^{-2}$ in Sweden ;) $\endgroup$ – user1415066 Nov 22 '16 at 21:24

This is a nonlinear model, even if you linearize $\sin\theta$ to $\theta$. If you want to linearize it further you would also have to assume that $\theta$ stays close to one value and use that value to define $B$. But if you want to solve this differential equation you could use separation of variables,

$$ \int\!\!\!\int \frac{1}{\sin\theta} d\theta\,d\theta = \int\!\!\!\int \frac{g + u(t)}{L}dt\,dt, $$

but the left hand side does not yield a nice solution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.