0
$\begingroup$

enter image description here

if we just examine the translational dynamics (X and Y direction), the dynamics are coupled, So if I want to design a PID controller to control the position, the outcome will be unpredictable because of the coupling.I some research papers , I have seen people assume (psi =0 ) at the equilibrium point to remove the coupling in x and y dynamics. But I would like to design a controller along with the couple system? Is there any solution like decoupling the dynamics equations or something.

$\endgroup$
  • $\begingroup$ Do you have a reference for the paper where these equations came from? $\endgroup$ – Paul Apr 23 '18 at 23:13
  • $\begingroup$ yea you can find a lot of papers on internet,almost everyone end up in these equations $\endgroup$ – robin chacko Apr 24 '18 at 7:59
  • 2
    $\begingroup$ I suggest you to make the definitions of the inputs appear. $\endgroup$ – N. Staub Apr 24 '18 at 8:38
1
$\begingroup$

You can linearize and decouple the system about equilibrium under the following assumptions:

  • Small angle approximation: $\sin(x) \approx x$ and $\cos(x) \approx 1$

  • Angular acceleration due to actuators only: $\dot{p} = \frac{U_2}{I_{xx}}$, $\dot{r} = \frac{U_3}{I_{yy}}$, $\dot{p} = \frac{U_4}{I_{zz}}$

$\endgroup$
0
$\begingroup$

Quadrotors are coupled systems because you have 4 inputs for 6D motions.

Usually people are using cascade controllers to regulate both orientation and position and assume that desired orientation is achieved when considering the control of position.

see this famous RAM tutorial paper to get familiar with quadrotors: https://ieeexplore.ieee.org/document/6289431/?arnumber=6289431

Also, quadrotors are flat, in the control theory sens, for position and yaw, this means that the internal orientation dynamics are stable, if the flat output are stabilized. That is the reason why you can easily fly them controlling the position.

$\endgroup$
  • $\begingroup$ What do you mean by “flat” in this context? Can you elaborate moe in your answer? $\endgroup$ – Paul Apr 24 '18 at 15:21

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.