3
$\begingroup$

I'm trying to implement with simulink a PD controller for my quadcopter. I use a simplified model, and for the roll case I have $ I_x * \phi = L $, where L is the roll torque. So, the transfer function of my model is $ Ix/s^2 = L $in continous time. My real system needs that the output of the transfer function is a torque, that is converted (with the other torques) in the PWM frequencies for the 4 motors. In order to implement my simulation and close the loop (that in reality is closed by the IMU unit), I have expressed the same previous equation in terms of $\phi$ to get the obtained angle and close the loop. When I'm trying to tune my PD automatically with matlab, it doesn't find a stable configuration of the parameters. I'm pretty sure that I'm doing something wrong, but I'm not very familiar with control. I attach two pictures of my control loop (continous and discrete, even if the final system will be discrete).

Continous Loop Discrete Loop

Someone can tell me what is the problem? Thanks

$\endgroup$
4
$\begingroup$

Your goal is to stabilize the roll angle $\varphi$.
Its dynamics are described via $$ \ddot{\varphi}I_{xx} = u $$ where $u$ is the control input (motor torque in your case).

The 1$^{\text{st}}$ flaw I found is that you're using four integrators instead of two. Assuming that the measured output that is used for an output feedback control is $$y = \int\int\ddot{\varphi}\: \text{d}t\text{d}t= \varphi,$$

only two states are required (denoted phi and dphi in the figure). Stated as a SL model this would be

Closed loop roll angle stabilization

The 2$^{\text{nd}}$ problem I noted is the choice of your controller. Applying a PD controller with the transfer function

$$ K_{PD}(s) = \frac{Vs}{1+Ts} $$

renders one of your critically stable poles ($s=0$) uncontrollable due to a pole-zero cancellation in the open loop

$$ L(s) = \frac{Vs}{1+Ts} \frac{1}{s^2} = \frac{V}{s(1 + Ts)}. $$

An output-feedback controller $K$ that does the job of stabilizing the $\varphi$ can be achieved via open loop shaping. The main design goal is to retain a reasonable phase reserve of about 60°. This can be achieved by a lead/lag controller design

$$ K_{LL}(s) = V_{LL}\frac{1+T_{lead}s}{1+T_{lag}s}$$

where I chose for your nominal system ($I_{xx} = 1$) the following parameters:

$$ V_{LL} = 100, \qquad T_{lead} = \frac{1}{5}, \qquad T_{lag} = \frac{1}{50}. $$

This choice yields the following Bode plot which shows a phase reserve of around 55° at a breakthrough frequency of $\omega_{c} \approx 18 \frac{\text{rad}}{\text{s}}$.

Bode plot of lead/lag open loop

The corresponding step response gives a good idea of the performance of the controller. Note that the integrators of your model covers a vanishing control error $\lim\limits_{t\rightarrow \infty}e = 0$ for the chosen output feedback loop but I don't know if you're feeling comfortable with an overshoot of $20$%+.

enter image description here

In case your IMU has the capability of measuring roll velocity as well you could switch over to a state feedback controller design which has the advantage of pole placement. This means that you can freely choose thoroughly new dynamics of your roll angle behaviour (limited by input constraints and measuring noise, of course..) at the cost of losing the structural property of vanishing control error.

The Ackermann algorithm for doing so is $$ k^T = [0, 1]S^{-1}(A,b)(p_0 + p_1A + p_2A^2), $$ where $k^T$ is the feedback vector, $S^{-1}(A,b)$ is the inverse of the controllability matrix, $A$ is your system matrix and $p_i$ are the coefficients of your desired characteristic polynomial.

Your system has the following parameters: $$ A = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right), \qquad b = \left( \begin{array}{c} 0 \\ \frac{1}{I_{xx}} \\ \end{array} \right), \qquad S(A,b) = \left( b, Ab \right). $$

I.e. your nominal system yields a feedback vector $k^T = [k_1, k_2] = [400, 40]$, given desired eigenvalues of $\lambda_1 = \lambda_2 = -20$.

It can be implemented in the following fashion:

QuadCop state feedback as Simulink model

The step response needs output scaling, though:

Step response of state feedbacked nominal system

Don't forget to reassure that you don't run into any input constraints when designing the controller.

$\endgroup$

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.