Quadrotor 2D Dynamics

Question

I am trying to understand the control of the quadrotor in 2 dimensions from the Penn course on aerial robotics. The attached image describes the linearization of the control efforts assuming that the quadrotor is near the hover position. Everything seems fine except Equation 7 to me. The control effort u1 makes sense as u1 = mg + mz¨c. The original equation z¨ = − g + u1*m is solved for u1, and z¨c (the actual z acceleration) is used.

However, the equation for u2 goes from φ¨ =u2/Ixx to u2 =Ixx*φ¨T (t), where φ¨T (t) is the desired value of φ acceleration instead of the actual value of φ acceleration.

Can anyone explain this to me?

EDIT: For further clarity, let me try rephrasing this question.

We start with three simplified, linearized equations:

$$ \ddot{y} = -g * \phi $$ $$ \ddot{z} = -g + \frac{u_1}{m} $$ $$ \ddot{\phi} = \frac{u_2}{I_{xx}} $$

For $u_1$, we solve to get: $$ u_1 = m\ddot{z} +mg $$

and since our control effort is defined as:

$$ \ddot{r_c} = \ddot{r_T}(t) k_{v}(\dot{r_T}-\dot{r}) + k_{p}(r_T-r) + $$

we plug in the Z part of this control effort and get

$$ u_1 = m\ddot{z} +mg $$ $$ u_1 = \ddot{z_T}(t) + k_{p,z}(\dot{z_T}(t)-z) + k_{v,z}(\dot{z_T}(t)-\dot{z}) +mg $$

In this case, when we took the Z part of the PD control effort, we replaced $\ddot{z}$ with the formula for $\ddot{z_c}$ (our control effort)

However, that is not what happens when solving for u2. We start off

$$ u_2 = I_{xx}*\ddot{\phi} $$

This time, when we replace $\ddot{\phi}$, I would think it would be replaced the same way as above, using $\ddot{\phi_c}$ our control effort. So it would look like this: $$ u_2 = I_{xx}*\ddot{\phi_c} $$ $$ u_2 = I_{xx} [\ddot{\phi_T}(t) + k_{v,\phi}(\dot{\phi_T}(t)-\dot{\phi}) + k_{p,\phi}(\phi_T(t)-\phi)] $$

Instead, we do this:

$$ u_2 = I_{xx}*\ddot{\phi_T} $$ $$ u_2 = I_{xx} [\ddot{\phi_c}(t) + k_{v,\phi}(\dot{\phi_c}-\dot{\phi}) + k_{p,\phi}(\phi_c-\phi)] $$

If we go back to our basic PD control effort formula, this contradicts it. With this replacement, we are saying that: $$ \ddot{\phi_T} = \ddot{\phi_c}(t) + k_{v,\phi}(\dot{\phi_c}-\dot{\phi}) + k_{p,\phi}(\phi_c-\phi) $$ which is not true because $$ \ddot{r_c} = \ddot{r_T}(t) + k_{v}(\dot{r_T}-\dot{r}) + k_{p}(r_T-r) $$ which means that the real solution for $\ddot{\phi_T}$ would be $$ \ddot{\phi_T} = \ddot{\phi_c}(t) - k_{v,\phi}(\dot{\phi_c}-\dot{\phi}) - k_{p,\phi}(\phi_c-\phi) $$ That is my confusion!

Answer 1

Remember, the $y$ position of the quadrotor depends on the angle $\phi$, therefore, in order to control $y$, we need to control $\phi$ in a way that yields the desired $y$. To control the angle $\phi$, we start from the original equation: $$ \ddot{\phi} = \frac{u_2}{I_{xx}} $$ Now the goal is to design a controller by which the angle is controlled. The PD-computed torque controller has this form:

$$ u_2 = I_{xx} [\ddot{\phi_c} + k_{v,\phi}(\dot{\phi_c}-\dot{\phi}) + k_{p,\phi}(\phi_c-\phi)] $$

Now plug $u_2$ inside the original function of the angle, hence

$$ \begin{align} \ddot{\phi} &= \frac{u_2}{I_{xx}} \\ &= \frac{I_{xx} [\ddot{\phi_c} + k_{v,\phi}(\dot{\phi_c}-\dot{\phi}) + k_{p,\phi}(\phi_c-\phi)]}{I_{xx}} \\ &= \ddot{\phi_c} + k_{v,\phi}(\dot{\phi_c}-\dot{\phi}) + k_{p,\phi}(\phi_c-\phi) \end{align} $$

Notice that the term $I_{xx}$ is eliminated deliberately. The notion is trivial. Relying on the mathematical model, we can design a controller that compensates for linear or nonlinear terms (e.g. as the case with $I_{xx}$).