# feedback linearization for a quadrotor

I am trying to develop a controller for a quadrotor using dynamic feedback linearization technique, but I am not sure I am doing thigs correctly. I know that for designing such controller I have to keep derivating until I see the input and it appears in a nonsingular way.

In my case, if I do this for example just for the $$z$$ coordinate I obtain that if I derivate two times I see the input in the expression of the second derivative, so after derivating two times I have:

$$\ddot{z} = -cos(\theta )cos(\varphi )\frac{T}{m}+g$$

where we can see that the thrust $$T$$ appears. Now we can derive a control law using a dynamic controller of the form :

$$J(\xi ,\zeta )^{-1}(-l(\xi ,\zeta)+v)$$

where $$\zeta = \begin{pmatrix} T\\ \dot{T} \end{pmatrix}$$, and I have understand well, we have do design such controller such that $$\ddot{z} = v$$ .

Now, if I apply this to the quadrotor, I obtain:

$$T = \frac{m(g-\ddot{z}))}{cos(\theta )cos(\gamma )}$$

where $$\ddot{z} = \ddot{z_{d}} -K_{d}(\dot{z_{d}} - \dot{z}) - K_{p}(z_{d}-z)$$ and with $$z_{d}$$ as desired value for the $$z$$.

The problem is that it looks too much as a PID controller to me, so I don't see the difference between the two types of controller.

[EDIT] moreover, what are the benefits of using feedback linearization and what is the difference between this tecnhnique and using backstepping or PID controllers?

• What do you mean by "the two types of controller"? You mean feedback linearization (or input-output linearization) and PID? Namely feedback linearization alone isn't a full control law yet, but only your equation for $T$ with $\ddot{z}$ set equal to some to be determined new virtual input. – fibonatic Oct 14 '19 at 0:26