UAV-Autonomous-control

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3D UAV simulation and autonomous control for path tracking

• [x] Build a controller to follow a trajectory
• [x] Follow hardcoded waypoints
• [x] From hardcoded waypoints, generate an optimal trajectory (minimum snap)
• [x] From hardcoded waypoints and hardcoded obstacles, generate an optimal trajectory (minimum snap + collision-free)
• [x] Generate waypoints from a path planning algorithm (RRT*), given hardcoded obstacles and generate an optimal trajectory (minimum snap + collision-free)
• [ ] Localize the UAV using a Map
• [ ] Localize the UAV and build a map (SLAM)
• [ ] Handle dynamic obstacles

Result

https://github.com/Mdhvince/UAV-Autonomous-control/assets/17160701/6dff700c-9ed4-4d6f-b976-8f42022d5d80

https://github.com/Mdhvince/UAV-Autonomous-control/assets/17160701/2dcca69f-2981-4cef-a6e1-313b70ec2882

Vanilla RRT

RRT*

Controller response

1. Control

source: A platform for aerial robotics research and demonstration: The Flying Machine Arena
Sergei Lupashin, Markus Hehn, Mark W. Mueller, Angela P. Schoellig, Michael Sherback, Raffaello D’Andrea

2. Trajectory Planner with minimum snap

Finding a path (waypoints) to a goal location is one of many important steps for an autonomous robot. But this path must satisfy some conditions such that a controller can handle it. Some of these conditions can be summarized as follows:

• The path must be feasible
• The path must be collision-free
• The path must be differentiable
• The path must be smooth enough

The thing is that a system as a quadrotor can be represented (in this case) as a 4th-order system, so a path for this kind of system must be differentiable at least 4 times: Let's denote k as the order of derivative:

• k=1 : velocity
• k=2 : acceleration
• k=3 : jerk
• k=4 : snap

So a good trajectory for this system can be thought of as a minimum snap trajectory, hence a trajectory that minimizes the snap criterion. So we need to find the optimal path

$$ \boxed{ x^{*}(t) = argmin_{x(t)} = \int_{0}^{T} \mathcal{L}(\ddddot{x}, \dddot{x}, \ddot{x}, \dot{x}, x, t) dt = \int_{0}^{T} \ddddot{x}^{2} dt } $$

where $\mathcal{L}$ is the Lagrangian of the system and $\ddddot{x}$ is the snap of the trajectory.

We can find the optimal path by solving the Euler-Lagrange equation:

$$ \boxed{ x(t) = c_{7}t^7 + c_{6}t^6 + c_{5}t^5 + c_{4}t^4 + c_{3}t^3 + c_{2}t^2 + c_{1}t + c_{0} } $$

Why 8 coefficients? Because we have 8 boundary conditions to respect. The boundary conditions are:

• Position at t=0, and t=T
• Velocity at t=0, and t=T
• Acceleration at t=0, and t=T
• Jerk at t=0, and t=T

By applying these conditions, we can find the 8 coefficients, hence the optimal path that minimizes the snap criterion.

Differentiating this equation gives the velocity/acceleration/jerk/snap constraints and so on...

$$ \dot{x}(t) = 7c_{7}t^6 +6 c_{6}t^5 + 5c_{5}t^4 + 4c_{4}t^3 + 3c_{3}t^2 + 2c_{2}t + c_{1} $$

what we are interested in is finding the coefficients c0, c1, c2, c3, c4, c5 that satisfy all the constraints (boundary conditions) mentioned above. _note: If I have another constraint to respect, I will have to find one more coefficient.

Each of the conditions gives an equation, so we can represent them in a Matrix $A$. We can write the equation in terms of unknown constants and boundary conditions. Solving for these constants (coefficients) is a linear problem.

To respect the position constraint:

$$x(t) = c_{7}t^7 + c_{6}t^6 + c_{5}t^5 + c_{4}t^4 + c_{3}t^3 + c_{2}t^2 + c_{1}t + c_{0}$$

$$x(0) = c_{0}$$

$$x(T) = c_{7}T^7 + c_{6}T^6 + c_{5}T^5 + c_{4}T^4 + c_{3}T^3 + c_{2}T^2 + c_{1}T + c_{0}$$

Matrix form at t=0

$$ \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} c_{7} \ c_{6} \ c_{5} \ c_{4} \ c_{3} \ c_{2} \ c_{1} \ c_{0} \end{bmatrix} = p_0 $$

Matrix form at t=T

$$ \begin{bmatrix} T^7 & T^6 & T^5 & T^4 & T^3 & T^2 & T^1 & T^0 \end{bmatrix} \cdot \begin{bmatrix} c_{7} \ c_{6} \ c_{5} \ c_{4} \ c_{3} \ c_{2} \ c_{1} \ c_{0} \end{bmatrix} = p_T $$

To respect the velocity constraint we differentiate the position equation:

$$\dot{x}(t) = 7c_{7}t^6 +6 c_{6}t^5 + 5c_{5}t^4 + 4c_{4}t^3 + 3c_{3}t^2 + 2c_{2}t + c_{1}$$

$$\dot{x}(0) = c_{1}$$

$$\dot{x}(T) = 7c_{7}T^6 +6 c_{6}T^5 + 5c_{5}T^4 + 4c_{4}T^3 + 3c_{3}T^2 + 2c_{2}T + c_{1}$$

Matrix form at t=0

$$ \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} c_{7} \ c_{6} \ c_{5} \ c_{4} \ c_{3} \ c_{2} \ c_{1} \ c_{0} \end{bmatrix} = v_0 $$

Matrix form at t=T

$$ \begin{bmatrix} 7T^6 & 6T^5 & 5T^4 & 4T^3 & 3T^2 & 2T^1 & T^0 & 0 \end{bmatrix} \cdot \begin{bmatrix} c_{7} \ c_{6} \ c_{5} \ c_{4} \ c_{3} \ c_{2} \ c_{1} \ c_{0} \end{bmatrix} = v_T $$

To respect the acceleration constraint we differentiate the velocity equation:

$$\ddot{x}(t) = 42c_{7}t^5 + 30c_{6}t^4 + 20c_{5}t^3 + 12c_{4}t^2 + 6c_{3}t + 2c_{2}$$

$$\ddot{x}(0) = 2c_{2}$$

$$\ddot{x}(T) = 42c_{7}T^5 + 30c_{6}T^4 + 20c_{5}T^3 + 12c_{4}T^2 + 6c_{3}T + 2c_{2}$$

Matrix form at t=0

$$ \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} c_{7} \ c_{6} \ c_{5} \ c_{4} \ c_{3} \ c_{2} \ c_{1} \ c_{0} \end{bmatrix} = a_{0} $$

Matrix form at t=T

$$ \begin{bmatrix} 42T^5 & 30T^4 & 20T^3 & 12T^2 & 6T^1 & 2T^0 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} c_{7} \ c_{6} \ c_{5} \ c_{4} \ c_{3} \ c_{2} \ c_{1} \ c_{0} \end{bmatrix} = a_{T} $$

To respect the jerk constraint we differentiate the acceleration equation:

$$\dddot{x}(t) = 210c_{7}t^4 + 120c_{6}t^3 + 60c_{5}t^2 + 24c_{4}t + 6c_{3}$$

$$\dddot{x}(0) = 6c_{3}$$

$$\dddot{x}(T) = 210c_{7}T^4 + 120c_{6}T^3 + 60c_{5}T^2 + 24c_{4}T + 6c_{3}$$

Matrix form at t=0

$$ \begin{bmatrix} 0 & 0 & 0 & 0 & 6 & 0 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} c_{7} \ c_{6} \ c_{5} \ c_{4} \ c_{3} \ c_{2} \ c_{1} \ c_{0} \end{bmatrix} = j_{0} $$

Matrix form at t=T

$$ \begin{bmatrix} 210T^4 & 120T^3 & 60T^2 & 24T^1 & 6T^0 & 0 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} c_{7} \ c_{6} \ c_{5} \ c_{4} \ c_{3} \ c_{2} \ c_{1} \ c_{0} \end{bmatrix} = j_{T} $$

All 8 constraints can be written as an 8x8 matrix to find the coefficients of the polynomial (coefficients of the trajectory). The full matrix is the following:

$$ A = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ T^7 & T^6 & T^5 & T^4 & T^3 & T^2 & T & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 7T^6 & 6T^5 & 5T^4 & 4T^3 & 3T^2 & 2T & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 \ 42T^5 & 30T^4 & 20T^3 & 12T^2 & 6T & 2 & 0 & 0 \ 0 & 0 & 0 & 0 & 6 & 0 & 0 & 0 \ 210T^4 & 120T^3 & 60T^2 & 24T & 6 & 0 & 0 & 0 \ \end{bmatrix} $$

So $A$ is the matrix of the constraints, and we need to find the vector of the coefficients of the polynomial $c$ from $A$ and $b$ the vector of the constraints:

$$\boxed{c = A^{-1}b}$$

Define $b$ vector

$b$ is a column vector of 8 elements, each element is a constraint.
Let's say we want a trajectory that starts from $x=2$, and ends at $x=5$, with an average velocity of $v=1$.

$$ b = \begin{bmatrix} 2 \ 5 \ 1 \ 1 \ 0 \ 0 \ 0 \ 0 \ \end{bmatrix} $$

Find the coefficients $c$ of the polynomial

$$ c = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ T^7 & T^6 & T^5 & T^4 & T^3 & T^2 & T & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 7T^6 & 6T^5 & 5T^4 & 4T^3 & 3T^2 & 2T & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 \ 42T^5 & 30T^4 & 20T^3 & 12T^2 & 6T & 2 & 0 & 0 \ 0 & 0 & 0 & 0 & 6 & 0 & 0 & 0 \ 210T^4 & 120T^3 & 60T^2 & 24T & 6 & 0 & 0 & 0 \ \end{bmatrix}^{-1} \cdot \begin{bmatrix} 2 \ 5 \ 1 \ 1 \ 0 \ 0 \ 0 \ 0 \ \end{bmatrix} $$

:rocket: