Q4: Differential Models – Fundamentals of Sensing, Tracking, and Autonomy 2

Question 1¶

Consider a Dubins car model with the following equations of motion:

\dot{x} = u_s \cos\theta, \quad \dot{y} = u_s \sin\theta, \quad \dot{\theta} = \frac{u_s}{L} \tan(u_φ), \\\

and the following constraints:

The car can only move forward, so 𝑢ₛ = 1 (constant forward speed).
The steering input 𝑢_φ is constrained to −π/4 ≤ 𝑢_φ ≤ π/4.
A maximum steering input (𝑢_φ = ±π/4) produces a circle of radius 𝜌 = 1.
The initial configuration is 𝑞_init = (0, 0, 0) (the car starts at the origin facing along the positive 𝑥-axis).

Note: For a Dubins car model, when a steering input 𝑢_φ is applied, the car follows a circular arc of radius:

\rho = L / \tan(u_φ). \\\

Define the following motion primitives:

S(𝑡): straight motion (𝑢_φ = 0) for 𝑡 seconds.
L(𝑡): maximum left turn (𝑢_φ = π/4) for 𝑡 seconds.
R(𝑡): maximum right turn (𝑢_φ = −π/4) for 𝑡 seconds.

Which of the following are true?

For the given Dubins car model and constraints, the length of the car (distance between its axles) is 𝐿 = 10.

If 𝐿 = 1, then 𝜌 = 1 / tan(𝑢_φ), and 𝑢_φ = π/4 gives 𝜌 = 1.

Applying 𝑢_φ = 0 for 1 second followed by 𝑢_φ = −π/4 for 4 seconds moves the car to (10, 10, π).

Applying S(1), R(π) from (0, 0, 0) returns the car to the 𝑥-axis with heading 0.

The configuration (2, 2, 0) can be reached from (0, 0, 0) using L(π/2), R(π/2).

The configuration (2, 2, 0) can also be reached from (0, 0, 0) using exactly three motion primitives (with no repeated primitive in consecutive steps).

The configuration (4, 0, 0) can be reached from (0, 0, 0) using R(π/2), L(π), R(π/2).

The configuration (4, 0, 0) can also be reached from (0, 0, 0) using exactly one motion primitive.

The configuration (0, 4, 0) can be reached from (0, 0, 0) using L(π), R(π).

The configuration (0, 4, 0) can also be reached from (0, 0, 0) using exactly three motion primitives (with no repeated primitive in consecutive steps)

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Question 2¶

Consider the same Dubins car model, with the same constraints and initial configuration as in Question 1.

Your task is to provide a sequence of exactly three motion primitives that reaches the target configuration (2, 2, 0) from (0, 0, 0).

Additional requirements:

You must use exactly one of the following sequence types: LSL, RSR, LRL, or RLR.
Do not use any other sequence (for example, LSR or RSL are not allowed).
Question 2: Only LRL or RLR are allowed.
Questions 3–4: Only LSL or RSR are allowed.

Submit your answer in the following strict format:

Use exactly three primitives.
Write primitives in the form L(value), S(value), R(value), separated with commas.
value cannot be 0.
For L and R, value must be within the interval (0, 2π).
Write values using numbers, pi, and standard mathematical expressions (e.g., pi/2, pi, 3*pi/2, 1, atan(3/4) + sqrt(3)).

Example of a valid submission:

L(pi/2),S(10),L(pi/2)

Question 3¶

Consider the same Dubins car model, with the same constraints and initial configuration as in Question 1.

Your task is to provide a sequence of exactly three motion primitives that reaches the target configuration (0, 0, π/2) from (0, 0, 0).

Submit your answer in the same format as in Question 2.

Question 4¶

Consider the same Dubins car model, with the same constraints and initial configuration as in Question 1.

Your task is to provide a sequence of exactly three motion primitives that reaches the target configuration (3, 2, π) from (0, 0, 0).

Submit your answer in the same format as in Question 2.

Question 5¶

Consider a point robot in the plane with configuration 𝑞 = (𝑥, 𝑦). The robot's velocity is: 𝑞̇ = (𝑥̇, 𝑦̇).

Now consider the model in which the acceleration of the robot is directly controlled:

\ddot{x} = u_x, \quad \ddot{y} = u_y. \\\

This model is called the double integrator. The state is (𝑥, 𝑦, 𝑥̇, 𝑦̇).

Which of the following are true?

If 𝑥̈ = 𝑦̈ = 0, then the robot must move at constant velocity (which could be zero).

If 𝑥̈ = 𝑦̈ = 0, then the robot is not moving.

If 𝑥̈ > 0 and 𝑦̈ = 0 and the initial velocity is (0, 0), then the robot moves along the positive 𝑥-axis with constant speed.

If the initial state is (0, 0, 1, 1) and 𝑥̈ = 𝑦̈ = 0, then the robot moves along a straight line at 45° forever with constant speed.

If ‖𝑞̈‖ ≤ 1, then the robot cannot stop (i.e., it can never reach 𝑥̇ = 𝑦̇ = 0).

If 𝑥̈ = 1 and 𝑦̈ = 1, the robot follows a circular path.

If 𝑥̈ = 1 and 𝑦̈ = 1 with initial velocity (0, 0), the robot follows a straight line at 45° with increasing speed.

Under the constraint ‖𝑞̈‖ = 1, the robot follows a circular path regardless of initial velocity.

If the initial state is (0, 0, −4, −4) and constant acceleration (𝑢_𝑥, 𝑢_𝑦) = (2, 2) is applied, the robot can never reach zero velocity.

If the initial state is (0, 0, −4, −4) and constant acceleration (𝑢_𝑥, 𝑢_𝑦) = (2, 2) is applied, the robot reaches zero velocity at 𝑡 = 2 and then reverses direction.

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Question 6¶

Consider the 1D double integrator from LaValle (Section 14.2.1). The continuous model is 𝑞̈ = 𝑢, with 𝑢 ∈ {−1, 0, 1}. The state is 𝑥 = (𝑞, 𝑞̇) ∈ ℝ², and the continuous solution for constant 𝑢 is:

\dot{q}(t) = \dot{q}(0) + ut, \quad q(t) = q(0) + \dot{q}(0)\,t + \tfrac{1}{2}u\,t^2. \\\

A discrete lattice is formed by applying constant 𝑢 ∈ {−1, 0, 1} for unit time steps Δ𝑡 = 1. The discrete update rule is:

\dot{q}' = \dot{q} + u, \quad q' = q + \dot{q} + \tfrac{1}{2}u. \\\

Which of the following are true?

If (𝑞, 𝑞̇) ∈ ℤ² and 𝑢 ∈ ℤ, the next state is also in ℤ². The lattice is integer.

Applying 𝑢 = 0 produces a straight line in the state space, regardless of the current state.

Applying constant 𝑢 = 1 from (0, 0) produces positions 𝑞 = 0, 1, 2, 3, … — a straight line.

Applying constant 𝑢 = 1 from (0, 0) produces positions 𝑞 = 0, 1/2, 2, 9/2, 8, 25/2, … — points on a parabola.

The control sequence [−1, +1, +1, −1] from (0, 0) traces a closed loop: (0,0) → (−1/2,−1) → (−1,0) → (−1/2,1) → (0,0).

Applying [+1, +1, −1, −1] from (0, 0) stops the car at 𝑞 = 4.

The 20-step control sequence [−1, −1, −1, +1, +1, +1, +1, +1, +1, +1, −1, −1, −1, −1, −1, −1, 0, +1, 0, +1] starting from (0, 0) ends at (−9, 0).

The 20-step control sequence [−1, −1, −1, +1, +1, +1, +1, +1, +1, +1, −1, −1, −1, −1, −1, −1, 0, +1, 0, +1] starting from (0, 0) traces a large loop reaching 𝑞 = −9 and 𝑞 = 7, and returns exactly to (0, 0) with zero velocity.

For any points in the state space, there are infinitely many paths that would bring the point back to the same state space forming a closed loop in the state space.

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Question 7¶

Consider a system in which the configuration space 𝒞 has dimension 𝑛. The state space 𝑋 is obtained by including velocity information needed for the equations of motion.

Which of the following are true?

If all 𝑛 configuration variables are controlled by acceleration (double integrator), then the state space has dimension 2𝑛.

If all 𝑛 configuration variables are controlled by acceleration, then the state space has dimension 𝑛.

If 𝑘 of the 𝑛 variables are controlled by acceleration and the remaining 𝑛 − 𝑘 are controlled directly by velocity, the state space dimension is 𝑛 + 𝑘.

The state space dimension is always exactly 2𝑛, regardless of the control model.

For the Dubins car with configuration (𝑥, 𝑦, 𝜃), the configuration space has dimension 3, and the state space also has dimension 3 because the inputs directly control velocity-level quantities.

For the Dubins car, the state space has dimension 6 because each of the 3 configuration variables needs a velocity variable.

For a point robot in the plane (𝑞 = (𝑥, 𝑦)) controlled by acceleration, the state is (𝑥, 𝑦, 𝑥̇, 𝑦̇), which has dimension 4.

For a point robot in the plane controlled by acceleration, the state space has dimension 2.

Adding acceleration control to a variable increases the state space dimension by 1 per variable, because the velocity becomes part of the state.

The state space dimension depends on the number of obstacles, not the control model.

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Question 8¶

Consider the Dubins car model with an added nature parameter 𝜓 that perturbs the heading. The equations become:

\dot{x} = u_s \cos(\theta + \psi), \quad \dot{y} = u_s \sin(\theta + \psi), \quad \dot{\theta} = \frac{u_s}{L} \tan(u_φ), \\\

in which 𝜓 ∈ [−𝜓_max, 𝜓_max] is chosen by nature (it models wind or disturbance). The controls are 𝑢_φ ∈ [−𝜋/4, 𝜋/4], and 𝑢ₛ = 1, 𝐿 = 1 as in Question 1.

Which of the following are true?

The nature parameter 𝜓 is chosen by the robot to help it reach the goal faster.

If 𝜓_max = 0, the car cannot move because nature provides no input.

If 𝜓_max = 0, then the model reduces to the standard Dubins car with no disturbance.

The reachable set from a configuration grows larger as 𝜓_max increases, because nature can push the car in more directions.

Increasing 𝜓_max shrinks the reachable set because the disturbance opposes the robot's motion.

For a worst-case (adversarial) nature, planning must ensure the goal is reached for every possible 𝜓 ∈ [−𝜓_max, 𝜓_max], not just one.

It is sufficient to plan for 𝜓 = 0 and the plan will work for all other values of 𝜓.

With 𝜓_max > 0 and with 𝜓 dependent on time, the car can deviate from a pure circular arc even when 𝑢_φ is held constant, because the effective heading is 𝜃 + 𝜓(t).

The nature parameter only affects the 𝜃̇ equation and has no effect on the car's position trajectory.

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Lovelace

QUIZ 4: Differential Models¶

Question 1¶

Question 2¶

Question 3¶

Question 4¶

Question 5¶

Question 6¶

Question 7¶

Question 8¶

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