Quiz 3 Solutions

SOLUTIONS: QUIZ 3 CONTROL OF MOBILE ROBOTS

1 This question investigates how to obtain linearized models of non-linear dynamical systems. A controlled pendulum with friction can be described by mℓφ¨ = −mg sin φ − kℓφ˙ + τ, where φ is the angle of the pendulum and τ is the input torque applied at the base of the pendulum. m is the mass of the pendulum, ℓ its length, g is the gravitational constant, and k is a friction coefficient. Linearize this model around the two equilibrium points, i.e., when the pendulum is hanging straight ˙ and u = τ , giving the linearized down (φ = 0) and balancing straight up (φ = π). Let x1 = φ, x2 = φ, system x˙ = Ax + Bu, where   0 B= . 1 mℓ

But what is A? Which of the following A-matrices give the correct linearizations around the two equilibrium points?

SOLUTION Let x1 = φ, x2 = φ˙ and u = τ , which gives the dynamics x˙ 1 = x2 x˙ 2 = − gℓ sin(x1 ) −

k m x2

+

1 mℓ u.

The linearization is given by x˙ = Ax + Bu =

∂f (xe , ue ) ∂f (xe , ue ) x+ u, ∂x ∂u

where the partial derivatives are ∂f = ∂x



0 1 k − gℓ cos(x1 ) − m



,

∂f = ∂u



˙ = (0, 0) and (φ, φ) ˙ = (π, 0) gives Plugging in the values (π, φ)   0 1 φ=0: A= k −g −m   ℓ 0 1 , φ=π: A= g k −m ℓ which is the answer. 1

0 1 mℓ



2 Which of the following statements is correct?

SOLUTION Let’s go through the answers one by one. The stability properties of x˙ = Ax is determined by A’s eigenvalues. We saw in class that this is indeed the case, with asymptotic stability requiring that the real part of all the eigenvalues be strictly negative, while the presence of a single eigenvalue with positive real part renders the system unstable. So this statement is true. Output-feedback is more general than state-feedback. Output feedback means that u = −Ky, but since y = Cx we can write this as u = −KCx or u = −K ′ x for a particular choice of K ′ . As such, output feedback can be written as a state feedback controller, while the opposite is not true. As such, output feedback is not more general, and this statement is false. The unicycle model has a useful linearization. In class we saw that the linearization of the unicycle around the point (x, y, φ) = (0, 0, 0) gave us a system where we could move in the x-direction and turn the robot, but not move in the y-direction. This means that the linearization is clearly not telling the whole story since an actual robot can indeed move in the y-direction as well. As such, this linearization is not particularly useful and the statement is false. Stability is not a control objective when designing controllers. This statement is clearly false. Without stability, we cannot do anything so not only is it a control objective, it is by far the most important objective. Most systems are already linear so linearizations are not really needed. This statement is not true. Most systems are nonlinear but the linearization provides a tight way of describing their behaviors (at least around operating points).

3 This question involves the computation of eigenvalues and how these eigenvalues relate to the stability properties of the underlying nonlinear system. Given a linear system   α β x˙ = x. 0 α Which of the options below captures all the values of α and β for which the system is asymptotically stable? Hint: The following fact might come in handy: The determinant of a 2 × 2 matrix is given by a b c d = ac − bd.

SOLUTION The eigenvalues to the A matrix are given by  λ−α 0 = det(λI − A) = det 0

−β λ−α

2



= (λ − α)2 ⇒ λ = α,

i.e., both eigenvalues are given by α. This means that β does not matter, i.e., it can be whatever, while for asymptotic stability we need that the real part (in this case λ is real so it is equal to its real part) to be negative. As such, the answer is: α < 0, β whatever.

4 Let the output of a third order system (x˙ = Ax, y = Cx) be given in the figure below: 1

y(t)

0.5

0

−0.5

0

1

2

3

4

5

6

7

8

9

10

t

The options below show the possible placements of the eigenvalues to the A matrix, with the axis being the real parts and the imaginary parts of the eigenvalues. Which option corresponds to the system used to generate the figure?

SOLUTION There are a few observation we should make right away: First of all, it is a third order system so there needs to be three eigenvalues. Moreover, as the system is stable (it clearly does not blow up), we know that the real parts of all three eigenvalues must be negative. This leaves three possible candidates. We also see that in the plot, the output (and hence also the state) seems to oscillate quite a bit before settling down. This implies that at least one of the eigenvalues has a non-zero imaginary part. But, we also know that there is no way we can have just one complex eigenvalue. They always show up in complex conjugate pairs, i.e., if σ + jω is an eigenvalue then so is σ − jω. This leaves only one possible choice, and the solution is given below:

3

Im

× Re

× ×

5 Recall the quadrotor model from Module 1 ¨ = cv − g, h where h is the altitude of the quadrotor and v is the input. The purpose of this question is to revisit PID design but in the context of state-feedback to see that there really isn’t anything magical/special about PID. So, please keep reading and follow along in the math... First, if we directly compensate for gravity by introducing a new control input v = u + g/c we get the ¨ = cu. much simpler system h Now, let x1 = h, x2 = h˙ and the resulting linear system is     0 1 0 x˙ = x+ u. 0 0 c ˙ i.e., As seen, a PD regulator is u = KP e + KD e˙ = KP (r − h) + KD (r˙ − h),   r − KP D x, u = KP D r˙ where KP D = [KP , KD ], i.e., it is just a state-feedback controller, and the closed-loop system becomes   r . x˙ = (A − BKP D )x + BKP D r˙ So doing state-feedback is exactly the same as PD control for this system. But what about the I-part? This requires some more work. For this we actually need to introduce another, extra state. Let the new state be Z t x1 (τ )dτ. x3 (t) = 0

The resulting system has three states instead of two and if we let   x1 x ˜ =  x2  , x3

4

we get the closed-loop PID dynamics 

 r ˜ P ID )˜ ˜ P ID  r˙  , x˜˙ = (A˜ − BK x + BK R r

where KP ID = [KP , KD , KI ] and where



 0 ˜ =  c . B 0 ˜ (Write out the dynamics of the new, third order But, which of the options below give the correct A? system in order to figure out the new A-matrix.)

SOLUTION Since x3 =

Z

t

x1 (τ )dτ,

0

we directly see that x˙ 3 = x1 . As such, the new A˜ matrix has to be  A˜ = 

A 1

0

  0 0 0 = 0 1 0

which is the answer.

5

 1 0 0 0 , 0 0