0₁

1) Consider the single link with flexible joint example in the lecture.

Assume that this time there is one more rigid element connected

between the motor and the rod. Therefore, this system involves three

rigid bodies connected in series via two springs: In this assignment, Motor shaft

you will make a closed-loop control simulation (via Matlab/Simulink)

on which you will practice two types of controllers. First, obtain the

state-space model of the system and prepare your simulated plant in

Simulink. Please use fixed time-step with ode45 solver.

In your simulation, the goal is to raise the link from down-right

position to 120 degrees in the CCW direction, as shown in the figure

on the left corner. The figures present the system from the side and

behind. Jm, Jc, and Jr are the rotational inertias of the bodies. Mr. is

the mass of the rod of which the center of mass is apart from its hinge

by L. k₁ and k₂ are the stiffness coefficients of the springs. The rod is

being actuated by a torque applied by the motor on its shaft.

k₁

k₁

100 Nm/rad

k₂ = 100 Nm/rad

Jm = 2 Nms²/rad

Jc = 1 Nms²/rad

Jr = 4 Nms²/rad

Mr = 12 kg

Flexible joints

0₂

coupler

120 deg

k2

L = 0.5 m

rod

03

a) Derive an input-state linearized controller and implement it on your simulation

b) Derive the control law, using the back-stepping control method and implement it.

c) Compare the performance of each of the controllers above. You can choose the controller gains as you please.

d) This time assume that our estimation of some of the model parameters are off to a certain degree. Derive the control

laws above assuming:

Jr = 2 Nms² /rad and k₁,2 = 120 Nm/rad

Which of the above methods perform better under these conditions? Note: The true parameters of the system model

should be those given in the figure above.)