Inverted Pendulum System The inverted pendulum system shown in Figure Q3.1 consists of a pole and a trolley on which the pole is hinged. The trolley moves on the rail tracks to its right or left, depending on the force exerted on the trolley. The control goal is to balance the pole starting from nonzero conditions by applying appropriate force to the trolley. Our control goal here is to balance the pole without regard to the trolley position and velocity, with x₁ = 0 and x2 = as the angular displacement and angular velocity of the pole. The relevant equation of motion is given by gsino + cose -ml mc+m 1 ė²sino + 9 = mc + m ml - cos20 mc + m Assume that trolley mass mc = 1.0 kg, pole mass m = 0.1 kg, half-length of pole / = 0.5 m, gravity acceleration g = 9.81m/s and F is the applied force in Newtons. From the above equation of motion, the state equations of this inverted pendulum system can be derived as x₁ = X2 x₂ = gsinx₁+cosx₁(-a₁x2² sinx₁ + a₂F) b₁ - b₂cos²x1 where a₁ = ml mc+m = 0.0455, a₂ mc+m ; = 0.9091, b₁ = 1 = 0.6667, ml b₂ = = 0.0455. mc+m Assuming that the sampling time T = 0.02 sec, and using backward difference discretisation, the dynamics of the inverted pendulum system can be approximated by x₁(k + 1) = x₁(k) + Tx₂(k) (gsinx₁ x2(k+1)=x2(k) +T (k)+cosx₁(k)(-a₁[x₂(k)]²sinx₁(k) + α₂F)` b₁-b₂ [cosx1(k)]² The task here is to design a control system, whose inputs are xie[-0.2,0.2] rad, x2 = [-1.0, 1.0] rad/s, and whose output is Fe[-10, 10] N such that the final states will be x1=0 and x2=0. Fuzzy logic is required for the control of this inverted pendulum system. In this simple fuzzy logic controller, a set of linguistic variables is chosen to represent 5 degrees of angular position xi [-0.2,-0.1, 0, 0.1, 0.2], 5 degrees of angular velocity x2 [-1.0, -0.5, 0, 0.5, 1.0], and 5 degrees of control force F [-10, -5, 0, 5, 10] as shown in Figure Q3.2. The generic rule set in the form of "Fuzzy Associate Memories" is shown in Figure Q3.3. The initial states of this inverted pendulum system are given to xi(1) 0.15 rad and x2(1) = -0.4 rad/s. 3.1 If the Centre of Area (COA) defuzzification strategy is used with the fire strength ai of the i-th rule calculated from min(ux₁(x1), Hx(x2)) determine the defuzzified control force F(1) and the next state vector [x1(2), x2(2)]. [20 marks] 3.2 If Mean of Maximum (MOM) defuzzification strategy is used with the fire strength ai of the i-th rule calculated from determine the defuzzified control force F(1) and the next state vector [x1(2), x2(2)] Figure Q3.1 An inverted pendulum system [20 marks] Control Force (F) Displacement (x1) Velocity (x2) NM NS ZE PS PM -0.2 -0.15 -0.1 -0.05 ° 0.05 0.1 0.15 0.2 NM NS ZE PS PM 1 0.5 OT .... -0.8 -0.6 -0.4 -0.2 10 0.2 0.4 0.6 0.8 1 NM NS ZE PS PM 1 0.5 -10 -8 -6 -2 6 8 10 Figure Q3.2 Membership functions of an inverted pendulum system Angular Velocity (✗2) Displacement (X1) NM NS ZE PS PM NM PM PM PM PS ZE NS PM PM PS ZE NS ZE PM PS ZE NS NM PS PS ZE NS NM NM PM ZE NS NM NM NM Figure Q3.3 Generic Fuzzy Associative Memories 10