QUESTION THREE

Inverted Pendulum System

Fuzzification Combined fuzzy inference with rule tables Defuzzified control angle Next state vector Calculation/software code

Answer the questions please

3.1 If the Centre of Area (COA) defuzzification strategy is used with the fire strength i of the i-th rule calculated from = min(1 (1 ), 2 (2 ) ) determine the defuzzified control force F(1) and the next state vector [x1(2), x2(2)

3.2 If Mean of Maximum (MOM) defuzzification strategy is used with the fire strength i of the i-th rule calculated from = 1 (1 ) 2 (2 ) determine the defuzzified control force F(1) and the next state vector [x1(2), x2(2)] [20 marks]

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 x = 0 and x = 0 as the angular displacement and angular velocity of the pole. The relevant equation of motion is given by gsino + cose o sing + -ml m. + m ml cos28 me + m ! Assume that trolley mass me = 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 *1 = x2 gsinx: + cosx(-a,x sinx+azF) b. - b2cos2x1 = 0.9091, b = 1 = 0.6667, mem ml where a = = 0.0455, az mem ml b2 = = 0.0455. mem 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 X:(k+1) = x2(k) + Tx,(k) gsinx (k) + cosx1(k)(-a,[x2(k)] sinx,(k) + azF) xz(k+1) = x2(k)+10 b - bz [cosx, (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 X: (-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, 10as 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 xx(1)=-0.15 rad and xz(1) = -0.4 rad's. 3.1 If the Centre of Area (COA) defuzzification strategy is used with the fire strength di of the i-th rule calculated from di = min(4x)(x). Axz/(x2)) determine the defuzzified control force F(1) and the next state vector [x(2), xz(2)]. [20 marks] 3.2 If Mean of Maximum (MOM) defuzzification strategy is used with the fire strength di of the i-th rule calculated from Qi = ux, (x2) - Hx2, (x2) determine the defuzzified control force F(1) and the next state vector [x(2), x2(2)] [20 marks) Figure Q3.1 An inverted pendulum system NM ZE PS PM Deplacement (1) 0.5 2 0.15 0.06 0.15 NM NS ZE PM Velocity (12) 5 -0.8 0.2 04 0.6 NM NS ZE PM Control Force 2 10 Figure Q3.2 Membership functions of an inverted pendulum system Displacement (x) Angular Velocity (x) 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 03.3 Generic Fuzzy Associative Memories 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 x = 0 and x = 0 as the angular displacement and angular velocity of the pole. The relevant equation of motion is given by gsino + cose o sing + -ml m. + m ml cos28 me + m ! Assume that trolley mass me = 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 *1 = x2 gsinx: + cosx(-a,x sinx+azF) b. - b2cos2x1 = 0.9091, b = 1 = 0.6667, mem ml where a = = 0.0455, az mem ml b2 = = 0.0455. mem 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 X:(k+1) = x2(k) + Tx,(k) gsinx (k) + cosx1(k)(-a,[x2(k)] sinx,(k) + azF) xz(k+1) = x2(k)+10 b - bz [cosx, (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 X: (-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, 10as 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 xx(1)=-0.15 rad and xz(1) = -0.4 rad's. 3.1 If the Centre of Area (COA) defuzzification strategy is used with the fire strength di of the i-th rule calculated from di = min(4x)(x). Axz/(x2)) determine the defuzzified control force F(1) and the next state vector [x(2), xz(2)]. [20 marks] 3.2 If Mean of Maximum (MOM) defuzzification strategy is used with the fire strength di of the i-th rule calculated from Qi = ux, (x2) - Hx2, (x2) determine the defuzzified control force F(1) and the next state vector [x(2), x2(2)] [20 marks) Figure Q3.1 An inverted pendulum system NM ZE PS PM Deplacement (1) 0.5 2 0.15 0.06 0.15 NM NS ZE PM Velocity (12) 5 -0.8 0.2 04 0.6 NM NS ZE PM Control Force 2 10 Figure Q3.2 Membership functions of an inverted pendulum system Displacement (x) Angular Velocity (x) 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 03.3 Generic Fuzzy Associative Memories