AV243 lab 1: Compensator design for a typical electromechanical engine gimbal control (EGC) system and Linear system performance assessment using MATLAB/SIMULINK Karun Mathews Manoj (SC19B120) March 12, 2021 Objective: of (proportional + rate feedback), (Proportional +pensators rateand feedback + Integral) andDesign (Proportional + Integral + Derivative) type ofand comfor typical engine gimbal systemmodel of launch vehicles linear system performance assessment using control SIMULINK SIMULINK LTIa Viewer.electromechanical Theory: The block diagram representation of a typical electromechanical engine gimbal control system using proportional plus rate feedback controller is shown in Fig.1. Figure 1: Block diagram representation proportional + rate feedback controller of a typical electromechanical EGC system with The parameter values are given below: 1 J = Jm B=B + K .K /R m t b a +( Nm Nl +( Nm )2B Nl l (1) )2 B K = K /R l t a (2) (motor torque constant) Kt = 0.181Nm/A (3) (motor back emf constant) Kb = 0.181 V/(rad/sec) (4) (MI of motor rotating assembly) Jm = 1.1694 10−4 kg.m2 ∗ (5) (MI of engine gimbal) Jl = 12.753 kg.m2 Nm θl 1 (6) (Ball screw gear ratio)Nl = θm= 398 (Viscous damping coefficient of motor shaft) Bm = 2.943 10−4Nm/(rad/sec) (7) ∗ (8) (Viscous damping coefficient of the engine gimbal) Bl = 58.86Nm/(rad/sec) (9) (Armature resistance) Ra = 8.6 ohms (10) (Position sensor scale factor) Kp = 0.36V/rad (11) (Tacho generator scale factor) Ktg = 0.1 V/(rad/sec) (12) (13) Controller gain design equations In order to get a closed loop system with transfer equation: w2n Gd(s) = ; (14) n s2 + 2ζ.wn.s + w2 2 the design equations for K1 and K2 are given by: wn2.J K1 = = 2ζ.wn.J−K.K B K2 K tgp.K (15) (16) (17) Specifications for closed system (-3 dB bandwidth) wb = 5 ∗ 2 ∗ πrad/sec; (Damping factor) ζ = 0.6; (18) Use the following formula for finding wn, thenaturalfrequencyoftheclosedloopsystem : √ . wb = wn (1 − 2ζ 2 ) + 4ζ 4 − 4ζ 2 + 2 Procedure for controller design and simulation: (19) 1. Find outwhich the (k1) values natural frequency (wn), sencontroller andof rateundamped feedback controller gain (k2) using a proportional MATprogram usespecifications. the parameter values of the dynamics, sor LAB scale factors andgain closed Write down theplant values below: 2. 3. 4. wn = 27.36 rad/sec; (20) k1 = 19.5077; (21) k2 = 0.9538; (22) Draw the Simulink fortime the LTI closed system and find out M-peak, the following performances SIMULINK -3fordBstep bandwidth, peak overshoot, riseusing time,model settling (the Viewer: lastloop 3 are response) Add an integral controller Ki/s, and in parallel withmeasurement the pro- portional controller gain ‘K1’, with(2). awith gainG(s) Ki = =K1/10 repeat the of the parameters given in step Add a derivative controller with G(s) = Kd.s/(1+s/300), where Kd = K2*KTG/Kp in parallel with the PI controller, and remove the rate feed- back controller. Repeat the exercise in step (2). 5. Make a performance comparison table as given below: 3 Performance parameter P + Rate comparison feedback table P + I + rate feedback Table 1: Performance -3 dB Bandwidth (Hz) 5.9046 (@ -3.03 dB) 5.9046 (@ -3.02 dB) M peak (dB) 2.38 2.43 Peak Overshoot (%) 23.5% 24% Rise time (ms) (%) 54.6 54.5 6. Settling Give your performance comparison. timecomments (ms) (%) on the 307 304 The PID controller outperforms the other two controllers in the two as-the pects of peak overshoot and settling significantly other two. This implies that thetime PID- having controlled system lower wouldvalues havethan much better stability. However, the PID controller provides it will provide a weaker output signal.much lesser gain than the other two, hence The P + Rate feedback and P+I+Rate feedback controllers have very similar performances. the system haswould aθslightly smaller as compared to the other sive two. In my PID opinion, make the bandwidth PID control system less responto varying inputsthis (like c = A.sin(wt)). 4 P+I+D 5.5704 (@ -3.02 dB) 0.592 10.7% 58.3 202