Question

Design a PI controller to drive the step-response error to zero for the negative unity feedback system shown in Fig. 1, where G(s) S+1(s+3)(s+10) The system operates with a damping factor of 0.4. * Design a PI controller whose compensator zero located at -0.1 Use MATLAB or any other computer program to simulate the step response * to closed-loop system

Answer 1

Given that

we have to Design a PI controller whose compensator zero located at -0.1

The gain function for the unity feedback system is, K G1+3)G-10) (s)= The damping ratio of the system is, 0.4. Let us first analyze the performance of the uncompensated system t drawing root locus. The MATLAB code to draw the root locus is, G-zpk(, 1 -3-101,1) rlocus (G) sgrid(0.4,0); axis (I-14 0-s 5)) The root locus of the uncompensated system is, Root Locus -3 14 12 10 -8 -6 -4 Real Axis (seconds' For given damping ratio the dominant pole pair location is,

For given damping ratio the dominant pole pair location is, Root Locus System: G Gain: 0 Pole:-10 Damping:1 Overshoot (%): 0 Frequency (rad/s): 10 System: G Gain: 106 Pole:-1.36 3.191 Damping: 0.392 Overshoot (%): 26 Frequency (rad/s): 3.47 .2 .3 Real Axis (seconds) From the root locus, the pole pair for the given damping ratio of 0.4 is, 4,2 -1.36 j3.19 . The corresponding gain of the uncompensated system is, K = 106 and higher order pole for open loop systemis at, s=-10. So, the second order approximation is valid since higher order pole is much left to the s-plane. For zero steady state error, put PI controller, having one pole at origin and a zero nearby origin into left half s-plane so that the introduced pole does not affect the uncompensated system root locus very much Let the transfer function of the compensator be G, (s)- The MATLAB code to draw the root locus of the compensated system is, G =zpk([], [-1-3-10], 1); Gc-zpk (-0.1, [0,1) rlocus (Gc G) sgrid (0.4,0) axis ([-14 0-5 5]) The root locus of the compensated system using MATLAB is,

Root Locus .3 12 10 .8 Real Axis (seconds For given damping ratio the dominant pole pair location is, Root Locus Svstem: untitledi Galn: 2.15 Pole: -10 Damping: 1 Overshoot ():0 Frequency (rad/s): 10 Svstem: untitled1 Gain: 100 Pole: -1.35 3.07 Damping: 0.402 Overshoot (%): 25.2 Frequency (rad/s): 3.36 3 10 .S Real Axis (seconds) 12

The new dominant pole pair is at same location 1.353.07 and the gain also remains almost the same at K=100 and higher order pole for the open loop system is also at s=-10. So, the second order approximation is valid. The MATLAB code to plot the step responses of the compensated and uncompensated systems is, syms t t=0:0.0001 :60; G-zpk (,-1-3-10],106); Gc-zpk (-0.1], [0 -13 -10,100) step (feedback (G,1),t) hold orn step (feedback (Gc, 1), t) grid xlabel t ylabel output The step responses of the compensated and uncompensated systems are Step Response 0.9 0.8 0.7 0.6 0.4 0.3 0.2 0.1 10 20 30 t (seconds) 50 60 The Steady state error for step imput of the system is 106 106 K, lim G (s)-lim 3.534 s+1 s+3)s+1030

The roots are 1.36tj3.07 ,--1 1.88 -0.78 The comparison of uncompensated and compensated closed loop systems is given in the following table: Plant and compensator (s+1)(s+3)(s+10) (s+1)(s+3)(s+10) 103 100 e() 0.22 3.534 Dominant second order pole --1.36 j3.19 -1.36t j3.07 Third pole 1125 -11.88 Fourth None None -0.78 -0.1