Question

Design a PI controller to drive the step response error to zero for the unity feedback system shown in Figure P9.1, where G(s) s1) (s +3) (s 10) The system operates with a damping ratio of 0.5. Compare the specifications of the uncompensated and compensated systems. [Section: 9.2] C(s) FIGURE P9.1

Answer 1

Hi there,
Please find the answer attached as under. Please give a Thumbs UP if you find the answer useful. Have a rocking day Ahead:)

Refer to unity feedback system in Figure P9.1 in the textbook. The open loop transfer function is, G(s) = (s+1)(8+3)(s+10) The damping ratio of the system is, < =0.5 First analyze the performance of the uncompensated system by drawing root locus. Write the MATLAB code to draw the root locus. G=zpk([],[-1 -3 -10], 1); rlocus (G); sgrid(0.5, 0); axis ([-14 0 -5 5])

The root locus of the uncompensated system is shown in Figure 1. Root Locus 0.5 Imaginary Axis (seconds) 0.5 -12 4 2 o -10 1 -8 6 Real Axis (seconds 1) Figure 1

The dominant pole pair location at damping ratio of 0.5 is shown in Figure 2. Root Locus System: G Gain: 0 - Pole:-10 6 Damping: 1 Overshoot (%): 0 Frequency (rad/s): 10 Imaginary Axis (seconds) System: G Gain: 72.1 Pole: -1.52 - 2.64i Damping: 0.5 Overshoot (%): 16.3 Frequency (rad/s): 3.04 "P 214 -12 -10 -4 2 -8 -6 Real Axis (seconds-1) Figure 2

From Figure 2 of the root locus, the dominant pole pair for the damping ratio of 0.5 is, S2 =-1.52 j2.64 The corresponding gain of the uncompensated system is, K = 72.1 The higher order pole for open loop system is 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 Pl 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 is, G-K(s+0.1)

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(0.5, 0); axis ([-14 0-5 5]). The root locus of the compensated system using MATLAB is, Root Locus System: untitled1 Gain: 0 - Pole:-10 6 Damping: 1 Overshoot (%): 0 Frequency (rad/s): 10 Imaginary Axis (seconds) System: untitled1 Gain: 71.5 Pole: -1.5 - 2.591 Damping: 0.5 Overshoot (%): 16.3 Frequency (rad/s): 2.99 1 -12 -10 4 2 -8 6 Real Axis (seconds 4) Figure 3

The new dominant pole pair is at same location $1,2 = -1.5512.59 and the gain also remains almost the same K = 71.5. From Figure 3, the value of frequency is, o = 3 rads/s 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, symst t=0:0.0001:60; G=zpk([],[-1 -3 -10],72.4); Gc=zpk ([-0.1], [0 -1 -3 -10], 71.7); step (feedback (G,1),t) hold on step (feedback (Gc,1),t) grid xlabel t ylabel output

The step responses of the compensated and uncompensated systems are Step Response 0.9 --- - output - Uncompensated system Compensated system - 10 20 40 50 60 30 t(seconds) Figure 4

The static error constant of the uncompensated system is, K = lim G(s) ->0 =lim 72.1 5-0 (s+1)(3+3)(s +10) _72.1 30 = 2.4 The Steady state error for step input of the uncompensated system is, és 1+K, 1+2.4 = 0.294

The static error constant of the compensated system is, K, = lim G(s) 71.5(s +0.1) **s(s+1)(s+3)(s +10) 71.5(0+0.1) (0) (0+1)(0+3)(0+10) The Steady state error for step input of the compensated system is, 1+K, II 1+00 -1

Determine the percentage overshoot for both compensated and uncompensated system. %OS = e V1-7 (100%) (0.5) = V140.5) (100%) = e.$14 (100%) = (0.163)(100%) = 16.3 Determine settling time for both compensated and uncompensated system. T. =- 50 (0.5)(3) = 2.66 s

==0 The poles of the closed loop system of the uncompensated system are the roots of the characteristic equation is, 1+G(s)=0 72.1 (s+1)(8+3)(8 +10) (s+1)(s +3)(s +10) + 72.1 = 0 s +14s +43s +102.1 = 0) The roots are: S =-1.54 ;2.64 s, = -10.9

The poles of the closed loop system of the compensated system are the roots of the characteristic equation. 1+G(S)=0 71.5(s +0.1) = 0 s(s+1)(8+3)(s +10) s(s+1)(8+3)(s +10)+71.5(s +0.1) = 0) $* +145 +43s +114.7s +7.15 = 0 The roots are: S =-1.5 12.6 s =-10.9 s=-0.726

72.1 The comparison of uncompensated and compensated closed loop systems is shown in the table 1. Table 1: Parameter Uncompensated Compensated Plant K K(s +0.1) compensator (s+1)(s +3)(s+4) s(s+1)(s +3)(s +4) 71.5 0.294 %OS 16.3 16.3 2.66 s 2.66 s 2.4 Dominant pole -1.52 + |2.64 -1.5+ j2.59 Third pole -10.9 -10.9 Fourth pole None None zero None None Therefore, the specifications between compensated and uncompensated systems are compared. Kp

******************************************************************************************************************************
PS: Please do not forget the thumbs up!