Question

1 CONTROL SYSTEM ANALYSIS & DESIGN Spring 2019 HW 7 Due 4/4/2019, Thursday, 11:59pm 1. Design a lead compensator for the closed-loop (CL) system whose open loop transfer function is given below. Design objectives: reduce the time constant by 50% while maintaining the same value of the damping ratio for the dominant poles. Please note that H(s)-1. Please use the method based on root locus plot. G(s) 2 [s(s+2)] Please include detailed step Obtain the location of the desired dominant poles. 2) Draw the root loci of the CL system without the lead compensator 3) Calculate the angle deficiency 4 Determine the location of the pole associated with the lead compensator if the zero is chosen to be at (-2.5, 0) 5) Obtain the gain value of the compensator (Hint: use the Magnitude Condition) 6) Write the complete expression of the lead compensator The following requires the use of Matlab (please write a Matlab script file): 7) Upon finishing the design, please use Matlab to plot the root loci of CL systems with and without the lead compensator. Mark the locations of the desired dominant poles. Please make sure that the root loci of CL systems with the lead compensator pass through the desired dominant poles. 8) Simulate and compare the unit step responses of the uncompensated and compensated system. Please observe from the plots and comment if the design objectives have been achieved. The plot and the Matlab code with proper documentation should be included in the homework.

Answer 1

(1) Consider the open-loop transfer function 2 G(s) s(s+2) (1) Determine the closed-loop transfer function. 2 G(s) C(s) s(s2 R(s) 1+G(s 2 + s(s+2) 2 s2s2 The dominant poles are at s 1+ j. (2) The system has no zeros and has two complex conjugate poles at The root locus originates at the pole and terminates at the zero. As the system has no zeros, the poles terminate at zeros at infinity. Determine the angle of asymptotes 1800(2k1) ;k =0,1...P - Z-1 Z=± P-Z P 2;Z 0 Z=±90°,2700 Determine the centroid of the asymptotes. Σpoles-Σ zeros P-Z -1+j-- =-1 2 0 Consider the characteristic equation of the system 1+KG(s)= 0 2 = 0 1 K s(s+2) s2 2s 2K=0

Construct the Routh table 2K 2 0 0 s 2K For the system to be stable, all elements in the first column of the Routh array should be greater than zero 2K 0 K0 The root locus intersects the imaginary axis when K is equal to 0. Consider the characteristic equation with this value of K s22s 0 s -20 Draw the root locus ja 1 900 1.5 -0.5 0 -1 2700 -1

(3) Consider the characteristic equation of the closed loop system s2s 2 0 Compare the characteristic equation with the desired characteristic equation s2,s =0 1 2C 2 2 1.4 rad/s 1 1 14 5 =0.714 Consider the closed-loop transfer function C(s) R(s) Determine the time constant of the uncompensated system. T20.5T 2 2 1 2 s2s2 2(0.5s2s10.5s2 +s+1 0.707 S 1 25r. 1r 0.7 S S 24 Time constant should be reduced by 50%. Determine the damping ratio for the new time constant 25(0.5x0.7) 1 51.428

The real part of the desired pole location is, -Sa,1.428 x1.4 =-2 Determine the angle of the uncompensated pole. 1 +90° 1350 tan Determine the imaginary part of the desired pole location. -2 tan (135) = 2 jo Desired Compensated dominant pole 2+2 135 xOpen-loop poles x Closed-loop pole Assume a compensator zero at s = -2.5 jo Desired Compensated dominant pole -2 j2 1350 1 -2.5-2 x Open-loop poles x Closed-loop pole

Determine the sum of the angles of the poles and zeros. 2 -90° -135° =-149° 0.5 tan The angle contribution or angle deficiency required from the compensator pole is, e1800 149° O 31° (4) Determine the location of the compensator pole, pc. 2 tan(310) р. — 2 2 0.6 р. — 2 Pe 2 3.33 Pe=5.333 (5) s2.5 К S5.33s(s+2) 2 G. (s)G(s) A Consider the magnitude condition s 2.5 K S5.33s(s+2)|_,+i2. 2 = 1 -2+ j2+ 2.5 K 2 1 -2 j2 5.33(-2 j2)(-2+ j2 2) 2.06 K 3.88 (2.828)(2 2 = 1 K 5.3 (6) Thus, the lead compensator is S2.5 G (s) 5.3 s5.33

(7) Draw the root locus of the uncompensated system using MATLAB num=2; den=[1 2 2]; sys=tf (num, den); rlocus (sys) Root Locus -0.5 -1.5 Real Axis (seconds Draw the root locus of the compensated system using MATLAB. num-2*5.3* [1 2.5]; den-conv (1 5.33], [1 2 2]); sys-tf (num, den); rlocus (sys) Root Locus 15 10 -10 -4 -3 -2 Real Axis (seconds) Imaginary Axis (seconds') Imaginary Axis (seconds)

(8) Draw the step responses of the compensated and uncompensated systems num=2; den [1 2 2]; sys=tf(num, den); step (sys) figure num-2*5.3* [1 2.5]; den=conv ([1 5.33],[1 2 2]); sys-tf(num, den); sysf-feedback (sys,1) step (sysf) Step Response Step Response 1.4 0.9 0.8 1.2 0.7 0.6 0.8 ,5 0.6 .4 0.3 0.4 0.2 0.2 0.1 2 5 7 0 0.5 1.5 Time (seconds) 2.5 3 3.5 Time (seconds) Observe that the settling time is reduced to half. Amplitude Amplitude