Question

Problem 2: (20 points) A DC Motor with negligible armature inductance is used in a position control applications. The open-loop transfer function is 50 G(s) = s(s/5+1) Using a Bode Plot of the system and Frequency Domain Methods, design an analog compensator to meet the following specifications: The bandwidth of the compensated system is no less than that of the uncompensated system. • The step response overshoot is less than 20%. • The steady-state error to a unit-ramp input is less than 1/200.

Answer 1

(a) Write the open loop transfer function. 50 G(s)=- +1 250 s(s+5) Write the MATLAB code to calculate the phase margin of the uncompensated system. num=[250] den=[1 5 0); sys=tf(num, den);

Bode Diagram Gm = Inf dB (at Inf rad/s), Pm= 18 deg (at 15.4 rad/s) 60 40F 201 Magnitude (dB) 0 System: sys Frequency (rad/s): 63.6 Magnitude (dB): -24.2 -20 Phase (deg) -135 -180 10-1 10° 10 10 Frequency (rad/s) Figure 1

Consider a compensator, D(s) of gain K. The steady state error of the compensated system is, e = lims 1+KD()G(s) |R(s) how [+ 5-0 Here, R(s) is ramp input. 1 R(S) e = lims 5-0 250 1+KD(S) s(s+5) = lim 250 s+sKD(s) s(s+5) 250 KD(0) (0+5) 50KD(0) From the specification the steady state error to a unit ramp input is less than 1 200 < 1 50KD(0) 200 50K > 200 K >4 Therefore, choose the value of K as 10.

Write the relation between overshoot and damping. overshoot = e From the specification, the overshoot should be less than 20% <0.2 > 0.5 VI-52 5>0.5(1-5) 5 > 0.5

From the approximation, PM 100 PM 0.5 100 PM = 50° Therefore, the phase margin should be greater than or equal to 50°

The phase margin of the uncompensated system is 18° hence, an additional phase is needed to make the system stable. Therefore, lead compensator is used. The transfer function of lead compensatoris, Ts +1 D(S)=K ats +1' To maintain the same low frequency gain and adding a compensator zero would increase the cross over frequency Therefore, to be safe, design a compensator such that it adds an extra phase margin of 55° a<1 From Figure 6.54 in the textbook for a phase margin of 55°, the value of a should be 1 10 From the bode plot shown in Figure 1, the maximum cross over frequency is, 63 rad sec The zero at is, т 1 ſa T 63 0.1 = 20 $+1 D(s)=10-20 =1020 -, a<1 5+1 200 KK S + 20 S+ 200 = 10

Write the MATLAB code for compensated system. num1=[250] den=conv([1].[15]); sys1=tf(num1, dent); num2=10*[1 201 den2= [1 200] sys2=tf(num2 den2); sys=sys1*sys2; margin(sys) sysf=feedback(sys,1) margin(sys) step(sysf)

The matlab plot for step response is, Step Response 1.5 1 System: syst Peak amplitude: 1.22 Overshoot (%): 21.9 At time (seconds): 0.171 1 Amplitude 0.5 0 0.6 0.2 0.4 Time (seconds) Figure 2 Therefore the step response of the overshoot is almost less than 20%.