1. Design system with Bode Hany Ferdinando Dept. of Electrical Eng. Petra Christian University

2. Design System with Bode - Hany Ferdinando2 General Overview  Bode vs Root Locus design  Information from open-loop freq. response  Lead and lag compensator

3. Design System with Bode - Hany Ferdinando3 Bode vs Root Locus  Root Locus method gives direct information on the transient response of the closed-loop system  Bode gives indirect information  In control system, the transient response is important. For Bode, it is represented indirectly as phase and gain margin, resonant peak magnitude, gain crossover, static error constant

4. Design System with Bode - Hany Ferdinando4 Open-loop Freq. Response  Low-freq. region indicates the steady-state behavior of the closed-loop system  Medium-freq. region indicates the relative stability  High-freq. region indicates the complexity of the system

5. Design System with Bode - Hany Ferdinando5 Lead and Lag Compensator  Lead Compensator:  It yields an improvement in transient response and small change in steady-state accuracy  It may attenuate high-freq. noise effect  Lag Compensator:  It yields an improvement in steady-state accuracy  It suppresses the effects of high-freq. noise signal

6. Design System with Bode - Hany Ferdinando6 Lead Compensator (0 <  < 1) The minimal value of a is limited by the construction of lead compensator. Usually, it is taken to be about 0.05

7. Design System with Bode - Hany Ferdinando7 Lead Compensator 1. Define K c  = K then then 2. With gain K, draw the Bode diagram and evaluate the phase margin Determine gain K to satisfy the requirement on the given static error constant

8. Design System with Bode - Hany Ferdinando8 Lead Compensator 3. Determine the phase-lead angle to be added to the system (  m ), add additional 5-12 o to it 4. Use to determine the attenuation factor . Find  c in G 1 (s) as |G 1 (s)| = - 20 log(1/ √  ) and  c is 1/(√  T) 5. Find zero (1/T) and pole (1/  T)

9. Design System with Bode - Hany Ferdinando9 Lead Compensator 6. Calculate K c = K/  7. Check the gain margin to be sure it is satisfactory

10. Design System with Bode - Hany Ferdinando10 Lead Compensator - example  It is desired that the Kv is 20/s  Phase margin 50 o  Gain margin at least 10 dB

11. Design System with Bode - Hany Ferdinando11 Lead Compensator - example K = 10

12. Design System with Bode - Hany Ferdinando12 Lead Compensator - example

13. Design System with Bode - Hany Ferdinando13 Lead Compensator - example  Gain margin is infinity, the system requires gain margin at least 10 dB.  Phase margin is 18 o, the system requires 50 o, therefore there is additional 32 o for phase margin  It is necessary to add 32 o with 5-12 o as explained before…it is chosen 5 o  37 o

14. Design System with Bode - Hany Ferdinando14 Lead Compensator - example  Sin 37 o = 0.602, then  = 0.25  |G 1 (s)| = -20 log (1/ √  )  c = 8.83 rad/s  new crossover freq.

15. Design System with Bode - Hany Ferdinando15 Lead Compensator - example   c = 1/( √  T), then 1/T =  c √   zero = 4.415  Pole = 1/(  T) =  c / √  = 17.66  K c = K/  = 10/0.25 = 40

16. Design System with Bode - Hany Ferdinando16 Lead Compensator - example

17. Design System with Bode - Hany Ferdinando17 Lead Compensator - example clear; K = 10; num = 4; den = [1 2 0]; margin(K*num,den) [Gm,Pm] = margin(K*num,den) new_Pm = 50 - Pm + 5; new_Pm_rad = new_Pm*pi/180; alpha = (1 - sin(new_Pm_rad))/(1 + sin(new_Pm_rad)) Wc = 8.83; zero = Wc*sqrt(alpha); pole = Wc/sqrt(alpha); Kc = K/alpha; figure; margin(conv(4*Kc,[1 zero]),conv([1 2 0],[1 pole])) Clear variable Gain from Kv Plot margin Get Gm & Pm New Pm in deg New Pm in rad alpha From calculation Get zero & pole Compensator gain Plot final result Original system

18. Design System with Bode - Hany Ferdinando18 Lag Compensator (  > 1) With b > 1, its pole is closer to the origin than its zero is

19. Design System with Bode - Hany Ferdinando19 Lag Compensator 1. Assume K c  = K K c  = K Calculate gain K for required static error constant or you can draw it in Bode diagram

20. Design System with Bode - Hany Ferdinando20 Lag Compensator 2. if the phase margin of KG(s) does not satisfy the specification, calculate  m ! It is  m = 180 – required_phase_margin (don’t forget to add 5-12 to the required phase margin). Find  c for the new  m ! 3. Choose  = 1/T (zero of the compensator) 1 octave to 1 decade below  c.

21. Design System with Bode - Hany Ferdinando21 Lag Compensator 4. At  c, determine the attenuation to bring the magnitude curve down to 0 dB. This attenuation is equal to -20 log (  ). From this point, we can calculate the pole, 1/(  T) 5. Calculate the gain K c as K/ 

22. Design System with Bode - Hany Ferdinando22 Lag Compensator - example  It is desired that the Kv is 5/s  Phase margin 40 o  Gain margin at least 10 dB

23. Design System with Bode - Hany Ferdinando23 Lag Compensator - example K = K c  With Kv = 5, K = 5

24. Design System with Bode - Hany Ferdinando24 Lag Compensation - example

25. Design System with Bode - Hany Ferdinando25 Lag Compensation - example  The required phase margin is 40 o, therefore, fm = -180 o + 40 o + 10 o = -130 o  Angle of G 1 (s) is -130 o,  c = 0.49 rad/s   = 1/T = 0.2  c = 0.098 rad/s  At  c, the attenuation to bring down the magnitude curve to 0 dB is -18.9878 dB (rounded to -19 dB)  From -20log(  ) = -19,  is 8.9125

26. Design System with Bode - Hany Ferdinando26 Lag Compensation - example  Pole of the compensator is 1/(  T), with 1/T = 0.098, the pole is 0.011  Kc = K/b, and Kc = 0.561

27. Design System with Bode - Hany Ferdinando27 Lag Compensation - example K = 5; num = 1; den = conv([1 1 0],[0.5 1]); margin(K*num,den) [Gm,Pm] = margin(K*num,den); new_Pm = (-180 + 40 + 10)*pi/180; %rad wc = 0.49; att = -(20*log10(5/abs(i*wc*(i*wc+1)*(i*0.5*wc+1)))) beta = 10^(att/-20) zero = 0.2*wc pole = zero/beta Kc = K/beta Gain Plot margin Get gain and phase margin Find new phase margin