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

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

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

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

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

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

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

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)

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

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

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

Design System with Bode - Hany Ferdinando12 Lead Compensator - example

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

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.

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

Design System with Bode - Hany Ferdinando16 Lead Compensator - example

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

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

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

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.

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/ 

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

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

Design System with Bode - Hany Ferdinando24 Lag Compensation - example

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 = rad/s  At  c, the attenuation to bring down the magnitude curve to 0 dB is dB (rounded to -19 dB)  From -20log(  ) = -19,  is

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

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 = ( )*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