Design system with Bode Hany Ferdinando Dept of

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

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

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 Ferdinando 3

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 Ferdinando 4

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 Ferdinando 5

Lead Compensator (0 < a < 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 Ferdinando 6

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

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

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

Lead Compensator - example § It is desired that the Kv is 20/s § Phase margin 50 o § Gain margin at least 10 d. B Design System with Bode - Hany Ferdinando 10

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

Lead Compensator - example Design System with Bode - Hany Ferdinando 12

Lead Compensator - example § Gain margin is infinity, the system requires gain margin at least 10 d. B. § 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 Ferdinando 13

Lead Compensator - example § Sin 37 = 0. 602, then a = 0. 25 § |G 1(s)| = -20 log (1/√a) o wc = 8. 83 rad/s new crossover freq. Design System with Bode - Hany Ferdinando 14

Lead Compensator - example § wc = 1/(√a. T), then 1/T = wc√a = 4. 415 zero = 4. 415 § Pole = 1/(a. T) = wc/√a = 17. 66 § Kc = K/a = 10/0. 25 = 40 Design System with Bode - Hany Ferdinando 15

Lead Compensator - example Design System with Bode - Hany Ferdinando 16

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

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

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

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

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

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

Lag Compensator - example K = Kcb With Kv = 5, K = 5 Design System with Bode - Hany Ferdinando 23

Lag Compensation - example Design System with Bode - Hany Ferdinando 24

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, wc = 0. 49 rad/s § w = 1/T = 0. 2 wc = 0. 098 rad/s § At wc, the attenuation to bring down the magnitude curve to 0 d. B is -18. 9878 d. B (rounded to -19 d. B) § From -20 log(b) = -19, b is 8. 9125 Design System with Bode - Hany Ferdinando 25

Lag Compensation - example § Pole of the compensator is 1/(b. T), with 1/T = 0. 098, the pole is 0. 011 § Kc = K/b, and Kc = 0. 561 Design System with Bode - Hany Ferdinando 26

Lag Compensation - example Gain K = 5; num = 1; den = conv([1 1 0], [0. 5 1]); Plot margin Get gain and phase margin Find new phase margin(K*num, den) [Gm, Pm] = margin(K*num, den); new_Pm = (-180 + 40 + 10)*pi/180; %rad wc = 0. 49; att = -(20*log 10(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 Design System with Bode - Hany Ferdinando 27