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

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

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

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

Design system with BodeDesign system with BodeHany FerdinandoHany FerdinandoDept. of Electrical Eng.Dept. of Electrical Eng.

Petra Christian UniversityPetra Christian University

Page 2: Design system with Bode Hany Ferdinando Dept. of Electrical Eng. Petra Christian University.

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General OverviewGeneral Overview

Bode vs Root Locus designBode vs Root Locus design Information from open-loop freq. responseInformation from open-loop freq. response Lead and lag compensatorLead and lag compensator

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Bode vs Root LocusBode vs Root Locus

Root Locus method gives direct Root Locus method gives direct information on the transient response of information on the transient response of the closed-loop systemthe closed-loop system

Bode gives indirect informationBode gives indirect information In control system, the transient response In control system, the transient response

is important. For Bode, it is represented is important. For Bode, it is represented indirectly as phase and gain margin, indirectly as phase and gain margin, resonant peak magnitude, gain crossover, resonant peak magnitude, gain crossover, static error constantstatic error constant

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Open-loop Freq. ResponseOpen-loop Freq. Response

Low-freq. regionLow-freq. region indicates the steady-state indicates the steady-state behavior of the closed-loop systembehavior of the closed-loop system

Medium-freq. regionMedium-freq. region indicates the relative indicates the relative stabilitystability

High-freq. regionHigh-freq. region indicates the complexity indicates the complexity of the systemof the system

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Lead and Lag CompensatorLead and Lag Compensator

Lead Compensator:Lead Compensator: It yields an improvement in transient response It yields an improvement in transient response

and small change in steady-state accuracyand small change in steady-state accuracy It may attenuate high-freq. noise effectIt may attenuate high-freq. noise effect

Lag Compensator:Lag Compensator: It yields an improvement in steady-state It yields an improvement in steady-state

accuracyaccuracy It suppresses the effects of high-freq. noise It suppresses the effects of high-freq. noise

signalsignal

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Lead CompensatorLead Compensator

Ts

Ts

KTs

TsKsG ccc

1

1

1

1)(

(0 < (0 < < 1) < 1)

The minimal value of a is limited by the The minimal value of a is limited by the construction of lead compensator. Usually, it construction of lead compensator. Usually, it is taken to be about 0.05is taken to be about 0.05

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Lead CompensatorLead Compensator

1.1. Define KDefine Kcc = K then then = K then then

2.2. With gain K, draw the Bode diagram and With gain K, draw the Bode diagram and evaluate the phase marginevaluate the phase margin

1

1)(

Ts

TsKsGc

)(1

1)(

1

1)(

1

1)()( 1 sG

Ts

TssKG

Ts

TssG

Ts

TsKsGsGc

Determine gain K to satisfy the Determine gain K to satisfy the requirement on the given static error requirement on the given static error constantconstant

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Lead CompensatorLead Compensator

3.3. Determine the phase-lead angle to be Determine the phase-lead angle to be added to the system (added to the system (mm), add additional ), add additional

5-125-12oo to it to it

4.4. Use to determine the Use to determine the attenuation factor attenuation factor . Find . Find cc in G in G11(s) as (s) as |G|G11(s)| = -(s)| = -20 log(1/20 log(1/√√) and ) and cc is is 1/(√1/(√T)T)

5.5. Find zero (1/T) and pole (1/Find zero (1/T) and pole (1/T)T)

1

1sin m

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Lead CompensatorLead Compensator

6.6. Calculate KCalculate Kcc = K/ = K/

7.7. Check the gain margin to be sure it is Check the gain margin to be sure it is satisfactorysatisfactory

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Lead Compensator - exampleLead Compensator - example

It is desired that the Kv is 20/sIt is desired that the Kv is 20/s Phase margin 50Phase margin 50oo Gain margin at least 10 dBGain margin at least 10 dB

)2(

4)(

sssG

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Lead Compensator - exampleLead Compensator - example

)(1

1)(

1

1)(

1

1)()( 1 sG

Ts

TssKG

Ts

TssG

Ts

TsKsGsGc

202)2(

4

1

1lim

)(1

1lim)(

1

1lim)()(lim

0

01

00

Kss

Ks

Ts

Ts

sKGTs

TsssG

Ts

TsssGssG

s

ssc

s

K = 10K = 10

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Lead Compensator - exampleLead Compensator - example

-50

0

50M

agni

tude

(dB

)

10-1

100

101

102

-180

-135

-90

Pha

se (

deg)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 18 deg (at 6.17 rad/sec)

Frequency (rad/sec)

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Lead Compensator - exampleLead Compensator - example

Gain margin is infinity, the system requires Gain margin is infinity, the system requires gain margin at least 10 dB.gain margin at least 10 dB.

Phase margin is 18Phase margin is 18oo, the system requires , the system requires 5050oo, therefore there is additional 32, therefore there is additional 32oo for for phase marginphase margin

It is necessary to add 32It is necessary to add 32oo with 5-12 with 5-12oo as as explained before…it is chosen 5explained before…it is chosen 5oo 37 37oo

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Lead Compensator - exampleLead Compensator - example

Sin 37Sin 37oo = 0.602, then = 0.602, then = 0.25 = 0.25

|G|G11(s)| = -20 log (s)| = -20 log (1/(1/√√))

5.002.6)2(

40

25.0

1log20

)2(

40

dBjj

jj

cc

cc

cc = 8.83 rad/s = 8.83 rad/s new crossover freq. new crossover freq.

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Lead Compensator - exampleLead Compensator - example

cc = 1/( = 1/(√√T), then 1/T = T), then 1/T =

cc√√zero = 4.415zero = 4.415

Pole = 1/(Pole = 1/(T) = T) = cc//√√ = 17.66 = 17.66

KKcc = K/ = K/ = 10/0.25 = 40 = 10/0.25 = 40

)2(

4

66.17

415.440)()(

)2(

41

1

)()(

sss

ssGsG

ssT

s

Ts

KsGsG

c

cc

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Lead Compensator - exampleLead Compensator - example

-100

-50

0

50M

agni

tude

(dB

)

10-1

100

101

102

103

-180

-135

-90

Pha

se (

deg)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 49.6 deg (at 8.83 rad/sec)

Frequency (rad/sec)

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Lead Compensator - exampleLead Compensator - exampleclear;

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

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Lag CompensatorLag Compensator

Ts

Ts

KTs

TsKsG ccc

1

1

1

1)(

(( > 1) > 1)

With b > 1, its pole is closer to the origin than With b > 1, its pole is closer to the origin than its zero isits zero is

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Lag CompensatorLag Compensator

1.1. Assume Assume

KKcc = K = K

Ts

Ts

KTs

TsKsG ccc

1

1

1

1)(

)(1

1)(

1

1)(

1

1)()( 1 sG

Ts

TssKG

Ts

TssG

Ts

TsKsGsGc

Calculate gain K for required static error Calculate gain K for required static error constant or you can draw it in Bode constant or you can draw it in Bode diagramdiagram

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Lag CompensatorLag Compensator

2.2. if the phase margin of KG(s) does not if the phase margin of KG(s) does not satisfy the specification, calculate satisfy the specification, calculate mm! It is ! It is

mm = 180 – required_phase_margin (don’t = 180 – required_phase_margin (don’t

forget to add 5-12 to the required phase forget to add 5-12 to the required phase margin). Find margin). Find cc for the new for the new mm!!

3.3. Choose Choose = 1/T (zero of the = 1/T (zero of the compensator) 1 octave to 1 decade compensator) 1 octave to 1 decade below below cc..

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Lag CompensatorLag Compensator

4.4. At At cc, determine the attenuation to bring , determine the attenuation to bring

the magnitude curve down to 0 dB. This the magnitude curve down to 0 dB. This attenuation is equal to -20 log (attenuation is equal to -20 log (). From ). From this point, we can calculate the pole, this point, we can calculate the pole, 1/(1/(T)T)

5.5. Calculate the gain KCalculate the gain Kcc as K/ as K/

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Lag Compensator - exampleLag Compensator - example

It is desired that the Kv is 5/sIt is desired that the Kv is 5/s Phase margin 40Phase margin 40oo Gain margin at least 10 dBGain margin at least 10 dB

)15.0)(1(

4)(

ssssG

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Lag Compensator - exampleLag Compensator - example

)15.0)(1(

1

1

1

)15.0)(1(

1

1

1)()(

sssTs

TsK

sssTs

TsKsGsG cc

K = KK = Kcc

)15.0)(1(

1

1

1)()(lim

0

sssTs

TssKsGssGK c

sv

With Kv = 5, K = 5With Kv = 5, K = 5

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Lag Compensation - exampleLag Compensation - example

-150

-100

-50

0

50

100M

agni

tude

(dB

)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Pha

se (

deg)

Bode DiagramGm = -4.44 dB (at 1.41 rad/sec) , Pm = -13 deg (at 1.8 rad/sec)

Frequency (rad/sec)

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Lag Compensation - exampleLag Compensation - example

The required phase margin is 40The required phase margin is 40oo, , therefore, fm = -180therefore, fm = -180oo + 40 + 40oo + 10 + 10oo = -130 = -130oo

Angle of GAngle of G11(s) is -130(s) is -130oo, , cc = 0.49 rad/s = 0.49 rad/s

= 1/T = 0.2= 1/T = 0.2cc = 0.098 rad/s = 0.098 rad/s

At At cc, the attenuation to bring down the , the attenuation to bring down the

magnitude curve to 0 dB is -18.9878 dB magnitude curve to 0 dB is -18.9878 dB (rounded to -19 dB)(rounded to -19 dB)

From -20log(From -20log() = -19, ) = -19, is 8.9125 is 8.9125

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Lag Compensation - exampleLag Compensation - example

Pole of the compensator is 1/(Pole of the compensator is 1/(T), with 1/T T), with 1/T = 0.098, the pole is 0.011= 0.098, the pole is 0.011

Kc = K/b, and Kc = 0.561Kc = K/b, and Kc = 0.561

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Lag Compensation - exampleLag Compensation - exampleK = 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

GainGain

Plot marginPlot margin

Get gain and phase marginGet gain and phase margin

Find new phase marginFind new phase margin