Understanding and Implementation the Bode Plot

Bode Plot

1. Introduction
•Frequencyresponseisthesteady-stateresponseofa
systemtoasinusoidalinput.
•Infrequency-responsemethods,thefrequencyof
theinputsignalisvariedoveracertainrangeandthe
resultingresponseisstudied.
System

2. The Concept of Frequency Response
•Inthesteadystate,sinusoidalinputstoalinear
systemgeneratesinusoidalresponsesofthe
samefrequency.
•Eventhoughtheseresponsesareofthesame
frequencyastheinput,theydifferinamplitude
andphaseanglefromtheinput.
•Thesedifferencesarefunctionsoffrequency.

3. The Concept of Frequency Response
•Sinusoidscanberepresentedascomplexnumbers
calledphasors.
•Themagnitudeofthecomplexnumberisthe
amplitudeofthesinusoid,andtheangleofthe
complexnumberisthephaseangleofthesinusoid.
•Thus canberepresentedas
wherethefrequency,ω,isimplicit.)cos(tM
1 11M

4. The Concept of Frequency Response
•Asystemcausesboththeamplitudeandphaseangle
oftheinputtobechanged.
•Therefore,thesystemitselfcanberepresentedbya
complexnumber.
•Thus,theproductoftheinputphasorandthesystem
functionyieldsthephasorrepresentationofthe
output.

5. The Concept of Frequency Response
•Considerthemechanicalsystem.
•Iftheinputforce,f(t),issinusoidal,thesteady-stateoutput
response,x(t),ofthesystemisalsosinusoidalandatthesame
frequencyastheinput.

6. The Concept of Frequency Response
•Assume that the system is represented by the complex number
•Theoutputisfoundbymultiplyingthecomplexnumber
representationoftheinputbythecomplexnumber
representationofthesystem.)()(M )()(M

7. The Concept of Frequency Response
•Thus, the steady-state output sinusoid is
•M
o(ω)isthemagnituderesponseandΦ(ω)isthephaseresponse.
Thecombinationofthemagnitudeandphasefrequency
responsesiscalledthefrequencyresponse.)()(M )]()([)()()()( 
iioo MMM 

8. Frequency Domain Plots
•Bode Plot
•Nyquist Plot
•Nichol’s Chart

9. Bode Plot
•ABodediagramconsistsoftwographs:
–Oneisaplotofthelogarithmofthemagnitudeof
asinusoidaltransferfunction.
–Theotherisaplotofthephaseangle.
–Bothareplottedagainstthefrequencyona
logarithmicscale.

11. Basic Factors of a Transfer Function
•Thebasicfactorsthatveryfrequentlyoccurin
anarbitrarytransferfunctionare
1.GainK
2.IntegralandDerivativeFactors(jω)
±1
3.FirstOrderFactors(jωT+1)
±1
4.QuadraticFactors))((
)(
)(
251
1320
2



ssss
s
sG

12. Basic Factors of a Transfer Function
1.GainK
•Thelog-magnitudecurveforaconstantgainKisahorizontal
straightlineatthemagnitudeof20log(K)decibels.
•ThephaseangleofthegainKiszero.
•TheeffectofvaryingthegainKinthetransferfunctionisthat
itraisesorlowersthelog-magnitudecurveofthetransfer
functionbythecorrespondingconstantamount,butithasno
effectonthephasecurve.

13. -15
-5
5
15
Magnitude (decibels)
Frequency (rad/sec)
0.1 1 10 1005KIf db)((K) 1452020 loglog Then 10
3
10
4
10
5 10
6
10
7
10
8 10
9

14. -90
o
-30
0
30
o
90
o
Phase (degrees)
Frequency (rad/sec)
0.1 1 10 1005KIf 0 )
5
0
(tan)
Re
Im
(tan Then
1-1-
 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
o

15. Basic Factors of a Transfer Function
2.IntegralandDerivativeFactors(jω)
±1jswheressG  ,)( )log()( 20jG 
90
0
1


)(tan)(

jG
Derivative Factor
Magnitude
Phase
ω
0.1 0.2 0.4 0.5 0.7 0.8 0.9 1
db
-20 -14 -8 -6 -3 -2 -1 0
Slope=20db/decade
Slope=6b/octave

16. -30
-10
10
30
Magnitude (decibels)
Frequency (rad/sec)
0.1 1 10 100decade
db20 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
-20

17. -180
o
-60
0
60
o
180
o
Phase (degrees)
Frequency (rad/sec)
0.1 1 10 100
90 )
0
(tan
1-
 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
o
90
0

18. Basic Factors of a Transfer Function
2.IntegralandDerivativeFactors(jω)
±1
•Whenexpressedindecibels,thereciprocalofanumber
differsfromitsvalueonlyinsign;thatis,forthenumberN,)log()log(
N
N
1
2020  )log()( 

 20
1

j
jG
Magnitude
•Therefore,forIntegralFactortheslopeofthemagnitudelinewould
besamebutwithoppositesign(i.e-6db/octaveor-20db/decade).
90
0
1


)(tan)(

jG
Phase

19. -30
-10
10
30
Magnitude (decibels)
Frequency (rad/sec)
0.1 1 10 100decade
db20 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
20

20. -180
o
-60
0
60
o
180
o
Phase (degrees)
Frequency (rad/sec)
0.1 1 10 100
90 )
0
(tan
1-
 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
o
-90
0

21. Basic Factors of a Transfer Function
3.FirstOrderFactors(jωT+1)
–ForLowfrequenciesω<<1/T
–Forhighfrequenciesω>>1/T)log()( TjM   120 )log()(
22
120 TM   0120  )log()(M )log()( TM 20 )()()( 1
3
1
3  sssG T T
1

22. Basic Factors of a Transfer Function
3.FirstOrderFactors(jωT+1))(tan
-1
T)( 
000  )(tan when
-1
)(, 
451
1
 )(tan when
1-
)(,
T 
90 )(tan when
-1
)(,

23. -30
-10
10
30
Magnitude (decibels)
Frequency (rad/sec)
0.1 1 10 100 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
20)()()( 1
3
1
3  sssG
ω=3
20 db/decade

24. -90
o
-30
0
30
o
90
o
Phase (degrees)
Frequency (rad/sec)
0.1 1 10 100 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
o
45
o

25. Basic Factors of a Transfer Function
3.FirstOrderFactors(jωT+1)
-1
–ForLowfrequenciesω<<1/T
–Forhighfrequenciesω>>1/T)log()( TjM   120 )log()(
22
120 TM   0120  )log()(M )log()( TM 20 )(
)(
3
1


s
sG

26. Basic Factors of a Transfer Function
3.FirstOrderFactors(jωT+1)
-1)(tan
-1
T )( 
000  )(tan when
-1
)(, 
451
1
 )(tan when
1-
)(,
T 
90 )(tan when
-1
)(,

27. -30
-10
10
30
Magnitude (decibels)
Frequency (rad/sec)
0.1 1 10 100 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
-20)(
)(
3
1


s
sG
ω=3
-6 db/octave
-20 db/decade

28. -90
o
-30
0
30
o
90
o
Phase (degrees)
Frequency (rad/sec)
0.1 1 10 100 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
o
-45
o

29. Example#1
•DrawtheBodePlotoffollowingTransferfunction.)(
)(
10
20


s
s
sG
Solution:).(
)(
110
2


s
s
sG
•Thetransferfunctioncontains
1.GainFactor(K=2)
2.DerivativeFactor(s)
3.1
st
OrderFactorindenominator(0.1s+1)
-1

30. Example#1).(
)(
110
2


s
s
sG
1.GainFactor(K=2)dbK
db
6220  )log(
2.DerivativeFactor(s)db/decade 2020  )log(
db
s
3.1
st
OrderFactorindenominator(0.1s+1)0120
110
1
10 

 )log(
.
,
db
j
 when db/dec when 201020
110
1
10 

 ).log(
.
, 


db
j

31. -30
-10
10
30
Magnitude (decibels)
Frequency (rad/sec)
0.1 1 10 100 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
-20)(
)(
10
20


s
s
sG
K=2
20 db/decade
-20 db/decade

32. -30
-10
10
30
Magnitude (decibels)
Frequency (rad/sec)
0.1 1 10 100 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
-20)(
)(
10
20


s
s
sG
20 db/decade
-20 db/decade+20db/decade

33. Example#1).(
)(
110
2





j
j
jG ).()( 1102   jjjG ).(tan)(tan)(tan)( 

 10
02
0
111 
jG ).(tan)(  1090
1


jG
ω
0.1 1 5 10 20 40 70 1001000 ∞
Φ(ω)89.484.263.445 26.514 8 5.7 0.5 0

34. -90
o
-30
0
30
o
90
o
Phase (degrees)
Frequency (rad/sec)
0.1 1 10 100 10
3
10
4
10
5 10
6
10
7
10
8 10
9
0
o
-45
o
ω 0.1 1 5 10 20 40 70 1001000 ∞
Φ(ω)89.484.263.4 45 26.5 14 8 5.7 0.5 0

35. -20
-10
0
10
20
30
Magnitude (dB)
10
-1
10
0
10
1
10
2
10
3
0
45
90
Phase (deg)
Bode Diagram
Frequency (rad/sec)

36. Example#2))((
)(
)(
10020
320



sss
s
sG
Solution:).)(.(
).(.
)(
10101050
1330030



sss
s
sG