Bode Plots step by step guide

8/15/2019 Bode Plots step by step guide

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Control Systems

EE

-

475

Lec Sara

Bode plots

Frequency Response

Analysis

ں د � ودى و ں رے دں ى ں

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Frequency Response Analysis

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ں د ے ا ا ا ر ذ ں ا ر ا د

ر ہ ر ا و س ہ

ر و ر ہ ا ر ل ں ا ا ل د

ا

ں ك د ا

ج ا د ا و

ن

ب

ك ے د ا ت ر ف

م

ا � م ى ا ز 3


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1

. Introduction

Steady State Response of System to a Sinusoidalinput. (Effect of input frequency variation onoutput).

Three common frequency response analysismethods are◦ Bode Diagrams

◦ Nyquist Plots

◦ Nicholas Charts

If input is sin wave than output will also be

wave but with different amplitude and phaseangle.

By replacing = we can shift from Laplace tofrequency domain (Fourier)

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Example

Let a System

For sin input put

As it’s a complex number so it s

magnitude and angle can be determined

as

( )1

K G j

jT ω

ω =

+

2 2( )

1

K G j

T ω

ω =

+( )1( ) tanG j T φ ω ω −= ∠ = −

s jω =

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Bode Diagrams/Logarithmic Plots

Consists of two plots magnitude and

phase angle plotted against frequency on

logarithmic scale.

Standard Logarithmic magnitude for base10 is represented as 20 log

Curves are drawn on semi graph

paper using log scale for frequency andlinear scale for magnitude in dB and phase

angle in degrees.

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Cont…

Followings are the basic factors that

frequently occur in arbitrary transfer

functions.

◦ Gain K

◦ Integral and Derivative Factors

◦ First order Factors

◦ Quadratic Factors

( ) 1

jω ±

( ) 1

1 j T ω ±

+ 12

1 2n n

j jω ω ζ

ω ω

± + +

7

ل � s2

s jw

in second order Transfer function

real part imaginary part


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Cont…

Once we become familiar with plots of

these basic factors we will be able to

construct plot of any general transfer

function by adding individual curvesgraphically.

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Gain Factor K

Gain is some constant value so log-

magnitude curve for gain is horizontal

straight line curve at

its phase angle will be zero.

Effect of variation in Gain is that it raise

or lower the log-magnitude curve of

transfer function and have no effect onthe phase curve.

( )1020log K

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Cont…

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Integral and Derivative Factors

For integral part Logarithmic magnitude is

Phase angle is

So log-magnitude plot will be straight linewith a slope of -20db per decade and phaseangle will be straight horizontal line at

( ) 1 1 j jω

ω

−

=

( )10 101

20log 20log dB j

ω ω

= −

90degφ = −

90o−

11

if (jw)-2 then -40log(w),phi=-180...


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Cont…

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Cont…

For derivative part Logarithmic magnitude is

Phase angle is

So log-magnitude plot will be straight linewith a slope of +20db per decade and phaseangle will be straight horizontal line at

+90deg

( ) jω

( )10 1020log 20log j dBω ω =

90degφ =

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Cont…

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First Order Factors

For

Logarithmic magnitude is

For lower frequencies

magnitude is zero

For higher frequencies

magnitude is . = − 20

( )1

1 j T ω +

( )2 2

10 10

120log 20log 1

1T dB

j T ω

ω = − +

+

1

T ω

1

T ω

( )1020log T dbω −

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Cont…

So magnitude plot for this factor canbe approximated by two straight lines

asymptotes.

Horizontal line of 0db for Straight line of slope for

Frequency at which two asymptotes

meets is called Corner/Break frequencythat in this case will be

10 T ω < <

1T

ω < < ∞

1

T ω =

20 dBdecade

−

log

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Cont…

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Cont…

Phase angle will be

At zero frequency

At corner frequency

At infinity

( )1Tan T φ ω −= −

1(0) 0degTanφ −= − =

1

45deg

T

Tan T φ

− = − = −

1

T ω =

1( ) 90degTanφ −= − ∞ = −

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Cont…

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Cont…

Error due to use of asymptotes can becalculated as

At corner frequency

At frequency one octave below

At frequency one octave above

1

T ω =

( ) ( ) ( )10 10 1020 log 1 20 log 1 1 10 log 2 3.03db− + = − −=1

2T ω =

2T

ω =

( )10 10 101 5

20 log 1 20 log 1 20 log 0.97

4 2

db

− + = − = −

( ) ( )10 10 105

20 log 2 20 log 1 4 20 9log 0. 72

db

− + = − =

−

20

octave=2*w and decade = 10 * w


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Cont…

So asymptotic plot can be corrected toexact by making correction of 3 at

corner frequency and 1 at one octave low

and one octave high. Variation in T only shifts the corner freq

to the left or right but no effect on the

shape of plot. So .( )

1

1 j T ω +

21

ر ۔

و س ہ ز omegavaluefactor

and this what a low pass filter does, it just allows lower frequencies


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Cont…

To Draw the Phase curve Accuratelylocate different points on curve

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Cont…

23

These are octaves, NOT decades


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Cont…

For

Logarithmic magnitude is

For lower frequencies

magnitude is zero

For higher frequencies

magnitude is

( )1 j T ω +

( )2 2

10 10

20 log 1 20 log 1 j T T dBω ω + = +

1

T ω

1

T ω

( )1020log T dbω

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Cont…

Horizontal line of 0db for

Straight line of slope +20db for

10T

ω < <1

T ω < < ∞

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Cont…

Phase angle will be

At zero frequency

At corner frequency

At infinity

( )1Tan T φ ω −=

1(0) 0degTanφ −= =

1

45deg

T

Tan T φ

− = =

1

T ω =

1( ) 90 degTanφ −= ∞ =

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Cont…

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Quadratic Factor

For quadratic factor

if it will be product of two first order

factors.(Because it’ll be over-damped system will two –real poles) If it will be product of two complex

conjugate factors. (Because it’ll be under-damped system willtwo – complex conjugate poles)

Magnitude and phase of quadratic factordepends upon corner freq. and dampingratio

12

1 2n n

j jω ω ζ ω ω

−

+ +

1ζ >

0 1ζ < <

( )n and ω ζ

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Cont…

Magnitude will be

For

Magnitude will be

For

Magnitude will be

22 2

10 102

120 log 20 log 1 2

1 2 n n

n n

j j

ω ω ς

ω ω ω ω ς ω ω

= − − +

+ +

nω ω >

2

10 10220log 40log

n n

dbω ω

ω ω

− = −

?

29

� ہ


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Cont…

So log-magnitude plot can be estimated bytwo straight line asymptotes.

◦ Horizontal line of 0db for

◦ Straight line with − 40

slope for

The frequency Is corner frequency in thiscase.

Near the corner freq.() a resonant peak occurs.

Damping ratio determines the magnitude of that peak . Asymptotes are independent of the damping ratio.

n

ω ω < < ∞

0 nω ω < <

nω

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Cont…

Phase Angle will be

At zero frequency

At corner frequency

At infinity

1

2

2

1

n

n

Tan

ω ξ

ω φ

ω

ω

−

= − −

0degφ =

90degφ = −

180degφ = −

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Cont…

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Contd.

For quadratic factor

Magnitude will be

2

1 2n n

j jω ω ζ ω ω

+ +

2 2 2

10 1020 log 1 2 20 log 1 2n n n n

j jω ω ω ω ξ ξ

ω ω ω ω

+ + = − +

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Cont…

For

Magnitude will be

For

Magnitude will be

nω ω >

2

10 10220 log 40 log

n n

dbω ω

ω ω

=

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Cont…

So log-magnitude plot can be estimatedby two straight line asymptotes.

◦ Horizontal line of 0db for

◦ Straight line with 40db/decade slope forn

ω ω < < ∞

0n

ω ω < <

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Cont…

Phase Angle will be

At zero frequency

At corner frequency

At infinity

1

2

2

1

n

n

Tan

ω ξ

ω φ

ω

ω

−

= −

0degφ =

90degφ =

180degφ =

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Example#1 Bode Plots

Let a System

Put =

Split into Sub Systems

( )( )( )

50000

5 500

sG s

s s=

+ +

( )( )( )

50000 20

5 5001 1

5 500

j jG j

j j j j

ω ω ω

ω ω ω ω = =

+ + + +

1 ( ) 20G jω =

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Cont…

Draw bode plot for each factor

for

( )

( )

( )

2

1

3

1

4

15

1 500

G j j

jG j

jG j

ω ω

ω ω

ω ω

−

−

=

= +

= +

1( ) 20G jω =

101( ) 20 log (20) 26.02G jω = =

0degφ =

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Cont…

25

25.5

26

26.5

27

27.5

M

a g n i t u d e ( d B )

100

101

-1

-0.5

0

0.5

1

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

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Cont…

for ( )2G j jω ω =

2 10( ) 20log ( )G jω ω =

90degφ =

-5

0

5

10

15

20

M a g n i t u d e ( d B )

100

101

89

89.5

90

90.5

91

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

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Cont…

For ( )1

3 15

jG j ω ω

−

= +

3 10

( ) 20 log5

G j for higher freqω

ω

= −

1

5Tan

ω φ −

= −

3 ( ) 0G j for lower freqω =

0 deg 0

45 deg 5

90deg

for

for

for

φ ω

φ ω

φ ω

= =

= − =

= − = ∞

( )1 15

1

51

5

j j T

T

cutt off freqT

ω ω

+ = +

=

= =

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Cont…

-30

-20

-10

0

10

M a g n

i t u d e ( d B )

Bode Diagram

Fre uenc rad/sec

10-1

100

101

102

-90

-45

0

P

h a s e ( d e g )

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Cont…

For ( )1

4 1500 jG j ω ω

−

= +

4 10

( ) 20log500

G j for higher freqω

ω

= −

1

500Tan

ω φ −

= −


0 deg 0

45 deg 500

90deg

for

for

for

φ ω

φ ω

φ ω

= =

= − =

= − = ∞

( )1 1500

1

5001

500

j j T

T

cutt off freqT

ω ω

+ = +

=

= =

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Cont…

Bode Diagram

Fre uenc rad/sec

-30

-20

-10

0

10

M

a g n i t u d e ( d B )

101

102

103

104

-90

-45

0

P h

a s e ( d e g )

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Cont…

Now Plot All of them on Single GraphPaper.

-50

0

50

M a g n i t u d e

( d B )

10-1

100

101

102

103

104

-90

-45

0

45

90

P h a s e ( d e g )

Bode Diagram

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Cont…

Now add the graphs to get final plot

10-1

100

101

102

103

104

-90

-45

0

45

90

P h a s e ( d

e g )

Bode Diagram

Fre uenc rad/sec

-50

0

50

M a g n i t u

d e ( d B )

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Cont…

0

20

40

60


10-1

100

101

102

103

104-90

-45

0

45

90

P h a s e ( d e g )

Bode Diagram

Fre uenc rad/sec

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Cont…

Draw bode plot for each factor

for

( ) ( )

( ) ( )

1

2

12 6

3

1 .01

100 10

G j j

G j j

ω ω

ω ω ω

−

−

= +

= − + +

71( ) 10G jω =

7

1 10( ) 20 log (10 ) 140G jω = =

0degφ =

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Cont…

139

139.5

140

140.5

141


100

101

-1

-0.5

0

0.5

1

P

h a s e ( d e g )

Bode Diagram

Fre uenc rad/sec

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Cont…

For ( ) ( ) 12 1 0.01G j jω ω −= +

( )2 10( ) 20log .01G j for higher freqω ω = −

( )1 .01Tanφ ω −= −


0 deg 0

45deg 100

90deg

for

for

for

φ ω

φ ω

φ ω

= =

= − =

= = ∞

( ) ( )1 1 .01

.01

1 100

j T j

T

cutt off freqT

ω ω + = +

=

= =

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Cont…

-40

-30

-20

-10

0

10


100 101 102 103 104-90

-45

0

P h a s e ( d e g )

Bode Diagram

Frequency (rad/sec)

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Cont…

For ( ) ( ) 1

2 63 100 10G j jω ω ω

−

= − + +

( )3 10

( ) 40log .001G j for higher freqω ω = −


0deg 0

90deg 1000

180deg

for

for

for

φ ω

φ ω

φ ω

= =

= − =

= − = ∞

2 610

1000

2 100

0.05

n

n

n

ω

ω

ξω

ξ

=

=

=

=

1

2

2

1

n

n

Tan

ω ξ

ω φ

ω

ω

−

= − −

21 2 997r n

ω ω ξ = − =

2

110

2 1r M

ξ ξ = =

−

( ) 12

90 sin 87.13

1

r G j

ξ ω

ξ

−

∠ = − + = −

−

Do apply 20 log to get

magnitude of in dB

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Cont…Do apply 20 log to get

magnitude of in dB.20 log 10 = 20

This has been verified in

Matlab

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Cont…

-60

-40-20

0

20

40


100

101

102

103

104

-270

-225

-180

-135

-90

-45

0

45

P

h a s e ( d e g )

Bode Diagram

Fre uenc rad/sec

Bode Plots step by step guide

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