Case StudyCase Study - Purdue Engineering Notes/Frequency Response.pdf · Case StudyCase Study...

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School of Mechanical EngineeringPurdue University

ME375 Frequency Response - 1

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POWER WIRE

Electric train derives power through a pantograph, which contacts the powerwire, which is suspended from a catenary. During high-speed runs between NewHaven, CT and New York City, the train experiences intermittent power loss at42 k /h d 100 k /h



42 km/hr and 100 km/hr.

Case StudyCase Study –– Pantograph ModelPantograph ModelCase Study Case Study Pantograph ModelPantograph Model

F (t) 0b b k k b k

m2

z2

Fc(t) 1 1 1 2 1 1 2 1 2 2 2 2

2 2 2 2 2 2 2 1 2 1

0( )c

mz b b z k k z b z k zm z b z k z b z k z F t

k2

mz1

b2 For m1 = 23.0 kg, b1 = 150 N/(m/s), k1 = 9600 N/m, m2 = 11.5 kg, b2 = 75 N/(m/s), and k2 = 9580 N/m:

k1

m1

b1

22

4 3 2( ) 0.087( 9.78 834)( ) 16.3 1709 8155 347,705c

Z s s sF s s s s s



Case StudyCase Study –– Frequency ResponseFrequency ResponseCase Study Case Study Frequency ResponseFrequency Response

-70

-60)

-90

-80

Mag

nitu

de (d

B)

-110

-100

135

180

45

90

135

Phas

e (d

eg)

100

101

102

0

Frequency (rad/sec)



Frequency ResponseFrequency ResponseFrequency ResponseFrequency Response

•• Forced Response to Sinusoidal InputsForced Response to Sinusoidal Inputs•• Frequency Response of LTI SystemsFrequency Response of LTI Systems•• Frequency Response of LTI SystemsFrequency Response of LTI Systems•• Bode PlotsBode Plots



Forced Response to Sinusoidal InputsForced Response to Sinusoidal InputsEx:Ex: Let’s find the forced response of a stable first order system:Let’s find the forced response of a stable first order system:

Forced Response to Sinusoidal InputsForced Response to Sinusoidal Inputs

5 10y y u

to a sinusoidal input:to a sinusoidal input:–– Forced response:Forced response:

( ) sin(2 )u t t

( ) ( ) ( )Y s G s U s

where ( ) and ( ) sin(2 )

( )

G s U s L t

Y s

–– PFE:PFE:31 2 ( )

AA AY s

–– Compare coefficients to find Compare coefficients to find AA11, , AA22 and and AA33 ::



Forced Response to Sinusoidal InputsForced Response to Sinusoidal InputsEx:Ex: (cont.)(cont.)

–– Use ILT to find Use ILT to find yy((tt) :) :

Forced Response to Sinusoidal InputsForced Response to Sinusoidal Inputs

1 1 ( ) ( ) y t L Y s L

U f l F l A t B t A B ti ( ) ( ) i ( ) 2 2

–– Using this formula, the forced response can be represented byUsing this formula, the forced response can be represented by

Useful Formula: Where atan2( ) = ( )

A t B t A B tB A A jB

sin( ) cos( ) sin( ),

5( ) sin(2 )ty t e t



Forced Response of 1st Order SystemForced Response of 1st Order SystemForced Response of 1st Order SystemForced Response of 1st Order System

2Input is sin(2t)

1

1.5

Output

Input

0

0.5

-1

-0.5

Res

pons

e

0 2 4 6 8 10 12 14-2

-1.5



0 2 4 6 8 10 12 14Time (sec)

Forced Response to Sinusoidal InputsForced Response to Sinusoidal InputsForced Response to Sinusoidal InputsForced Response to Sinusoidal Inputs

Ex:Ex: Given the same system as in the previous example, find the forced response Given the same system as in the previous example, find the forced response (( ) i () i ( ))to to uu((tt) = sin(10 ) = sin(10 tt).).

( ) ( ) ( ) where ( ) and ( ) sin(10 )

Y s G s U s

G s U s L t

( ) ( ) ( )

( )Y s



Forced Response of 1st Order SystemsForced Response of 1st Order SystemsForced Response of 1st Order SystemsForced Response of 1st Order SystemsInput is sin(10t)

OO

0.6

0.8

1OutputOutput

0

0.2

0.4

Res

pons

e

-0.6

-0.4

-0.2

R

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-1

-0.8

InputInput



Time (sec)

Frequency ResponseFrequency ResponseFrequency ResponseFrequency ResponseEx:Ex: Let’s revisit the same example where Let’s revisit the same example where

5 10y y u

and the input is a general sinusoidal input: sin(and the input is a general sinusoidal input: sin(tt).).

2 210 10( ) ( ) ( )

5 5 ( )( )Y s G s U s

j j

2 2

31 2

( ) ( ) ( )5 5 ( )( )

( )5

s s s j s jsAA AY s

s s j s j

–– Instead of comparing coefficients, use the Instead of comparing coefficients, use the residue formula residue formula to find to find AAii’s:’s:

1 2 255

10( 5) ( ) ( 5)( 5)s

sA s Y s s

s s

2 2 2

5

( ) ( ) ( ) ( )

s

s js j

A s j Y s s j G ss



3 2 2( ) ( ) ( ) ( )s js j

A s j Y s s j G ss

Frequency ResponseFrequency ResponseFrequency ResponseFrequency ResponseEx: Ex: (Cont.)(Cont.)

1 2 210

5A

2

3

1 10 1 1( )2 5 2 2

1 10 1 1( )

A G jj j j j

A G j

The steady state response The steady state response YYSSSS((ss) is:) is:

3 ( )2 5 2 2

A G jj j j j

32( )SSAAY s

12 3

( )

( ) ( )

SS

j t j tSS SS

Y ss j s j

y t L Y s A e A e



( ) ( ) sin( ) where ( )SSy t G j t G j

Frequency ResponseFrequency ResponseFrequency ResponseFrequency Response•• Frequency ResponseFrequency Response



In Class ExerciseIn Class ExerciseIn Class ExerciseIn Class ExerciseFor the current example,For the current example,

5 10y y u

Calculate the magnitude and phase shift of the steady state response when the Calculate the magnitude and phase shift of the steady state response when the system is excited by (i) sin(2system is excited by (i) sin(2tt) and (ii) sin(10) and (ii) sin(10tt). Compare your result with the ). Compare your result with the steady state response calculated in the previous examplessteady state response calculated in the previous examples

y y

steady state response calculated in the previous examples.steady state response calculated in the previous examples.Note:Note: 10 10( ) ( )

5 5G s G j

s j

2 2

10( ) and ( ) atan2( ,5)5

G j G j



Frequency ResponseFrequency ResponseFrequency ResponseFrequency Response••Frequency response Frequency response is used to study the steady state output is used to study the steady state output yySSSS((tt) of a stable system ) of a stable system due to sinusoidal inputs at different frequencies.due to sinusoidal inputs at different frequencies.

In general, given a stable system:In general, given a stable system:( ) ( 1) ( ) ( 1)

1 1 0 1 1 01

n n m mn n m m

m m

a y a y a y a y b u b u b u b u

If the input is a sinusoidal signal with frequency If the input is a sinusoidal signal with frequency , , i.e. i.e.

11 1 0 1 2

11 21 1 0

( )( ) ( )( )( )( ) ( )( ) ( )

m mm m m m

n nn nn n

b s b s b s b b s z s z s zN sG sD s a s p s p s pa s a s a s a

( ) i ( )A

then the steady state output then the steady state output yySSSS((tt) is also a sinusoidal signal with the same frequency as the ) is also a sinusoidal signal with the same frequency as the input signal but with different magnitude and phase:input signal but with different magnitude and phase:

( ) sin( )uu t A t

where where GG((jj) is the complex number obtained by substitute ) is the complex number obtained by substitute jjfor for ss in in GG((ss) , i.e.) , i.e.

( ) ( ) sin( ( ))SS uy t G j A t G j

11 1 0( ) ( ) ( )( ) ( )

m mm mb j b j b j bG G



1 1 01

1 1 0

( ) ( ) ( )( ) ( )( ) ( ) ( )

m mn ns j

n n

b j b j b j bG j G sa j a j a j a


LTI SystemInput u(t) Output y(t)LTI System

G(s)

u 2/ ySS 2/

Input u(t)

U(s)

Output y(t)

Y(s)

t t

( ) ( ) i ( ( ))t G j A t G j( ) i ( )t A t ( ) ( ) sin( ( ))SS uy t G j A t G j ( ) sin( )uu t A t

A different perspective of the role of the transfer function:

Amplitude of the steady state sinusoidal output( )Amplitude of the sinusoidal input

( ) Phase difference (shift) between ( ) and the sinusoidal inputSS

G j

G j y t




G

Input u(t)

Output y(t)

G



In Class ExerciseIn Class Exercise

Ex: Ex: 1st Order System1st Order SystemTh i f i i li d bTh i f i i li d b

In Class ExerciseIn Class Exercise

(2) Calculate the steady state output of the (2) Calculate the steady state output of the t h th i t it h th i t iThe motion of a piston in a cylinder can be The motion of a piston in a cylinder can be

modeled by a 1st order system with force modeled by a 1st order system with force as input and piston velocity as output:as input and piston velocity as output:

system when the input is system when the input is

Input f(t) Steady State Output v(t) sin( t) G(j) sin( t + )

f(t)

( ) (j ) ( )sin(0t) sin(0t +

sin(10t) sin(10t +

sin(20t) sin(20t +The EOM is:The EOM is:

(1) Let(1) Let M =M = 0 10 1 kg andkg and BB = 0 5 N/(m/s)= 0 5 N/(m/s)( )Mv Bv f t

vsin(20t) sin(20t +

sin(30t) sin(30t +

sin(40t) sin(40t + (1) Let (1) Let M M 0.10.1 kg and kg and B B 0.5 N/(m/s), 0.5 N/(m/s), find the transfer function of the system:find the transfer function of the system: sin(50t) sin(50t +

sin(60t) sin(60t +



In Class ExerciseIn Class ExerciseIn Class ExerciseIn Class Exercise(3) Plot the frequency response plot

1 6

1.8

2

20

-10

0

1

1.2

1.4

1.6

e ((

m/s

)/N)

-50

-40

-30

-20

e (d

eg)

0 4

0.6

0.8

1

Mag

nitu

de

-80

-70

-60

-50

Pha

se

0 10 20 30 40 50 60 700

0.2

0.4

Frequency (rad/sec)0 10 20 30 40 50 60 70

-90

80

Frequency (rad/sec)



Frequency (rad/sec) Frequency (rad/sec)

ExampleExample -- Vibration Absorber (I)Vibration Absorber (I)Example Example Vibration Absorber (I)Vibration Absorber (I)Without vibration absorber:Without vibration absorber: EOM:EOM:

( )M B K f t

Let Let MM11 = 10 kg, = 10 kg, KK11 = 1000 N/m, = 1000 N/m, BB11 = 4 N/(m/s).= 4 N/(m/s).Find the steady state response of the system for Find the steady state response of the system for ff((tt) = ) =

M1

z11 1 1 1 1 1 ( )M z B z K z f t

y p yy p y ff(( ))(a) sin(8.5(a) sin(8.5tt) (b) sin(10) (b) sin(10tt) (c) sin(11.7) (c) sin(11.7tt).).

K1 B1f(t)

Input f(t) Steady State Output z1(t)

TF (from TF (from ff((tt) to ) to zz11):):

sin( t) G(j) sin( t + ) sin(8.5t) sin(8.5t +

sin(10t) sin(10t +( ) (

sin(11.7t) sin(11.7t +



ExampleExample -- Vibration Absorber (I)Vibration Absorber (I)

0 005

0.01f(t) = sin(8.5 t)

Example Example Vibration Absorber (I)Vibration Absorber (I)

-0.01

-0.005

0

0.005

z 1(m

)

0

0.02

0.04

z 1(m

)

f(t) = sin(10 t)

-0.04

-0.02

0.005 f(t) = sin(11.7 t)

-0.005

0

z 1(m

)



0 5 10 15 20 25 30 35 40 45 50Time (sec)

ExampleExample -- Vibration Absorber (II)Vibration Absorber (II)Example Example Vibration Absorber (II)Vibration Absorber (II)With vibration absorber:With vibration absorber: EOM:EOM:

z2 1 1 1 2 1 1 2 1 2 2 2 2( ) ( )M z B B z K K z B z K z f t

Let Let MM11 = 10 kg, = 10 kg, KK11 = 1000 N/m, = 1000 N/m, BB11 = 4 N/(m/s), = 4 N/(m/s), MM2 2 = 1 kg, = 1 kg, KK2 2 = 100 N/m, and = 100 N/m, and BB2 2 = 0.1 N/(m/s). Find = 0.1 N/(m/s). Find

M2

K2 B2

2 2 2 2 2 2 2 1 2 1 0M z B z K z B z K z

the steady state response of the system for the steady state response of the system for ff((tt) =) =(a) sin(8.5(a) sin(8.5tt) (b) sin(10) (b) sin(10tt) (c) sin(11.7) (c) sin(11.7tt).).

K

M1

B

z1

Input f(t) Steady State Output z1(t)

TF (fromTF (from ff((tt) to) to zz11):):

K1 B1f(t) Input f(t) Steady State Output z1(t)sin( t) G(j) sin( t + ) sin(8.5t) sin(8.5t +

i (10 ) i (10t +TF (from TF (from ff((tt) to ) to zz11):): sin(10t) sin(10t +

sin(11.7t) sin(11.7t +



ExampleExample -- Vibration Absorber (II)Vibration Absorber (II)

0.02

0.04f(t) = sin(8.5 t)

Example Example Vibration Absorber (II)Vibration Absorber (II)

-0.04

-0.02

0

0.02

z 1(m

)

0

0.002

0.004

z 1(m

)

f(t) = sin(10 t)

-0.004

-0.002

0 01

0.02 f(t) = sin(11.7 t)

0 5 10 15 20 25 30 35 40 45 50-0.02

-0.01

0

0.01

z 1(m

)



0 5 10 15 20 25 30 35 40 45 50

Time (sec)

ExampleExample -- Vibration Absorber (II)Vibration Absorber (II)Example Example Vibration Absorber (II)Vibration Absorber (II)Take a closer look at the poles of the transfer function:Take a closer look at the poles of the transfer function:Th h t i ti tiTh h t i ti tiThe characteristic equationThe characteristic equation

4 3 2

1 2

10 5.1 2100.4 500 100000 0 Poles: 0.1 8.5

s s s sp j

What part of the poles determines the rate of decay for the transient response?What part of the poles determines the rate of decay for the transient response?

1,2

3,4 0.155 11.7

p j

p j

(Hint: when (Hint: when p = p = jjthe response is the response is eet t e e jjtt ))



ExampleExample -- Vibration AbsorbersVibration AbsorbersExample Example Vibration Absorbers Vibration Absorbers Frequency Response PlotFrequency Response PlotNo absorber addedNo absorber added

Frequency Response PlotFrequency Response PlotAbsorber tuned at 10 rad/sec addedAbsorber tuned at 10 rad/sec addedNo absorber addedNo absorber added Absorber tuned at 10 rad/sec addedAbsorber tuned at 10 rad/sec added

0 015

0.025

0.02

e (m

/N)

0 015

0.02

0.025

e (m

/N)

0 2 4 6 8 10 12 14 16 18 20

0.005

0.015

0

0.01

Mag

nitu

de

0 2 4 6 8 10 12 14 16 18 200

0.005

0.01

0.015

Mag

nitu

de-45

0

deg)

Frequency (rad/sec)0 2 4 6 8 10 12 14 16 18 20

-45

0

deg)

Frequency (rad/sec)

0 2 4 6 8 10 12 14 16 18 20-180

-135

-90

Phas

e (d

0 2 4 6 8 10 12 14 16 18 20-180

-135

-90Ph

ase

(d



Frequency (rad/sec) Frequency (rad/sec)

ExampleExample -- Vibration AbsorbersVibration AbsorbersExample Example Vibration Absorbers Vibration Absorbers

Bode PlotBode Plot Bode PlotBode PlotNo absorber addedNo absorber added Absorber tuned at 10 rad/sec addedAbsorber tuned at 10 rad/sec added

-50-40-30

50-40-30

agni

tude

(dB

)

-90-80-70-60-50

agni

tude

(dB

)

-90-80-70-60-50

Phas

e (d

eg);

M -100

-45

0

Phas

e (d

eg);

Ma

-100

-45

0

010 110 210-180

-135

-90

100 101 102-180

-135

-90



Frequency (rad/sec) Frequency (rad/sec)10 10 10

Bode Diagrams (Plots)Bode Diagrams (Plots)•• Bode Diagrams (Plots)Bode Diagrams (Plots)

A i f l i h f f iA i f l i h f f i GG((jj ) f) f ff

Bode Diagrams (Plots)Bode Diagrams (Plots)

A unique way of plotting the frequency response function, A unique way of plotting the frequency response function, GG((jj), w.r.t. frequency ), w.r.t. frequency of of systems.systems.Consists of two plots:Consists of two plots:

Magnitude PlotMagnitude Plot : plots the magnitude of: plots the magnitude of GG((jj) in decibels w r t logarithmic frequency) in decibels w r t logarithmic frequency–– Magnitude PlotMagnitude Plot : plots the magnitude of : plots the magnitude of GG((jj) in decibels w.r.t. logarithmic frequency, ) in decibels w.r.t. logarithmic frequency, i.e.i.e.

–– Phase PlotPhase Plot : plots the linear phase angle of : plots the linear phase angle of GG((jj) w.r.t. logarithmic frequency, i.e.) w.r.t. logarithmic frequency, i.e.

G j G j( ) log ( ) dB 10 vs log 20 10

To plot Bode diagrams, one needs to calculate the magnitude and phase of the To plot Bode diagrams, one needs to calculate the magnitude and phase of the corresponding transfer function.corresponding transfer function.

G j( ) vs log 10

Ex: Ex: G s s

s s( )

1

102



Bode DiagramsBode DiagramsBode DiagramsBode DiagramsRevisit the previous example:Revisit the previous example:

2

1 ( ) 1( ) ( )10 ( 10)

s jG s G js s j j

30

50

10 ( 10) ( )

( )

s s j j

G j

G j

B)

-100

10

( )G j

g); M

agni

tude

(dB

-50

-30

100

G j( ) 20 10log ( )G j G j( )0.10.2

Phas

e (d

eg

50

1000.20.512

-50

05102050



10-1 100 101 102

Frequency (rad/sec)

-10050100

Bode DiagramsBode DiagramsBode DiagramsBode DiagramsRecall that ifRecall that if

G b s b s b s b b s z s z s zm m

( ) ( )( ) ( ) 1

11 0 1 2

ThenThen

G s b s b s b s ba s a s a s a

b s z s z s za s p s p s p

m m

nn

nn

m m

n n

( ) ( )( ) ( )( )( ) ( )

1 1 0

11

1 0

1 2

1 2

G j b j z j z j zj j j

m m( ) ( )( ) ( )( )( ) ( )

1 2

a j p j p j p

ba j p j p j p

j z j z j z

n n

m

n nm

( )( ) ( )

( ) ( ) ( )( ) ( ) ( )

1 2

1 21 2

1 1 1

G j ba j p j p

j

m

n n

log ( ( ) ) log log( )

log( )

log (

FHGIKJ

FHG

IKJ

FHG

IKJ

10 10 101

10

10

20 20 20 1 20 1

20

z j z1 10 120) log ( )c h c h

and and jlog ( 1020 z j z1 10 120) log ( )c h c h

G j b j z j z j za j p j p j p

m m

n n

( ) ( )( ) ( )( )( ) ( )

1 2

1 2



j z j z j z j p j p j pm n( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2

ExampleExampleExampleExampleEx:Ex: Find the magnitude and the phase of the following transfer function:Find the magnitude and the phase of the following transfer function:

G s s s( ) 3 12 93 2

G ss s s

( )(

2 22 76 803 2 )( )



Bode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building Blocks•• 1st Order Real Poles1st Order Real Poles

Transfer Function:Transfer Function: -30

Frequency Response:Frequency Response:

G ssp11

10( )

,

B)

-20

G jj

G j

p11

10

1

( )

( )

R|

,

g); M

agni

tude

(dB

-40

G j

G j

p

p

1 2 2

11

11

( )

( ) ( , )tan

S|T|

atan2a f

Phas

e (d

eg

-45

0

QQ:: By just looking at the Bode diagram, can you By just looking at the Bode diagram, can you determine the time constant and the steady determine the time constant and the steady state gain of the systemstate gain of the system ??

-90



state gain of the systemstate gain of the system ??

Frequency (rad/sec)0.01/ 0.1/ 1/ 10/ 100/

ExampleExampleExampleExample•• 1st Order Real Poles1st Order Real Poles

Transfer Function:Transfer Function: 20

Plot the straight line approximation Plot the straight line approximation ff GG(( )’ B d di)’ B d di

G ss

( ) 50

5

)

5

10

15

of of GG((ss)’s Bode diagram: )’s Bode diagram:

); M

agni

tude

(dB

0

Phas

e (d

eg)

-45

0

-90

1 0 1 2



Frequency (rad/sec)10

-110

010

110

2

Bode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building Blocks•• 1st Order Real Zeros1st Order Real Zeros

Transfer Function:Transfer Function:

40

Frequency Response:Frequency Response:1( ) 1 , 0zG s s

G j j 1 0( )

20

B)G j j

G j

G j

z

z

z

1

12 2

11

1 0

1

1

( )

( )

( ) ( , )t

RS|T|

,

atan2a f

30

g); M

agni

tude

(d

1tan a f

45

90Ph

ase

(deg

0

45



Frequency (rad/sec)0.01/ 0.1/ 1/ 10/ 100/

ExampleExampleExampleExample•• 1st Order Real Zeros1st Order Real Zeros

Transfer Function:Transfer Function:20


Plot the straight line approximation Plot the straight line approximation of of GG((ss)’s Bode diagram: )’s Bode diagram:

G s s( ) . . 0 7 0 7

dB) 5

10

15

(( ) g) g

eg);

Mag

nitu

de (d 0

90

Phas

e (d

e

45

90

0

-1 0 1 2



Frequency (rad/sec)10 10 10 10

ExampleExampleExampleExample•• Lead CompensatorLead Compensator


Plot the straight line approximation Plot the straight line approximation ff GG(( )’ B d di)’ B d di

G s ss

( )

35 355

dB) 0

10

20

of of GG((ss)’s Bode diagram: )’s Bode diagram:

eg);

Mag

nitu

de (d -10

90

Phas

e (d

e

0

90

2

-90

-1 0 1



Frequency (rad/sec)10 10 10 10

1st Order Bode Diagram Summary1st Order Bode Diagram Summary1st Order Bode Diagram Summary1st Order Bode Diagram Summary•• 1st Order Poles1st Order Poles •• 1st Order Zeros1st Order Zeros

11( ) , 0

1pG s

1( ) 1 , 0zG s s

–– Break FrequencyBreak Frequency

M Pl t A i tiM Pl t A i ti

–– Break FrequencyBreak Frequency1p s

1 rad/sb

1( ) ,z

1 rad/sb –– Mag. Plot ApproximationMag. Plot Approximation

0 dB from DC to 0 dB from DC to bb and a straight line with and a straight line with 20 dB/decade slope after 20 dB/decade slope after bb..

–– Phase Plot ApproximationPhase Plot Approximation

–– Mag. Plot ApproximationMag. Plot Approximation0 dB from DC to 0 dB from DC to bb and a straight line and a straight line with 20 dB/decade slope after with 20 dB/decade slope after bb..Ph Pl t A i tiPh Pl t A i ti

pppp0 deg from DC to . Between and 0 deg from DC to . Between and 1010b b , a straight line from 0 deg to , a straight line from 0 deg to 90 deg 90 deg (passing (passing 45 deg at 45 deg at bb). For frequency ). For frequency higher than 10higher than 10 straight line onstraight line on 90 deg90 deg

–– Phase Plot ApproximationPhase Plot Approximation0 deg from DC to . Between 0 deg from DC to . Between and 10and 10b b , a straight line from 0 deg to 90 , a straight line from 0 deg to 90 deg (passing 45 deg at deg (passing 45 deg at bb). For frequency ). For frequency

110b

110b 1

10b1

10b

higher than 10higher than 10bb , straight line on , straight line on 90 deg.90 deg.higher than 10higher than 10bb , straight line on 90 deg., straight line on 90 deg.

Note:Note: By looking at a Bode diagram you should be able to determine By looking at a Bode diagram you should be able to determine the relative order of the systemthe relative order of the system, its , its break frequencybreak frequency, and , and DC (steadyDC (steady--state) gainstate) gain. This process should also be reversible, i.e. given a . This process should also be reversible, i.e. given a

f f b bl l h l d d df f b bl l h l d d d



transfer function, be able to plot a straight line approximated Bode diagram.transfer function, be able to plot a straight line approximated Bode diagram.

Bode Diagram Building BlocksBode Diagram Building Blocks•• Integrator (Pole at origin)Integrator (Pole at origin)


Bode Diagram Building BlocksBode Diagram Building Blocks

0

20


G ssp01( )

de (d

B)

-40

-20

0

G jj

G j

p01

1

( )

( )

R|

e (d

eg);

Mag

nitu

d

-60

0G j

G j

p

p

0

0 90

( )

( )

S|T|

Phas

e

-90

-45

Frequency (rad/sec)

0.1 1 10 100 1000-135



Bode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building Blocks•• Differentiator (Zero at origin)Differentiator (Zero at origin)


60


G s sz0( )

de (d

B)

0

20

40

G j j

G jz

z

0

0

( )

( )

RS|| e

(deg

); M

agni

tud

-20

135

G jz0 90( ) ST|

Phas

e45

90

Frequency (rad/sec)

0.1 1 10 100 1000

0



Frequency (rad/sec)

ExampleExampleExampleExample•• Combination of SystemsCombination of Systems


Plot the straight line approximation of Plot the straight line approximation of GG((ss)’s Bode diagram: )’s Bode diagram:

G s ss s

( )( )

35 355

B)

0

10

20

); M

agni

tude

(dB

-10

Phas

e (d

eg)

0

90

-90



2

Frequency (rad/sec)10

-110

010

110




G ss s s

( )( )

250055 2502

dB)

-40

0

eg);

Mag

nitu

de (d

-120

-80

Phas

e (d

e

-90

0

-270

-180



Frequency (rad/sec)

10-1 100

101

102

103

Bode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building Blocks•• 2nd Order Complex Poles2nd Order Complex Poles


20


G ss sp

n

n n2

2

2 220( )

, 1

1 (dB

)

60

-40

-20

G j

jp

n n

2 2

2

1

2 1

1 deg)

; Mag

nitu

de

-80

-60

0

G jp

n n

22 2

2

2

2

2

1

4 1

Phas

e (d

-90

G jp n n2

1 2 1 2

2

tan

n n n n n

-180



Frequency (rad/sec)

ExampleExampleExampleExample•• SecondSecond--Order SystemOrder System

Transfer Function:Transfer Function:2 00

40


G ss s

( )

250010 25002

(dB

)

80

-40

0

(( ) g) g

eg);

Mag

nitu

de (

-120

-80

Phas

e (d

-90

0

0 1 2 3 4

-180



Frequency (rad/sec)

100 101 102 10310

4

Bode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building BlocksBode Diagram Building Blocks•• 2nd Order Complex Zeros2nd Order Complex Zeros


80


22

2 22( ) , 1 0n n

zn

s sG s

B)

20

40

q y pq y p

G j jzn n

2

2

22 1

G G1 g); M

agni

tude

(dB

-20

0

G j G jz p

n n

2 2

2 2

2

2

2

2

1

4 1

Phas

e (d

eg

90

180

n n

G j G jz p

n

2 2

1 2 1 2

2

tan

0



n n Frequency (rad/sec)n n n n n

Bode Diagrams of Poles and ZerosBode Diagrams of Poles and ZerosBode Diagrams of Poles and Zeros Bode Diagrams of Poles and Zeros Bode Diagrams of stable complex zeros are Bode Diagrams of stable complex zeros are the mirror images of the Bode diagrams of the mirror images of the Bode diagrams of

40

B)the identical stable complex poles w.r.t. the the identical stable complex poles w.r.t. the 0 dB line and the 0 deg line, respectively.0 dB line and the 0 deg line, respectively.

Let G s( ) 120

0

20

Mag

nitu

de (d

B

Let G sG s

G jG j

pz

pz

( )( )

( )( )

RS|

1

-40

-20

180

G j G j

G j G j

p z

p z

( ) ( )

log ( ) log ( )

ST|

R

S|20 2010 10

e j c h 0

80

hase

(deg

)

G j G jp z( ) ( )

ST|

-180

Ph



Frequency (rad/sec)

n nn

2nd Order Bode Diagram Summary2nd Order Bode Diagram Summary2nd Order Bode Diagram Summary2nd Order Bode Diagram Summary•• 2nd Order Complex Poles2nd Order Complex Poles

2

( ) 1 0nG

•• 2nd Order Complex Zeros2nd Order Complex Zeros22 2s s

–– Break FrequencyBreak Frequency

2 22( ) , 1 02

np

n n

G ss s

rad/sb n –– Break FrequencyBreak Frequency

2 22( ) , 1 0n n

zn

s sG s

d/–– Mag. Plot ApproximationMag. Plot Approximation

0 dB from DC to 0 dB from DC to nn and a straight line with and a straight line with 40 dB/decade slope after 40 dB/decade slope after nn. Peak value . Peak value occurs at:occurs at:

b n

–– Mag. Plot ApproximationMag. Plot Approximation0 dB from DC to 0 dB from DC to nn and a straight line with and a straight line with 40 dB/decade slope after40 dB/decade slope after

rad/sb n

occurs at:occurs at:

–– Phase Plot ApproximationPhase Plot Approximation

40 dB/decade slope after 40 dB/decade slope after nn..–– Phase Plot ApproximationPhase Plot Approximation

0 deg from DC to . Between 0 deg from DC to . Between and and nn , a straight line from 0 deg to 180 , a straight line from 0 deg to 180

15 n

15 n

2

2

1 2 1( )

2 1

r n

p r MAXG j

pppp0 deg from DC to . Between 0 deg from DC to . Between and and n n , a straight line from 0 deg to , a straight line from 0 deg to 180 180 deg (passing deg (passing 90 deg at 90 deg at nn). For frequency ). For frequency hi h hhi h h i h lii h li 180 d180 d

15 n

15 n

n n , g g, g g

deg (passing 90 deg at deg (passing 90 deg at nn). For frequency ). For frequency higher than higher than nn , straight line on 180 deg., straight line on 180 deg.



higher than higher than nn , straight line on , straight line on 180 deg.180 deg.

2nd Order System Frequency Response2nd Order System Frequency Response2nd Order System Frequency Response2nd Order System Frequency Response

A Closer Look:A Closer Look: 2

Frequency Response Function:Frequency Response Function:

2 22( ) , 1 02

np

n n

G ss s

2 1G

Magnitude:Magnitude: Phase:Phase:

2 22 2

2

1( ) 2 ( ) 2 1

np

n n

n n

G jj j

j

2 22 2

2

1

2 1

pG j

2

2

2

2

22

2

1

atan2 , 1

nnpG j j

The maximum value of |The maximum value of |GG((jj)| occurs at the )| occurs at the Peak (Resonant) FrequencyPeak (Resonant) Frequency r r ::n n 2,

n n

22

11 2 and ( )G j



21 2 and ( )2 1

r n p rG j

2nd Order System Frequency Response2nd Order System Frequency Response2nd Order System Frequency Response2nd Order System Frequency Response

20

40

(dB

) -20

0

eg);

Mag

nitu

de

-60

-40

Pha

se (d

e

-90

-45

0

-180

-135

0 1 10



Frequency (rad/sec)0.1n n 10n

2nd Order System Frequency Response2nd Order System Frequency Response2nd Order System Frequency Response2nd Order System Frequency ResponseA Few Observations:A Few Observations:•• ThreeThree differentdifferent characteristic frequencies:characteristic frequencies:Three Three differentdifferent characteristic frequencies:characteristic frequencies:

–– Natural FrequencyNatural Frequency ((nn))–– Damped Natural FrequencyDamped Natural Frequency ((dd): ):

R t (P k) FR t (P k) F (( )) 21 2

21d n –– Resonant (Peak) FrequencyResonant (Peak) Frequency ((rr):): 21 2r n

r d n

•• When the damping ratio When the damping ratio , there is , there is no peakno peak in the Bode magnitude plot. in the Bode magnitude plot. DO NOT confuse this with the condition for overDO NOT confuse this with the condition for over--damped and underdamped and under--damped damped systems: when systems: when the system is underthe system is under--damped (has overshoot) and when damped (has overshoot) and when th t ith t i d d ( h t)d d ( h t)the system is overthe system is over--damped (no overshoot).damped (no overshoot).

•• As As , , r r nn and and GG((jj)) increases; also the phase transition from increases; also the phase transition from 0 deg to0 deg to 180 deg becomes sharper180 deg becomes sharper



0 deg to 0 deg to 180 deg becomes sharper.180 deg becomes sharper.


Transfer Function:Transfer Function: G s s s( ) ( )

2000 252


G ss s s s

( )( )( ) 200 10 25002



ExampleExample

20

40

ExampleExample

60

-40

-20

0

Mag

nitu

de (d

B)

-80

-60

180

0

90

180

se (d

eg)

270

-180

-90Pha

s



10 -1 100 101102 10 3

-270

Frequency (rad/sec)

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