6 - frequency response (read-only)

06/03/2017

© 2017 University of the West of England, Bristol. 1

UWE Bristol

Industrial ControlUFMF6W-20-2

Control Systems EngineeringUFMEUY-20-3

Lecture 6: Frequency ResponseBode Plots and Nyquist Diagrams

Today’s Lecture

• Frequency Response• Amplitude Ratio and Phase• System Analysis using

– Bode Plot– Nyquist Diagram

• System Identification

Frequency Response

• Up to now – Time response– How the system behaves to a specific input

with respect to time• Now – Frequency Response

– System response to a sinusoidal input– Range of frequencies used – Used for system identification– Used for stability analysis (Next week)

Frequency Response

• System behaviour determined from the steady state response to a sinusoidal input in the form:

• Sine wave used:– easy to analyse– easy to generate– easy to measure experimentally

tAR wsin=

Frequency Response

• Sinusoidal applied to LINEAR system:– output will be sinusoidal– output amplitude proportional to input

amplitude– harmonic input produces harmonic output at

same frequency• Variation in Amplitude and Phase• Function of Frequency

Frequency Response

G(s)Input Output

tA wsin

06/03/2017


Frequency Response

G(s)Input Output

tA wsin ( )fw +tBsin

Frequency Response

• Amplitude Ratio (AR)

AB

G(s) Output

( )fw +tBsinInput

tA wsin

AB

=AR [dimensionless]

Frequency ResponseG(s) Output

( )fw +tBsinInput

tA wsin!360One Period =

Frequency Response

• Phase Shift (angle)• If output follows input à LAG à negative

G(s) Output

( )fw +tBsinInput

tA wsin

f [degrees]

f

Frequency Response

• Data gained experimentally• Can be gained from OL transfer function

– Input:– R can be represented as: – Take first derivative:

– Replace s terms in TF with jω• Moving from s-domain to frequency domain

tAR wsin=

tjAeR w= ( )1-=j

tjAjRsdtdR ww==

tAR wsin=

Frequency Response

• First order system:

• Second order system:

( ) ( )wt

wt j

jGs

sG+

=®+

=11

11

( ) ( ) 22

2

22

2

22 nn

n

nn

n

jjG

sssG

wwzwwww

wzww

++-=®

++=

Note: ω is driving frequency, ωn is natural frequency

06/03/2017


Frequency Response

• Consider second order transfer function:

• Substitute for s132

5)( 2 ++=

sssG

( ) ( ) 1325)( 2 ++

=ww

wjj

jG

1325)( 2 ++-

=ww

wj

jG( ) ( )ww

w321

5)( 2 jjG

+-=

Frequency Response• Complex quantity that depends on frequency:

• Multiply by complex conjugate to separate real and imaginary parts:

( ) ( )www

3215)( 2 j

jG+-

=

( ) ( )( ) ( )( ) ( )úû

ùêë

é----

úû

ùêë

é+-

=wwww

www

321321

3215)( 2

2

2 jj

jjG

( ) ( )[ ]( ) ( )( ) ( )( ) ( )( ) ( )( )wwwwwww

www3321321321

3215)(2222

2

jjjjjjG

-+---+-

--=

( ) ( )[ ]242

2

94413215)(wwwwww

++---

=jjG

15415

154105)( 2424

2

++-

++-

=www

wwww jjG

Frequency Response• Complex number in form:

• Real part:

• Imaginary part:

15415

154105)( 2424

2

++-

++-

=www

wwww jjG

154105Re 24

2

++-

=www

15415Im 24 ++

-=www

ImRe j+

Frequency Response• From complex number theory:• Amplitude ratio:

• Phase angle:

• Used to generate a table of results

22 ImRe +=AR

÷øö

çèæ= -

ReImtan 1f

Frequency Response

Frequency(Rad/s) Real Imaginary AR Phase0.01 4.9965 –0.1499 4.9988 –1.7187

0.02 4.9860 –0.2994 4.9950 –3.4364

0.03 4.9686 –0.4480 4.9888 –5.1520

0.05 4.9135 –0.7407 4.9690 –8.5730

0.1 4.6649 –1.4280 4.8786 –17.0205

0.2 3.8130 –2.4867 4.5522 –33.1113

0.3 2.7658 –3.0356 4.1066 –47.6630

0.5 1.0000 –3.0000 3.1623 –71.5651

1 -0.5000 –1.5000 1.5811 –108.4349

2 -0.4118 –0.3529 0.5423 –139.3987

3 -0.2297 –0.1216 0.2599 –152.1027

5 -0.0933 –0.0286 0.0976 –162.9795

10 -0.0246 –0.0037 0.0248 –171.4270

Representing Data

• Some methods– Nyquist Diagram– Bode Plot

06/03/2017


Nyquist Diagram

• Polar plot on Argand diagram of open loopfrequency locus (i.e. G(jω)H(jω))

• Amplitude ratio is radial coordinate.• Phase angle is the angular coordinate.

Nyquist Diagram

• Example154

15154

105)( 2424

2

++-

++-

=www

wwww jjG

Frequency(Rad/s) AR Phase

0.01 4.9988 –1.7187

0.02 4.9950 –3.4364

0.03 4.9888 –5.1520

0.05 4.9690 –8.5730

0.1 4.8786 –17.0205

0.2 4.5522 –33.1113

0.3 4.1066 –47.6630

0.5 3.1623 –71.5651

1 1.5811 –108.4349

2 0.5423 –139.3987

3 0.2599 –152.1027

5 0.0976 –162.9795

10 0.0248 –171.4270

Nyquist Diagram

• Example154

15154

105)( 2424

2

++-

++-

=www

wwww jjG


0.01 4.9988 –1.7187

0.02 4.9950 –3.4364

0.03 4.9888 –5.1520

0.05 4.9690 –8.5730

0.1 4.8786 –17.0205

0.2 4.5522 –33.1113

0.3 4.1066 –47.6630

0.5 3.1623 –71.5651

1 1.5811 –108.4349

2 0.5423 –139.3987

3 0.2599 –152.1027

5 0.0976 –162.9795

10 0.0248 –171.4270

Nyquist Diagram

• Example154

15154

105)( 2424

2

++-

++-

=www

wwww jjG


0.01 4.9988 –1.7187

0.02 4.9950 –3.4364

0.03 4.9888 –5.1520

0.05 4.9690 –8.5730

0.1 4.8786 –17.0205

0.2 4.5522 –33.1113

0.3 4.1066 –47.6630

0.5 3.1623 –71.5651

1 1.5811 –108.4349

2 0.5423 –139.3987

3 0.2599 –152.1027

5 0.0976 –162.9795

10 0.0248 –171.4270

Bode Plots154

15154

105)( 2424

2

++-

++-

=www

wwww jjG

• Two logarithmic plots– Amplitude ratio vs frequency– Phase angle vs frequency

Bode Plots

• Amplitude measured in dB, converted by:( ) ( )ARAR log20dB =Frequency

(Rad/s) AR AR(dB)

0.01 4.9988 13.98

0.02 4.9950 13.97

0.03 4.9888 13.96

0.05 4.9690 13.93

0.1 4.8786 13.77

0.2 4.5522 13.16

0.3 4.1066 12.27

0.5 3.1623 10.00

1 1.5811 3.98

2 0.5423 –5.32

3 0.2599 –11.70

5 0.0976 –20.21

10 0.0248 –32.11

06/03/2017


Bode Plots

• Phase measured in degrees:Frequency(Rad/s) Phase

0.01 –1.7187

0.02 –3.4364

0.03 –5.1520

0.05 –8.5730

0.1 –17.0205

0.2 –33.1113

0.3 –47.6630

0.5 –71.5651

1 –108.4349

2 –139.3987

3 –152.1027

5 –162.9795

10 –171.4270

Bode Plots

• Constructing Bode plots– Sections of TF can be represented as straight

lines = asymptotic approximations– Example:

Bode Plots

• Each part of TF has certain type of frequency response:

• Building blocks:– Gain– Differentiator– Integrator– First order lead/lag– Second order lead/lag

K

s

s1

st+11

st+1

121

2

2 ++ ssnn wz

w

122

2 ++ ss

nn wz

w

Bode Plots

0

dB

0

phase

frequency

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

1 Rad/s

+90

1 Rad/s

-90

1/t

1/t

wn

+20 dB/decade

+20 dB/decade

-20 dB/decade

-20 dB/decade

+40 dB/decade

0 dB/decade

0 dB/decade

0 dB/decade

0 dB/decade

+90

-90

1/t

1/t

0.1/t 1/10t

wn

+180+90

0

dB

0

phase

frequency

wn

-40 dB/decade

0 dB/decade wn

-180-90

Gain: G(s) = KAR(db) = 20 log KPhase = 0°

Differentiator: G(s) = s+20 dB/decade risePhase = +90°

Integrator: G(s) = 1/s–20 dB/decade fallPhase = –90°

SEE HANDOUT ON BLACKBOARD

Bode Plots

0

dB

0

phase

frequency

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

1 Rad/s

+90

1 Rad/s

-90

1/t

1/t

wn

+20 dB/decade

+20 dB/decade

-20 dB/decade

-20 dB/decade

+40 dB/decade

0 dB/decade

0 dB/decade

0 dB/decade

0 dB/decade

+90

-90

1/t

1/t

0.1/t 1/10t

wn

+180+90

0

dB

0

phase

frequency

wn

-40 dB/decade

0 dB/decade wn

-180-90

1st order lead: G(s) =Corner frequency: 1/tTrue curve 3dB above break point

1st order lag:

G(s) =

Corner frequency: 1/tTrue curve 3dB below break point

st+1

st+11


Bode Plots

0

dB

0

phase

frequency

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

0

dB

0

phase

frequency

1 Rad/s

+90

1 Rad/s

-90

1/t

1/t

wn

+20 dB/decade

+20 dB/decade

-20 dB/decade

-20 dB/decade

+40 dB/decade

0 dB/decade

0 dB/decade

0 dB/decade

0 dB/decade

+90

-90

1/t

1/t

0.1/t 1/10t

wn

+180+90

0

dB

0

phase

frequency

wn

-40 dB/decade

0 dB/decade wn

-180-90

2st order lead:

G(s) =

Corner frequency: wnTrue curve at break point depends on z

2st order lag:

G(s) =

Corner frequency: wnTrue curve at break point depends on z

122

2 ++ ss

nn wz

w

121

2

2 ++ ssnn wz

w


06/03/2017


Bode Plot

• Example:• Draw the Bode diagram for the following

transfer function:

( ) ( )( ) ( ) ( )( )wwww

jjjj

EB

ssss

EB

05.012.015

5.012.015

++=®

++=

( ) ( ) ( )22 05.01log202.01log20log205log20dB www +-+--=AR

( ) ( )wwf 05.0tan2.0tan90 11 -- ---= !

Bode Plot Example

• Take each term separately:

dB145log20(a) 1 ==R

14R1

wlog20(b) 2 -=R

01=w

slope = -20 dB/dec

R2

Bode Plot Example2

3 )2.0(1log20(c) w+-=R

rad/s52.011atpointbreak ===

tw

dB/dec0sloperad/s5for =<w

dB/dec20sloperad/s5for -=>w

05=w

slope = -20 dB/dec

R3

Bode Plot Example

24 )05.0(1log20(d) w+-=R

rad/s2005.01atpointbreak ==w

dB/dec0sloperad/s20for =<w

dB/dec20sloperad/s20for -=>w

020=w

slope = -20 dB/dec

R4

(e) combining plots:

0

slopes = -20 dB/decR

R1

R2 R3 R4

1 5 20

Bode Plot Example

0

slope = -20 dB/dec

R

R1

R2

1 5 20

rad/s5for <w

Bode Plot Example

06/03/2017


0

slope = -20 dB/dec

R

1 5 20

rad/s20for <w

slope = -40 dB/dec

Bode Plot Example

0

slope = - 20 dB/dec

R

1 5 20

rad/s20for >w

slope = - 40 dB/dec

slope = - 60 dB/dec

Bode Plot Example

Phase lag given by:

321 ffff ++=

ω – rad/s Φ1 - deg Φ2 - deg Φ3 - deg Φ - deg1 -90 -11 -3 -104

5 -90 -45 -7 -149

10 -90 -63 -27 -18020 -90 -76 -45 -211

50 -90 -84 -68 -242

Bode Plot Example

( ) ( )wwf 05.0tan2.0tan90 11 -- ---= !

-20

.1 1 10 100Frequency – rad/s

0

40

20

-40

-60-100°

-140°

-180°

-220°

R -d

B

phas

e

5 20

R

Φ

Bode Plot Example

System Identification

• Can use Bode Plot to estimate transfer function

• Use same rules as before• Example


• Step 1: Plot amplitude and phase data on Bode plot

• Step 2: Fit asymptotic approximations to amplitude data– slopes must be 0 dB/dec, ±20 dB/dec,

±40 dB/dec etc.– approximations must pass through the data

points at the lowest and highest frequencies• Step 3: Use relationships to determine

values for transfer function

06/03/2017


System IdentificationFollowing experimental data obtained from a frequency response test undertaken on control system

Freq - (rad/s) R - dB Phase - deg2 8 -1126 -2.5 -21212 10 -15230 26 -20290 52 -245

200 72 -260


-20

1 10 100 1000Frequency – rad/s

0

10

-10

-40

-60

-100°

-140°

-180°

-260°

R -d

B

phas

e

-30

-80

-50

-70 -220°

20

0 dB/dec

-20 dB/dec-40 dB/dec

-20 dB/dec

-40 dB/dec -60 dB/dec

6 dB


• Amplitude: Asymptotic slopes– low frequency = -20dB/decade– high frequency = -60dB/decade– no intermediate asymptotes

• Low frequency amplitude and phase– -20dB/decade and phase à -90°

Indicates a integrator term 1𝑠

System Identification• High frequency amplitude and phase

– -60 dB/decade slope and phase à 270°

• Break point– Only one internal break point at intersection of low and

high frequency asymptotes• slope change = -40dB/dec• break point frequency = 20 rad/s• difference between point of intersection and true curve = -6dB

Indicates a 3rd order system

Indicates two coincident 1st order lags


• System transfer function:– integrator and two coincident first order lags:

• Transfer function terms– Break point frequency:

TF =𝐾

𝑠 1 + 𝜏𝑠 )

𝜔𝑏𝑝 =-.→ 𝜏 = -

012= -

)3= 0.05 sec


• Transfer function terms– Integral term supposed to cross zero at 1

rad/s– In this case, integral term crosses at 5 rad/s à integrator is 14 dB higher

– Therefore, gain: ( )( )

01.5107.0logdB14log20

7.0 ==

==

KKK

TF =5.01

𝑠 1 + 0.05𝑠 )

06/03/2017


Today’s Lecture

• Introduction to Frequency response• Determining Amplitude Ratio and Phase• Representing diagrammatically

– Nyquist– Bode

• Plotting frequency response• System Identification

6 - frequency response (read-only)

Documents