Baaode Irwin8e

Variable-Frequency Response AnalysisNetwork performance as function of frequency.Transfer function

Sinusoidal Frequency AnalysisBode plots to display frequency response data

Filter NetworksNetworks with frequency selective characteristics:low-pass, high-pass, band-pass

VARIABLE-FREQUENCY NETWORKPERFORMANCE

LEARNING GOALS

SINUSOIDAL FREQUENCY ANALYSIS

)(sH

Circuit represented bynetwork function

++

)cos(0

)(0

tBeA tj ( )

++

+

)(cos|)(|)(

0

)(0

jHtjHBejHA tj

)()()(

)()(|)(|)(

jeMjH

jHjHM

===

Notation

stics.characteri phase and magnitudecalledgenerally are offunctionas of Plots ),(),(M

)(log)(

))(log2010

10

PLOTS BODE vs(M

. of function a as function network theanalyzewefrequency theof functionaasnetworkaofbehavior thestudy To

)( jH

HISTORY OF THE DECIBEL

Originated as a measure of relative (radio) power

1

2)2 log10(| PPP dB =1Pover

21

22

21

22)2

22 log10log10(|

II

VVP

RVRIP dB ==== 1Pover

By extension

||log20|||log20|||log20|

10

10

10

GGIIVV

dB

dB

dB

===

Using log scales the frequency characteristics of network functionshave simple asymptotic behavior.The asymptotes can be used as reasonable and efficient approximations

Poles and Zeros and Transfer Functions

Transfer Function: A transfer function is defined as the ratio of the Laplacetransform of the output to the input with all initial conditions equal to zero. Transfer functions are definedonly for linear time invariant systems.

Considerations: Transfer functions can usually be expressed as the ratioof two polynomials in the complex variable, s.

Factorization: A transfer function can be factored into the following form.

)(...))(()(...))(()(

21

21

n

m

pspspszszszsKsG +++

+++=

The roots of the numerator polynomial are called zeros.

The roots of the denominator polynomial are called poles.

wlg

Poles, Zeros and the S-Plane

An Example: You are given the following transfer function. Show thepoles and zeros in the s-plane.

)10)(4()14)(8()( ++

++=ssssssG

S - plane

xxoxo0-4-8-10-14

origin

axis

j axis

wlg

Poles, Zeros and Bode Plots

Characterization: Considering the transfer function of the previous slide. We note that we have 4 differenttypes of terms in the previous general form:These are:

)1/(,)1/(

1,1, ++ zspssKB

Expressing in dB: Given the tranfer function:

)1/)(()1/()( +

+=pjwjwzjwKjwG B

|1/|log20||log20|)1/(|log20log20|(|log20 +++= pjwjwzjwKjwG B

wlg


Mechanics: We have 4 distinct terms to consider:

20logKB

20log|(jw/z +1)|

-20log|jw|

-20log|(jw/p + 1)|

wlg


Mechanics: The gain term, 20logKB, is just so manydB and this is a straight line on Bode paper,independent of omega (radian frequency).

The term, - 20log|jw| = - 20logw, when plottedon semi-log paper is a straight line sloping at -20dB/decade. It has a magnitude of 0 at w = 1.

0

20

-20

=1

-20db/dec

wlg


Mechanics: The term, - 20log|(jw/p + 1), is drawn with the following approximation: If w < p we use theapproximation that 20log|(jw/p + 1 )| = 0 dB,a flat line on the Bode. If w > p we use the approximation of 20log(w/p), which slopes at-20dB/dec starting at w = p. Illustrated below.It is easy to show that the plot has an error of-3dB at w = p and 1 dB at w = p/2 and w = 2p.One can easily make these corrections if it is appropriate.

0

20

-20

-40

= p

-20db/dec

wlg


0

20

-20

-40

= z

+20db/dec

Mechanics: When we have a term of 20log|(jw/z + 1)| weapproximate it be a straight line of slop 0 dB/decwhen w < z. We approximate it as 20log(w/z)when w > z, which is a straight line on Bode paperwith a slope of + 20dB/dec. Illustrated below.

wlg

Example 1:

Given: 50,000( 10)( )( 1)( 500)

jwG jwjw jw

+= + +First: Always, always, always get the poles and zeros in a form such that

the constants are associated with the jw terms. In the above example we do this by factoring out the 10 in the numerator and the 500 in thedenominator.

50,000 10( /10 1) 100( /10 1)( )500( 1)( / 500 1) ( 1)( / 500 1)

x jw jwG jwjw jw jw jw

+ += =+ + + +Second: When you have neither poles nor zeros at 0, start the Bode

at 20log10K = 20log10100 = 40 dB in this case.

wlg

Example 1: (continued)

Third: Observe the order in which the poles and zeros occur.This is the secret of being able to quickly sketch the Bode.In this example we first have a pole occurring at 1 whichcauses the Bode to break at 1 and slope 20 dB/dec.Next, we see a zero occurs at 10 and this causes aslope of +20 dB/dec which cancels out the 20 dB/dec,resulting in a flat line ( 0 db/dec). Finally, we have apole that occurs at w = 500 which causes the Bodeto slope down at 20 dB/dec.

We are now ready to draw the Bode.

Before we draw the Bode we should observe the rangeover which the transfer function has active poles and zeros.This determines the scale we pick for the w (rad/sec)at the bottom of the Bode.

The dB scale depends on the magnitude of the plot and experience is the best teacher here. wlg

1 1 1 1 1 1

(rad/sec)

dB Mag Phase (deg)

1 1 1 1 1 1

(rad/sec)

dB Mag Phase (deg)

Bode Plot Magnitude for 100(1 + jw/10)/(1 + jw/1)(1 + jw/500)

0

20

40

-20

-60

60

-60

0.1 1 10 100 1000 10000

wlg

Using Matlab For Frequency Response

Instruction: We can use Matlab to run the frequency response forthe previous example. We place the transfer functionin the form:

]500501[]500005000[

)500)(1()10(5000

2 +++=++

+ss

ssss

The Matlab Program

num = [5000 50000];den = [1 501 500];Bode (num,den)

wlg

In the following slide, the resulting magnitude and phase plots (exact)are shown in light color (blue). The approximate plot for the magnitude(Bode) is shown in heavy lines (red). We see the 3 dB errors at thecorner frequencies.

Frequency (rad/sec)

P

h

a

s

e

(

d

e

g

)

;

M

a

g

n

i

t

u

d

e

(

d

B

)

Bode Diagrams

-10

0

10

20

30

40From: U(1)

10-1 100 101 102 103 104-100

-80

-60

-40

-20

0

T

o

:

Y

(

1

)

1 10 100 500

)500/1)(1()10/1(100)(

jwjwjwjwG ++

+=Bode for:

wlg

Phase for Bode Plots

Comment: Generally, the phase for a Bode plot is not as easy to drawor approximate as the magnitude. In this course we will usean analytical method for determining the phase if we want tomake a sketch of the phase.

Illustration: Consider the transfer function of the previous example.We express the angle as follows:

)500/(tan)1/(tan)10/(tan)( 111 wwwjwG =

We are essentially taking the angle of each pole and zero.Each of these are expressed as the tan-1(j part/real part)

Usually, about 10 to 15 calculations are sufficient to determinea good idea of what is happening to the phase.

wlg

Bode PlotsExample 2: Given the transfer function. Plot the Bode magnitude.

2)100/1()10/1(100)(

ssssG +

+=Consider first only the two terms of

jw100

Which, when expressed in dB, are; 20log100 20 logw.This is plotted below.

1

0

20

40

-20

The isa tentative line we use until we encounter the first pole(s) or zero(s)not at the origin.

-20db/dec

wlg

dB

(rad/sec)

1 1 1 1 1 1

(rad/sec)

dB Mag Phase (deg)0

20

40

60

-20

-40

-601 10 100 10000.1

2)100/1()10/1(100)(

ssssG +

+=

(rad/sec)

dB Mag Phase (deg)0

20

40

60

-20

-40

-601 10 100 10000.1

Bode Plots

Example 2: (continued)

-20db/dec

-40 db/dec

2)100/1()10/1(100)(

ssssG +

+=

wlg

The completed plot is shown below.

1 1 1 1 1 1

(rad/sec)

dB Mag

Bode PlotsExample 3:

Given:3

3 2

80(1 )( )( ) (1 / 20)

jwG sjw jw

+= +

10.1 10 100

40

20

0

60

-20 .

20log80 = 38 dB

-60 dB/dec

-40 dB/dec

wlg

1 1 1 1 1 1

(rad/sec)

dB Mag Phase (deg)0

20

40

60

-20

-40

-601 10 100 10000.1

Bode Plots

-40 dB/dec

+ 20 dB/dec

Given:

Sort of a lowpass filter

Example 4:

2

2

10(1 / 2)( )(1 0.025 )(1 / 500)

jwG jwj w jw

= + +

wlg

1 1 1 1 1 1

(rad/sec)

dB Mag Phase (deg)0

20

40

60

-20

-40

-601 10 100 10000.1

Bode Plots

22

22

)1700/1()2/1()100/1()30/1()(

jwjwjwjwjwG ++

++=

-40 dB/dec

+ 40 dB/dec

Given:

Sort of a lowpass filter

Example 5

wlg

Bode Plots

Given: problem 11.15 text

)11.0()()101.0)(1(64

)10()()101.0)(1(640)(

22 +++=+

++=jwjw

jwjwjwjw

jwjwjwH

0.01 0.1 1 10 100 1000

0

20

40

-20

-40

dB mag

.

.

.

.

.

-40dB/dec

-20db/dec

-40dB/dec

-20dB/dec

Example 6

wlg

Bode Plots

Design Problem: Design a G(s) that has the following Bode plot.

dB mag

rad/sec

0

20

40

0.1 1 10 100 100030 900

30 dB

+40 dB/dec -40dB/dec

? ?

Example 7

wlg

Bode PlotsProcedure: The two break frequencies need to be found. Recall:

#dec = log10[w2/w1]

Then we have:

(#dec)( 40dB/dec) = 30 dB

log10[w1/30] = 0.75 w1 = 5.33 rad/sec

Also:

log10[w2/900] (-40dB/dec) = - 30dB

This gives w2 = 5060 rad/sec

wlg

Bode PlotsProcedure:

2 2

2 2

(1 / 5.3) (1 / 5060)( )(1 / 30) (1 / 900)

s sG ss s

+ += + +Clearing: 2 2

2 2

( 5.3) ( 5060)( )( 30) ( 900)s sG ss s+ += + +

Use Matlab and conv:

2 2 71 ( 10.6 28.1) 2 ( 10120 2.56 )N s s N s s xe= + + = + +

N = conv(N1,N2)

N1 = [1 10.6 28.1] N2 = [1 10120 2.56e+7]

1 1.86e+3 2.58e+7 2.73e+8 7.222e+8

s4 s3 s2 s1 s0

wlg

Bode PlotsProcedure: The final G(s) is given by;

Testing: We now want to test the filter. We will check it at = 5.3 rad/secAnd = 164. At = 5.3 the filter has a gain of 6 dB or about 2.At = 164 the filter has a gain of 30 dB or about 31.6.We will check this out using MATLAB and particularly, Simulink.

)29.7022.5189.91860()194.7716.2571.26.10130()(

872234

882834

esesessesesesssG ++++

++++=

wlg

Matlab (Simulink) Model:

wlg

Filter Output at = 5.3 rad/sec

Produced from Matlab Simulinkwlg

Filter Output at = 70 rad/sec

Produced from Matlab Simulinkwlg

Reverse Bode PlotRequired:

From the partial Bode diagram, determine the transfer function(Assume a minimum phase system)

dB

20 db/dec

20 db/dec

-20 db/dec30

1 110 850

68

Not to scale

wlg

Example 8

Reverse Bode Plot

Not to scale

100 dB

w (rad/sec)

50 dB

0.5

-40 dB/dec

-20 dB/dec

40

10 dB

300

-20 dB/dec

-40 dB/dec

Required:From the partial Bode diagram, determine the transfer function

(Assume a minimum phase system)

wlg

Example 9

Appendix

]...)()(21)[1(]...)()(21)[1()()( 2

233310

bbba

N

jjjjjjjKjH +++

+++=

General form of a network function showing basic terms

Frequency independent

Poles/zeros at the origin

First order terms Quadratic terms for complex conjugate poles/zeros

..|)()(21|log20|1|log20

...|)()(21|log20|1|log20

||log20log20

21010

233310110

10010

+++++++++

=

bbba jjj

jjj

jNK

DN

DN

BAAB

loglog)log(

loglog)log(

=+=

|)(|log20|)(| 10 jHjH dB=

212

1

2121

zzzz

zzzz

=+=

...)(1

2tantan

...)(1

2tantan

900)(

211

23

3311

1

+++=

b

bba

NjH

Display each basic termseparately and add theresults to obtain final answer

Lets examine each basic term

Constant Term

Poles/Zeros at the origin

==

90)(

)(log20|)(|)( 10Nj

Njj NdB

NN

linestraight a is thislogisaxis-xthe 10

Simple pole or zero j+1 =+

+=+

1

210

tan)1()(1log20|1|

jj dB

asymptotefrequency low 0|1| + dBj(20dB/dec)asymptotefrequency high 10log20|1| + dBj

frequency)akcorner/bre1whenmeet asymptotestwoThe (=Behavior in the neighborhood of the corner

FrequencyAsymptoteCurvedistance to asymptote Argument

corner 0dB 3dB 3 45

octave above 6dB 7db 1 63.4

octave below 0dB 1dB 1 26.6

1=2=5.0=

+ 0)1( j+ 90)1( j

1

Asymptote for phase

High freq. asymptoteLow freq. Asym.

Simple zero

Simple pole

Quadratic pole or zero ])()(21[ 22 jjt ++= ])()(21[ 2 += j( ) ( )222102 2)(1log20|| +=dBt 212 )(1 2tan = t

1 asymptote freq. high 2102 )(log20|| dBt 1802t

1= )2(log20|| 102 =dBt = 902tCorner/break frequency221 = 2102 12log20|| =dBt

21

221tan = t

22Resonance frequency

Magnitude for quadratic pole Phase for quadratic pole

dB/dec40

These graphs are inverted for a zero

LEARNING EXAMPLE Generate magnitude and phase plots

)102.0)(1()11.0(10)( ++

+=

jjjjGvDraw asymptotes

for each term1,10,50 :nersBreaks/cor

40

20

0

20

dB

90

90

1.0 1 10 100 1000

dB|10

decdB /20

dec/45

decdB /20

dec/45

Draw composites

asymptotes

LEARNING EXAMPLE Generate magnitude and phase plots

)11.0()()1(25)( 2 +

+=

jjjjGv 101,:(corners)Breaks

40

20

0

20

dB

90

270

90

1.0 1 10 100

Draw asymptotes for each

dB28

decdB /40

180

dec/45

45

Form composites

( )21020 0)( KjK

dB

==

Final results . . . And an extra hint on poles at the origin

decdB40

decdB20

decdB40

LEARNING EXTENSION Sketch the magnitude characteristic

)100)(10()2(10)(

4

+++=

jj

jjGformstandardinNOTisfunctiontheBut

10010,2,:breaks

Put in standard form)1100/)(110/(

)12/(20)( +++=

jj

jjG We need to show about 4 decades

40

20

0

20

dB

90

90

1 10 100 1000

dB|25


2)(102.0(100)(

j

jjG +=origin theat pole Double

50at breakformstandardinisIt

40

20

0

20

dB

90

270

90

1 10 100 1000

Once each term is drawn we form the composites

Put in standard form

)110/)(1()( ++=

jjjjG


)10)(1(10)( ++=

jjjjG

10 1, :breaksorigin theat zero

formstandardinnot

40

20

0

20

dB

90

270

90

1.0 110

100Once each term is drawn we form the composites

decdB /20decdB /20

LEARNING EXAMPLE A function with complex conjugate poles

[ ]1004)()5.0( 25)( 2 +++= jjj jjGPut in standard form [ ]125/)10/()15.0/( 5.0)( 2 +++= jjj jjG

40

20

0

20

dB

90

90

01.0 1.0 1 10 100270

1= )2(log20|| 102 =dBt

2.01.025/12 =

==

])()(21[2

2 jjt ++=

dB8

Draw composite asymptote

Behavior close to corner of conjugate pole/zerois too dependent on damping ratio.Computer evaluation is better

Evaluation of frequency response using MATLAB

[ ]1004)()5.0( 25)( 2 +++= jjj jjG num=[25,0]; %define numerator polynomial den=conv([1,0.5],[1,4,100]) %use CONV for polynomial multiplicationden =

1.0000 4.5000 102.0000 50.0000 freqs(num,den)

Using default options

Evaluation of frequency response using MATLAB User controlled

>> clear all; close all %clear workspace and close any open figure>> figure(1) %open one figure window (not STRICTLY necessary)>> w=logspace(-1,3,200);%define x-axis, [10^{-1} - 10^3], 200pts total

[ ]1004)()5.0( 25)( 2 +++= jjj jjG

>> G=25*j*w./((j*w+0.5).*((j*w).^2+4*j*w+100)); %compute transfer function>> subplot(211) %divide figure in two. This is top part>> semilogx(w,20*log10(abs(G))); %put magnitude here

>> grid %put a grid and give proper title and labels>> ylabel('|G(j\omega)|(dB)'), title('Bode Plot: Magnitude response')

>> semilogx(w,unwrap(angle(G)*180/pi)) %unwrap avoids jumps from +180 to -180>> grid, ylabel('Angle H(j\omega)(\circ)'), xlabel('\omega (rad/s)')>> title('Bode Plot: Phase Response')

Evaluation of frequency response using MATLAB User controlled Continued

Repeat for phase

No xlabel here to avoid clutter

USE TO ZOOM IN A SPECIFIC REGION OF INTEREST

Compare with default!


]136/)12/[()1(2.0)( 2 ++

+=

jjjjjG 6/136/12

12/1==

=

])()(21[ 22 jjt ++=

40

20

0

20

dB

90

270

90

1.0 110

100

1= )2(log20|| 102 =dBt

decdB /20

decdB /40

decdB /0

12

dB5.9=

]136/)12/[()1(2.0)( 2 ++

+=

jjjjjG

num=0.2*[1,1]; den=conv([1,0],[1/144,1/36,1]); freqs(num,den)

DETERMINING THE TRANSFER FUNCTION FROM THE BODE PLOT

This is the inverse problem of determining frequency characteristics. We will use only the composite asymptotes plot of the magnitude to postulatea transfer function. The slopes will provide information on the order

A

A. different from 0dB.There is a constant Ko

B

B. Simple pole at 0.11)11.0/( +j

C

C. Simple zero at 0.5

)15.0/( +j

D

D. Simple pole at 3

1)13/( +j

E

E. Simple pole at 20

1)120/( +j)120/)(13/)(11.0/(

)15.0/(10)( ++++=

jjj

jjG

20|

00

0

1020|dBK

dB KK ==

If the slope is -40dB we assume double real pole. Unless we are given more data

LEARNING EXTENSIONDetermine a transfer function from the composite magnitude asymptotes plot

A

A. Pole at the origin. Crosses 0dB line at 5

j5

B

B. Zero at 5

C

C. Pole at 20

D

D. Zero at 50

E

E. Pole at 100

)1100/)(120/()150/)(15/(5)( ++

++= jjj

jjjG

Sinusoidal

(non-inverting op-amp)DESIGN EXAMPLE BASS-BOOST AMPLIFIER DESIRED BODE PLOT

OPEN SWITCH

(6dB)

5002P

f =

Switch closed??

DESIGN EXAMPLE TREBLE BOOSTOriginal player response Desired boost

Proposed boost circuit

Non-inverting amplifier

Design equations

Filters

Documents

Baaode Irwin8e