Frequency Response (I&N Chap 12) - UBC ECE |...

22
Slide 3.1 Frequency Response (I&N Chap 12) Introduction & TFs Decibel Scale & Bode Plots Resonance Scaling Filter Networks Applications/Design Frequency response; based on slides by J. Yan

Transcript of Frequency Response (I&N Chap 12) - UBC ECE |...

Slide 31

Frequency Response

(IampN Chap 12)

bull Introduction amp TFs

bull Decibel Scale amp Bode Plots

bull Resonance

bull Scaling

bull Filter Networks

bull ApplicationsDesign

Frequency response based on slides by J Yan

Slide 32

RC Circuit AC Response Letrsquos motivate this topic with a simple RC circuit example

If R=1Ω and C=1F determine the forced response

Time Domain Freq Domain

Input vs(t) Output vo(t) (steady-state)

cos(t) V

5cos(t-20ordm) V

1 V

cos(10t) V

cos(100t) V

Frequency response based on slides by J Yan

Slide 33

RC Example Comments bull From AC analysis if the input frequency is known only the

magnitude and phase of the output need to be computed

bull Using linearity and time-invariance only the magnitude gain and phase difference need to be computed This is evident when comparing the outputs to the 1st and 2nd inputs The gain in both cases is 2-frac12asymp0707 and the relative phase is -45ordm

bull Both the magnitude gain and phase difference vary with frequency For which frequency is the capacitor voltage identical to that of the source input and why

Frequency response based on slides by J Yan

Slide 34

Frequency Response The frequency response of a circuit is the variation in its output behaviour with change in input signal frequency These variations occur in the magnitude gain and phase difference

bull Info about both quantities (magnitude gain and phase difference) can be computed directly from the relevant transfer function by evaluating it at s=jω What is the TF for our example on slide 32

bull The frequency response is often represented graphically by plots of these quantities on a graph using frequency as the independent variable

Frequency response based on slides by J Yan

Slide 35

Freq Resp for Passive Elements

CjsCZC ω

11 == LjsLZL ω==RZR =Transfer Function

Magnitude Plot

Phase Plot

Resistor Capacitor Inductor

Impedance is the TF relating voltage to current Z(s)=V(s)I(s) Consider the impedance for each passive element

Frequency response based on slides by J Yan

Slide 36

More on Transfer Functions We defined the transfer function as a ratio of two s-domain signals in a system assuming the system is ldquorelaxedrdquo (ie ICs are zero)

φωω jejHjHsD

sN

sX

sYsH )()(

)(

)(

)(

)()( ===

bull H(jω)is the magnitude gain while φ is the phase difference of Y(s)relative to X(s) for frequency ω

bull The TFs in EE253 can all be expressed as a rational polynomial in s (NB A pure delay is a practical example not having this property)

bull A value of s that makes N(s)=0 is a zero of H(s) bull A value of s that makes D(s)=0 is a pole of H(s) bull Express TFs in terms of s rather than ω This makes it easier to perform

algebraic manipulation and to determine poles and zerosbull By our definition of the TF the s-domain input signal neednrsquot be part of

the TF however if it does appear it is in the denominator of the TF In this case D(s)=0 is the system characteristic equation

Frequency response based on slides by J Yan

Slide 37

RC Circuit Revisited Revisiting the RC circuit example

1

1

)1(

1)( 0

+=

+==

sCRsCR

sC

V

VsH

s

Frequency response based on slides by J Yan

Slide 38

Example Find H(s)=Vo(s)Ii(s) the poles and zeros Sketch the freq resp

Frequency response based on slides by J Yan

Slide 39

Logarithm Properties

Logarithms will be used to more quickly and systematically sketch the frequency response graphs Itrsquos worthwhile to review a few of their properties

1 log(P1P2)=log(P1)+log(P2)2 log(P1P2)=log(P1)-log(P2)3 log(Pn)=nlog(P)4 log(01)=-1 log(1)=0 log(10)=1 log(100)=2 hellip

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 32

RC Circuit AC Response Letrsquos motivate this topic with a simple RC circuit example

If R=1Ω and C=1F determine the forced response

Time Domain Freq Domain

Input vs(t) Output vo(t) (steady-state)

cos(t) V

5cos(t-20ordm) V

1 V

cos(10t) V

cos(100t) V

Frequency response based on slides by J Yan

Slide 33

RC Example Comments bull From AC analysis if the input frequency is known only the

magnitude and phase of the output need to be computed

bull Using linearity and time-invariance only the magnitude gain and phase difference need to be computed This is evident when comparing the outputs to the 1st and 2nd inputs The gain in both cases is 2-frac12asymp0707 and the relative phase is -45ordm

bull Both the magnitude gain and phase difference vary with frequency For which frequency is the capacitor voltage identical to that of the source input and why

Frequency response based on slides by J Yan

Slide 34

Frequency Response The frequency response of a circuit is the variation in its output behaviour with change in input signal frequency These variations occur in the magnitude gain and phase difference

bull Info about both quantities (magnitude gain and phase difference) can be computed directly from the relevant transfer function by evaluating it at s=jω What is the TF for our example on slide 32

bull The frequency response is often represented graphically by plots of these quantities on a graph using frequency as the independent variable

Frequency response based on slides by J Yan

Slide 35

Freq Resp for Passive Elements

CjsCZC ω

11 == LjsLZL ω==RZR =Transfer Function

Magnitude Plot

Phase Plot

Resistor Capacitor Inductor

Impedance is the TF relating voltage to current Z(s)=V(s)I(s) Consider the impedance for each passive element

Frequency response based on slides by J Yan

Slide 36

More on Transfer Functions We defined the transfer function as a ratio of two s-domain signals in a system assuming the system is ldquorelaxedrdquo (ie ICs are zero)

φωω jejHjHsD

sN

sX

sYsH )()(

)(

)(

)(

)()( ===

bull H(jω)is the magnitude gain while φ is the phase difference of Y(s)relative to X(s) for frequency ω

bull The TFs in EE253 can all be expressed as a rational polynomial in s (NB A pure delay is a practical example not having this property)

bull A value of s that makes N(s)=0 is a zero of H(s) bull A value of s that makes D(s)=0 is a pole of H(s) bull Express TFs in terms of s rather than ω This makes it easier to perform

algebraic manipulation and to determine poles and zerosbull By our definition of the TF the s-domain input signal neednrsquot be part of

the TF however if it does appear it is in the denominator of the TF In this case D(s)=0 is the system characteristic equation

Frequency response based on slides by J Yan

Slide 37

RC Circuit Revisited Revisiting the RC circuit example

1

1

)1(

1)( 0

+=

+==

sCRsCR

sC

V

VsH

s

Frequency response based on slides by J Yan

Slide 38

Example Find H(s)=Vo(s)Ii(s) the poles and zeros Sketch the freq resp

Frequency response based on slides by J Yan

Slide 39

Logarithm Properties

Logarithms will be used to more quickly and systematically sketch the frequency response graphs Itrsquos worthwhile to review a few of their properties

1 log(P1P2)=log(P1)+log(P2)2 log(P1P2)=log(P1)-log(P2)3 log(Pn)=nlog(P)4 log(01)=-1 log(1)=0 log(10)=1 log(100)=2 hellip

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 33

RC Example Comments bull From AC analysis if the input frequency is known only the

magnitude and phase of the output need to be computed

bull Using linearity and time-invariance only the magnitude gain and phase difference need to be computed This is evident when comparing the outputs to the 1st and 2nd inputs The gain in both cases is 2-frac12asymp0707 and the relative phase is -45ordm

bull Both the magnitude gain and phase difference vary with frequency For which frequency is the capacitor voltage identical to that of the source input and why

Frequency response based on slides by J Yan

Slide 34

Frequency Response The frequency response of a circuit is the variation in its output behaviour with change in input signal frequency These variations occur in the magnitude gain and phase difference

bull Info about both quantities (magnitude gain and phase difference) can be computed directly from the relevant transfer function by evaluating it at s=jω What is the TF for our example on slide 32

bull The frequency response is often represented graphically by plots of these quantities on a graph using frequency as the independent variable

Frequency response based on slides by J Yan

Slide 35

Freq Resp for Passive Elements

CjsCZC ω

11 == LjsLZL ω==RZR =Transfer Function

Magnitude Plot

Phase Plot

Resistor Capacitor Inductor

Impedance is the TF relating voltage to current Z(s)=V(s)I(s) Consider the impedance for each passive element

Frequency response based on slides by J Yan

Slide 36

More on Transfer Functions We defined the transfer function as a ratio of two s-domain signals in a system assuming the system is ldquorelaxedrdquo (ie ICs are zero)

φωω jejHjHsD

sN

sX

sYsH )()(

)(

)(

)(

)()( ===

bull H(jω)is the magnitude gain while φ is the phase difference of Y(s)relative to X(s) for frequency ω

bull The TFs in EE253 can all be expressed as a rational polynomial in s (NB A pure delay is a practical example not having this property)

bull A value of s that makes N(s)=0 is a zero of H(s) bull A value of s that makes D(s)=0 is a pole of H(s) bull Express TFs in terms of s rather than ω This makes it easier to perform

algebraic manipulation and to determine poles and zerosbull By our definition of the TF the s-domain input signal neednrsquot be part of

the TF however if it does appear it is in the denominator of the TF In this case D(s)=0 is the system characteristic equation

Frequency response based on slides by J Yan

Slide 37

RC Circuit Revisited Revisiting the RC circuit example

1

1

)1(

1)( 0

+=

+==

sCRsCR

sC

V

VsH

s

Frequency response based on slides by J Yan

Slide 38

Example Find H(s)=Vo(s)Ii(s) the poles and zeros Sketch the freq resp

Frequency response based on slides by J Yan

Slide 39

Logarithm Properties

Logarithms will be used to more quickly and systematically sketch the frequency response graphs Itrsquos worthwhile to review a few of their properties

1 log(P1P2)=log(P1)+log(P2)2 log(P1P2)=log(P1)-log(P2)3 log(Pn)=nlog(P)4 log(01)=-1 log(1)=0 log(10)=1 log(100)=2 hellip

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 34

Frequency Response The frequency response of a circuit is the variation in its output behaviour with change in input signal frequency These variations occur in the magnitude gain and phase difference

bull Info about both quantities (magnitude gain and phase difference) can be computed directly from the relevant transfer function by evaluating it at s=jω What is the TF for our example on slide 32

bull The frequency response is often represented graphically by plots of these quantities on a graph using frequency as the independent variable

Frequency response based on slides by J Yan

Slide 35

Freq Resp for Passive Elements

CjsCZC ω

11 == LjsLZL ω==RZR =Transfer Function

Magnitude Plot

Phase Plot

Resistor Capacitor Inductor

Impedance is the TF relating voltage to current Z(s)=V(s)I(s) Consider the impedance for each passive element

Frequency response based on slides by J Yan

Slide 36

More on Transfer Functions We defined the transfer function as a ratio of two s-domain signals in a system assuming the system is ldquorelaxedrdquo (ie ICs are zero)

φωω jejHjHsD

sN

sX

sYsH )()(

)(

)(

)(

)()( ===

bull H(jω)is the magnitude gain while φ is the phase difference of Y(s)relative to X(s) for frequency ω

bull The TFs in EE253 can all be expressed as a rational polynomial in s (NB A pure delay is a practical example not having this property)

bull A value of s that makes N(s)=0 is a zero of H(s) bull A value of s that makes D(s)=0 is a pole of H(s) bull Express TFs in terms of s rather than ω This makes it easier to perform

algebraic manipulation and to determine poles and zerosbull By our definition of the TF the s-domain input signal neednrsquot be part of

the TF however if it does appear it is in the denominator of the TF In this case D(s)=0 is the system characteristic equation

Frequency response based on slides by J Yan

Slide 37

RC Circuit Revisited Revisiting the RC circuit example

1

1

)1(

1)( 0

+=

+==

sCRsCR

sC

V

VsH

s

Frequency response based on slides by J Yan

Slide 38

Example Find H(s)=Vo(s)Ii(s) the poles and zeros Sketch the freq resp

Frequency response based on slides by J Yan

Slide 39

Logarithm Properties

Logarithms will be used to more quickly and systematically sketch the frequency response graphs Itrsquos worthwhile to review a few of their properties

1 log(P1P2)=log(P1)+log(P2)2 log(P1P2)=log(P1)-log(P2)3 log(Pn)=nlog(P)4 log(01)=-1 log(1)=0 log(10)=1 log(100)=2 hellip

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 35

Freq Resp for Passive Elements

CjsCZC ω

11 == LjsLZL ω==RZR =Transfer Function

Magnitude Plot

Phase Plot

Resistor Capacitor Inductor

Impedance is the TF relating voltage to current Z(s)=V(s)I(s) Consider the impedance for each passive element

Frequency response based on slides by J Yan

Slide 36

More on Transfer Functions We defined the transfer function as a ratio of two s-domain signals in a system assuming the system is ldquorelaxedrdquo (ie ICs are zero)

φωω jejHjHsD

sN

sX

sYsH )()(

)(

)(

)(

)()( ===

bull H(jω)is the magnitude gain while φ is the phase difference of Y(s)relative to X(s) for frequency ω

bull The TFs in EE253 can all be expressed as a rational polynomial in s (NB A pure delay is a practical example not having this property)

bull A value of s that makes N(s)=0 is a zero of H(s) bull A value of s that makes D(s)=0 is a pole of H(s) bull Express TFs in terms of s rather than ω This makes it easier to perform

algebraic manipulation and to determine poles and zerosbull By our definition of the TF the s-domain input signal neednrsquot be part of

the TF however if it does appear it is in the denominator of the TF In this case D(s)=0 is the system characteristic equation

Frequency response based on slides by J Yan

Slide 37

RC Circuit Revisited Revisiting the RC circuit example

1

1

)1(

1)( 0

+=

+==

sCRsCR

sC

V

VsH

s

Frequency response based on slides by J Yan

Slide 38

Example Find H(s)=Vo(s)Ii(s) the poles and zeros Sketch the freq resp

Frequency response based on slides by J Yan

Slide 39

Logarithm Properties

Logarithms will be used to more quickly and systematically sketch the frequency response graphs Itrsquos worthwhile to review a few of their properties

1 log(P1P2)=log(P1)+log(P2)2 log(P1P2)=log(P1)-log(P2)3 log(Pn)=nlog(P)4 log(01)=-1 log(1)=0 log(10)=1 log(100)=2 hellip

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 36

More on Transfer Functions We defined the transfer function as a ratio of two s-domain signals in a system assuming the system is ldquorelaxedrdquo (ie ICs are zero)

φωω jejHjHsD

sN

sX

sYsH )()(

)(

)(

)(

)()( ===

bull H(jω)is the magnitude gain while φ is the phase difference of Y(s)relative to X(s) for frequency ω

bull The TFs in EE253 can all be expressed as a rational polynomial in s (NB A pure delay is a practical example not having this property)

bull A value of s that makes N(s)=0 is a zero of H(s) bull A value of s that makes D(s)=0 is a pole of H(s) bull Express TFs in terms of s rather than ω This makes it easier to perform

algebraic manipulation and to determine poles and zerosbull By our definition of the TF the s-domain input signal neednrsquot be part of

the TF however if it does appear it is in the denominator of the TF In this case D(s)=0 is the system characteristic equation

Frequency response based on slides by J Yan

Slide 37

RC Circuit Revisited Revisiting the RC circuit example

1

1

)1(

1)( 0

+=

+==

sCRsCR

sC

V

VsH

s

Frequency response based on slides by J Yan

Slide 38

Example Find H(s)=Vo(s)Ii(s) the poles and zeros Sketch the freq resp

Frequency response based on slides by J Yan

Slide 39

Logarithm Properties

Logarithms will be used to more quickly and systematically sketch the frequency response graphs Itrsquos worthwhile to review a few of their properties

1 log(P1P2)=log(P1)+log(P2)2 log(P1P2)=log(P1)-log(P2)3 log(Pn)=nlog(P)4 log(01)=-1 log(1)=0 log(10)=1 log(100)=2 hellip

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 37

RC Circuit Revisited Revisiting the RC circuit example

1

1

)1(

1)( 0

+=

+==

sCRsCR

sC

V

VsH

s

Frequency response based on slides by J Yan

Slide 38

Example Find H(s)=Vo(s)Ii(s) the poles and zeros Sketch the freq resp

Frequency response based on slides by J Yan

Slide 39

Logarithm Properties

Logarithms will be used to more quickly and systematically sketch the frequency response graphs Itrsquos worthwhile to review a few of their properties

1 log(P1P2)=log(P1)+log(P2)2 log(P1P2)=log(P1)-log(P2)3 log(Pn)=nlog(P)4 log(01)=-1 log(1)=0 log(10)=1 log(100)=2 hellip

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 38

Example Find H(s)=Vo(s)Ii(s) the poles and zeros Sketch the freq resp

Frequency response based on slides by J Yan

Slide 39

Logarithm Properties

Logarithms will be used to more quickly and systematically sketch the frequency response graphs Itrsquos worthwhile to review a few of their properties

1 log(P1P2)=log(P1)+log(P2)2 log(P1P2)=log(P1)-log(P2)3 log(Pn)=nlog(P)4 log(01)=-1 log(1)=0 log(10)=1 log(100)=2 hellip

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 39

Logarithm Properties

Logarithms will be used to more quickly and systematically sketch the frequency response graphs Itrsquos worthwhile to review a few of their properties

1 log(P1P2)=log(P1)+log(P2)2 log(P1P2)=log(P1)-log(P2)3 log(Pn)=nlog(P)4 log(01)=-1 log(1)=0 log(10)=1 log(100)=2 hellip

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 310

Decibel ScaleHistorically the ldquogainrdquo G in communication systems was the ratio of power levels between 2 signals measured in bels (B)

The decibel (dB) is a tenth of a bel and is more convenient to use Since the power in a signal is proportional to the square of the voltage or to the square of the current

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 311

Decibel Scale1 If power ratio is known use

10log(P2P1) If voltage or current ratio is known use 20log(V2V1) or 20log(I2I1)

bull Only the voltage or current magnitudes are used in the calculations Negative signs and angles are handled by the phase

bull When specifying the TF gain the dB absolute value is only meaningful for voltage or current gains

Magnitude H 20logH (dB)

0001

001

01

05

2-05

1

205

2

10

100

1000

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 312

Bode Plots

)log())(log())(log(

)()(

ejjHjH

ejHjH j

times+=

=

φωωωω φ

Plotting the frequency response on a linear scale as done in slides 37 amp 38 is somewhat laborious In the late 1930s Hendrick W Bode developed standard Bode plots which take advantage of logarithm properties to allow for quick and easy plots that conveniently span a large frequency range

ωωφωω

log vs)( (semilog) sticCharacteri Phase

log vs)(log20 log)-(log sticCharacteri Magnitude

Plots Bode

jH

jHH dB

ang==

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 313

( ) ( ) cpzszn

issn

is

n

issn

isN

n

i i

m

i i

nnn

mm

nnNms

K

ps

zsK

ssasaa

sbsbb

sD

sN

sX

sYsH

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz 2 where1)1(

1)1(

)(

)(

)(

)(

)(

)()(

1

22

1

1

22

1

1

11

1110

10

++=+++

+++=

minus

minus=

+++++++==

=

prodprodprodprod

prodprod

==

==

=

=

minusminus

ωωζ

ω

ωωζ

ω

L

L

Transfer FunctionThe TF is commonly written in one of the following forms

Characteristicpolynomial

zeros

poles

of simple zeros

of zeros

of complex pairs of zeros

of zeros at origin(assumes Ngt0)

This form most convenient for frequency response plots (itrsquos not as ugly as it looks)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 314

Computation for Bode Diagram

Thus we have 4 types of terms to consider in the contribution to the magnitude and phase characteristics

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

ωωζ

ω

ωωζ

ω

11

11loglog20)(log20

1

2

1

2

11

2

2

2

2

sumsum

sumsum

==

==

+minusminus+minus+

+minus+++==rArr

cpp

icpp

icpp

icpp

cpz

icpz

icpz

icpz

sp

ip

sz

iz

n

i

n

i

n

i

jn

i

jdB

jj

NKjHH

ωωζ

ωω

ωωζ

ωω

ωω

ωωωω L

tantan

tantan900 if 180

0 if 0

1

21

1

21

1

1

1

1

2222 sumsum

sumsum

= minusminus

= minusminus

=minus

=minus

minus+

minus+degsdot+

ltdegplusmngedeg

=rArr

cpp

icpp

icppicppcpz

icpz

icpzicpz

sp

ip

sz

iz

n

i

n

i

n

i

n

iN

K

K

ωω

ωωζ

ωω

ωωζ

ωω

ωωφ L

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 315

Constant TermK

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 316

PolesZeros at Origin

N poles at origin contribute -20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and ndashN90degto the phase

N zeros at origin contribute 20N dBdecade to the magnitude (passing through 1 rads at 0 dB) and N90deg to the phase

( ) ( )( )

degtimesplusmn=

timesplusmn=rArr=

plusmn

plusmnplusmn

=

plusmn

90

)log(20

Njangle

Njjs

NdB

NN

js

N

ω

ωωω

ω

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 317

Simple Zeros

degasymp+

asymp+rArrltlt

01

011 asymptote freq Low

i

i

i jdB

j

angle ωω

ωω

ωω

( ) ( )

=+

+=+rArr+=+

minus=ii

ii

ii jdB

jj

js

s

angle ωω

ωω

ωω

ωω

ωω

ωω 1

2

tan1

1log20111

degasymp+

asymp+rArrgtgt

901

log2011 asymptote freq Hi

i

ii

i jdB

j

angle ωω

ωω

ωω

ωω

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 318

Simple Zeros

Magnitude characteristic approximated by 2 straight line asymptotes Horizontal line at 0dB ldquobreaks awayrdquo at ω=ωi with slope 20dBdecade At this corner or break frequency Where is this approximation worst and what is the error value

Phase characteristic approximated by 3 straight line segments 2 horizontal lines at 0deg and 90degconnected by 45degdecade line passing through ωi at 45deg

ω101ω1 10ω1

ω101ω1 10ω1

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 319

ω

ω

Simple poles

( ) ( ) ( )( ) ( )

( )

minus=+

+minus=+rArr+=+=+

minusminus

minus

minusminus

=

minus

ii

ii

ii j

dB

j

ij

js

s

anglej

ωω

ωω

ωω

ωω

ωω

ωω ωτ11

21

111

tan1

1log201111

Itrsquos easy to see that the frequency response to a simple pole is the mirror image of that for a simple zero reflected across the ω axis

ω101ω1 10ω1

ω101ω1 10ω1

HdB

φ

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 320

Quadratic Pole( ) ( )

( ) ( ) ( )( )

minus=+minus

+minusminus=+minusrArr

+minus=++

minusminusminus

minus

minus

=

minus

222

2

2

2

2

2

2

2

2

2

2112

22212

1212

tan1

1log201

11

ωωζωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωζω

ωω

ωωωζ

n

n

nn

nnnn

nnnn

jangle

j

j

dB

js

ss

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 321

Algorithm for Sketching Bode

Diagrams

( )( )prodprod

prodprod==

==

+++

+++=

cpp

icppicpp

icppsp

ip

cpz

icpzicpz

icpzsz

iz

n

issn

is

n

issn

isNs

KsH

1

2

1

1

2

1

2

2

2

2

1)1(

1)1()(

form standard in the Express 1 Step

ωωζ

ω

ωωζ

ω

Step 2 Use straight-line sketches to show the contribution to the magnitude and phase characteristics from each multiplicand

Step 3 Add lines together for complete frequency response

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan

Slide 322

Example

)10(

)2(5)(for plot Bode theDraw

++=

ss

ssH

10-1

100

101

102

103

-20

-10

0

10

20

Mag

nitu

de (

dB)

01 1 10 100 1000-90

-45

0

45

90

Pha

se (

deg)

Frequency response based on slides by J Yan