chapter 6 - frequency response

8/2/2019 Chapter 6 - Frequency Response

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11

Frequency ResponseFrequency Response(Bode Plot)(Bode Plot)

22

Frequency ResponseFrequency Response

Definition

The frequency response of a system is defined as thesteady state response of the system to a sinusoidalinput signal.

tsin A )t(u oω=

y(t) ?

2o

2

o

s

A )s(U

ω+

ω=

∏=

+

=n

1i

ips

)s(m)s(G

)s(G)s(U

33

Frequency ResponseFrequency ResponseResponse to sinusoidal input

)s(U)s(G)s( Y =22n

1i

i

s

A

ps

)s(m)s( Y

ω+

ω

+

=

∏=

o

*

o

o

n

n

1

1

js jspsps)s( Y o

ω−

α+

ω+

α+

+

α++

+

α= L

Taking the inverse Laplace transform:

( )φ+ωα+α++α= −− tsin A 2ee)t(y oo

tp

n

tp

1n1

L

α

α=φ −

)Re(

)Im(tan

o

o1

44


If the system is stable, the steady state response is

( )φ+ω= tsin AM)t(y o

Where

( ){ } ( ){ }2o

2

o jso ) j(GIm) j(GRe)s(G) j(GMo

ω+ω==ω=ω=

ω

ω=φ −

) j(GRe

) j(GImtan

o

o1

In polar form φ=ω jo Me) j(G

Both magnitude M and phaseangle φ depend on thefrequency of the sinusoidalinput.

Response to sinusoidal input



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Example:

Find the frequency response

y(t) ?)s(G

)s(U

u(t)=sin 10tWhere:1s

1)s(G+

=


66


u(t)=sin 10t1s

1)s(G

+=


77



The steady state response to sinusoidal input is alsosinusoidal, which differs from sinusoidal input only inamplitude and phase angle.

y(t) ?

T(s)Y(s)R(s)

)s(U)s(G

88


How to represent frequency response graphically?




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Bode Plot TechniquesBode Plot Techniques

Display of frequency response is a problem that hasDisplay of frequency response is a problem that hasbeen studied for a long time.been studied for a long time.

The most useful technique for hand plotting wasThe most useful technique for hand plotting wasdeveloped by H.W. Bode, so called as Bode Plot.developed by H.W. Bode, so called as Bode Plot.

The idea in BodeThe idea in Bode’’s method is to plot magnitudes method is to plot magnitude

curves using a logarithmic scale and phase curvescurves using a logarithmic scale and phase curvesusing a linear scale.using a linear scale.

1010


The Bode plots consists of two graphs:The Bode plots consists of two graphs:

Magnitude M (GainMagnitude M (Gain ||G(jG(jωω))||) is plotted against) is plotted againstlog(log(ωω) again is represented as logarithmic gain in) again is represented as logarithmic gain indBdB

Phase anglePhase angle ∠∠G(jG(jωω) is plotted against log() is plotted against log(ωω))

Phase is usually measured in degreesPhase is usually measured in degrees

Glog20Gdb

=

1111


Advantages of Bode Plots Advantages of Bode Plots

Dynamic compensator design can be basedDynamic compensator design can be based

entirely on Bode plots.entirely on Bode plots. Bode plots can be determined experimentallyBode plots can be determined experimentally

Bode plots of system in series simply add, whichBode plots of system in series simply add, whichis quite convenient.is quite convenient.

The use of a log scale permits a much widerThe use of a log scale permits a much widerrange of frequencies to be displayed on a singlerange of frequencies to be displayed on a singleplot than is possible with linear scales.plot than is possible with linear scales.

1212


• For the following first order system

2s

1)s(T

+=

• We substitute jω for s

2 j

1) j(T

+ω=ω

• The magnitude and phase are

4

1) j(T

2 +ω=ω

ω−=ωφ=ω∠ −

2tan)() j(T 1

Example:



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10-1

100

101

102

-50

-40

-30

-20

-10

0

Frequency rad/s

M a g n i t u d e d B

10-1

100

101

102

-100

-80

-60

-40

-20

0

Frequency rad/s

P h a s e d e g

Example:

1414


The transfer function of RC filter is

1s

1)s(G

+τ=

1 j

1) j(G

+ωτ=ω

How to make Bode diagrams for complex system?

decade

1515

Bode Plot TechniquesBode Plot TechniquesThe primary advantage of the logarithmic plot is theconversion of multiplicative factor into additive factorby virtue of the definition of logarithmic gain.

Consider the open-loop transfer function

( )

( ) ( ) ( )∏ ∏

∏

= =

=

ω+ωζ++

+

=M

1m

R

1k

2nk

2m

N

Q

1i

i

k nk s2spss

zs

K )s(G

1616


In working with frequency response, it is moreconvenient to replace s with jω and to write thetransfer function in the Bode-form:

( )

( ) ( ) ( )( )∏ ∏

∏

= =

=

ωω+ωωζ++ωτω

+ωτ=ω

M

1m

R

1k

2

nnk m

N

Q

1ii

o

k k / j j) /2(11 j j

1 jK ) j(G

?) j(Gdb

=ω ?) j(G =ω∠=φ



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( )∑=

ω−ωτ++=ωN

1i

N

iodbjlog20 j1log20K log20) j(G

Then:

∑=

ωτ+−M

1mm j1log20 ( )( )∑

=

ωω+ωωζ+−R

1k

2

nnk k k / j j) /2(1log20

Since the gain is plot in dB, we can treat each pole andzero individually, and then add up the results.

1818

Bode Plot TechniquesBode Plot TechniquesFurthermore the phase angle is

∑∑=

−

=

− ωτ−−ωτ+=ωφM

1mm

1N

1ii

1 tan)90(Ntan0)(

∑=

−

ω−

ω

ωζ−

R

1k

2

2n

nk 1

k

k 2

tan

Similarly, the phase angle due to each pole and zerocan be computed individually, and then the results isadded together.

1919


Given the response of each basic term in theGiven the response of each basic term in thetransfer function, the algebraic sum wouldtransfer function, the algebraic sum would

yield the total response.yield the total response.

If we can make an approximation of eachIf we can make an approximation of eachterm we should be able to quickly sketch theterm we should be able to quickly sketch theBode diagrams.Bode diagrams.

2020

Bode Plot SummaryBode Plot SummaryAll transfer function for kinds of systems we havetalked about, so far are composed of three classes ofterms

• Ko(jω)n

• (jωτ+1)±1

•

We can determine the logarithmic magnitude plot andphase angle for these four factors then utilize themto obtain a Bode diagram for any general form of atransfer function.

1

n

2

n

1 j

2 j

±

+

ω

ωζ+

ω

ω



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Bode Plot SummaryBode Plot Summary

1. Ko

(jωωωω)n

The logarithmic gain is

( )ω+=ω jlogn20Kolog20)(M

The phase angle is

090n)( =ωφ

The easiest way to draw the curve is to locate ω = 1and plot 20logKo at that frequency and then draw theline with slope 20n db/dec through that point.

2222


0.1 1 10

20

40

0

-20

-40

Kolog20 20 dB/dec

40 dB/dec

-20 dB/dec

-40 dB/dec

n=1

n=2

n=-1

n=-2

1. Ko

(jωωωω)n

2323


0.1 1 10

90

180

0

-90

-180

n=1

n=2

n=-1

n=-2

1. Ko(jωωωω)n

2424

Bode Plot SummaryBode Plot Summary2. (jωτωτωτωτ + 1)±1

The magnitude of this term approaches one asymptoteat very low frequencies and another asymptote at veryhigh frequency.

(a) For ω << 1/τ, (jωτ + 1)n

≅ 1(b) For ωτ >> 1/τ, (jωτ + 1)n ≅ (jω)±1 slope ±1x20 dB/dec

The phase curve can also be drawn easily by using thefollowing low- and high-frequencies asymptotes:

(a) For ω << 1/τ, φ = 0

(b) For ω >> 1/τ, φ = ±1x900

(c) For ω ≅ 1/τ, φ = ± 1x450

ωωωω = 1/ττττ is the break point

/ /



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Bode Plot SummaryBode Plot Summary(jωτωτωτωτ + 1)

(jωτωτωτωτ + 1)-1

2626


The magnitude of this term approaches one asymptoteat very low frequencies and another asymptote at veryhigh frequency.

(a) For ω << ωn, |G(jω)|db ≅ 0

(b) For ω >> ωn, |G(jω)|db ≅ (jω)±2 slope ±1x40 dB/dec

1

n

2

n

1 j

2 j

±

+

ω

ωζ+

ω

ω

The phase curve can also be drawn easily by using the

following low- and high-frequencies asymptotes:(a) For ω << ωn, φ = 0

(b) For ω >> ωn, φ = ±1x1800

(c) For ω ≅ ωn, φ = ± 1x900

ωωωωn = natural frequency

2727

Bode Plot SummaryBode Plot Summary1

n

2

n

1 j

2 j

±

+

ω

ωζ+

ω

ω

2828

Bode Plot TechniquesBode Plot TechniquesExample

Plot the Bode plots for the following transfer function

)1.0s(

)1s(10)s(G

+

+=

)100s4.0s(s

100)s(G

2 ++=

)1s(

)1.0s(10)s(G

+

+=

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Bode Plot TechniquesBode Plot TechniquesExample

Plot the Bode plots for the following transfer function

)5s)(10s(s

)5.0s(2000)s(G

++

+=

)4s4.0s(s

10)s(G

2 ++=

[ ]1)2 /s(02.04 /ss

)1s01.0s(01.0)s(G

22

2

++

++=

3030


)5s)(10s(s

)5.0s(2000)s(G

++

+=

3131

)4s4.0s(s

10)s(G

2 ++=

Bode Plot Techniques

3232


[ ]1)2 /s(02.04 /ss

)1s01.0s(01.0)s(G

22

2

++

++=

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3333

Gain and Phase MarginGain and Phase Margin

-5 -4 -3 -2 -1 0 1 2-3

-2

-1

0

1

2

3Root Locus

RealAxis

I m a g i n a r y

A x i s

2)1s(s

1

+

3434

Gain Margin (GM)Gain Margin (GM)

Gain margin (GM): difference (in dB) betweenGain margin (GM): difference (in dB) betweenunity gain condition (0 dB)unity gain condition (0 dB) andand magnitudemagnitude of of frequency responsefrequency response @ phase cross@ phase cross--overoverfrequencyfrequency

Phase crossPhase cross--over frequency,over frequency, ωωpp :: frequencyfrequency,,

at whichat which phase =phase = --180180oo

GM = 0GM = 0 -- |G(j|G(j ωωpp)| =)| = -- |G(j|G(j ωωpp)|)|

3535

Phase Margin (PM)Phase Margin (PM)

Phase margin (PM): difference (inPhase margin (PM): difference (indegrees) betweendegrees) between phase @ gain crossphase @ gain cross--over frequencyover frequency andand --180180oo

Gain crossGain cross--over frequency,over frequency, ωωgg:: frequencyfrequency,,at whichat which magnitude = 1 (0 dB)magnitude = 1 (0 dB)

PM =PM = ∠∠G(jG(j ωωgg) + 180) + 180oo

3636

Gain and Phase MarginGain and Phase Margin

-150

-100

-50

0

50

M

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

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

P h a s e ( d e g )

Bode DiagramGm = 26 dB (at 1 rad/sec) , Pm = 78.7 deg (at 0.099 rad/sec)

Frequency (rad/sec)

Phase cross-over frequency

Gain cross-over frequency

Gain margin}}}}

}

Phase margin

chapter 6 - frequency response

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