Frequency Response Methods - Department of …me.emu.edu.tr/ranjbar/Lecture9Control.pdf · • The important issue in frequency response methods is ... Bode Diagram • Advantage

Frequency Response Methods

• The frequency response

• Nyquist diagram – polar plots

• Bode diagram – magnitude and phase

• Frequency domain specifications


• The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input signal.

• The sinusoid is a unique input signal, and theresulting output signal for a linear system aswell as signals throughout the system, is sinusoidal in the steady state (the out of thesystem); it is differs from the input waveformonly in amplitude and phase angle.

• In preceding chapters the response and performance of a system have been described in terms of the complex frequency variable s and the location of the poles andzeros on the s-plane.

• A very practical and important alternative approach to the analysis and design of a system is the frequency response method.

Frequency Response• Consider the following

)cos(R)cos()(

ttRtuω

αω=

+=

)cos(B)cos()(

ttBtxss

ωφω

=+=

( )B( ) ( ) ( )( ) ( ) ( )R( ) ( )( )

jj

jj B e Bj G e Gj RR e

φ ωθ

αω ω ω φω ω ω θω ω αω

∠= = = = = ∠

∠G

is known as the gain and is known as the phase of the system.

)(ωG αφωθ −=)(

on

nn

n ajajaj

+++= −

− ....)()(1)( 1

1 ωωωG

For example for a system with dynamics described by the differential equation

uyadtdya

dtyda

dtyda n

n

nn

n

n =++++ −

−

− 011

1

1 ...

The transfer function from u to y is :

on

nn

n asasasG

+++= −

− ....1)( 1

1

This means a frequency dependent system gain (or transfer function):

Frequency Response

The frequency response can be obtained from the transfer function by substituting . Brief derivation:

Write u and y as and substitute into DE:

ωjs =tjtj Beyu ωω == ,Re

tjtjtjtjnn

tjnn

n

n

nn

n

n

BeaBejaBejaBeja

yadtdya

dtyda

dtyda

ωωωωω ωωω Re...)()(

...

011

1

011

1

1

=++++=

=++++

−−

−

−

−

on

nn

ntj

tj

ajajaBej

+++== −

− ....)()(1

Re)( 1

1 ωωω ω

ω

G

This means that the frequency dependent system gain (or transfer function) is :

Hence the substitution rule is justified.ωjs =

Frequency Response

11 1 0[ ( ) ( ) ... ] Ren n j t j t

n na j a j a a Be ω ωω ω −−= + + + + =

Frequency ResponseAlternatively, consider the system

Then in partial fraction form we have

Taking the inverse Laplace transform yields

where pi are assumed to be distinct poles.

where α and β are constants which are problem dependent.

1−l

Frequency Response

1−l

If the system is stable, then all pi are have positive nonzero real parts, (poles are − pi), and

since each exponential term decays to zero as t → ∞.

1−l

• Thus the steady-state output signal depends only on the magnitude and phase of T(jω) at a specific frequency ω.

• Notice that the steady state response as described the above is true only for stable systems, T(s).

Laplace vs. Fourier Transform

Laplace transform:

where

Fourier transform:

Setting s=jω in F(s) yields the Fourier transform of f (t)

(But Fourier Transform is often used for signals that exist for t<0)

Jean Baptiste Joseph Fourier

• Born in 1768 in Auxerre, France

• Died in 1830 in Paris• Was nearly guillotinedin 1794

• Was taught by Laplace,Lagrange and Monge

• Created Cairo Institute• Developed Fourier series

while prefect in Grenoble

Advantage of Frequency response method • The frequency response method is the readyavailability of sinusoid test signals forvarious ranges of frequencies and amplitudes.

• Thus the experimental determination of thefrequency response of a system is easily

accomplished and is the most reliable anduncomplicated method for the experimentalanalysis of a system.

• Furthermore the design of a system in thefrequency domain provides the designer withcontrol of the bandwidth of a system andsome measure of the response of the systemto undesired noise and disturbance.

Transfer Function in Frequency Response Method

• The transfer function representing the sinusoidal steady-state behavior is then a function of the complex

variable jω and is itself a complex function T(jω).

• The frequency response method is that the transfer function describing the sinusoid steady-state behaviorof a system can be obtained by replacing s with jω in the system transfer function T(s).

• Direct correlations between the frequency response and the corresponding transient response characteristics in the time domain are somewhat tenuous (very weak).


• The sinusoid is a unique input signal, and theresulting output signal for a linear system aswell as signals throughout the system, is sinusoidal in the steady state (the out of thesystem); it is differs from the input waveformonly in amplitude and phase angle.

• The important issue in frequency response methods is how to descript the amplitude and phase angle of the system. We will study different methods to represent amplitude and phase.

Frequency Response Plots

(Review Appendix G in textbook)

)Re()Im(tan

)(Im)(Re|)(|

1

22

ωωφ

ωωω

−=

+=G

Frequency Response Plots

)(tan1/tan

)Re()Im(tan)(

1

1111

ωωωω

ωωωφ −−− −=

−==

1st Ordersystem

Polar Plot or Nyquist Diagram

)1(tantan)Re()Im(tan)( 1

211

ωττωω

ωωωφ −−− =

−−

==KK

2rd system)1()(

+=

τssKsG

Harold Nyquist

• Born in 1889 in Sweden• Died in 1976, USA• Yale PhD, 1917• Career at Bell Labs• 138 patents• Nyquist diagram,

criterion, sampling theorem

• Laid the foundation forinformation theory, datatransmission andnegative feedback theory

Real Axis

Imag

inar

y A

xis

Nyquist Diagrams

-1 -0.5 0 0.5 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1G(s)=1/(s3+2s2+2s+1)

Frequency response diagrams

ωj

ωj

1221)( 23 +++

=sss

sG

1)(2)(2)(1)( 23 +++

=ωωω

ωjjj

jG

is plotted for ])0,((),0[ −∞∈∞∈ ωω

The Nyquist diagram of

Bode DiagramPlots of 20log10 | G(ω ) | and of Φ(ω) as a function of log ω

RC filter:

Gain in decibels (dB)

Hendrik Wade Bode

• 1905-1982, USA• PhD from Columbia

in 1935• Entire career at Bell

Labs• Invented magnitude

and phase frequencyplots in 1938

• Many other contributions in electrical engineeringand control

Bode Diagram-3dB

Break or corner frequency

Bode Diagram

• Advantage of logarithmic plot is that multiplicative factors are converted into additive ones

• We can then decompose a high order transfer function into a product of simple standard components to sketch the broad features of the Bode diagram

decade

Form of the Transfer Function

• We treat real and complex poles and zeros separately,and use one of two standard forms.

• We then plot magnitude (in db) and phase on a log scale of frequency

• By using db for magnitude, we can use addition tocombine the effects of each pole or zero since theproduct becomes a sum of log terms.

• The phase effect is already a linear combination• The log frequency scale allows piecewise linear

approximations with reasonable accuracy• Bode plots work best for real poles and zeros. The

complex case is less accurate and requires carefultreatment.

How to draw a Bode plot for a given transfer function?

There are only 8 types of factors in any transfer function - which are these?

Lswyn

j

w

knknkkj

y

m

ii

Lsyn

jj

y

m

ii

Ls

n

m

Lsn

nn

n

mm

mm

esspss

zsk

epss

zsk

epspspszszszsk

easasasabsbsbsbsG

−−−

= =

=

−−

=

=

−

−−

−

−−

∏ ∏

∏

∏

∏

+++

+

=

+

+

=

++++++

=

++++++++

=

2

1 1

22

1

1

1

21

21

011

1

011

1

)2()(

)(

)(

)(

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

........)(

ωωζ

Ljwyn

j

w

k nknk

kpj

y

m

izi

Ljyn

jpi

y

m

izi

Lj

pnpp

zmzz

Ljn

nn

n

mm

mm

ejjTjj

TjK

eTjj

TjK

eTjTjTjTjTjTjK

eajajaja

bjbjbjbj

ω

ω

ω

ω

ωωω

ωζωω

ω

ωω

ω

ωωωωωωωωωωωωω

−−−

= =

=

−−

=

=

−

−−

−

−−

∏ ∏

∏

∏

∏

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

+=

+

+=

++++++

=

++++++++

=

2

1 12

21

1

1

21

21

011

1

011

1

)(21)1()(

)1(

)1()(

)1(

)1)...(1)(1()1)....(1)(1(

....)()(....)(G

If we knew the look of the Bode plots for each of the 8 types, we could add up the Bode plots from them.

Bode plot analysis techniques

Ljwyn

j

w

k nknk

kpj

y

m

izi

ejjTjj

TjKj ω

ωωω

ωζ

ωω

ωω −

−−

= =

=

∏ ∏

∏

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

+

= 2

1 12

21

)(21)1()(

)1()(G

( )

( )

1

2

1 2

21

( 2 ) 21 1

2

2

Lm ( ) 20 log ( ) 20 log 20 log 1

20 log 1 .... 20 log 1 20 log

20 log 1 20 log 1 ....

220 log 1 20 log 1 ...

220 log 1

z

z zm

p p

p n y wn n

w

nw nw

j j K j T

j T j T y

j T j T

jj T j

jj

ω ω ω

ω ω ω

ω ω

ωζω ωω ω

ωζ ωω ω

− −

= = + + +

+ + + + − −

+ − + −

− + − + + −

− + +

G G

Factorization

Gain in dB :


⎟⎟⎠

⎞⎜⎜⎝

⎛++∠

−⎟⎟⎠

⎞⎜⎜⎝

⎛++∠−+∠

−−+∠−+∠−−+∠+++∠++∠+∠=∠

−−

2

2

21

2

1

1)2(

21

21

)(21

...)(21)1(

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

nwnw

w

nnwynp

pp

zmzz

jj

jjTj

TjTjTjTjTjKj

ωωω

ωζ

ωωω

ωζω

ωωπωωωωGPhase:

The laborious procedure of plotting the amplitude and the phase by means of substituting several values of can be avoided when drawing Bode diagrams, because we can use several short cuts. These short cuts are based on simplifying approximations, which allow us to represent the exact, smooth plots with straight-line approximations. The difference between actual curves and these asymptotic approximations is small, and can be

added as a correction.

ω

The 8 types of factors in a transfer function

Transport

delay

Constant

K

Lje ω−

Differentiators

Integrators

yj )(1ω

rj )( ω


First order lag terms(real poles)

First order Lead terms(real

zeros)

rziTj )1( ω+

rpjTj )1(

1ω+


Quadratic lead terms(complex

zeros)

Quadratic lag terms (complex poles

r

nknk

k jj ⎟⎟⎠

⎞⎜⎜⎝

⎛++ 2

2)(21

1

ωωω

ωζ

r

nknk

k jj ⎟⎟⎠

⎞⎜⎜⎝

⎛++ 2

2)(21ωωω

ωζ


Bode DiagramPlots of 20log10 | G(ω ) | and of Φ(ω) as a function of log ω

RC filter:

Gain in decibels (dB)


Ljwyn

j

w

k nknk

kpj

y

m

izi

ejjTjj

TjKj ω

ωωω

ωζ

ωω

ωω −

−−

= =

=

∏ ∏

∏

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

+

= 2

1 12

21

)(21)1()(

)1()(G

( )

( )

1

2

1 2

21

( 2 ) 21 1

2

2

Lm ( ) 20 log ( ) 20 log 20 log 1

20 log 1 .... 20 log 1 20 log

20 log 1 20 log 1 ....

220 log 1 20 log 1 ...

220 log 1

z

z zm

p p

p n y wn n

w

nw nw

j j K j T

j T j T y

j T j T

jj T j

jj

ω ω ω

ω ω ω

ω ω

ωζω ωω ω

ωζ ωω ω

− −

= = + + +

+ + + + − −

+ − + −

− + − + + −

− + +

G G

Factorization

Gain in dB :


⎟⎟⎠

⎞⎜⎜⎝

⎛++∠

−⎟⎟⎠

⎞⎜⎜⎝

⎛++∠−+∠

−−+∠−+∠−−+∠+++∠++∠+∠=∠

−−

2

2

21

2

1

1)2(

21

21

)(21

...)(21)1(

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

nwnw

w

nnwynp

pp

zmzz

jj

jjTj

TjTjTjTjTjKj

ωωω

ωζ

ωωω

ωζω

ωωπωωωωGPhase:

The laborious procedure of plotting the amplitude and the phase by means of substituting several values of can be avoided when drawing Bode diagrams, because we can use several short cuts. These short cuts are based on simplifying approximations, which allow us to represent the exact, smooth plots with straight-line approximations. The difference between actual curves and these asymptotic approximations is small, and can be

added as a correction.

ω

Detailed examination of the 8 factors

System type corresponds to integrators (for 0 type there is not integrator factor)Diagram of a constant

πdBlog20Lm KK =

K>0K<0

Ljwyn

j

w

k nknk

kpj

y

m

izi

ejjTjj

TjKj ω

ωωω

ωζ

ωω

ωω −

−−

= =

=

∏ ∏

∏

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

+

= 2

1 12

21

)(21)1()(

)1()(G


ωωωω

log20log201log20)(

1log20)(

1Lm yyjj yy −=−==⎟⎟

⎠

⎞⎜⎜⎝

⎛

yjj

yy 90)(1

)(1

−=∠−∠=⎟⎟⎠

⎞⎜⎜⎝

⎛∠ ω

ω

ωj1

Diagram of integrators


( ) ωωωω log20log20)(log20)(Lm yyjj yy ===

( ) yj y 90)( =∠ ω

Bode diagram of a differentiator

y=1


2)(1log20

1log201log201

1log201

1Lm

T

TjTjTj

ω

ωωω

+−=

+−=+

=⎟⎟⎠

⎞⎜⎜⎝

⎛+ TTj

Tjωω

ω1tan)1(1

11 −−=+∠−∠=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

∠

dB01log201

1Lm =≈⎟⎟⎠

⎞⎜⎜⎝

⎛+ Tjω

.1<<Tω

TTjTj

ωωω

log201log201

1Lm −=≈⎟⎟⎠

⎞⎜⎜⎝

⎛+

.1>>Tω

Bode diagram of a first order lag term


frequency error

Corner frequency (bandwidth) -3dB

At half the corner frequency -1dB

At a quarter of the corner frequency

-0.26 dB

Bode diagram of a first order lag term

Detailed examination of the 8 factors( )

2)(1log20

1log201log201Lm

T

TjTjTj

ω

ωωω

+=

+=+=+

( ) TTj ωω 1tan1 −=+∠.1<<Tω

( ) dB01log201Lm =≈+ Tjω

.1>>Tω

( ) TTjTj ωωω log20log201Lm =≈+

First order lead term


22 )(121

1

ωω

ωωζ jj

nn

++

1<ζ

22

2

2

22

22

21log20

)(121

1log20)(121

1Lm

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

++=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++

nn

nnnn

jjjj

ωζω

ωω

ωω

ωωζω

ωω

ωζ

221

22

/1/2tan

)(121

1

n

n

nn

jj ωωωζω

ωω

ωωζ −

−=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++∠ −

Quadratic (second order) Lag

dB01log20)(121

1Lm2

2

=−≈⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++ ωω

ωωζ jj

nn

For small ω

nn

nn

jj

ωω

ωω

ωω

ωωζ

log40log20

)(121

1Lm

2

2

22

−=−≈

≈

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++

For large ω

Detailed examination of the 8 factorsQuadratic (second order) Lag

2121

ζζ −=mM

221 ζωω −= nmFor there is a resonant peak at 707.0<ζ

with peak size



Lee LjLj ωωω −=∠= −− ,0Lm

Transport Lag

Non-Minimum Phase System

• A transfer function is called minimumphase if all its zeros lie in the left-handplane.

• It is called non-minimum phase if it haszeros in the right-hand plane.

331)( 2 ++

+=

ssssGH

Apply steps 1-4 and step 5 for breakaway point

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Axis

Imag

Axi

s

Root locus examples

331)( 2 ++−

=ss

ssGH

Apply steps 1-4,

Step 4 for crossing and

Step 5 for breakaway point

…

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Axis

Imag

Axi

s

Root locus examples

Non-Minimum Phase SystemNote that |G(jω)| is identical, but the phase is different

Drawing the Bode Diagram

20log5=14

40dB/De

-20dB

-40dB

?

?


20log5=14

40dB/De

-20dB

-40dB

?


20log ( ) 20log5 20logG jω ω= −

20log 1 0.5j ω− +

2 0 log 1 0 .1j ω+ +

220log 1 0.6( / 50) ( / 50)j ω ω− + +

(ω <1)

(ω >2 )

(ω >10)

(ω >50)



Using Matlab

Example

530)(+

=s

sG

)2.01)(41(10)(

ωωω

jjj

++=G

Example

))125.0(05.01)(21()5.01(4)( 2ωωωω

ωωjjjj

jj+++

+=G

Example

Performance Specifications in the Frequency Domain

• The basic disadvantage of the frequency response method for analysis and design is the indirect link between the frequency and the time domain.

• Then for given a set of time-domain (transient performance) specifications, how do we specify the frequency response?

• Direct correlations between the frequency response and the corresponding transient response characteristics are somewhat tenuous (very weak).

• However, we need to develop a method to evaluate the performance in the frequency response method.

• Like in the time-domain approach, we only consider the performance of a simple second order system to a step input.

1 21( ) 1 sin( cos ), 1tnny t e tζω ω β ζ β ζ

β− −= − + = −

%)2(4== δ

ζωforT

ns

21/100 ζζπ −−= ePO

nrT

ωζ 6.016.2

1+

≅

Settling time

Percentage overshoot

Peak time

Rise time (10% - 90%)

21 ζωπ−

=n

pT

Step response for second ordersystems-in the time domain


.2

)( 22

2

nn

nss

sTωζω

ω++

=

The closed-loop transfer function in the frequency domain:

Consider a second order system

The Bode diagram of the frequency response of this feedback system is shown in Fig

dB01log20)(121

1Lm2

2

=−≈⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++ ωω

ωωζ jj

nn

For small ω

nn

nn

jj

ωω

ωω

ωω

ωωζ

log40log20

)(121

1Lm

2

2

22

−=−≈

≈

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++

For large ω



rωωpM

• At the resonant frequency, , a maximum value of the frequency response, , is attained.

• The bandwidth, , is a measure of a system’s ability to faithfully reproduce an input signal.

Bω

• The bandwidth is the frequency, , at which thefrequency response has declined 3 dB from its low-frequency value.

Bω

.2

)( 22

2

nn

nss

sTωζω

ω++

=


.121 ζζπ −−+= eM tp

ωpMThe resonant peak indicates the relative stability of a system

The resonant frequency and –3dB bandwidth can berelated to the speed of the transient response. Thus as the bandwidth increase, the rise time of the step response of the system will decrease. Furthermorethe overshoot to a step input can be related to, through the damping ratio ζ. by

rω

Bω

ωpM


The bandwidth of a system as indicated on the frequency response can be approximately related to the natural frequency of the system. Figure 8.26 shows the normalized bandwidth, versusζ for the second-order system

Bω

,/ nB ωω


<ωpM

nζωτ /1=

Thus desirable frequency-domain specifications are as follows:

1. Relatively small resonance magnitude:

2. Relatively large bandwidths so that the system time constant is sufficiently small

1.5, for example.


• The usefulness of these frequency response specifications and their relation to the actual transient performance depend upon the approximation of the system by a second-order pair of complex poles, that is the dominant roots.

• If the frequency response is dominated by a pair of complex poles, the relationships between the frequency response and the time response discussed in this section will be valid.

• Fortunately a large proportion of control system satisfied this dominant second-order approximation in practice.

Steady-state error constants

The steady-state error specification can also be related to the frequency response of a closed-loop system.

• As we knew, the steady-state error for a specific test input signal can be related to the gain and number of integrations (poles at the origin) of the open-loop transfer function, i.e.,the type of the system. • In frequency response method, the type of the system

determines the slop of the logarithmic gain curve at low frequency, since steady-state error is defined at

,0→s .0→ωji.e.,

Thus, information concerning the existence and magnitude of the steady-state error of a control system to a given input can be determined from the observation of the low-frequency region of the logarithmic gain curve.

Determine of static position error constants.

For type 0 system (N=0), we have

)(lim)(lim00

ωω

jGsGKjs

P→→

==

Consider the transfer function as follows

.)1()(

)1()(

1

1

∏

∏

=

=

+

+= Q

kk

N

M

ii

jj

jKjG

ωτω

ωτω

For type 0 (N=0) system, at the low frequency, we have

KjG ≈)( ω or KjGKj

P ==→

)(lim0

ωω

∏

∏

∏

∏

=

=

=

=

+

+=

+

+= Q

kk

M

ii

Q

kk

M

ii

j

jK

jj

jKjG

1

1

1

0

1

)1(

)1(

)1()(

)1()(

ωτ

ωτ

ωτω

ωτω

Determine of static position error constants.

KjGKj

P ==→

)(lim0

ωω

Hence, we can determine the steady-state position error by measurethe value from its logarithmic gain curve (let 20logK=c),

KKcpK log)20/)log20()20/( 101010 ===

Determine of static velocity error constant


)(lim)(lim00

ωωω

jGjssGKjs

v→→

==


.)1()(

)1()(

1

1

∏ +

∏ +=

=

=Q

kk

N

M

ii

jj

jKjG

ωτω

ωτω

According to the definition, we have

KjGjKj

v ==→

)(lim0

ωωω

(at the low frequency ).

)1()(

)1()(

1

1

1

ωωτω

ωτω

jK

jj

jKjG Q

kk

M

ii

≈+

+=

∏

∏

=

=

Determine of static velocity error constant

||log20log201

vj

v KjK

==ωω

Also, we can find out Kv using the fact that the intersection of the initial –20dB/decade segment (or its extension) with the 0dBline has a frequency numerically equal to Kv

.

1=ωj

Kv or 1ω=vK

At the intersection of the initial –20dB/decade segment (or its extension) with the 0-dB line, the horizontal coordinate, i.e., the frequency isnumerically equal to the. vK

Determine of static acceleration error constant

KjGjKja ==→

)()(lim 2

0ωω

ω



)()(lim)(lim 20

20

ωωω

jGjsGsKjs

a→→

==

.)1()(

)1()(

1

1

∏

∏

=

=

+

+= Q

kk

N

M

ii

jj

jKjG

ωτω

ωτω

(at the low frequency ).)()1()(

)1()( 2

1

2

1

ωωτω

ωτω

jK

jj

jKjG Q

kk

M

ii

≈+

+=

∏

∏

=

=

Determine of static acceleration error constant

||log20)(

log201

2 aj

a KjK

==ωω

aω

aK

01log20)(

log20 2 ==ωjKa

aa K=ω .2aaK ω=

The frequency at the intersection of the initial-40db/decade segment (or its

extension) with the 0-dB line gives the square root of numerically.

or which yields

Design Example: Engraving Machine

The goal is to select an appropriate gain K, utilizing frequency response method, so that the time response to step commands is acceptable


To represent the frequency response of the system,we will first obtain the open-loop and closed-loop Bode diagram.

)2)(1(1)(

++=

sssjG ω


.223

2)( 23 +++=

ssssT

)8.0(.78.1 == rpMor ωω

Then we use the closed-loop Bode diagram to predict the time response of the system and check the predicted result with the actual result

20log|T|=5 dB at 8.0=rω

.

5log20 =ωpM

ωωωω

ω jsj

jT =−+−

=)2()32(

2)(22


)8.0(.78.1 == rpMor ωω

5log20 =ωpM

If we assume that the system has dominant second-order roots, we can approximate the system with a second-order frequency response of the form shown in Fig.

29.08.0 == ζωr


78.1=ωpM

29.08.0 == ζω r

nr ωω / =0.91.

.88.091.08.0==nω

.774.051.0

774.02

)(222

2

++=

++=

sssssT

nn

n

ωζω

ω

Since we are now approximating T(s) as a second-order system, we have


.sec7.1588.0)29.0(

44 ondsTn

s ===ζω

The overshoot to a step input as 37% for 29.0=ζ

The settling time to within 2% of the final value is estimated as

.774.051.0

774.02

)(222

2

++=

++=

sssssT

nn

n

ωζω

ω


• The actual overshoot for a step input is 34%, and the actual settling time is 17 seconds.

• We see that the second-order approximation is reasonable in this case and can be used to determine suitable parameters on a system.

• If we require a system with lower overshoot, we would reduce K to 1 and repeat the procedure.

Disk Drive Read System

• The disk drive uses a flexure suspension to hold the reader head mount, as shown in Fig

• we will include the effect of thespringy flexure within the model of the motor-load system.

• We model the flexure with the mounted head as a mass M, a spring k, and a sliding friction b.


.)/()/2(1

12

)()()(

222

23

nnnn

nssss

sGsUsY

ωωζωζωω

++=

++==

3.0=ζHzfn 3000= 31085.18 ×=nω

The transfer function of a spring-mass-damper was developed in Chapter 2, where

A typical flexure and head has and natural resonance at or


|,)()()()1(|log20 321 ωωωω jGjGjGjK +

The sketch is a plot of the magnitude characteristics for the open-loop Bode diagram ,or

K=400

Note the resonance at nω

We wish to avoid exciting

This resonance.


msTn

s 5.24≈=

ζω

8.0≈ζwhere

,400≤KAs long as the resonance is outside the bandwidth of the system.

Plots of the magnitude of the open-loop Bode diagram andthe closed-loop Bode diagram are shown in following

sec/2000radBn =≈ωω

Mpω

Determine the transfer function of the system that has the following frequency response:

15+s

Problems: Experimental determination of transfer function of a system based on its frequency response

Determine the transfer function of the system that has the following frequency response:

)1)(101.0()10(1.0

)1)(101.0(11.0

+++

=++

+ss

sss

s

Problems: Experimental determination of transfer function of a system based on its frequency response

Summary

• In this chapter we have considered the representation of a feedback control system by its frequency response characteristics.

• The frequency response of a system was defined as the steady-state response of the system to a sinusoidal input signal.

• Several alternative forms of frequency plots were considered, including the polar plot of the frequency response of a system G(jω) and logarithmic plots, often called Bode plots, and the value of the logarithmic measure was illustrated.

• The ease of obtaining a Bode plot for the various factors of G(jω) was noted, and an example was considered in detail.

• The asymptotic approximation for sketching the Bode diagram simplifies the computation considerably

Summary of fifteen typical Bode plots

1.

2.

3.


5.


7.

9.


12.

11.

10.


14.

13.

Documents

Frequency Response Methods - Department of …me.emu.edu.tr/ranjbar/Lecture9Control.pdf · • The important issue in frequency response methods is ... Bode Diagram • Advantage