IntroCtrlSys_Chapter3.pdf

Lecture NotesLecture Notes

Introduction to Control SystemsIntroduction to Control Systems

Instructor: Dr. Huynh Thai HoangDepartment of Automatic Control

Faculty of Electrical & Electronics EngineeringHo Chi Minh City University of Technology

Email: [email protected]@yahoo.com

Homepage: www4.hcmut.edu.vn/~hthoang/

24 September 2011 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1

Chapter 3Chapter 3

SYSTEM DYNAMICSSYSTEM DYNAMICSSYSTEM DYNAMICSSYSTEM DYNAMICS

24 September 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 2

The concept of system dynamics

ContentContent

The concept of system dynamicsTime responseFrequency responseq y p

Dynamics of typical componentsDynamics of control systems


The concept of system dynamicsThe concept of system dynamics


The concept of system dynamicsThe concept of system dynamics

System dynamics is the study to understanding the behaviour ofSystem dynamics is the study to understanding the behaviour of complex systems over time.

Systems described by similar mathematical model will exposeSystems described by similar mathematical model will expose similar dynamic responses.

T t d th d i i t i l ll h tTo study the dynamic responses, input signals are usually chosen to be basic signals such as Dirac impulse signal, step signal, or sinusoidal signal.

Time responseImpulse responseStep response

Frequency response


Impulse responseImpulse response

G(s)U (s) Y (s)

Impulse response: behavior of a system to a Dirac impulse)()().()( sGsGsUsY == (due to U(s) = 1)

{ } { } )()()()( 11 tgsGsYty === −− LL⇒ Impulse response is the inverse Laplace transform of the transfer function

Impulse response is also referred as weighting function.It is possible to calculate the response of a system to a arbitrarily input by taking convolution of the weighting function and the input.

∫t

dtttt )()()(*)()(


∫ −== dtugtutgty0

)()()(*)()( τττ

Step responseStep response

G(s)U (s) Y (s)

Step response: behavior of a system to a step input

ssGsGsUsY )()().()( == (because U(s) = 1/s)

∫⎫⎧ tsG )({ } ∫=⎭⎬⎫

⎩⎨⎧== −− dg

ssGsYty

0

11 )()()()( ττLL

⇒ The step response is the integral of the impulse response⇒ The step response is the integral of the impulse responseThe step response is also referred as the transient function of the system.


Impulse and step response exampleImpulse and step response example

Calculate the impulse response and step response of the system

G(s)U (s) Y (s)

Calculate the impulse response and step response of the system described by the transfer function:

)5(1)( +

=ssG ( )

)5()(

+ss

{ } ⎬⎫

⎨⎧

+⎬⎫

⎨⎧ + −−− 411)()( 111 sGt LLL

Impulse response:

{ }⎭⎬

⎩⎨ +

+=⎭⎬

⎩⎨ +

==)5(55)5(

)()(ssss

sGtg LLL

tetg 5

54

51)( −+=⇒

55

4141)( ⎫⎧ +⎫⎧ ssGStep response:

)5(254

51

254

)5(1)()( 22

11

+−+=

⎭⎬⎫

⎩⎨⎧

++

=⎭⎬⎫

⎩⎨⎧= −−

ssssss

ssGth LL

441)( 5− th


25255)( 5 +−= − tetth⇒

Frequency responseFrequency response

Observe the response of a linear system at steady state when theinput is a sinusoidal signal.


Frequency response definitionFrequency response definition

It can be observed that for linear system if the input is a sinusoidalIt can be observed that, for linear system, if the input is a sinusoidalsignal then the output signal at steady-state is also a sinusoidalsignal with the same frequency as the input, but different amplitude

d hand phase.

HTU (j ) ( )

u (t)=Umsin (jω) y (t)=Ymsin (jω+ϕ)

Definition: Frequency response of a system is the ratio between the

U (jω) Y (jω)

Definition: Frequency response of a system is the ratio between thesteady-state output and the sinusoidal input.

)( ωjYF)()(

ωjUj

=responseFrequency

It is proven that: )()(responseFrequency ωjGsG ==


It is proven that: )()(responseFrequency ωω

jGsGjs==

=

Magnitude response and phase responseMagnitude response and phase response

In general G(jω) is a complex function and it can be represented inIn general, G(jω) is a complex function and it can be represented inalgebraic form or polar form.

)().()()()( ωϕωωωω jeMjQPjG =+=

where:

)()()()( 22 ωωωω QPjGM +== Magnitude response)()()()( ωωωω QjG Magnitude response

⎥⎦

⎤⎢⎣

⎡=∠= −

)()()()( 1

ωωωωϕ

PQtgjG Phase response⎥

⎦⎢⎣ )(ωP

Physical meaning of frequency response:The magnitude response provides information about the gain ofthe system with respect to frequency .The phase response provides information about the phase shift


p p p pbetween the output and the input with respect to frequency .

Graphical representation of frequency responseGraphical representation of frequency response

Bode diagram: is a graph of the frequency response of a linearBode diagram: is a graph of the frequency response of a linearsystem versus frequency plotted with a log-frequency axis. Bodediagram consists of two plots:

Bode magnitude plot expresses the magnitude response gainL(ω) versus frequency ω .

)(l20)( ML

Bode phase plot expresses the phase response ϕ(ω) versusf

)(lg20)( ωω ML = [dB]

frequency ω.

Nyquist plot: is a graph in polar coordinates in which the gain andphase of a frequency response G(jω) are plotted when ω changingfrom 0→+∞.


Graphical representation of frequency response (cont’)Graphical representation of frequency response (cont’)

Bode diagram Nyquist plotBode diagram Nyquist plotBode diagram Nyquist plotBode diagram Nyquist plot

Gain margin

Gain marginGain margin

Phase margin

Phase margin

Phase margin


Crossover frequencyCrossover frequency

Gain crossover frequency(ω ): is the frequency where the amplitudeGain crossover frequency(ωc): is the frequency where the amplitudeof the frequency response is 1 (or 0 dB).

1)( =cM ω 0)( =cL ω⇔)( c c

Phase crossover frequency (ω−π): is the frequency where phase shiftof the frequency response is equal to −1800 (or equal to −π radian).

0180)( −=−πωϕ rad )( πωϕ π −=−⇔


Stability marginStability margin

G i i (GM)Gain margin (GM):

)(1ω

=M

GM ⇔ )( πω−−= LGM [dB] )( πω−M

Phase margin (ΦM))(1800

cM ωϕ+=Φ )( cϕ

The phase margin is the amount of additional phase lag at the gain crossover frequency required to bring the system to the stability boundary.


Dynamics of basic factorsDynamics of basic factors


The proportional gainThe proportional gain

Transfer function: KsG )(Transfer function: KsG =)(

Time response:

Impulse response:

Step response:

)()( tKtg δ=

)(1)( tKth =

Frequency response:

M it d

KjG =)( ω

KM )( KL l20)(⇒Magnitude response: KM =)(ω KL lg20)( =ω

0)( =ωϕ

⇒

Phase response:


The proportional gain The proportional gain –– Time responseTime response

(a) Weighting function (b) Transient function( ) g g ( )


The proportional gain The proportional gain –– Frequency responseFrequency response


Bode diagram Nyquist plot

Integral factorIntegral factor

T f f ti G 1)(Transfer function:s

sG )( =

Time response:

Impulse response:

Step response:

)(1)( tKtg =

)(1)( tKtth =

Frequency response:ωω

ω 11)( jj

jG −==

p p )()(

Magnitude response:

ωωj

ωω 1)( =M ωω lg20)( −=L⇒

ω090)( −=ωϕPhase response:


Time response of integral factor Time response of integral factor

(a) Weighting function (b) Transient function(a) Weighting function (b) Transient function


Frequency response of integral factor Frequency response of integral factor



Derivative factorDerivative factor

Transfer function: ssG =)(Transfer function: ssG =)(

Time response:

Impulse response:

Step response:

)()( tKtg δ&=

)()( tKth δ=

Frequency response: ωω jjG =)(

Magnitude response: ωω =)(M ωω lg20)( =L

090)(Phase response:

⇒

090)( =ωϕPhase response:


Time response of derivative factorTime response of derivative factor

Transient function


Frequency response of derivative factorFrequency response of derivative factor



FirstFirst--order lag factororder lag factor

1Transfer function:1

1)(+

=Ts

sG

Time response:

Impulse response: )(111

1)( 1 teTTs

tg Tt

−− =⎭⎬⎫

⎩⎨⎧

+=L

Step response: )(1)1()1(

1)( 1 teTss

th Tt

−− −=⎭⎬⎫

⎩⎨⎧

+= L

⎭⎩


Time response of firstTime response of first--order lag factororder lag factor



Frequency response of firstFrequency response of first--order lag factororder lag factor

1)(Frequency response:

Magnitude response:

11)(+

=ω

ωTj

jG

1)(ωM =Magnitude response:

⇒

221)(

ωω

TM

+=

221lg20)( ωω TL +−=

)()( 1 ωωϕ Ttg−−=Phase response:

g)(

Approximation of the Bode diagram by asymptotes:

: the asymptote lies on the horizontal axisT/1<ω : the asymptote lies on the horizontal axis

: the asymptote has the slope of −20dB/dec

T/1<ω

T/1>ω


Frequency response of firstFrequency response of first--order lag factor (cont’)order lag factor (cont’)

corner frequencycorner frequency



FirstFirst--order lead factororder lead factor

T f f i 1)( TGTransfer function: 1)( +=TssG

Time response:

Step response: )(1)()1()( 1 ttTs

Tsth +=⎭⎬⎫

⎩⎨⎧ +

= − δL

)()()()( ttTthtg δδ +== &&Impulse response:


Time response of firstTime response of first--order lead factororder lead factor

Transient function


Frequency response of firstFrequency response of first--order lead factororder lead factor

Frequency response:

Magnitude response:

1)( += ωω TjjG

221)( ωω TM +=Magnitude response:

⇒

1)( ωω TM +=

221lg20)( ωω TL +=

)()( 1 ωωϕ Ttg−=Phase response:

Approximation of the Bode diagram by asymptotes:: the asymptote lies on the horizontal axisT/1<ω

: the asymptote has the slope of +20dB/decT/1>ω


Frequency response of firstFrequency response of first--order lead factor (cont’)order lead factor (cont’)

corner frequency



SecondSecond--order oscillating factororder oscillating factor

1Transfer function:

Time-domain dynamics:12

1)( 22 ++=

TssTsG

ξ)10( << ξ

Impulse response: [ ]tetg n

tn

n

)1(sin1

)( 2

2ξω

ξ

ω ξω

−−

=−

Step response:

ξ

[ ]θξωξ

ξω

+−−

−=−

teth n

tn

)1(sin1

1)( 2

2ξ1


Time response of secondTime response of second--order oscillating factororder oscillating factor

(a) Weighting function (b) Transient function


Frequency response of secondFrequency response of second--order oscillating factororder oscillating factor

1Frequency response:

M i d

121)( 22 ++−

=ωξω

ωTjT

jG

1)(MMagnitude response:

⇒

222222 4)1()(

ωξωω

TTM

+−=

222222 4)1(l20)( ξ TTL +

Phase response:

⇒ 222222 4)1(lg20)( ωξωω TTL +−−=

⎟⎠⎞

⎜⎝⎛−= −

221

12)( ωξωϕ

TTtg

Approximation of the Bode diagram by asymptotes:

⎠⎝ − 221 ωT

: the asymptote lies on the horizontal axis

: the asymptote has the slope of −40dB/dec

T/1<ω

T/1>ω


Frequency response of secondFrequency response of second--order oscillating factororder oscillating factor

Corner frequency



Time delay factorTime delay factor

TTransfer function:

Time-domain dynamics:

TsesG −=)(

Impulse response: { } )()( 1 Ttetg Ts −== −− δL

Step response: )(1)( 1 Tts

ethTs

−=⎭⎬⎫

⎩⎨⎧

=−

−L⎭⎩


Time response of time delay factorTime response of time delay factor



Frequency response of time delay factorFrequency response of time delay factor

Frequency response:

M it d

ωω TjejG −=)(

1)(M 0)(LMagnitude response:

ωωϕ T−=)(Phase response:

⇒1)( =ωM 0)( =ωL


Frequency response of time delay factor (cont’)Frequency response of time delay factor (cont’)



Dynamics of control systemsDynamics of control systems


Time response of control systemsTime response of control systems

Consider a control system which has the transfer function G(s):Consider a control system which has the transfer function G(s):

nnmm

mm

asasasabsbsbsbsG

++++++++

= −−

−

11

10

11

10)(L

L

nn asasasa ++++ −110

Laplace transform of the transient function:1

sasasasasbsbsbsb

ssGsH

nnnn

mmmm

)()()(

11

10

11

10

++++++++

==−

−−

−

L

L


Remarks on the time response of control systemsRemarks on the time response of control systems

If G( ) d t t i id l i t l d i ti f t thIf G(s) does not contain a ideal integral or derivative factor then:Weighting function decays to 0

1 ⎞⎛ ++++ −mm bsbsbsb 0lim)(lim)(1

110

110

00=⎟⎟

⎠

⎞⎜⎜⎝

⎛++++++++

==∞−

−−

→→nn

nnmm

ss asasasabsbsbsbsssGg

L

L

Transient function approaches to non-zero value at steady state:

0.1lim)(lim)( 11

110 ≠=⎟⎟

⎞⎜⎜⎛ ++++

==∞ −−

mmmmm bbsbsbsbsssHh L 0.lim)(lim)(

11

1000

≠⎟⎟⎠

⎜⎜⎝ ++++

∞−

−→→nnn

nnss aasasasassssHh

L


Remarks on the time response of control systems (cont’)Remarks on the time response of control systems (cont’)

If G( ) t i id l i t l f t ( 0) thIf G(s) contain a ideal integral factor (an = 0) then:Weighting function has non-zero steady-state:

1 ⎞⎛ ++++ − bsbsbsb mm

0lim)(lim)(1

110

110

00≠⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

++++==∞

−−

−

→→ sasasabsbsbsbsssGg

nnn

mm

ss L

L

Transient function approaches infinity at steady-state

∞=⎟⎟⎞

⎜⎜⎛ ++++

==∞ −− bsbsbsbsssHh mm

mm1

1101lim)(lim)( L

∞=⎟⎟⎠

⎜⎜⎝ +++

==∞−

−→→ sasasassssHh

nnnss

11

1000

.lim)(lim)(L



If G( ) t i id l d i ti f t (b 0) thIf G(s) contain a ideal derivative factor (bm = 0) then:Weighting function approaches zero at steady-state.

1 ⎞⎛ +++ −mm sbsbsb 0lim)(lim)(1

110

110

00=⎟⎟

⎠

⎞⎜⎜⎝

⎛++++

+++==∞

−−

−

→→nn

nnm

ss asasasasbsbsbsssGg

L

L

Transient function approaches zero at steady-state.

01lim)(lim)( 11

10 =⎟⎟⎞

⎜⎜⎛ +++

==∞ −−

mmm sbsbsbsssHh L 0.lim)(lim)(

11

1000

=⎟⎟⎠

⎜⎜⎝ ++++

==∞−

−→→nn

nnss asasasassssHh

L



If G( ) i ( ≤ ) th h(0) 0If G(s) is proper (m ≤ n) then h(0) = 0.

0.1lim)(lim)0( 11

110 =⎟⎟

⎠

⎞⎜⎜⎝

⎛ +++== −

−−

nnm

mm sbsbsbsHh L

11

10⎟⎠

⎜⎝ ++++ −

∞→∞→nn

nnss asasasas L

If G(s) is strictly proper (m < n) then g(0) = 0.( ) y p p ( ) g( )

0lim)(lim)0(1

110

11

10 =⎟⎟⎠

⎞⎜⎜⎝

⎛++++

+++==

−−

−−

∞→∞→nn

nnm

mm

ss asasasasbsbsbsGg

L

L

110 ⎠⎝ nn


Frequency response of control systemFrequency response of control system

Consider a control system which has the transfer function G(s)Consider a control system which has the transfer function G(s).Suppose that G(s) consists of basis factors in series:

∏l

GG )()( ∏=

=i

i sGsG1

)()(

∏l

Frequency response: ∏=

=i

i jGjG1

)()( ωω

Magnitude response: ∏=l

MM )()( ωω ∑=l

LL )()( ωω⇒

Phase response: ∑l

)()( ωϕωϕ

Magnitude response: ∏=

=i

iMM1

)()( ωω ∑=

=i

iLL1

)()( ωω⇒

⇒ The Bode diagram of a system consisting of basic factors in series

Phase response: ∑=

=i

i1

)()( ωϕωϕ


⇒ The Bode diagram of a system consisting of basic factors in seriesequals to the summation of the Bode diagram of the basic factors.

Approximation of Bode diagramApproximation of Bode diagram

S th t th t f f ti f th t i f th fSuppose that the transfer function of the system is of the form:K)()()()( 321 sGsGsGKssG α=

(α>0: the system has ideal derivative factor(s)

Step 1: Determine all the corner frequency ω =1/T and sort them

(α>0: the system has ideal derivative factor(s)α<0: the system has ideal integral factor(s))

Step 1: Determine all the corner frequency ωi =1/Ti , and sort themin ascending order ω1 <ω2 < ω3 …

Step 2: The approximated Bode diagram passes the point A havingStep 2: The approximated Bode diagram passes the point A havingthe coordinates:

⎨⎧ = 0ωω

⎩⎨ ×+= 0lg20lg20)( ωαω KL

ω0 is a frequency satisfying ω0 < ω1 . If ω1 > 1 then it is possible to


chose ω0 =1.

Approximation of Bode diagram (cont’)Approximation of Bode diagram (cont’)

Step 3: Through point A draw an asymptote with the slope:Step 3: Through point A, draw an asymptote with the slope:(− 20 dB/dec × α) if G(s) has α ideal integral factors(+ 20 dB/dec × α) if G(s) has α ideal derivative factors

The asymptote extends to the next corner frequency.

Step 4: At the corner frequency ωi =1/Ti , the slope of the asymptote is S ep : e co e eque cy ωi / i , e s ope o e asy p o e sadded with:

(−20dB/dec × βi) if Gi(s) is a first-order lag factor (multiple βi)(+20dB/dec × β ) if G (s) is a first-order lead factor (multiple β )(+20dB/dec × βi) if Gi(s) is a first-order lead factor (multiple βi)(−40dB/dec × βi) if Gi(s) is a second-order oscillating factor (multiple βi)(+40dB/dec × βi) neáu Gi(s) is a second-order lead factor (multiple βi)

The asymptote extends to the next corner frequencyThe asymptote extends to the next corner frequency.

Step 5: Repeat the step 4 until the asymptote at the last corner


p p p y pfrequency is plotted.

Approximation of Bode diagram Approximation of Bode diagram –– Example 1Example 1

Plot the Bode diagram using asymptotes:Plot the Bode diagram using asymptotes:

)101,0()11,0(100)(

++

=ssssG

Based on the Bode diagram, determine the gain cross frequency ofthe system.

Solution:Corner frequencies:

(rad/sec) 10001,011

22 ===

Tω(rad/sec) 10

1,011

11 ===

Tω

The Bode diagram pass the point A at the coordinate:

⎨⎧ =1ω


⎩⎨ === 40100lg20lg20)( KL ω

Approximation of Bode diagram Approximation of Bode diagram –– Example 1 (cont’)Example 1 (cont’)

L(ω), dB

A

20dB/dec40

−20dB/dec

20dB/dec

0dB/dec20

−20dB/dec

ωc0

ω

lgω

100 10110-1

10-1 2

102

3


In the Bode diagram, the gain crossover frequency is 103 rad/sec.

Example 2 Example 2 –– Bode diagram to transfer function Bode diagram to transfer function

D i h f f i f h hi h h hDetermine the transfer function of the system which has theapproximation Bode diagram as below:

L(ω), dB

60 0dB/dec

20dB/d40

60 0dB/dec54

A

D E

−20dB/dec40

20 0dB/dec26

B C

0lgω10-1 21.301


gωg1 ωg2 ωg3

Example 2 Example 2 –– Bode diagram to transfer function (cont’)Bode diagram to transfer function (cont’)

2654 (dB/dec) 40301.122654

+=−−The slope of segment CD:

The corner frequencies:

(rad/sec) 510 7.01 ==gω

e co e eque c es:

7.020

26400lg 1 =−

+=gω ⇒

301.1lg 2 =gω ⇒ (rad/sec) 2010 301.12 ==gω

2lg 3 =gω ⇒ (rad/sec) 1001023 ==gω

The transfer function has the form: 23

221

)1()1)(1()(

+++

=sTs

sTsTKsG

3g

3 )(100 40lg20 =⇒= KK

0.2111 ===T 0.0511

2 ===T 0.01113 ===T


0.2 51

1g

Tω

0.05202

2g

Tω

0.011003

3g

Tω

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