IntroCtrlSys_Chapter5

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/

11 October 2011 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1

Chapter 5Chapter 5

ANALYSIS OF CONTROL SYSTEM ANALYSIS OF CONTROL SYSTEM PERFORMANCEPERFORMANCE

11 October 2011 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 2

Performance criteria

ContentContent

Performance criteriaSteady state errorTransient responsepThe optimal performance indexRelationship between frequency domain performances and timedomain performances.


Performance criteriaPerformance criteria


Performance criteria: Steady state errorPerformance criteria: Steady state error

(t)yfb(t)

r(t)ess

G(s)+_R(s) Y(s)E(s)

t0

e(t) essH(s)Yfb(s)

Error: is the difference between the set-point (input) and the feedback

0

Error: is the difference between the set point (input) and the feedback signal.

)()()( tytrte fb−= )()()( sYsRsE fb−=⇔

Steady-state error: is the error when time approaching infinity.

)(li )(li E


)(lim teetss ∞→

= )(lim0

ssEesss →

=⇔

Overshoot: refers to an output exceeding its steady state value

Performance criteria Performance criteria –– Percent of Overshoot (POT)Percent of Overshoot (POT)

Overshoot: refers to an output exceeding its steady-state value.

y(t)overshoot

y(t)

yss

overshoot

yss

ymaxymax− yss

t0

No overshoot t0

yss

Percent of Overshoot (POT) is an index to quantify the overshoot oft POT i l l t da system, POT is calculated as:

%100max ×−

= ssyyPOT


ssy

S ttli ti (t ) i th ti i d f th f t t

Performance criteria Performance criteria –– Settling time and rise timeSettling time and rise time

Settling time (ts): is the time required for the response of a system toreach and stay within a range about the steady-state value of sizespecified by absolute percentage of the steady-state value (usually2% or 5%)Rise time (tr): is the time required for the response of a system torise from 10% to 90% of its steady-state valuerise from 10% to 90% of its steady state value.

y(t) y(t)

(1+ε)yss

(1−ε)yss

yss yss0.9yss

t0

t0.1yss


0 ts

0tr

SteadySteady--state errorstate error


SteadySteady--state errorstate error

R( ) Y( )E( )G(s)+_

R(s) Y(s)E(s)

H(s)Yfb(s)

)()(1)()(

sHsGsRsE

+=Error expression:

)()(1)(lim)(lim

00 sHsGssRssEe

ssss +==

→→Steady-state error:

)()(100 sHsGss +→→

Remark: Steady-state error not only depends on the structure andt f th t b t l d d th i t i l


parameters of the system but also depends on the input signal.

SteadySteady--state error to step inputstate error to step input

Step input: ssR /1)( =Step input: ssR /1)( =

⇒ Steady-state error: ss Ke

+=

11

with )()(lim0

sHsGKsp →

= (position constant)

pK+1

yfb(t) yfb(t)

1 1

0t

G(s)H(s) does not have0

t

G(s)H(s) has at least 1


G(s)H(s) does not have deal integral factor

G(s)H(s) has at least 1 ideal integral factor

SteadySteady--state error to ramp inputstate error to ramp input

Ramp input: 2/1)( ssRRamp input: /1)( ssR =

ss Ke 1

= with )()(lim0

sHssGKsv →

= (velocity constant) ⇒vK 0s→

yfb(t)

r(t)

e ≠ 0

yfb(t)

r(t)

yfb(t)

r(t)

0

e(t)→∞

ess ≠ 0 ess = 0

G(s)H(s) does not have

0t

G(s)H(s) has 1 ideal

0t

G(s)H(s) has at least 2

0t


G(s)H(s) does not have deal integral factor

G(s)H(s) has 1 ideal integral factor

G(s)H(s) has at least 2 ideal integral factors

SteadySteady--state error to parabolic inputstate error to parabolic input

P b li i 3Parabolic input: 3/1)( ssR =

e 1= with )()(lim 2 sHsGsKa = (acceleration ⇒

ass K

e with )()(lim0

sHsGsKsa → constant )

⇒

yfb(t)

r(t)

e = 0ess≠0

yfb(t)

r(t)

yfb(t)

r(t)

e(t)→∞ess 0ss

G(s)H(s) has less than 20

t

G(s)H(s) has 2 ideal

0t

G(s)H(s) has more than 20

t


G(s)H(s) has less than 2 ideal integral factors

G(s)H(s) has 2 ideal integral factors

G(s)H(s) has more than 2 ideal integral factors

Transient responseTransient response


FirstFirst--order systemorder system

1+TsK Y(s)R(s)

Transfer function:1

)( =T

KsG1

)(+Ts

First order system has 1 real pole: T

p 11 −=

T1

Transient response: 1

.1)()()(+

==Ts

Ks

sGsRsY1+Tss

⇒ )1()( /TteKty −−=


FirstFirst--order system (cont’)order system (cont’)

Im s y(t)

(1+ε).K

(1−ε).KRe s

K

0−1/T0.63K

ts

Pole zero plotT

t0

Transient responsePole – zero plot of a first order system

Transient response of the first order

)1()( /TteKty −=


)1()( eKty −=

FirstFirst--order system order system –– Remarks Remarks

First order system has only one real pole at (−1/T), its transientresponse doesn’t have overshoot.

Ti T i h i i d f h f hTime constant T: is the time required for the step response of thesystem to reach 63% its steady-state value.

The further the pole ( 1/T) of the system is from the imaginaryThe further the pole (−1/T) of the system is from the imaginaryaxis, the smaller the time constant and the faster the time responseof the system.

Settling time of the first order system is:

⎟⎞

⎜⎛ 1lTt ⎟

⎠⎞

⎜⎝⎛=ε

lnTts

where ε = 0 02 (2% criterion) or ε = 0 05 (5% criterion)


where ε 0.02 (2% criterion) or ε 0.05 (5% criterion)

FirstFirst--order systemorder system

The relationship between the pole and the time responseThe relationship between the pole and the time responseThe relationship between the pole and the time responseThe relationship between the pole and the time response

The further the pole of the system is from the imaginary axis, thesmaller the time constant and the faster the time response of the

Im s y(t)

smaller the time constant and the faster the time response of thesystem.

Re sK

0

Pole zero plot Transient response

t0


Pole – zero plot of a first order system

Transient response of the first order

SecondSecond--order oscillating systemorder oscillating system

1222 ++ TssTKξ

Y(s)R(s)

The transfer function of the second-order oscillating system:2

)( nKKsG ω== )101( <<= ξω

The system has two complex conjugate poles:

2222 212)(

nnssTssTsG

ωξωξ ++=

++= )10 ,( <<= ξω

Tn

Transient response: 22

2

.1)()()( nKsGsRsY ω==

22,1 1 ξωξω −±−= nn jp

Transient response: 22 2.)()()(

nnssssGsRsY

ωξω ++

⇒ [ ]⎪⎬⎫⎪⎨⎧

+−−=−

θξωξω

teKtytn

)1(sin1)( 2 )(cos ξθ =


⇒ [ ]⎪⎭⎬

⎪⎩⎨ +

−θξω

ξtKty n )1(sin

11)(

2)(cos ξθ

SecondSecond--order oscillating system (cont’)order oscillating system (cont’)

y(t)Im s

(1+ε).K

(1−ε).KK

Re s

21 ξω −njωn

cos θ= ξ

( )Re s0−ξωn

21 ξω −− j

θ

tqñ

t0

1 ξωnj

Transient response of a second order oscillating system

Pole – zero plot of a second order oscillating system


SecondSecond--order oscillating system order oscillating system –– RemarkRemark

A second order oscillation system has two conjugated complexA second order oscillation system has two conjugated complexpoles, its transient response is a oscillation signal.

If ξ 0 i iIf ξ = 0, transient response isa stable oscillation signal atthe frequency ωn ⇒ ωn is

ξ = 0ξ = 0.2

called natural oscillationfrequency.If 0<ξ<1, transient response

ξ = 0.4

ξ , pis a decaying oscillationsignal ⇒ ξ is called dampingconstant, the larger the value ξ = 0.6, gξ , (the closer the poles are tothe real axis) the faster theresponse decays.

ξ = 0.6


response decays.

SecondSecond--order oscillating system order oscillating system –– OvershootOvershoot

Transient response of the second order oscillating system has

%100⎟⎞

⎜⎛ ξπPOT

Transient response of the second order oscillating system hasovershoot.

The percentage of o ershoot: %100.1

exp2 ⎟⎟⎠

⎜⎜⎝ −−=

ξξPOTThe percentage of overshoot:

The larger the value ξ, (the closer the poles are to the real axis) the smaller the (%

)

ea a s) e s a e ePOT.The smaller the value ξ, (the closer the poles are to

POT

(

(the closer the poles are to the imaginary axis) the larger the POTξ

The relationship


The relationship between POT and ξ

SecondSecond--order oscillating system order oscillating system –– Settling timeSettling time

Settling time:3

n

tξω

3=s5% criterion:

n

tξω

4=s2% criterion:



Relationship between pole location and transient responseRelationship between pole location and transient responseRelationship between pole location and transient responseRelationship between pole location and transient responseThe second order systems that have the poles located in the samerays starting from the origin have the same damping constant, thenh f h h h f h h l

Im s y(t)

the percentage of overshoots are the same. The further the polesfrom the origin, the shorter the settling time.

Im s

R

y(t)

Kcosθ = ξ Re s

0θ

cosθ = ξ

Pole – zero plot of a second Transient response of a second

t0


Pole zero plot of a second order oscillating system



Relationship between pole location and transient response (cont’)Relationship between pole location and transient response (cont’)Relationship between pole location and transient response (cont )Relationship between pole location and transient response (cont )The second order systems that have the poles located in the same distance from the origin have the same natural oscillation

Im s y(t)

frequency. The closer the poles to the imaginary axis, the smaller the damping constant, then the higher the percentage of overshoot.

s

R

y(t)

Kω Re s0

ωn

t0






Relationship between pole location and transient response (cont’)Relationship between pole location and transient response (cont’)Relationship between pole location and transient response (cont )Relationship between pole location and transient response (cont )The second order systems that have the poles located in the samedistance from the imaginary axis have the same ξωn, then the settling

Im s y(t)

time are the same. The further the poles from the real axis, the smallerthe damping constant, then the higher the percentage of overshoot.

s

Re s

y(t)

KRe s

0−ξωn

t0





Transient response of high order systemTransient response of high order system

High order systems are the system that have more than 2 polesHigh-order systems are the system that have more than 2 poles.If a high order system have a pair of poles located closer to theimaginary axis than the others then the high order system can beapproximated to a second order system. The pair of poles nearest tothe imaginary axis are called the dominated poles.

Im s y(t) f hi hs

Re s

y(t) Response of highorder system

Re s

0 Response of second order system with the

dominated poles

High order systems A high order system can be approximated0

dominated poles

t


High order systems have more than 2 poles

A high order system can be approximated by a dominated-pole second order systems

Performance indicesPerformance indices


Integral performance indicesIntegral performance indices

IAE criterionIAE criterion(Integral of the Absolute Magnitude of the Error )

∫+∞

∫=0

)( dtteJIAE

ISE criterionISE criterion(Integral of the Square of the Error)

∫+∞

2 )( dJ

ITAE criterion

∫=0

2 )( dtteJISE

(Integral of Time multiplied by the Absolute Value of the Error)

∫+∞

= )( dttetJ


∫=0

)( dttetJITAE

Optimum systemsOptimum systems

A control system is optimum when the selected performance index isA control system is optimum when the selected performance index is minimized

min→IAEJ when 707.0→ξiJ 50ξ

Second order system:min→ISEJ when 5.0→ξmin→ITAEJ when 707.0→ξ

y(t) ξ=0.3

ξ=0.5

ξ=0.707

0t

ξ=0.9


Transient response of second order systems0

ITAE optimal controlITAE optimal control

ITAE is usually used in design of control systemITAE is usually used in design of control systemAn n-order system is optimal according to ITAE criterion if thedenominator of its transfer function has the form:


ITAE optimal control (cont’)ITAE optimal control (cont’)

Optimal response according to ITAE criterionOptimal response according to ITAE criterion

y(t)1st d1st order system

2nd order system

3rd order system

4th order system

0t


Relationship between frequency domain Relationship between frequency domain p q yp q yperformances and time domain performancesperformances and time domain performances


Relationship between frequency response and steady state errorRelationship between frequency response and steady state error

R(s) Y(s)R(s)G(s)+−

Y(s)

)()(lim)()(lim00

ωωω

jHjGsHsGKsp →→

==

)()(lim)()(lim00

ωωωω

jHjGjsHsGsKsv →→

==

)()()(lim)()(lim 2

0

2

0ωωω

ωjHjGjsHsGsK

sa →→==


Relationship between frequency response and steady state errorRelationship between frequency response and steady state error

R( ) Y( )R(s)G(s)+

−Y(s)

Steady state error of the closed loop system depends on theSteady state error of the closed-loop system depends on themagnitude response of the open-loop system at low frequencies butnot at high frequencies.

The higher the magnitude response of the open-loop system at lowfrequencies, the smaller the steady-state error of the closed-loops stemsystem.In particular, if the magnitude response of the open-loop system isinfinity as frequency approaching zero, then the steady-state error of


y q y pp g , ythe closed-loop system to step input is zero.

Relationship between frequency response and transient responseRelationship between frequency response and transient response


Y(s)

In the frequency range ω <ω because then:1)( >ωjGIn the frequency range ω <ωc , because then:1)( >ωjG

1)()(

)(1)(

)( =≈+

=ωω

ωω

ωjGjG

jGjG

jGcl )()( ωω jGjG

In the frequency range ω >ωc , because then:1)( <ωjG)()( ωω jGjG

⇒ Bandwidth of the closed loop system is approximate the gain

)(1

)()(1

)()( ω

ωω

ωω jG

jGjG

jGjGcl =≈

+=


⇒ Bandwidth of the closed-loop system is approximate the gaincrossover frequency of the open-loop system.


Bode plot of a open-loop system Bode plot of the corresponding closed-loop system




Y(s)

The higher the gain crossover frequency of open loop system theThe higher the gain crossover frequency of open-loop system, thewider the bandwidth of closed-loop system ⇒ the faster theresponse of close-loop system, the shorter the settling time.

cqd

c

tωπ

ωπ 4

<<

The higher the phase margin of the open-loop system, the smallerthe POT of closed-loop system. In most of the cases, if the phasemargin of the open loop system is larger than 600 then the POT of


margin of the open-loop system is larger than 600 then the POT ofthe closed-loop system is smaller than 10%.

Example of relationship between gain crossover frequency and settling timeExample of relationship between gain crossover frequency and settling time

R(s) Y(s)

)1080)(110(10)(

++=

ssssG

R(s)G(s)+−

Y(s)

)108.0)(11.0( ++ sss


Example of relationship between gain crossover frequency and settling timeExample of relationship between gain crossover frequency and settling time


Y(s)

)110(50)(+

=ss

sG)11.0( +ss


Example of relationship between phase margin and POTExample of relationship between phase margin and POT

R(s) Y(s)

)1080)(110(6)(

++=

ssssG

R(s)G(s)+−

Y(s)

)108.0)(11.0( ++ sss


Example of relationship between phase margin and POT (cont’)Example of relationship between phase margin and POT (cont’)

R(s) Y(s)

)110(6)(+

=ss

sG

R(s)G(s)+−

Y(s)

)11.0( +ss


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