performance of feedback control systems (stachowicz, 2010)

7/29/2019 Performance of Feedback Control Systems (Stachowicz, 2010)

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Performance of Feedback Control

Systems

Prof. Marian S. StachowiczLaboratory for Intelligent Systems

ECE Department, University of Minnesota Duluth

February 25 March 2, 2010

ECE 3151 - Spring 2010


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Outline

Introduction

Test Input Signals

Performance of a second-order system Effects of a Third Pole and a Zero on the Second-

Order System Response

Estimation of the Damping Ratio

The s-plane Root Location and the TransientResponse

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References for reading

1. R.C. Dorf and R.H. Bishop, Modern Control Systems,

11th Edition, Prentice Hall, 2008,Chapter 5.1 - 5.12

2. J.J. DiStefano, A. R. Stubberud, I. J. Williams,Feeedback and Control Systems, Schaum's OutlineSeries, McGraw-Hill, Inc., 1990Chapter 9

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Test Input Signal

Since the actual input signal of the systemis usually unknown, a standard test input

signal is normally chosen. Commonlyused test signals include step input, rampinput, and the parabolic input.

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General form of the standard test

signals

r(t) = tn

R(s) = n!/sn+1

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Test signals r(t) = A tn

n = 0 n = 1 n = 2

r(t) = A r(t) = At r(t) = At2R(s) = 2A/s3R(s) = A/s R(s) = A/s2

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Table 5.1 Test Signal Inputs

Test Signal r(t) R(s)

Step

position

r(t) = A, t > 0

= 0, t < 0

R(s) = A/s

Ramp

velocity

r(t) = At, t > 0

= 0, t < 0

R(s) = A/s2

Parabolic

acceleration

r(t) = At2, t > 0

= 0, t < 0

R(s) = 2A/s3

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Test inputs vary with target type

parabola

ramp

step

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Steady-state error

Is a difference between input and the

output for a prescribed test input as

t

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Application to stable systems

Unstable systems represent loss ofcontrol in the steady state and are

not acceptable for use at all.

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Steady-state error:a) step input, b) ramp input

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Time response of systems

c(t) = ct(t) + css(t)

The time response of a control system is dividedinto two parts:

ct(t) - transient response

css(t) - steady state response

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Transient response

All real control systems exhibit transientphenomena to some extend before steady

state is reached.

lim ct(t) = 0 for t

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Steady-state response

The response that exists for a long timefollowing any input signal initiation.

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Poles and zeros of a first order system

Css(t) Ct(t

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A pole on the real axis generate an exponential response

of the form Exp[-t] where - is the pole location on real axis.

The farther to the left a pole is on the negative real axis,

the faster the exponential transit response will decay to zero.

Effect of a real-axis pole upon transientresponse

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Evaluating response using poles

C(s) K1

s

K2

s 2

K3

s 4

K4

s 5

Css(t) Ct(t

)

c(t) K1 K2e2t K3e

4 t K4e5t

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c(t) 1 eat

First order system

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C(s) R(s) G(s) as(s a)


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First-order system response to a unit step

Transient response specification:

1. Time-constant, 1/a

2. Rise time, Tr

3. Settling time, Ts

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Transient response specification

for a first-order system

1. Time-constant, 1/aCan be described as the time for (1 - Exp[- a t])

to rise to 63 % of initial value.

1. Rise time, Tr = 2.2/aThe time for the waveform to go from 0.1 to 0.9of its final value.

3. Settling time, Ts = 4/aThe time for response to reach, and stay within,2% of its final value

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Transfer function via laboratory testing

G(s) K

(s a)

C(s) K

s(s a)

Ka

s

Ka

(s a)

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Identify K and a from testing

The time for amplitude to reach 63% of its final value:63 x 0.72 = 0.45, or about 0.13 sec , a = 1/0.13 = 7.7

From equation, we see that the forced response reachesa steady-state value of K/a =0.72 .K= 0.72 x 7.7= 5.54

G(s) = 5.54/(s+7.7) .

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Exercise

A system has a transfer function

G(s)= 50/(s+50).

Find the transit response specifications

such as Tc, Tr, Ts.

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Steady-state response

If the steady-state response of the output does notagree with the steady-state of the input exactly, thesystem is said to have a steady-state error.

It is a measure of system accuracy when a specific typeof input is applied to a control system.

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Y(s) = R(s) G(s)

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Steady-state error

T(s) = 9/(s + 10) Y(s) = 9/s(s+10)

y(t) = 0.9(1- e

-10t

)

y() = 0.9

E(s) = R(s) - Y(s)

ess = lim s 0 s E(s) = 0.1

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E(s)

R(s)

1K

E(s)

R(s)

1 Ks

e(t) 1

1K e(t) eK t

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Performance of a second-order system

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Numerical example of the second-

order system

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Overdamped

C(s) 9

s(s2 9s 9)

9

s(s 7.854)(s 1.146)

c(t) 1 0.171e7.854t 1.17e1.146t

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Underdamped

C(s) 9

s(s

2

2s 9)

9

s(s 1 j 8)(s 1 j 8)

c(t) 1 et(cos 8t8

8sin 8t)

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c(t) 1 et(cos 8t8

8sin 8t)

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Undamped

C(s) 9s(s

2 9)

c(t) 1cos3t

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Critically damped

C(s)

9

s(s2 6s 9)

9

s(s 3)2

c(t) 1 3te3t e3t

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Step response for second order systemdamping cases

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fig_04_11

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Summary

Overdamped

Poles: Two real at - 1, - 2

Underdamped

Poles: Two complex at - d + jd, - d - jd

Undamped

Poles: Two imaginary at + j1, - j1

Critically dampedPoles: Two real at - 1,

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Performance of a second-order system

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Response to unit step input

Y(s) G(s)

1 G(s)R(s)

)(2)( 22

2

sRsssY nn

n

)2()(

22

2

nn

n

ssssY

)sin(1

1)(

tety ntn

21

1cos

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Natural frequency n - the frequency of naturaloscillation that would occur for two complexpoles if the damping were equal to zero

Damping ratio - a measure of damping forsecond-order characteristic equation

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s2 2ns n 2 0

s1 n n 2 1

s2

n

n

2 1

Characteristic equation

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Finding nand for a second-order

system

G(s) n2

s2 2nsn2

n2

362n 4.2

n 6

0.35

s2 2ns n2 0

s1 n n 2 1

s2 n n 2 1

G(s) 36

s2 4.2s 36

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Second-order responses for

underdamped

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Transit response

For step input as

a function of

For step input asa function of and nt

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Unit impulse response

Y(s) n

2

(s2 2nsn2

)R(s)

y(t) n

ent sinnt

R(s)=1 T(s)=Y(s)

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Constant real part

Constant imaginary part

Constant damping ratio


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Standard performance measures

Ts(s) 44

n

21

n

pT

Peak time

Mpt1 e

12

P.O. 100 e

12

Settling time

Percent overshoot

Peak response

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fig_04_14

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Settling time

The settling time is defined as the timerequired for a system to settle within acertain percentage of the input amplitude.

Ts(s) 44

n

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Settling time

Ts(s) 44

n

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Rise time

The time it takes for a signal to go from10% of its value to 90% of its final value

Tr(s) 2.16 0.60

n0.3 0.8

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Rise time

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Peak time

Peak time is the time required by a signalto reach its maximum value.

21

n

pT

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Peak time

21

n

pT

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Percent overshoot

Percent Overshoot is defined as:

P.O. = [(Mpt fv) / fv] * 100%

Mpt = The peak value of the time response

fv = Final value of the response

P.O. 100 e

12

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Percent overshoot

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Percent overshoot and normalizedpeak time versus

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Finding transient response

G(s) 25

s(s 5)

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T(s) 25

s2 5s 25

s2 5s 25

s2 2nsn2

n 25 5

2n 5, 0.5

Tp

n 12

0.726 sec

P.O. 100 e

12 16.3%

Ts(s) 4n

1.6 sec

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Gain design for transient response

G(s) K

s(s 5)

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T(s) K

s2 5sK

s2 5sK

s2 2nsn2

n K

2n 5, 5

2 K

for P.O.10%

P.O.100e

12 10.0%

0.591, K17.9

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Performance Indices

Elevator

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Simplified description of a control


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Simplified description of a controlsystem

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Elevator input and output

When the fourth floor button is pressed on the firstfloor, the elevator rises to the fourth floor with aspeed and floor level accuracy designed for

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Push of the fourth-floor button is an input thatrepresent a desired output, shown as a stepfunction.

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Transient response

Passenger comfort and passenger patience are

dependent upon the transient response.If this response is too fast, passenger comfortis sacrificed; if too slow, passenger patience issacrificed.

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Steady-state error

Passenger safety and convenience would be

sacrificed if the elevator is not properly level.

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Performance Indices

A performance index is a quantitativemeasure of the performance of a

system and is chosen so thatemphasis is given to the importantsystem specifications.

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Response of the system

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ISE - Integral of Square of Error

I1 e2(t)0

T

dt

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The Integral Squared Error

I1 e2(t)

0

T

dt

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IAE - Integral of the AbsoluteMagnitude of the Error

I2 e(t)0

T

dt

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ITAE - Integral of Time Multiplied byAbsolute Error

I3 t e(t)0

T

dt

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ITSE - Integral of Time Multiplied bySquared Error

I4 te2 (t)0

T

dt

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General form of the performance


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General form of the performanceintegral

I f[e(t), r(t),c((t), t]0

T

dt

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Section 5.9

T

dtteISE

0

2)(

T

dtteIAE

0

|)(|

T

dttetITAE

0

|)(| T

dttteITSE

0

2)(

T

dtttytrtefI

0

)),(),(),((

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Performance criteria

T(s) 1

s2 2s 1

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Optimum system

A control system is optimum when theelected performance index is minimized.

The optimum value of the parametersdepends directly upon the definition ofoptimum, that is, the performance index.

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The coefficients that will minimize theITAE performance criterion for a stepinput and a ramp input have beendetermined for the general closed-looptransfer function.

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General closed-loop T(s)

T(s) Y(s)

R(s)

b0

sn

bn1sn1

... b1s b0

The T(s) has n poles and no zeros.

This T(s) has a steady-state error equal zero for a step input.

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T bl Th O i


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Table 5.6 The OptimumCoefficients of T(s) Based on the

ITAE Criterion for a Step Input

s + n

s2 + 1.4ns + n2

s3 + 1.75n s2 + 2.15n2s+ n3

s4 + 2.1 n s3 + 3.4 n

2s2 + 2.7 n3s + n

4

s5 + 2.8 n s4 + 5.0 n

2s3 + 5.5 n3s2 + 3.4n

4s +n5

s6 + 3.25ns5 + 6.60 n

2s

4 + 8.60n3s3 + 7.45 n

4s2 + 3.95n5s

+

n

6

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Step response for optimum coefficients


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Step response for optimum coefficients

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Table 5.7 The Optimum Coefficientsof T(s) Based on the ITAE Criterion

for a Ramp Input

s2 + 3.2ns + n2

s3

+ 1.75

n s2

+ 3.25

n

2

s

+

n

3

s4 + 2.41 n s3 + 4.93 n

2s2 + 5.14 n3s + n

4

s5 + 2.19 n s4 + 6.50 n

2s3 + 6.30 n3s2 + 5.24n

4s +n5

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T(s) Y(s)

R(s)

b1s b0s

n bn1sn1 ... b1s b0

T(s) has a steady-state error equal to zero for aramp input.

T(s) has two or more pure integrations asrequired to provide zero steady-state error.

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A: Simplification of linear system

G(s) K

s(s 2)(s 30)

G(s) K/30s(s 2)

We can neglect the impact of the pole at s = - 30 ,however we must retain the steady-state responseand reduce the system to

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Impulse response

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B: Simplification of linear systems

H(s) Kams

m am1sm1 a1s 1

bnsn an1s

n1 b1s 1, m n

L(s) Kcps

p c1s 1

dgsg d1s 1

, p g n

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)()()(

sMds

dsM

k

kk

)()()(

sds

ds

k

kk

)!2(!

)0()0()1()2(2

02

kqk

MMM

kqkqkq

kq

M2q 2q q 1,2...

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Simplified model

3223)6/1()6/11(1

1

6116

6)(ssssss

sH

L(s) 6

1 d1s d2s2

2

211)( sdsdsM

32 )6/1()6/11(1)( ssss

2

21

)0(1)( sdsdsM M(0)(0) 1

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Example 5.9

sddsdsdds

dsM k 21

2

21

)(2)1()(

1)0()0( M

1

)1()0( dM

2)2( 2)0( dM

1)0()0(

6/11)0()1(

2)0()2(

1)0()3( 0)0()3( M

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Example 5.9

2

)0()0()1(

1

)0()0(

2

)0()0()1(

021120

2

MMMMMMM

2

122

2

122 2 dddddM

36

49

2

)0()0()1(

1

)0()0(

2

)0()0()1(

021120

2

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Example 5.9

18

72 d

22584.260.1

60.1

625.0165.11

1)(

sssssL

36

492

2

12 dd

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Example 5.9

6116

6)(

23

ssssH

L(s) 1.60

1.60 2.584s s2

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Impulse response

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Dominant poles of transfer function

It has been recognized in practice and inthe literature that if the magnitude of thereal part of a pole is at least 5 to 10 times

of a dominant pole or pair of complexdominant poles, than the pole may beregarded as insignificant insofar as thetransient response is concerned.

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Thank You.

performance of feedback control systems (stachowicz, 2010)

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