performance and stability test

8/7/2019 performance and stability test

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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 system

is usually unknown, a standard test input

signal is normally chosen. Commonlyused test signals include step input, ramp

input, 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 = 2r(t) = A r(t) = At r(t) = At2

R(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

tpg

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Application to stable systems

Unstable systems represent loss of

control 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 transient

phenomena to some extend before steady

state is reached.

lim ct(t) = 0 for tpg

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Steady-state response

The response that exists for a long time

following any input signal initiation.

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Poles and zeros of a first order system

Css(t) Ct(t

) 15

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Poles and zeros

1. A pole of the input function generates the form of the

forced response ( that is the pole at the origin generated

a step function at the output).

2. A pole of the transfer function generate the form of the

exponential response

3. The zeros and poles generate the amplitudes for both the

transit and steady state responses ( see A, B in partial

fraction extension)

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A pole on the real axis generate an exponential response

of the form Exp[-Et] where -E 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 transient

response

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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) !a

s(s a)


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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)!

Kas

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 reaches

a 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 sp0 s E(s) = 0.1

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(s)!

R(s)

1 K

E(s) !R(s)

1K

s

e(t) !1

1 K 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(s

2

9s 9)

!

9

s(s 7.854)(s 1.146)

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

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Underdamped

C(s) !9

s(s2

2s 9)

!9

s(s1 j 8)(s1 j 8)

c(t) ! 1 e t

(cos 8t8

8sin 8t)

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c(t) ! 1 e t(cos 8t

8

8sin 8t)

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Undamped

C(s) ! 9s(s 9)

c(t) !1 cos3t37Control Systems


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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 system

damping cases

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fig_04_11

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Summary

Overdamped

Poles: Two real at - W1, - W2

UnderdampedPoles: Two complex at - Wd + j[d, - Wd - j[d

Undamped

Poles: Two imaginary at + j[1, - j[1

Critically dampedPoles: Two real at - W1,

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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

sRss

sYnn

n

[^[

[

!

)2()(

22

2

nn

n

ssssY

[^[

[

!

)sin(1

1)( UF[F

^[ ! tety ntn

21 ^F !

^U 1cos!

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Natural frequency [n - the frequency of natural

oscillation that would occur for two complex

poles if the damping were equal to zero

Damping ratio ^ - a measure of damping for

second-order characteristic equation

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Finding [n and ^ for a second-order

system

G(s) ! [n2

s2 2^[ns[n2

[n2

! 362^[n ! 4.2

[n ! 6

^! 0.35

s2 2^[ ns [ n

2! 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 2^[ns[n2) R(s)

y(t) ![n

Fe\[ntsin[nFt

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) ! 4X !4

^[n

21 ^[

T

! npT

Peak time

Mpt!1 e

^T

1^ 2

P.O.!100e

^T

1^ 2

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 time

required for a system to settle within a

certain percentage of the input amplitude.

Ts(s) ! 4X !4

^[n

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Settling time

Ts(s) ! 4X !4

^[n

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Rise time

The time it takes for a signal to go from

10% of its value to 90% of its final value

Tr(s) !2.16^ 0.60

[n0.3 e ^ e 0.8

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Rise time

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Peak time

Peak time is the time required by a signal

to reach its maximum value.

21 ^[

T

!

n

pT

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Peak time

21 ^[

T

!

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.!100e

^T

1^2

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Percent overshoot

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Percent overshoot and normalized

peak 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 2^[ns[n2

[n ! 25 ! 5

2^[n ! 5, ^! 0.5

Tp!T

[n

1 ^2! 0.726 sec

P.O.!100e

T

1 2 !16. %

Ts(s) ! 4^[n

!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 5s K

( ! s2 5s K

( ! s2 2^[ns[n2

[n ! K

2^[n ! 5, ^!5

2 K

for P.O.!10

P.O.!100e

^T

1^ 2 !10.0

^! 0.591, K!17.9

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Performance Indices

Elevator

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Simplified description of a control

system

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Elevator input and output

When the fourth floor button is pressed on the first

floor, the elevator rises to the fourth floor with a

speed and floor level accuracy designed for

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Transient response

Passenger comfort and passenger patience are

dependent upon the transient response.If this response is too fast, passenger comfort

is sacrificed; if too slow, passenger patience is

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Performance Indices

A performance index is a quantitative

measure of the performance of a

system and is chosen so thatemphasis is given to the important

system 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 Absolute

Magnitude of the Error

I2

! e(t)0

T

dt

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ITAE - Integral of Time Multiplied by

Absolute Error

I3 ! te(t)0

T

dt

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ITSE - Integral of Time Multiplied by

Squared Error

I4 ! te2(t)

0

T

dt

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General form of the performance

integral

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

T

dt

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Section 5.9

!T

dtteISE0

2 )( !T

dtteIAE0

|)(|

!

T

dttetITAE0

)(

!

T

dttteITSE0

2)(

!T

dtttytrtefI

0

)),(),(),((

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Performance criteria

T(s) !1

s2 2^s 1

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Optimum system

A control system is optimum when the

elected performance index is minimized.

The optimum value of the parameters

depends directly upon the definition of

optimum, that is, the performance index.

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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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Table 5.6 The Optimum

Coefficients of T(s) Based on theITAE Criterion for a Step Input

s + n

s2 + 1.4ns + n2

s3 + 1.75n s2 + 2.15n

2s + 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

2 s4 + 8.60n3s3 + 7.45 n

4s2 + 3.95n5s

+n6

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St f ti ffi i t


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Step response for optimum coefficients

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Table 5.7 The Optimum Coefficients

of T(s) Based on the ITAE Criterionfor a Ramp Input

s2 + 3.2ns + n2

s3 + 1.75n s2 + 3.25n2s + n3

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 b0sn bn1s

n1 ... b1s b0

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

ramp 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 response

and reduce the system to

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Impulse response

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B: Simplification of linear systems

H(s) ! Kams

m am1sm1 L a1s 1

bnsn an1s

n1 L b1s 1, m e n

L(s) ! Kc ps

p L c1s1

dgsg L d1s 1

, p e g n

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)()()( sMds

dsM

k

kk !

)()()(

s

ds

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)(

sssssssH

!

!

L(s) !6

1 d1s d2s2

2

211)( sdsdsM !

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

2

21

)0( 1)( sdsdsM ! M(0)(0) ! 198Control Systems


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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 2.584 s s2

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Impulse response

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Dominant poles of transfer function

It has been recognized in practice and in

the literature that if the magnitude of the

real part of a pole is at least 5 to 10 times

of a dominant pole or pair of complex

dominant poles, than the pole may be

regarded as insignificant insofar as the

transient response is concerned.

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Thank You.

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