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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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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.