B_lecture7 Time Responses of Prototype Systems Automatic control System
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Transcript of B_lecture7 Time Responses of Prototype Systems Automatic control System
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Time Response of a Prototype First-Order System
One of the simplest systems is that represented by a first-order differential equation. Such a system is known as a first-order system. There are many examples of first-order systems.
Speed control of a DC Motor.
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RCiv ov
iRvv oi
dt
dvCi o
Taking Laplace transforms, assuming zero initial
conditions, and eliminating i result in
ooi RCsVsVsV )()( )1()( RCsVsV oi
1
1
1
1
)(
)(
TsRCssV
sV
i
o
RCT Time constant
Electrical first-order system
-
Mathematical Model
RTs1
C
trsc
dt
tdcT
1
1
TssR
sCs
A Prototype First-Order System
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Impulse Response
1
1
Ts)(sR )(sC
If the input is a unit impulse, then 1)( sR
leading to 1
1)(
TssC
Taking the inverse Laplace transform gives
TteT
tc /1
)(
-
t0
TteT
tc /1
)(
T
1
Impulse Response
-
Unit-Step Response
ssR
1)(
Hence the output becomes
TssTs
T
sTsssC
/1
11
1
1
)1(
1)(
giving the time-domain response
1
1
Ts)(sR )(sC
In this case, for a unit-step input
Ttetc /1)(
1
0 0
1 1| |
tT
t t
dc te
dt T T
-
Ttetc /1)(
0%
Unit-Step Response
632.0
865.0950.0
982.0
0111 hess
3 (5%)
4 (2%)s
Tt
T
-
Ramp Response
If the input is the unit ramp ttr )(
then 21
)(s
sR
leading to Ts
T
s
T
sTsssC
/1
1
)1(
1)(
22
The time-domain response becomes
T
t
TeTttc
)(
1
1
Ts)(sR )(sC
-
Tt
TeTttc
)(
The response consists of a transient part and a
steady-state part, and in the steady state the output
lags the input by a time equal to the consist T .
Ramp Response
-
22( )
d x dxm f Kx F t
dt dt
m
k
f
x(t)
F(t)
Spring-mass-damper
second order system
masstheofpositiontherepresentstx
tF
)(
force )(
2
2
( ) ( ) ( ) ( )
( ) 1
( )
ms X s fsX s Kx s F s
X s
F s ms fs K
Assuming zero initial conditions,
taking Laplace transforms
-
A Prototype Second-Order System
)2(
2
n
n
ss
)(tr
)(sR
)(tc
)(sC22
2
2)(
)(
nn
n
sssR
sC
The characteristic equation
02 22 nnss
2
2 2
22
n n n
d c t dc tc t r t
dtdt Mathematical Model
Damping ratio Natural undamped frequency n
-
tsts
n
nn
n
ececsss
c
ss
cL
sssssL
sssLsCLth
21
21
2
2
1
11
21
21
22
211
11
1
2
121
2
1)( sss
C n
212
2
2)( sss
C n
2
1,2 1n ns
ss
sRssCnn
n 1
2 22
2
A Prototype Second-Order System
For a unit-step input s
sR1
)(
the output becomes
02 22 nnss
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Root
Root j
na
21 nd
n
cos
is the imaginary part of the roots
is the real part of the roots na
21 nd
A Prototype Second-Order System
2 2
1,2 1 1n n ns a
-
>1 (overdamped)
=1 (critically damped)
0
-
2s1s
j
0
(a) 1
21 ss
j
0
(b) 1
2s
1s
j
0
(c) 10
2s
1s
j
0
(d) 0
2s
1s
j
0
(e) 01
2s1s
j
0
(f) 1
n21 n
21 n
n
21 n
21 n
n
n
A Prototype Second-Order System
Roots of the characteristic equation of the prototype second-order system
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The unit-step response of
a prototype second-order system
arccos 1sin1
1)( 22
te
tc n
tn
)2()(
22
2
nn
n
ssssC
0
-
arccos 1sin1
1)( 22
te
tc n
tn
The unit-step response of
a prototype second-order system
0
-
Maximum Overshoot
21
nd
pt
21/1)(
etc p
arccos 1sin1
1)( 22
te
tc n
tn0
-
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
10
20
30
40
50
60
70
80
90
100
%
Maximum Overshoot
%100%21/
e
Percent overshoot as a function of damping ratio for the step response of
the prototype second-order system
0
-
te
tc d
tn
sin1
1)(2
boundupper 1
12
tne
bound-lower1
12
tne
21ln snt
05.01 2
tne
Settling time
)8.0( 5.3
n
stst
0
-
3) % %100)(
)()(%
c
ctc p
)10(
If
-
The unit-step response of
a prototype second-order system
3.5n
st
To keep % less than a fixed value, Must have (%).
To keep tr less than a fixed value,
Must have n n min.
To keep ts less than a fixed value,
Must have
-
The unit-step response of a prototype second-order system
Putting these constraints together will yield an
allowable region for the poles
Note: The allowable region is a guide.
After a system is designed, the
performance will have to be evaluated.
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The characteristic equation 0222 nnss
The two roots can be expressed as
The unit-step response of
a prototype second-order system
>1 (overdamped)
where
1 2
2 21 1 1
2 2
1 2
1 1 21 2
1 2
1
2
11
n n
n n
s t s t
c t L C s L Ls s s s s ss s
c cL c e c e
s s s s s
121
2
1)( sss
C n
212
2
2)( sss
C n
2
1,21
n ns
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Maximum Overshoot 0%
The unit-step response of
a prototype second-order system
>1 (overdamped)
=c( ) 5% 1(6.45 1.7)
0.7s
n
t
No overshoot .No oscillation
-
ns 2,1
2
2
1 1n
n
c s ss ss
1 1 1tnc t L C s t e
n
5% )c( 75.4 nst
=1 (critically damped)
The unit-step response of
a prototype second-order system
-
njs 2,1
1 cos nc t t
continuous oscillation
=0 (undamped)
The unit-step response of
a prototype second-order system
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The unit-step response of
a prototype second-order system
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Addition of a Zero
to the Forward-Path Transfer Function
)2(
2
n
n
ss
)(tr
)(sR
)(tc
)(sCsTd1
)(te )(tu
The improved damping ratio:
ndd T2
1
The closed-loop transfer function:
)0(T 2
)1(
)2
1(2
)1(
)(
)(d22
2
22
2
nnd
nd
nnnd
nd
ss
sT
sTs
sT
sR
sC
-
The closed-loop transfer function:
)0(T 2
)1(
)2
1(2
)1(
)(
)(d22
2
22
2
nnd
nd
nnnd
nd
ss
sT
sTs
sT
sR
sC
)(2
)(2
)(22
2
22
2
sRss
sTsRss
sCnnd
nd
nnd
n
dt
tdcTtctc d
)()()( 11
22
2
2 nnd
n
ss
sTd
)(sR )(1 sC )(sC
Addition of a Zero
to the Forward-Path Transfer Function
Since multiplying by s is the same as differentiating in
the time domain
-
ssssC
nn
n 1
2)(
22
2
0
ssTs
sTsC
nnnd
nd 1
)2
1(2
)1()(
22
2
)(tc
)(0 tc
dt
tdcTtctc d
)()()( 11
)(tc
)(1 tc
dt
tdcTd
)(1
ssTs
sC
nnnd
n 1
)2
1(2
)(22
2
1
Addition of a Zero
to the Forward-Path Transfer Function
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The effect of the speed feedback
)2(
2
n
n
ss
)(tr
)(sR
)(tc
)(sC
sKt
)(te )(tu
222
2
)2()(
)(
nntn
n
sKssR
sC
The improved damping ratio: ntt K 2
1