Transforming C(s) into C(t):
Negative Feedback Control with Proportional Only Controller
Jigsaw Team EstrogenStephanie WilsonAmanda NewmanJessica Raymond
(We Laplace Transforms)
Goal: Advantages and disadvantages of different t0 (dead time) values
System: Any FOPDT system in a negative feedback loop
with a Proportional-Only Controller
From Block Diagram Algebra…
stc
stc
stc
stc
KeKsKeK
sKeKsKeK
sRsC
0
0
0
0
)1(
11
1 )()(
CLTF:
Negative feedback control loop for
FOPDT system with P-only controller
With Pade’s Approximation…
)2
1(
)2
1()1(
)2
1(
)2
1(
)()(
0
0
0
0
st
st
KKs
st
st
KK
sRsC
c
c
)2
1(
)2
1(
0
0
0
st
st
e st
Pade’s
Approximation:
CLTF:
This Reduces to…
sKKstKKtst
stKKsC
cc
c 1
1)22
(2
)2
1()(
0020
0
)2
1()1)(2
1(
)2
1(
)()(
00
0
stKKsst
stKK
sRsC
c
c
Therefore…
CLTF:
Final Value Theorem
sKKstKKtst
stKKs
cc
c
s
1
1)22
(2
)2
1( lim
0020
0
0
Value S. S. )( lim0
sCssIn Laplace Domain
0
0
0KKc
KKc1
Now for C(t)…
sKKstKKtst
stKKsC
cc
c 1
1)22
(2
)2
1()(
0020
0
)1)22
(2
( 0020 KKstKKtstcc
First we must simplify the denominator…
for partial fraction decomposition
CE:
Now Complete the Square
0 2 mqs
mps
0 2 mqs
mps
mq
mp()
mps )
4
2( 2
22
mq
mp()
mps 22 )
2
2(
)44()
4
2( 2
2
22
mm
mq
mp()
mps
21 2)
mp(
21 2)
mp(
22
22
44)
4
2(
mmq
mp()
mps
)4
4 2
( 2
22
mmqp()
mps
1
44)
2(
)2
1()( 2
2
0
mpmq
mpsm
stKKsC
c
sC
mpmq
mpsm
BmpsA
44)
2(
)2
( 2
2
sKKstKKtst
stKKsC
cc
c 1
1)22
(2
)2
1()(
0020
0
m
mpmq
mPsmCBss
mPsAstKK c 4
4222
12
220 )()(
s
qpsmsCBssmPsAstKK c
220
221 )(
Simplify the last term
CqsCpBmApsCmAstKKKK cc
22
20
Distribute the left sideFactor the right
Substitute for m, p, q
KK
tKKttKKB
c
oc
o
oc 12
22(
2
KK
tKKA
c
oc
1
2
KKKKCc
c
1
KKstKKtstcc 1 )
22(
2 0020
m p q
Separate the terms
sC
mpmq
mpsm
B
mpmq
mpsm
mPsA
sC
44)
2(
44)
2(
)2
()( 2
22
2
sC
mpmq
mpsm
BmpsA
sC
44)
2(
)2
( )( 2
2
Manipulate…
sC
m
pqmmps
m
pmq
m
pmqm
B
24
242)2
(
24
24
24
2424m
2p4mq2
2mps
2mps
mAC(s)
sC
mpmq
mpsm
B
mpmq
mpsm
mPsA
sC
44)
2(
44)
2(
)2
()( 2
22
2
to Fit Page 15 Formulas…
Substitute A, B, C, m, p, q back into C(t)
KKKK
t
tKKtKKtte
c
ccc
tt
tKK
tc
122
12sin
220
200
022
0
00
220
200
00
00
0
220
200
0
22
0
0
2212
2
1222
22212
cos
2
12
)()(0
00
t
tKKtKKtt
KK
tKKttKK
tt
tKKtKKtet
KK
tKK
tutC
cc
c
c
ccct
t
tKK
t
c
c c
KKKKt
t
tKKtKKte
c
ccc
tt
tKK
tc
122
12sin
220
200
022
0
00
220
200
00
00
0
220
200
0
22
0
0
2212
2
1222
22212
cos
2
12
)()(0
00
t
tKKtKKtt
KK
tKKttKK
tt
tKKtKKte
tKK
tKK
tutC
cc
c
c
ccct
t
tKK
t
c
c c
Final Value Theorem
KKKK
c
c
1
Value S. S. )( lim
tCt
In Time Domain
0
0
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