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:

Substitute for Easier Math…

KKstKKtstcc 1 )

22(

2 0020

m p q

0

2

2

mqs

mps

qpsms

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

Solve for C

KKCq c

qKKC c

Solve for A

0CmA

qKmKA c

CmA

Solve for B

qp

qptKKB o

c 22

2

CpmAptKKB o

c 22

22o

ctKKCpB

mAp

qptKKB o

c 2

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…

Ctmpmqe

mpmqm

Bmtmpmqe

mAu(t)(C(s))

tmpt

mp

2

22

2

22

22

44sin

444

4cos-L

Inverse LaPlace Transform

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

CONCLUSION

Advantages of Increased t0

Reaches steady state faster Larger values of stable operating gain Bigger gain is better

Disadvantages of Increased t0

Increase in t0 results in a decrease in the ultimate 0 0gain of the loop (Kcu) Poor performance Limited values of stable operating gain