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

KeKs

KeK

s

KeKs

KeK

sR

sC0

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

sR

sC

c

c

)2

1(

)2

1(

0

0

0

st

st

e st

Pade’s

Approximation:

CLTF:

This Reduces to…

sKKst

KKt

st

st

KKsC

cc

c 1

1)22

(2

)2

1()(

0020

0

)2

1()1)(2

1(

)2

1(

)(

)(

00

0

st

KKsst

st

KK

sR

sC

c

c

Therefore…

CLTF:

Final Value Theorem

sKKst

KKt

st

st

KKs

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)…

sKKst

KKt

st

st

KKsC

cc

c 1

1)22

(2

)2

1()(

0020

0

)1)22

(2

( 0020 KKst

KKt

st

cc

First we must simplify the denominator…

for partial fraction decomposition

CE:

Substitute for Easier Math…

KKst

KKt

st

cc 1 )22

( 2

0020

m p q

0

2

2

m

qs

m

ps

qpsms

CE:

Now Complete the Square

0 2 m

qs

m

ps

0 2 m

qs

m

ps

m

q

m

p()

m

ps )

4

2(

2

22

m

q

m

p()

m

ps 22 )

2

2(

)4

4()

4

2( 2

2

22

m

m

m

q

m

p()

m

ps

2

1 2)m

p(

2

1 2)m

p(

22

22

4

4)

4

2(

m

mq

m

p()

m

ps

)4

4

2(

2

22

m

mqp()

m

ps

1

44

)2

(

)2

1()( 2

2

0

mpmq

mp

sm

st

KKsC

c

s

C

mpmq

mp

sm

Bmp

sA

44

)2

(

)2

( 2

2

sKKst

KKt

st

st

KKsC

cc

c 1

1)22

(2

)2

1()(

0020

0

m

m

pmq

m

PsmCBss

m

PsAs

tKK c 4

4

2221

2220 )()(

s

qpsmsCBssm

PsAs

tKK c

220

221 )(

Simplify the last term

CqsCpBm

ApsCmAs

tKKKK cc

22

20

Distribute the left sideFactor the right

Solve for C

KKCq c

q

KKC c

Solve for A

0CmA

q

KmKA c

CmA

Solve for B

q

p

q

ptKKB oc 2

2

2

Cpm

AptKKB oc

22

22o

c

tKKCpB

m

Ap

q

ptKKB oc 2

Substitute for m, p, q

KK

tKK

tt

KKBc

oc

o

oc 12

22(

2

KK

tKK

Ac

oc

1

2

KK

KKC

c

c

1

KKst

KKt

st

cc 1 )22

( 2

0020

m p q

Separate the terms

s

C

m

pmq

m

psm

B

m

pmq

m

psm

m

PsA

sC

4

4)

2(

4

4)

2(

)2

()( 2

2

2

2

s

C

mpmq

mp

sm

Bmp

sAsC

44

)2

(

)2

( )( 2

2

Manipulate…

sC

m

pqmmp

s

m

pmq

m

pmqm

B

24

242)2

(

24

24

24

2424m

2p4mq2

2mp

s

2mp

s

mA

C(s)

s

C

m

pmq

m

psm

B

m

pmq

m

psm

m

PsA

sC

4

4)

2(

4

4)

2(

)2

()( 2

2

2

2

to Fit Page 15 Formulas…

Ctm

pmqe

mpmq

m

Bmt

m

pmqe

m

Au(t)(C(s))

tm

pt

m

p

2

22

2

22

22

4

4sin

444

4cos-L

Inverse LaPlace Transform

Substitute A, B, C, m, p, q back into C(t)

KK

KK

t

tKK

tKKt

tec

ccc

tt

tKK

tc

122

12sin

220

2

000

22

0

00

220

2

000

0

00

0

220

2

000

22

0

0

2212

2

1222

22212

cos

2

12

)()(0

00

t

tKK

tKKt

t

KK

tKK

tt

KK

tt

tKK

tKKt

etKK

tKK

tutC

cc

c

c

ccct

t

tKK

t

c

c c

KK

KKt

t

tKK

tKKt

ec

ccc

tt

tKK

tc

122

12sin

220

2

000

22

0

00

220

2

000

0

00

0

220

2

000

22

0

0

2212

2

1222

22212

cos

2

12

)()(0

00

t

tKK

tKKt

t

KK

tKK

tt

KK

tt

tKK

tKKt

etKK

tKK

tutC

cc

c

c

ccct

t

tKK

t

c

c c

Final Value Theorem

KK

KK

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