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Process Control
CHAPTER III
LAPLACE TRANSFORM
The Laplace transform of a function f(t) is defined as;
In the application of Laplace transform variable time is eliminated and a new domain is introduced.
In the modeling of dynamic systems differential equations are solved by using Laplace transform.
dtetfsftfL st
0
)()()(
Properties of Laplace Transform
1. The Laplace transform contains no information about f(t) for t<0. (Since t represents time this is not a limitation)
2. Laplace transform is defined with an improper integral. Therefore the required conditions are
a. the function f(t) should be piecewise continuous
b. the integral should have a finite value; i.e., the function f(t) does not increase with time faster than e -st decreases with time.
3. Laplace transform operator transforms a function of variable t to a function of variable s. e.g., T(t) becomes T(s)
4. The Laplace transform is a linear operator.
5. Tables for Laplace transforms are available. In those tables inverse transforms are also given.
)()()()( 2121 tfbLtfaLtbftafL
)()(1 tfsfL
Transforms of Some Functions
1. The constant function
s
aaL
sL
ss
e
t
edtetfL
tf
t
t
stst
11
10)1()(
1)(0
00
2. The Step function
0
0
)()(
0,
0,0)(
1)1()(
0,1
0,0)(
s
adteatfL
ta
ttf
sdtetfL
t
ttf
st
st
3. The exponential function
as
AtfLAetfif
ase
as
dtedteetfL
te
ttf
at
t
t
tas
tasstat
at
)(,)(,
11
)(
0,
0,0)(
0
)(
0
)(
0
4. The Ramp function
svdtedv
dtdutu
duvvuudv
dttetfL
tt
ttf
stst
st
e,
,
,..
)(
0,
0,0)(
0
2
0
2
2
0
0
0
1)(
1
0
1
1
)(
sdttetfL
s
t
te
s
dtess
et
dttetfL
st
st
sttt
st
st
Laplace Transforms of Derivatives
)0()(
)().()(.0
))((0
)(
)(,
,eu
)(
0
00
st-
0
Fssf
sfsofe
dtestft
ttfedte
dt
df
tfvdtdt
dvdv
dtsedu
dtedt
dftf
dt
dL
ststst
st
st
)0('')0(')0()(
)0(')0()(
)0()(
233
3
22
2
fsffssfsdt
fdL
fsfsfsdt
fdL
fssfdt
dfL
Solution of Differential equations by Laplace Transform
For the solution of linear, ordinary differential equations with constant coefficients Laplace Transforms are applied.
The procedure involves:1. Take the Laplace Transform of both sides of the
equation.2. Solve the resulting equation for the Laplace
transform of the unknown function. i.e., evaluate x(s).
3. Find the function x(t), which has the Laplace Transform obtained in step 2.
Example:
Solve the following equation.
Solution:
2)0(,,03 xwherexdt
dx
tetx
sssx
ssx
sxxssx
32)(
3
1.2
3
2)(
2)3)((
0)(3)0()(
Inversion by Partial FractionsExample Solve the following equation.
Solution:
Partial fractions is to be applied;
0)0(,,1 xwherexdt
dx
)1(
1)(
1)1)((
1)()0()(
sssx
sssx
ssxxssx
1,010
1
0,11
1
1..
1)1(
1)(
AA
sfors
BsA
s
s
Bs
s
As
s
B
s
A
sssx
1)s(s
1s.
s with sides both multiply
I METHOD
1,0
1,
).1(
BB
sfor
s
1-
1
1s
B
s
1)A(s
1)s(s
11).(s
1),(s with sides both multiply
1
0
1
1
1)s(s
Bs
1)s(s
1)A(s
1s
A
)1(
1
II METHOD
B
BA
A
BsAAs
s
B
ss
1)1(
1)(
s
B
s
A
sssx
tetx
sssx
1)(
1
11)(
(by using Laplace Transform table)
Example Solve the following equation.
2
14
)(2
)0()(
)0()0()(2
)0()0()0()(
1)0(,0)0(,1)0(,
422
2
23
22
2
3
3
ss
sx
xssx
xsxsxs
xxsxssxs
xxxwhere
exdt
dx
dt
xd
dt
xd t
Apply Laplace transform to both sides of the equation.
1212)(
)1)(2)(1)(2(
896)(
)22)(2(
896)(
)2(
89622
2
1422)(
24
23
24
24223
s
E
s
D
s
C
s
B
s
Asx
sssss
ssssx
sssss
ssssx
ss
sssss
ssssssx
to determine multiply by Set s to results
A S 0 -2
B S-2 2 1/12
C S+1 -1 11/3
D S+2 -2 -17/2
E S-1 1 2/3
particularshomogeneou
22 )3/2()2/17()3/11()12/1(2)(
1
1.
3
2
2
1.
2
17
1
1.
3
11
2
1.
12
12)(
SSSolution
eeeetx
ssssssx
tttt
tttt
t
ttt
t
t
t
t
BeAecececSolution
BeAS
ecececS
er
er
er
rrr
rrr
exdt
dx
dt
xd
dt
xd
23
221
2particular
32
21shomogeneou
2
23
rt
22
2
3
3
,1
,2
,1
0)1)(1)(2(
022
e xsolution, shomogeneoufor
422
Example
Solve the following equation;
)45(
25)(
52
)45)((
2)(4)0()(5
245
1)0(,245
ss
ssY
sssY
ssYYssY
ydt
dyL
yydt
dy
t
t
tbtb
ety
ety
bbb
bsbs
bse
bb
bbe
bb
bbline
8.0
8.0
312
21
3
21
23
12
13
5.05.0)(
)1(08.0
04.0
8.00
8.04.0)(
4.0,8.0,0
))((,,11 21
Example; Find the solution of the equation with given laplace
transform.
tt eety
sssY
B
sss
Bs
s
As
ss
s
A
sss
Bs
s
As
ss
s
II
BABA
BA
sBsAs
s
sB
s
sA
ss
s
I
s
B
s
A
ss
s
ss
ssY
4
2
.3
1.
3
4)(
4
1.
3
1
1
1.
3
4)(
3
1
4),4(4
)4(1
)4()4)(1(
53
4
1),1(4
)1(1
)1()4)(1(
5
.
3
1,
3
4
45
1
)1()4(5
4
)1(
1
)4(
)4)(1(
5
.
41)4)(1(
5
45
5)(
Example;
321)6116(
1)(
1
)(6
)0()(11
)0()0()(6
)0()0()0()(
0)0()0()0(,16116
23
2
23
2
2
3
3
s
D
s
C
s
B
s
A
sssssY
s
sY
yssY
ysysYs
yysyssYs
yyyydt
dy
dt
yd
dt
yd
Solve the following equation.
to determine
multiply by Set s to results
A S 0 1/6
B S+1 -1 -1/2
C S+2 -2 1/2
D S+3 -3 1/6
ttt eeety
sssssY
32 .6
1.
2
1.
2
1
6
1)(
3
1.
6
1
2
1.
2
1
1
1.
2
11.
6
1)(
Example (Repeated factors)
Solve the equation for which Laplace transform is given.
Heaviside expansion rule can not be used for A, because the denominator of coefficient B becomes infinity.
s
C
s
B
s
A
sss
ssY
22 )2(2)44(
1)(
4
1,..
)2(.
2.
)2(
1
2
1,)2()2.(
)2()2.(
2)2.(
)2(
1
22
222
222
Css
Cs
s
Bs
s
As
ss
s
Bss
Cs
s
Bs
s
As
ss
s
4
1-A-C,AC.2,A.20
2)C2(sB2)A(2s1
2)C(sBs2)As(s1s
)2(2)44(
1
2
22
s
C
s
B
s
A
sss
s
Equalizing the derivatives gives
Taking derivative w.r.t. s gives
Further differentiation gives
s
sfdttfL
ssYty
ssYty
t
st
st
)()(
)(lim)(lim
)(lim)(lim
0
0
0
Final Value Theorem
Initial Value Theorem
Transform of an integral
Example:
Solve the following equation for x(t)
tt
t
eetx
ssssx
CBA
ssCssBssAs
s
C
s
B
s
A
sss
s
ss
s
s
s
s
ssx
s
s
s
ssx
ssssx
ss
sxxssx
xtdttxdt
dx
1)(
1
1
1
11)(
1,1,1
)1)(()1)(()1)(1(13
11)1)(1(
13
)1(
13
1.
13)(
131)(
13
1)(
1)()0()(
3)0(,)(
2
2
2
2
22
2
2
22
2
2
0
Example:
For a mixing tank, evaluate the change of composition w.r.t. time due to a step change in composition.
w1
x1 w
x
(Overflow system with constant volume, having a component balance)
ttexextx
sx
ss
xsx
s
xxssx
s
xsxxssx
xxdt
dxxx
dt
dx
w
V
wxwxdt
dxV
xVdt
dwxwx
1)0()1()(
1
1)0(
)1()(
)0(1)(
)()0()(
,
)(
1
1
1
1
11
1
1
Example: Considering Ex.2.1 for a sudden change in one of the input flow rates, evaluate
concentration change w.r.t. time.
dt
xVdwxxwxw
dt
Vdwww
)(
)(
2211
21
Total mass balance;
Component balance
w1, x1
w2, x2
w, x
For constant V and ρ ;
tt
c
eectx
ssscsx
xs
cssx
s
csxxssx
xcxdt
dx
xw
xwxw
dt
dx
w
V
wxxwxwdt
dxV
15.0)1(*)(
1
15.0
)1(
1*)(
)0(*
)1)((
*)()0()(
5.0)0(*,
*
2211
2211
τ
Example: Consider a CSTR and evaluate concentration change with respect to time.
t
A
t
AA
AA
A
AAA
AAAA
AA
A
AAAA
AAAA
AAAA
eCeCVkq
qsC
sC
ss
C
Vkq
qsC
s
C
Vkq
qCssC
s
C
Vkq
qsCCssC
s
C
Vkq
qC
dt
dC
Vkq
V
qCCVkqdt
dCV
qCVkCqCdt
dCV
VkCCCqdt
dCV
i
i
i
i
i
i
i
i
1)0()1()(
1
1)0(
)1()(
)0()1)((
)()0()(
)(
)(