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SOLVED PROBLEMS ON
LAPLACE TRANSFORM
Dr. Samir Al-Amer
November 2006
LAPLACE TRANSFORM
Many mathematical problems are solved using transformations. The idea is to transform the problem into another problem that is easier to solve. Once a solution is obtained, the inverse transform is used to obtain the solution to the original problem. The Laplace transform is an important tool that makes solution of linear constant coefficient differential equations much easier. The Laplace transform transforms the differential equations into algebraic equations which are easier to manipulate and solve. Once the solution is obtained in the Laplace transform domain is obtained, the inverse transform is used to obtain the solution to the differential equation. Laplace transform is an essential tool for the study of linear time-invariant systems. In this handout a collection of solved examples and exercises are provided. They are grouped into two parts: background material and Laplace transform. The background material is supposed to be covered in the prerequisite materials namely MATH 001, MATH002, MATH 101,MATH 102, MATH 201 and MATH 260. Most of the problems in this part are easy. They are listed here for quick review and to ensure that students can perfectly solve these problems. The second part is directly related to the material covered in this course. Read the text book for more details. The intension here is not to replace the textbook but rather to give supplementary material to improve the student understanding.
LEARNING OBJECTIVES
A list of learning objectives that students are expected acquire in the course is listed below. This problem set is not covering all of them. Read the textbook and make sure that you know all the items below.
1. To define (mathematically) the unit step and unit impulse. 2. To express some simple functions in terms of unit step and/or unit
impulse 3. To state and use the sampling property of the impulse. 4. To determine if a function is of exponential order or not. 5. To know basic integration rules (including integration by parts) 6. To be able to factor second order polynomials 7. To perform algebraic manipulation of complex numbers. 8. To state the definition of Laplace transform. 9. To give sufficient conditions for existence of Laplace transform.
10. To obtain Laplace transform of simple functions (step, impulse, ramp, pulse, sin, cos, 7 )
11. To obtain Laplace transform of functions expressed in graphical form.
12. To know the linear property of Laplace transform. 13. To know Laplace transform of integral and derivatives (first and high
orders derivatives. 14. To obtain inverse Laplace transform of simple function using the
Table of Laplace transform pairs. 15. To use the method of partial fraction expansion to express strictly
proper functions as the sum of simple factors (for the cases: simple poles, complex poles and repeated poles).
16. To perform long division and know the reason for using it in inverse Laplace transform.
17. To obtain inverse Laplace transform. 18. To solve constant coefficient linear ordinary differential equations
using Laplace transform. 19. To derive the Laplace transform of time-delayed functions. 20. To know initial-value theorem and how it can be used. 21. To know final-value theorem and the condition under which it can
be used.
BACKGROUND MATERIAL
Topics:
1. Factoring a Second Order Polynomial 2. Integration 3. Limits 4. Complex Number Manipulation 5. Long division 6. Useful identities
1. FACTORING A SECOND ORDER POLYNOMIAL
Factoring of polynomials are needed in solving problems related to Laplace Transform. Factoring high order polynomials may not be easy and computers may be needed to do the factorization. Factoring second order polynomials can be done manually.
EXAMPLE 1:
Factor the second order polynomial 65)( 2 ++= xxxP Answer:
This is a simple problem. Find two numbers such that their product is six and
their sum is five. They are two and three. )3)(2(65)( 2 ++=++= xxxxxP
EXAMPLE 2:
Factor the second order polynomial 52.29.0)( 2 −+= xxxP
Answer:
It is not easy to guess two numbers such that their product is 52.2− and
their sum is 9.0 . We can use the formula to find the roots then write the
factors.
)2.1)(1.2(25.29.0)(
1.22.1)1(2
)52.2)(1(49.09.0025.29.0
)(2
))((40
2
2
2
2
2
−+=−+=
−=−−±−
=−+
−±−=++
xxxxxP
andarexxofroots
a
cabbarecbxaxofroots
EXAMPLE 3:
Factor the second order polynomial 134)( 2 ++= xxxP
Answer:
We can use the formula to find the roots then write the factors.
)32)(32(134)(
32)1(2
)13)(1(4440134
2
2
2
jxjxxxxP
jarexxofroots
−+++=++=
±−=−±−
=++
EXAMPLE 4:
Express the second order polynomial 134)( 2 ++= xxxP in the sum of square
form Answer:
( ) ( )22
2222
22 32132
4
2
413
2
4
2
44134)( ++=+
−
+=+
−
++=++= xxxxxxsP
EXAMPLE 5:
Factor the second order polynomial 642)( 2 −+= xxxP
Answer:
( ) ( ) )1(32322642)( 22 −+=−+=−+= xxxxxxxP
EXAMPLE 6:
Factor the second order polynomial 16)2()( 2 ++= xxP
Answer:
This polynomial is expressed as the sum of squares. It has two complex poles. The real part of the roots is -2 and imaginary parts are 4± .
)42)(42(16)2()( 2 jxjxxxP −+++=++=
Exercise 1:Factor 3212)( 2 ++= xxxP
Exercise 2:Factor 9.14.4)( 2 ++= xxxP
Exercise 3:Factor 1122)( 2 ++= xxxP
Exercise 4:Factor 132)( 2 ++−= xxxP
Exercise 5: Express 172)( 2 ++= xxxP in sum of squares form.
INTEGRATION
Integration is a large topic. Here we will concentrates on integrals often used in problems related to Laplace transform. In particular, we
concentrate on integrals involving axe .
EXAMPLE 1:
Evaluate the integral ∫2
0
2 dxe x
Answer:
( )12
1
2
1
2222
4404
2
0
22
0
2 −=−=−==∫ eeeee
dxex
x
EXAMPLE 2:
Evaluate the integral ∫b
a
cxdxe where ba, and c are constants and c is not zero.
Answer:
( )acbc
b
a
cxb
a
cx eecc
edxe −==∫
1
EXAMPLE 3:
Evaluate the integral ∫∞
−
0
dxe cx where c is a positive constant.
Answer:
( ) ( )cc
eecc
edxe cc
cxcx 1
1011 0
00
=−−
=−−
=−
= −∞−
∞−∞−∫
EXAMPLE 4:
Evaluate the integral ∫∞
−
5
2 dte t .
Answer:
( ) ( )2
02
1
2
1
2
1010)5(2)(2
5
2
5
2−
−−∞−
∞−∞− =−
−=−
−=
−=∫
eeee
edte
tt
EXAMPLE 5:
Evaluate the integral ∫ −3
1
2 dtte t .
Answer:
Using integration by parts
( ).............22
3
22
2
23
1
623
1
23
1
3
1
3
1
2
22
−−
−−
=−
−−
=−=
−=⇒=
=⇒=
−−−−−
−−
∫∫∫ee
dte
te
vduvudtte
evdtedv
dtdutudefine
ttt
tt
EXAMPLE 6:
Evaluate the integral ∫ −3
1
22 dtet t .
Answer:
EXAMPLE 7:
Evaluate the integral ∫ −3
1
2)5sin( dtet t .
Answer:
EXAMPLE 8:
Evaluate the integral ∫∞
−
0
)5sin( dtet ct where c is a positive constant.
Answer:
EXAMPLE 9:
Evaluate the integral ∫10
0
)( dttf where
≤<+
≤≤=
10252
202)(
tt
tttf
Answer:
4452
2
2
2)52(2)()()(
10
2
22
0
210
2
2
0
10
2
2
0
10
0
=++=++=+= ∫∫∫∫∫ ttt
dtttdtdttfdttfdttf
LIMITS
EXAMPLE 1:
Evaluate the limit 1
3lim
2
2
−+
∞→ t
tt
t .
Answer:
∞∞
=−+
∞→ 1
3lim
2
2
t
tt
t Dividing by the highest power and taking the limit
101
01
11
31
lim1
3
lim1
3lim
222
2
22
2
2
2
=−+
=−
+=
−
+=
−+
∞→∞→∞→
t
t
tt
tt
t
t
t
t
tt
ttt
EXAMPLE 2:
Evaluate the limit t
tet 2)10sin(lim −
∞→ .
Answer:
0)10sin(lim 2 =−
∞→
t
tet
EXAMPLE 3:
Evaluate the limit t
tet 2lim −
∞→ .
Answer:
0lim 2 =−
∞→
t
tet
EXAMPLE 4:
Evaluate the limit t
tet 3lim
→∞ .
Answer:
∞=∞→
t
tet 3lim
EXAMPLE 5:
Evaluate the limit 1
3lim
2
2
0 −+
→ t
tt
t .
Answer:
01
0
1
3lim
2
2
0=
−=
−+
→ t
tt
t
EXAMPLE 6:
Evaluate the limit tt
tt
t −+
→ 2
2
0
3lim .
Answer:
0
03lim
2
2
0=
−+
→ tt
tt
t using L'Hopital's rule
31
3
12
32lim
3lim
02
2
0−=
−=
−+
=−+
→→ t
t
tt
tt
tt
COMPLEX NUMBER MANIPULATION
EXAMPLE 1:
Evaluate )3.04()32( jj +−++ .
Answer:
3.32)3.04()32( jjj +−=+−++
EXAMPLE 2:
Evaluate )3.04)(32( jj +−+ .
Answer:
4.119.8)3.04)(32( jjj −−=+−+
EXAMPLE 3:
Evaluate 24
32
j
j
−+
.
Answer:
8.01.024
32j
j
j+=
−+
EXAMPLE 4:
Let 2
1)(
−+
=s
ssF . Evaluate )32( jF + .
Answer:
jj
jjF −=
−+++
=+ 1232
132)32(
EXAMPLE 5:
Let )1)(2(
3)(
+++
=ss
ssF . Evaluate )22( jF +
Answer:
0.2077 j- 0.2615)122)(222(
322)22( =
++++++
=+jj
jjF
LONG DIVISION
EXAMPLE 1:
Let )1)(2(
39)(
2
++++
=ss
sssF . Perform long division and determine the quotient
and the remainder. Answer:
1
232 ++ ss 392 ++ ss
232 −−− ss
16 +s
23
161
23
39
)1)(2(
39)(
22
22
++
++=
++
++=
++++
=ss
s
ss
ss
ss
sssF
EXAMPLE 2:
Let 6116
1533173)(
23
23
+++
+++=
sss
ssssF . Perform long division and determine the
quotient and the remainder. Answer:
6116
33)(
23
2
+++
−−+=
sss
ssF
MISCELLANEOUS PROBLEMS
EXAMPLE 1:
Express the following function as a sum of shifted steps.
Answer:
)1()2(2)1()( −+−−−= tutututf
EXAMPLE 2:
Express the following function in terms of shifted steps.
Answer:
))2()1((1)1()(()(
0
]2,1[1
]1,0[
)(
2
2
−−−+−−=
∈
∈
=
tututttuttf
otherwise
t
tt
tf
1 2 3 4
1
-1
t2
1 2 3 4
1
-1
EXAMPLE 3:
Obtain an analytical expression of the saw function.
Answer:
<
+∈−
∈
=
00
)1,[
)1,0[
)(
t
nntnt
tt
tf
EXAMPLE 3:
Obtain an analytical expression of the function.
Answer:
1 2 3 4
1
-1
1 2 3 4
1
-1
EXAMPLE 4:
Obtain an analytical expression of the function.
Answer:
∈+−
∈∈
=
otherwise
tt
t
tt
tf
0
)3,2[62
)2,1[2
)1,0[2
)(
1 2 3 4
2
-2
USEFUL IDENTITIES
Often some problem can be made considerably easier if some identities are used.
( )
( )
))2cos(1(2
1)(sin
2
1)sin(
2
1)cos(
)sin()cos(
)sin()cos(
2 θθ
θ
θ
θθ
θθ
θθ
θθ
θ
θ
−=
−=
+=
−=
+=
−
−
−
jj
jj
j
j
eej
ee
je
je
identitiesuseful
EXAMPLE 1:
LAPLACE TRANSFORM
Computing the Laplace transform using definition Using Laplace transform Properties Inverse Laplace transform
DEFINITION OF THE LAPLACE TRANSFORM
The Laplace transform converts a function of real variable f(t) into a function of complex variable F(s). The Laplace transform is defined1 as
{ } ∫∞
−
−==0
)()()( dtetftfLsF at
The variable s is a complex variable that is commonly known as the Laplace operator.
EXAMPLE 1:
Find the Laplace transform of ∈
=otherwise
ttf
0
]2,0[2)(
Answer:
{ } satatat ess
dtedtedtetftfLsF 2
2
2
00
2202)()()( −∞ −−∞
−
− −=+=== ∫∫∫
EXAMPLE 2:
Find the Laplace transform of tetf 221)( −−=
Answer:
{ } ∫∫∫∫∞ −−∞ −∞ −−∞
−
−
+−=−=−===
0
2
00
2
0 2
212)21()()()(
ssdteedtedteedtetftfLsF attatattat
USING LAPLACE TRANSFORM PROPERTIES
Often it is possible to obtain Laplace Transforms of complicated functions using one or more of the Laplace transform properties. Some other problems can be made easy by applying trigonometric identities. In this section several examples are shown.
EXAMPLE 1:
Find the Laplace transform of )3cos()( 22 tettf x−=
Answer:
For convenience, let us define the following functions
32
2
2
2
2
22
13)+4s+(s
23)-4s+2)(s+2(s)()(
9)2(
2)(
9)(
)}({)()},({)()},({)(
)()(
)()3cos()(
)3cos()(
=
−−=
++
+=
+=
===
=
==
=−−
sHds
d
ds
dsF
s
ssH
fromTables
ssG
tfLsFthLsHtgLsGLet
thttfthen
tgeteth
ttg
xx
EXAMPLE 2:
Find the Laplace transform of [ ]2)3cos()( ttf =
Answer:
Using the trigonometric identity
( )
)36(22
1)}3({cos
)6cos(12
1)3(cos
2
2
2
++=
+=
s
s
stL
tt
EXAMPLE 3:
Find the Laplace transform of ( )22 1)( −= − tetf
Answer:
Using the trigonometric identity
ssseL
eee
t
ttt
1
2
2
4
1})1{(
12)1(
22
2422
++
−+
=−
+−=−
−
−−−
EXAMPLE 4:
Find the Laplace transform of )2cos()2sin()( tttf =
Answer:
Using the trigonometric identity
( )
)16(2
4)}2cos()2{sin(
)4sin(2
1)2cos()2sin(
2 +=
=
sttL
ttt
EXAMPLE 5:
Find the Laplace transform of 2)( ttf =
Answer:
Using the multiplication by t property
32
2
2
21}{
1}{
ssds
dtL
stL
=
−=
=
Exercises:
Find the Laplace transform of
1. [ ]3)3cos()( ttf =
2. )3(sin)( 2 ttf =
3. )43sin()( −= ttf
INVERSE LAPLACE TRANSFORM
The formula for computing the Laplace transform is well known. A simple way to compute the inverse transform is split the function as the sum of first and second order terms then obtain the inverse for each term alone and finally add them together. Standard Laplace transform tables can be used to obtain inverse transforms for simple terms.
EXAMPLE 1:
Find the inverse Laplace transform of ks
sF+
=2
)(
Answer:
{ } ktesFLtf −− == 2)()( 1
EXAMPLE 2:
Find the inverse Laplace transform of 23
2)(
2 ++=
sssF
Answer:
A first step in the solution is to factor the denominator
)1)(2(232 ++=++ ssss
Then F(s) can be expressed in terms of the unknown parameters A and B
1223
2)(
2 ++
+=
++=
s
B
s
A
sssF
Use the formula to determine A and B
1
2
2
2)(
22
2)()1(
21
2)()2(
11
22
++
+−
=
=+
=+=
−=+
=+=
−=−=
−=−=
sssF
ssFsB
ssFsA
ss
ss
{ } 022)()( 21 ≥+−== −−− tforeesFLtf tt
EXAMPLE 3:
Find the inverse Laplace transform of bsecs
sF −
+=
2)(
Answer:
{ }
{ } )(2)()(
2)()(2
)(
)(1
1
btueesGLtf
esGLtgcs
sGLet
btkbs
tc
−==
==⇒+
=
−−−−
−−
EXAMPLE 4:
Find the inverse Laplace transform of ( )321
)(+
=s
sF
Answer:
{ }== − )()( 1 sFLtf
EXAMPLE 5:
Find the inverse Laplace transform of 7
53)(
2 +
+=s
ssF
Answer:
{ }
( ) ( )( ) ( )tt
ss
sL
ss
sLsFLtf
7sin7
57cos3
7
7
7
5
73
7
5
7
3)()(
22
22
1
22
11
+=
++
+=
++
+==
−
−−
EXAMPLE 7:
Find the inverse Laplace transform of 9
5)(
2 −=s
sF
Answer:
{ }
tt
s
s
ee
sssB
sssA
s
B
s
AL
sLsFLtf
33
3
3
1
2
11
6
5
6
5
6
5
)3)(3(
5)3(
6
5
)3)(3(
5)3(
339
5)()(
−
−=
=
−−−
−=
−=
+−+=
=+−
−=
++
−=
−==
EXAMPLE 8:
Find the inverse Laplace transform of 32 )2(
1)(
+
+=
ss
ssF
Answer:
( )
( )
( )
????????????16
1
8
1
16
1
16
1)()2(
!2
1
0)()2(
4
1)()2(
16
1)(
8
1)(
)2()2()2()(
2
2
3
2
2
2
3
2
3
0
2
0
2
2
t
s
s
s
s
s
et
sFsds
dE
sFsds
dD
sFsC
sFsds
dB
sFsA
s
E
s
D
s
C
s
B
s
AsF
−
−=
−=
−=
=
=
++−
=
=+=
=+=
=+=
−==
==
++
++
+++=
EXAMPLE 9:
Find the inverse Laplace transform of 86
)3(3)(
2 ++
+=
ss
ssF
Answer:
{ } tt eesFLtf 241 5.15.1)()( −−− +==
EXAMPLE 10:
Find the inverse Laplace transform of )10005.34(
1)(
2 ++=
ssssF
Answer:
{ } )9.565.26sin(00119.0001.0)()( 25.171 ++== −− tesFLtf t
EXAMPLE 12:
Find the inverse Laplace transform of )23(
)(2
3
++=
−
sss
esF
s
Answer:
{ }
{ } { } )3(5.05.0)()(
5.05.0)(
1,5.0,5.0
12)()(
)()()23(
1)(
)3()3(21
2
11
3
2
−−+==
−+=
−===
++
++==
=⇒++
=
−−−−−
−−
−−
−
tueesFLtf
eetg
CBA
s
C
s
B
s
ALsGLtg
esGsFsss
sLetG
tt
tt
s
FINAL VALUE THEOREM
The final value theorem can provide the steady state value of a time-domain function from its Laplace transform without going through the inverse Laplace transform procedure. Conditions that must be satisfied:
No poles on imaginary axis (except simple pole at origin) No poles on having positive real part
EXAMPLE 1:
What is the steady state value of f(t) if you know that
)3)(2)(1(
2)(
+++=
sssssF
Answer:
A simple pole at origin and all other poles have negative real part.
3
1
)3)(2)(1(
2lim)(lim)(
00=
+++==∞
→→ ssssFsf
ss
EXAMPLE 2:
What is the steady state value of f(t) if you know that )2)(1(
1)(
+−=
sssF
Answer:
The final value theorem is not applicable because F(s) has a pole s = 1.
EXAMPLE 3:
What is the steady state value of f(t) if you know that )4()2(
1)(
2 ++=
sssF
Answer:
All poles have negative real part.
0)4()2(
lim)(lim)(200
=++
==∞→→ ss
ssFsf
ss
EXAMPLE 4:
What is the steady state value of f(t) if you know that )1(
10)(
2 +=
sssF
Answer:
The final value theorem is not applicable because F(s) has two poles at origin.
EXAMPLE 5:
What is the steady state value of f(t) if you know that )1)(1(
10)(
2 ++=
sssF
Answer:
The final value theorem is not applicable because F(s) has two poles on the
imaginary axis ( j and –j).
EXAMPLE 6:
What is the steady state value of f(t) if you know that
0,))(1(
)( >++
= awhereasss
bsF
Answer:
,))(1(
)()( limlim00 a
b
asss
bssFsf
ss
=++
==∞→→
SOLVING ODE
Solving ODE is one important application for Laplace transform. The general procedure is shown below.
1. Use Laplace transform to convert the ODE into algebraic equation.
2. Solve the algebraic equation for the unknown function
3. Use Partial fraction expansion to express the unknown function as the sum of first and second order terms
4. Use inverse Laplace transform to obtain the solution to the original problem
EXAMPLE 1:
0)0(;0)0(
1)(3)(4)(
==
=++
xx
txtxtx
&
&&&
Answer:
Step1: Convert to algebraic equation
{ } { }
{ } { }s
sXssXsXs
conditionsinitialngsubistiuti
ssXxssXxsxsXs
txtxtx
1)(3)(4)(
1)(3)0()(4)0()0()(
1)(3)(4)(
2
2
=++
=+−+−−
=++
&
&&&
Step2: Solve for X(s) { } { }
( )
( )34
1)(
1)(34
1)(3)(4)(
2
2
2
++=
=++
=++
ssssX
ssXss
ssXssXsXs
Step3: Use partial fraction expansion
( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) 2
1
)2(1
1
3
1
34
1)1(
6
1
)2(3
1
1
1
34
1)3(
3
1
34
1
34
1)(
1334
1)(
11
2
33
2
0
2
0
2
2
−=−
=+
=++
+=
=−−
=+
=++
+=
=++
=++
=
++
++=
++=
−=−=
−=−=
==
ss
ss
ss
ssssssC
ssssssB
ssssssA
s
C
s
B
s
A
ssssX
Step4: Inverse Laplace Transform
( ) ( ) ( )
02
1
6
1
3
1)(
1
2
1
3
6
1
3
1
34
1)(
3
2
≥−+=
+
−+
++=
++=
−− tforeetx
sssssssX
tt
EXAMPLE 2:
2)0(;1)0(
1)(3)(4)(
==
=++
xx
txtxtx
&
&&&
Answer:
Step1: Convert to algebraic equation
{ } { }
{ } { }s
sXssXssXs
conditionsinitialngsubistiuti
ssXxssXxsxsXs
txtxtx
1)(31)(42)(
1)(3)0()(4)0()0()(
1)(3)(4)(
2
2
=+−+−−
=+−+−−
=++
&
&&&
Step2: Solve for X(s)
{ } { }
( )
( )34
13)(
313
1)(34
1)(3)(4)(
2
2
22
2
++
++=
++=++=++
=++
sss
sssX
s
sss
ssXss
ssXssXsXs
Step3: Use partial fraction expansion
( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) 2
1
)2(1
1
3
13
34
13)1(
6
1
)2(3
1
1
13
34
13)3(
3
1
34
13
34
13)(
1334
1)(
1
2
1
2
2
3
2
3
2
2
0
2
2
0
2
2
2
=−−
=+
++=
++
+++=
=−−
=+++
=++
+++=
=++
++=
++
++=
++
++=
++=
−=−=
−=−=
==
ss
ss
ss
ss
ss
sss
sssC
ss
ss
sss
sssB
ss
ss
sss
sssA
s
C
s
B
s
A
ssssX
Step4: Inverse Laplace Transform
( ) ( ) ( )
02
1
6
1
3
1)(
1
2
1
3
6
1
3
1
34
1)(
3
2
≥++=
++
++=
++=
−− tforeetx
sssssssX
tt
EXAMPLE 3:
0)0(;0)0(
1)(4)(
==
=+
xx
txtx
&
&&
Answer:
Step1: Convert to algebraic equation
{ }
{ }s
sXsXs
conditionsinitialngsubistiuti
ssXxsxsXs
txtx
1)(4)(
1)(4)0()0()(
1)(4)(
2
2
=+
=+−−
=+
&
&&
Step2: Solve for X(s) { }
( )
( )41
)(
1)(4
1)(4)(
2
2
2
+=
=+
=+
sssX
ssXs
ssXsXs
Step3: Use partial fraction expansion
( ) ( )
( ) ( )( )
0,
24)(
4
1
4
1
4
1
44
1)(
2
0
2
0
2
22
=−=⇒
=+++
=+
=+
=
+
++=
+=
==
CBA
sAscBs
ssssA
s
CBs
s
A
sssX
ss
Step4: Inverse Laplace Transform
( ) ( )0)2cos(25.025.0)(
4
25.025.0
4
1)(
22
≥−=
+
−+=
+=
tforttx
s
s
ssssX
