3 – Laplace Transforms
ENGI 3424
3.
Page 3.01
Laplace Transforms
In some situations, a difficult problem can be transformed into an easier problem, whose
solution can be transformed back into the solution of the original problem. For example,
an integrating factor can sometimes be found to transform a non-exact first order first
degree ordinary differential equation into an exact ODE [Section 1.7]:
One type of first order ODE [not considered in this course] is the Bernoulli ODE,
y  P y  R y u , which, upon a transformation of the dependent variable, becomes a
linear ODE:
An initial value problem is an ordinary differential equation together with sufficient
initial conditions to determine all of the arbitrary constants of integration. A Laplace
transform will convert some initial value problems into much easier algebra problems.
The solution of the original problem is then the inverse Laplace transform of the solution
to the algebra problem.
Uses of Laplace transforms include:
1)
the solution of some ordinary differential equations
2)
the solution of some integro-differential equations, such as
y t   g t  
t
 0 h t  x  y  x  dx .
ENGI 3424
3 – Laplace Transforms
Page 3.02
Contents:
3.01
3.02
3.03
3.04
3.05
3.06
3.07
3.08
3.09
3.10
3.11
Definition, Linearity, Laplace Transforms of Polynomial Functions
Laplace Transforms of Derivatives
Review of Complex Numbers
First Shift Theorem, Transform of Exponential, Cosine and Sine Functions
Applications to Initial Value Problems
Laplace Transform of an Integral
Heaviside Function, Second Shift Theorem; Example for RC Circuit
Dirac Delta Function, Example for Mass-Spring System
Laplace Transform of Periodic Functions; Square and Sawtooth Waves
Derivative of a Laplace Transform
Convolution; Integro-Differential Equations; Circuit Example
Note: Sections 3.09, 3.10 and the latter part of 3.11 will be parts of this course only if
time permits.
3.01 – Definition; Polynomial Functions
ENGI 3424
3.01
Definition, Linearity, Laplace Transforms of Polynomial Functions
If f  t  is



defined on t  0 ,
piece-wise continuous on t  0
(that is, only a finite number of finite discontinuities) and
of exponential order
[ f  t   k e t for all t  0 and for some positive constants k and  ],
then
m
F  s   lim
m
e
 st
f  t  dt exists for all s  0 and
0
the Laplace transform of f  t  is

F s 

0
Example 3.01.1
Find L  1  .
e st f  t  dt  L  f  t 
Page 3.03
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3.01 – Definition; Polynomial Functions
Example 3.01.2
Find L  t  .
Linearity Property of Laplace Transforms
L  a f  t   b g  t   (where a, b are constants):
Page 3.04
ENGI 3424
Example 3.01.3
Find
L  2t  3  .
Example 3.01.4
Find
L tn  .
3.01 – Definition; Polynomial Functions
Page 3.05
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3.01 – Definition; Polynomial Functions
Example 3.01.5
Find


L t 4  2t 3  3 .
Page 3.06
3.02 – Laplace Transforms of Derivatives
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3.02
Page 3.07
Laplace Transforms of Derivatives
L  f  t   

e
 st
f   t  dt 
0
Continuing this pattern, we can deduce the Laplace transform for any higher derivative of
f t  :
or
This result allows us to find the Laplace transform of an entire initial value problem.
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3.02 – Laplace Transforms of Derivatives
Page 3.08
Example 3.02.1
Find the Laplace transform of the initial value problem
d2y
 6 y  6t 3 ,
2
dt
y  0  0 ,
y  0    1
and hence find the complete solution.
Let Y  s   L  y  t   , then
L  y  6 y  

L 6t 3

Therefore the Laplace transform of the initial value problem is
which is an algebra problem for Y  s  , (the Laplace transform of the complete solution).
Solving the algebra problem:
ENGI 3424
3.03
3.03 – Review of Complex Numbers
Page 3.09
Review of Complex Numbers
Before proceeding further with Laplace transforms, some familiarity with complex
numbers is required. A very brief review of complex numbers is provided here.
The set
of real numbers is closed under addition, subtraction, multiplication and (with
the exception of zero) division. However, the square root of any negative number is
not real. This generates the set of imaginary numbers, which are all real multiples of the
imaginary number j  1 . For example: 49  49  1  49  1  7 j .
Imaginary and real numbers can be combined by addition to form the set
numbers, which can be represented on an Argand diagram:
of complex
Example 3.03.1
Two complex numbers are defined by z  3  2 j and w   1  3 j :
The real parts of these numbers are Re  z   3 and Re  w   1
The imaginary parts of these numbers are Im  z   2 and Im  w  3
The sum (or difference) of any pair of complex numbers is calculated by taking the sum
(or difference) of the real parts and imaginary parts separately:
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3.03 – Review of Complex Numbers
Page 3.10
The real and imaginary parts of a complex number z  x  y j behave on the Argand
 x
diagram very like the Cartesian x and y components of a vector v    :
 y
Also like vectors in 2, complex numbers can be expressed in terms of their magnitude
and direction.
The modulus (or “magnitude” or “amplitude”) of z  x  y j
 x
z  x  y j  x2  y 2
(which is also the length of the vector v    ) is
 y
The argument (or “phase”) of z  x  y j
 x
(which is also the angle that the vector v    makes with the positive x axis) is ,
 y
y
where tan  
. When Re  z   x  0 , add or subtract  radians from the inverse
x
tangent in order to be in the correct quadrant (2nd or 3rd, not 4th or 1st).
The principal argument is usually taken to be in the range      or 0    2 .
In Example 3.03.1,
3.03 – Review of Complex Numbers
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Page 3.11
Multiplication and Division of Complex Numbers
Example 3.03.1 (continued)
zw   3  2 j  1  3 j  
It is generally true that
zw  z  w
and
arg  zw  arg  z   arg  w
Also, for division (where the divisor is not zero),
z
z

w
w
and
z
arg    arg  z   arg  w
 w
The argument of the product or quotient may need to be adjusted by the addition or
subtraction of 2 in order to bring it back inside the range of the principal argument.
3.03 – Review of Complex Numbers
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Page 3.12
In the eighteenth century Euler extended the definition of the exponential function to
non-real exponents:
e j  cos   j sin 
It then follows that
z  x  j y  r cos   j r sin   r e j
,
where r  z and   arg  z  .
Alternative rules for multiplication and division then follow immediately:

z1 z2  r1e
and
j1
 r e    r r e  e 
j
j 2
2
j 1 j 2
1 2
j
  r1r2  e
 r  j   
z1
re 1
r e 1
 1 j  1 j   1  e 1 2 ,
z2
r2 e 2
r2e 2
 r2 
j 12 
 r2  0 
Exponents and roots of complex numbers are best found in exponential form.
Example 3.03.2
For the complex numbers z  3  4 j, w   2  2 j , find zw ,
z
1
, z 3 and .
w
z
ENGI 3424
3.03 – Review of Complex Numbers
Example 3.03.2 (continued)
Definition: The complex conjugate of z  x  y j is z *  x  y j .
For any complex number z ,
zz*   x  y j  x  y j   x 2   y j   x 2  y 2  z
2
For any non-zero complex number,
Example 3.03.2(d) (again)
1

z
2
Page 3.13
3.03 – Review of Complex Numbers
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Page 3.14
Example 3.03.3
Find all distinct cube roots of the real number 8.
The only real cube root of 8 is 2. However, there is a pair of non-real roots also.
Method 1
Let z be a cube root of 8, then z 3  8

Method 2
 j 02n  
j 2n /3
z  8  2 e
, n  
 z  2e
  2e


 2n 
 2n 
 z  2cos 
  2 j sin 

 3 
 3 
n  0  z  2cos 0  2 j sin 0  2
 2 
 2 
n  1  z  2cos 
  2 j sin 
  1  j 3
 3 
 3 
 2 
 2 
n  1  z  2cos  
  2 j sin  
  1  j 3
 3 
 3 
and these are the only three distinct values.
n = ..., –7, –4, 2, 5, 8, ... return the same value for z as n = –1.
n = ..., –6, –3, 3, 6, 9, ... return the same value for z as n = 0.
n = ..., –5, –2, 4, 7, 10, ... return the same value for z as n = +1.
3
3
j  0 2n 
1/3
ENGI 3424
3.03 – Review of Complex Numbers
Page 3.15
Example 3.03.3 (continued)
On the Argand diagram the roots are equally spaced around a circle
of radius 2 centred on 0.
A similar graphical approach can be used to find the n distinct nth roots of any non-zero
complex number.
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3.03 – Review of Complex Numbers
Page 3.16
Example 3.03.4
On the Argand diagram sketch the locations of the five distinct fifth roots of the complex
number w. You may assume that the radius of the guide circle is
Also note that e j  cos   j sin 
and
e j  cos     j sin     cos   j sin 

and
5
w .
3.04 – First Shift Theorem, Cosine and Sine
ENGI 3424
3.04
First Shift Theorem, Transform of Exponential, Cosine and Sine Functions

L eat f  t 

The first shifting theorem then follows:
If
Example 3.04.1
Find


L t 3e5t .
Example 3.04.2
Find
Page 3.17
 
L eat .
L  f  t    F  s  , then

L eat f  t 
  F s  a
3.04 – First Shift Theorem, Cosine and Sine
ENGI 3424
Example 3.04.3
Find
L  cos  t  .
Example 3.04.4
Find
L  sin  t  .
It then follows that


s
L1  2
 cos  t
2 
 s  
and


1
sin  t
L1  2

2 

 s  
Page 3.18
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3.04 – First Shift Theorem, Cosine and Sine
Example 3.04.5
Find L  sin  t  directly, from the definition of a Laplace transform.
Example 3.04.6
Find L  sin  t  , using L  cos  t  .
Page 3.19
3.04 – First Shift Theorem, Cosine and Sine
ENGI 3424
Example 3.04.7
Find


L eat sin  t .
Using the first shift theorem, we have, immediately,
Similarly,

L eat cos  t
Example 3.04.8
Find
L  tan  t  .
Example 3.04.9
Find
L  sinh at  .
Similarly,
L  cosh at  
s
s  a2
2

sa
s  a
2
 2
Page 3.20
3.04 – First Shift Theorem, Cosine and Sine
ENGI 3424
Page 3.21
Summary so far:
Definition:
L  f t   


0
e st f  t  dt
Linearity:
L  a f  t   b g  t    a L  f t    b L  g t  
 a, b  constants 
Polynomial functions:
n!
t n1
 1 
L  t n   n1
 L 1  n  
 n  1 !
s
s 
First Shift Theorem:

L  f  t    F  s   L eat f  t 
  F s  a
 1 
at
and L 1 
  e
 sa 
Trigonometric Functions:




L eat sin  t
L eat cos  t
s  a
2
 2
sa
 s  a
2
 2
Hyperbolic Functions:

at
L e sinh bt


L eat cosh bt



1
eat sin  t


 L 1 


2
2


 s  a   



sa


at
 L 1 
  e cos  t
2
2
 s  a   



b
 s  a
2
 b2
sa
 s  a
 abt  e a bt
at


1

 e sinh bt e
 L 


2
2
b
2b
s

a

b






 abt  e a bt


sa
e
 at
1 
 L 
  e cosh bt 
2
2
2
 s  a  b 


1
2
 b2
Derivatives:
L  f   t    s L  f t    f  0
L  f   t    s 2 L  f  t    s f  0  f   0 
etc.
When trying to find the inverse Laplace transform of a rational function F  s  , one
usually attempts to break the function up into its partial fractions, with real linear and
quadratic denominators, so that one may read the inverses from the table on this page.
ENGI 3424
3.05
3.05 – Applications to Initial Value Problems
Applications to Initial Value Problems
Example 3.05.1
Solve the initial value problem
y  5 y  6 y  0 ,
y  0   1, y  0   0
Let Y  s   L  y  t   .
L  y  t   
L  y  t   
Taking the Laplace transform of the entire initial value problem:
L  y  5 y  6 y   L  0  
Check 1:
Page 3.22
ENGI 3424
3.05 – Applications to Initial Value Problems
Example 3.05.1 (continued)
Check 2:
Solution by Chapter 2 method:
y  5 y  6 y  0
Page 3.23
3.05 – Applications to Initial Value Problems
ENGI 3424
Example 3.05.2
Solve the initial value problem
y  4 y  13 y  26 e4t ,
y  0   5, y  0    29
Let Y  s   L  y  t   .
L  y  t   
L  y  t   

L 26 e4t

Taking the Laplace transform of the entire initial value problem:

L  y  4 y  13 y   L 26 e4t


Recall the three standard inverse Laplace transforms:
 1 
at
L 1 
  e
s

a



sa

L 1 
2
2

 s  a  


at
  e cos  t





L 1 
2
2

 s  a  


at
  e sin  t


Page 3.24
ENGI 3424
3.05 – Applications to Initial Value Problems
Page 3.25
Example 3.05.2 (continued)
We need to express Y  s  in partial fractions and to complete the square in the quadratic
denominator.
ENGI 3424
3.05 – Applications to Initial Value Problems
Example 3.05.2 (by a Chapter 2 method)
y  4 y  13 y  26 e4t ,
y  0   5, y  0    29
Page 3.26
3.06 – Laplace Transform of an Integral
ENGI 3424
3.06
Let
Page 3.27
Laplace Transform of an Integral
f t  

t
0
g   d , then
Example 3.06.1
Find the function f  t  whose Laplace transform is F  s  
1
.
s s  2 
2
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3.06 – Laplace Transform of an Integral
Page 3.28
Example 3.06.1 (Alternative solution, using partial fractions)
Example 3.06.2
Find the function f  t  whose Laplace transform is F  s  
Alternative method, using partial fractions:
1
.
s s  2 
2
2
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3.07
3.07 – Heaviside Function, Second Shift Theorem
Page 3.29
Heaviside Function, Second Shift Theorem; Example for RC Circuit
The Heaviside function H  t  a  (also known as the unit step function u  t  a  ) is
defined by

0
H t  a   

1
t  a 
t  a 
 a  0
The Laplace transform of the
Heaviside function is:
F  s   L  H t  a   
If

 g t 
f t   

 h t 
0  t  a
t  a 
,
then




f t   g t    h t   g t   H t  a 
 
 

ON
OFF


[The function h  t  is “switched on” and g  t  is “switched off” at time t = a.]
ENGI 3424
3.07 – Heaviside Function, Second Shift Theorem
Page 3.30
Example 3.07.1
Express the function

 2x
f  x  

 4
 x  2
 x  2
in a single line definition.
Answer:
f  x 
The Second Shift Theorem
The result of shifting the graph y  f  t  (defined only for t > 0) a units to the right, is
the graph y  f  t  a  H  t  a  :
The Laplace transform of the shifted function of t is:
L  H t  a  f t  a   


0
e st H  t  a  f  t  a  dt
L  H t  a  f t  a   
The second shift theorem (for a > 0) is
L  H  t  a  f  t  a    eas L  f  t  
ENGI 3424
3.07 – Heaviside Function, Second Shift Theorem
Another way to express the second shift theorem is:
L 1  F  s    f  t 

L 1  eas F  s    H  t  a  f  t  a 
Example 3.07.2


Find L H  t  4   t  4  .

3
L H t  4 t  4
3

Example 3.07.3
Find the function whose Laplace transform is
e 5 s
.
s2  4
Page 3.31
3.07 – Heaviside Function, Second Shift Theorem
ENGI 3424
Page 3.32
Example 3.07.4
In the RC circuit shown here, there is no
charge on the capacitor and no current
flowing at the time t = 0.
The input voltage Ein is a constant Eo
during the time t1  t  t2 and is zero at all
other times.
Find the output voltage Eout for this circuit.
Initial conditions:
q(0) = i(0) = 0
E
Ein   o
 0

dq 

 Note: dt  i 
 t1  t  t2 
 otherwise 
Ein 
Equating voltage drops around the circuit with Ein :
Let Q  s   L  q  t   .
Taking the Laplace transform of this initial value problem:
ENGI 3424
3.07 – Heaviside Function, Second Shift Theorem
Example 3.07.4 (continued)
Graph of the output voltage against time:
Page 3.33
3.07 – Heaviside Function, Second Shift Theorem
ENGI 3424
Example 3.07.5
Find the complete solution of the initial value problem
0
d2y
 4 y  f t   
2
dt
t
Let Y  s   L  y  t  
 0  t  3
 t  3
;
y  0   y  0   0 .
Page 3.34
ENGI 3424
3.07 – Heaviside Function, Second Shift Theorem
Example 3.07.5 (continued)
Page 3.35
3.08 – Dirac Delta Function
ENGI 3424
3.08
Dirac Delta Function, Example for Mass-Spring System
1

f t    

0
Let
Area under
graph


0
Page 3.36
f  t  dt


a  t  a   
 otherwise 
Area of
rectangle
1


 1
Also
f t  
Define the Dirac delta function to be
  t  a   lim f  t 
 0
then
L  f t   
 L   t  a   
Therefore  for a  0 
and
Also, for a  0 , the total area under the graph is 1 (even in the limit as   0+), so

0  t  a  dt
1
ENGI 3424
3.08 – Dirac Delta Function
Page 3.37
For g  t   any function that is continuous everywhere on  0,   ,

0
g  t    t  a  dt 

0
g  t    t  a  dt  g  a 
(which is the sifting property of the Dirac delta function).
Example 3.08.1
A damped mass-spring system, (with damping constant = c = 3m and spring modulus =
k = 2m, where m is the mass), is at rest in its equilibrium position, until an impulse of
5 Ns is applied at time t = 4 s. Find the response y  t  .
One can anticipate some features of the response. The forcing function is zero until
t  4 . Thus there is nothing to disturb the system until that instant. We can deduce that
y  t   0 during t  4 . The mass-spring system is set into motion abruptly by the arrival
of the impulse at the instant t  4 , but the forcing term is again zero thereafter. The
damping force will asymptotically restore the system to its equilibrium state. The only
question remaining is whether the system is under-damped or not (see the diagram on the
next page).
ENGI 3424
3.08 – Dirac Delta Function
Example 3.08.1 (continued)
Anticipated response:
The auxiliary equation is  2  3  2  0 . The discriminant is
Solution, using Laplace transforms:
y  3 y  2 y  5  t  4
y  0  y  0  0
Let Y  s   L  y  t  
Page 3.38
ENGI 3424
3.08 – Dirac Delta Function
Example 3.08.1 (continued)
 1 
 at
From the first shift theorem: L 1 
  e
 sa 
The second shift theorem states:
L 1  F  s    f  t   L 1  eas F  s    f  t  a  H  t  a 
Therefore
or, equivalently,
Page 3.39
3.09 – Periodic Functions
ENGI 3424
Page 3.40
3.09
Laplace Transform of Periodic Functions; Square and Sawtooth Waves
[only if time permits]
If the constant p  0 and f  t  p   f  t  for all t  0 , then
f  t  is a periodic function of t, with period p.
Example of a periodic function (with one finite discontinuity in each period):
Define a new set of functions g n  t  , each of which captures only a single period of f  t  :

 f  t   np  t   n  1 p 
gn  t   
 otherwise 

 0
Then
f t  

 gn  t 
n 0
Let Gn  s   L  g n  t   
 0 

 n1 p
np
e
 st


0
e st g n  t  dt
g n  t  dt  0 
Let F  s   L  f  t   , then

n1 p
np
e st f  t  dt
ENGI 3424
3.09 – Periodic Functions
Therefore, for a periodic function f  t  with fundamental period p,
L  f t  
1

1  e s p

p
0
e st f  t  dt
Example 3.09.1
Use the formula above to verify the Laplace transform of f  t   sin t .
Page 3.41
ENGI 3424
Example 3.09.2
3.09 – Periodic Functions
The Square Wave
The square wave is periodic, with fundamental period p  2a .
In the “zeroth” period  0  t  2a  ,

 1  0  t  a 
f  t   g0  t   

 1  a  t  2a 
The Laplace transform F  s  of the square wave f  t  then follows.
Page 3.42
ENGI 3424
3.09 – Periodic Functions
Example 3.09.3 The Saw-tooth Wave
The saw-tooth wave is periodic, with fundamental period p  a .
In the “zeroth” period  0  t  a  ,
bt
a
The Laplace transform F  s  of the saw-tooth wave f  t  then follows.
f  t   g0  t  
Page 3.43
ENGI 3424
3.09 – Periodic Functions
Page 3.44
Example 3.09.4
Find the function f  t  whose Laplace transform is F  s  
(where a is a constant).
1
 as 
tanh  
2
s
 2 
3.10 – Derivative of a Laplace Transform
ENGI 3424
3.10
Page 3.45
Derivative of a Laplace Transform [only if time permits]
Example 3.10.1
Find F  s   L  t cos t  .
Method via an ODE:
Let
f  t   t cos t
Then
f  t  
and
f   t  
Also
f  0 
f   0 
Taking the Laplace transform of this initial value problem,
F  s   L  f t   


0
e st f  t  dt
 Fs 

d    st
e f  t  dt 

ds  0


Using Leibnitz differentiation of an integral:


  st

F s  0  0 
e f  t  dt 
e st  t f  t   dt
0
0 s
Therefore
d
L  f  t     L  t f  t    L 1  F   s     t  L 1  F  s  
ds




ENGI 3424
3.10 – Derivative of a Laplace Transform
Example 3.10.1 (again)
Find F  s   L  t cos t  .
Method via the derivative of a transform:
Let g  t   cos t
Example 3.10.2
Find F  s   L  t sin t  .
Page 3.46
3.11 – Convolution
ENGI 3424
Page 3.47
3.11
Convolution; Integro-Differential Equations; Circuit Example
If
L  f t    F  s 
and
L  g t    G  s 
then the convolution of f  t  and g  t  is denoted by
 f  g  t 


t
0
 f  g  t  , is defined by
f   g  t    d
and the Laplace transform of the convolution of two functions is the product of the
separate Laplace transforms:
L   f  g  t    F  s  G  s 
An equivalent identity is
L 1  F  s  G  s    L 1  F  s   L 1  G  s  
Convolution is commutative:
g f 

t
0
f  t    g   d 

t
0
f   g  t    d  f  g
Example 3.11.1
Find the inverse Laplace transform of R  s  
1
.
s s  2 
2
2
ENGI 3424
3.11 – Convolution
Page 3.48
Example 3.11.1 (continued)
Note:
This inverse transform can also be found using partial fractions or the division of another
Laplace transform by s. Both of these alternatives are in example 3.06.2.
3.11 – Convolution
ENGI 3424
Page 3.49
Example 3.11.2

1

Find L 1 
2
2 2
  s   


 .

The alternative methods that were available in example 3.11.1 are not available here.
The only obvious way to proceed is via convolution.
Let
R s 
s
1
2
2 
2
3.11 – Convolution
ENGI 3424
Page 3.50
Transfer Function [only if time permits]
For a second order linear ordinary differential equation with constant coefficients, when
both initial conditions are zero, the complete solution can be expressed as a convolution.
y  b y  c y  r  t 
and
y  0   y  0   0

 s 2  bs  c Y  s   R  s 

Y  s   Q  s  R  s  , where Q(s) is the transfer function:
Q s 
1
s  bs  c
2


1
The complete solution is just y  t   q  t   r  t  , where q  t   L 1  2
 .
 s  bs  c 
Example 5.11.3
A unit impulse is delivered at time t  1 to an harmonic system (with c/m = 4) that is in
equilibrium until that instant. Find the response y  t  .
ENGI 3424
3.11 – Convolution
Some More Identities involving Convolution
1 y 
[only if time permits]
Page 3.51
ENGI 3424
3.11 – Convolution
Integro-Differential Equations
[only if time permits]
Example 3.11.4
Find the solution y  t  of the integro-differential equation
y t   t 2  1  9
t
0 t  x  y  x  dx
Page 3.52
ENGI 3424
Example 3.11.5
3.11 – Convolution
Page 3.53
[only if time permits]
Solve the system of simultaneous integral equations
t

4
i

12
 i1  i2  d  1
1

0


t
t
 i  2 i d  2  i  i  d
2
1
2
 2
0
0


Let







I1  s   L  i1  t   , I 2  s   L  i2  t   .
Taking the Laplace transform of the entire system of integral equations,
ENGI 3424
3.11 – Convolution
Example 3.11.5 (continued)
These are the currents in the circuit
with initial conditions i1  0   14 , i2  0   0 .
END OF CHAPTER 3
Page 3.54