LAPLACE TRANSFORMS

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Engineering Mathematics II
Laplace Tranforms
LAPLACE TRANSFORMS
Dr.A.T.Eswara Professor and Head Dept of
Mathematics PES College of Engineering Mandya -
571401
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Engineering Mathematics II
Laplace Tranforms

• INTRODUCTION
• Laplace transform is an integral transform
• employed in solving physical problems.
• Many physical problems when analysed assumes
  the form
• of a differential equation subjected to a set
  of initial
• conditions or boundary conditions.
• By initial conditions we mean that the
  conditions on the
• dependent variable are specified at a single
  value of the
• independent variable.
• If the conditions on the dependent variable are
  specified at
• two different values of the independent
  variable, the
• conditions are called boundary conditions.

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Laplace Tranforms

• The problem with initial conditions is referred
  to as the Initial
• value problem.
• The problem with boundary conditions is
  referred to as the
• Boundary value problem.

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Example 1 The problem of solving the equation
(0) 1
with conditions y(0)
is an initial value problem
Example 2 The problem of solving the equation
with y(1)1, y(2)3 is called Boundary value
problem.
Laplace transform is essentially employed to
solve initial value problems. This technique is
of great utility in applications dealing with
mechanical systems and electric circuits.
Besides the technique may also be employed to
find certain integral values also. The transform
is named after the French Mathematician P.S. de
Laplace (1749 1827).
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The subject is divided into the following sub
topics.
LAPLACE TRANSFORMS
Definition and Properties
Convolution theorem
Inverse transforms
Solution of differential equations
Transforms of some functions
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Laplace Tranforms
Definition Let f(t) be a real-valued function
defined for all t ? 0 and s be a parameter, real
or complex. Suppose the integral
exists (converges). Then this integral is called
the Laplace transform of f(t) and is denoted by
Lf(t).
Thus,
Lf(t)
(1)
We note that the value of the integral on the
right hand side of (1) depends on s. Hence Lf(t)
is a function of s denoted by F(s) or
.
Thus,
Lf(t) F(s)
(2)
Consider relation (2). Here f(t) is called the
Inverse Laplace transform of F(s) and is denoted
by L-1 F(s).
Thus,
L-1 F(s) f(t)
(3)
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Laplace Tranforms
Suppose f(t) is defined as follows
f1(t), 0 lt t lt a
f(t) f2(t), a lt t lt b
f3(t), t gt b
Note that f(t) is piecewise continuous. The
Laplace transform of f(t) is defined as
Lf(t)
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NOTE In a practical situation, the variable t
represents the time and s represents frequency.
Hence the Laplace transform converts the time
domain into the frequency domain. Basic
properties The following are some basic
properties of Laplace transforms 1. Linearity
property For any two functions f(t) and ?(t)
(whose Laplace transforms exist) and any
two constants a and b, we have
L a f(t) b ?(t) a L f(t) b L?(t)
Proof - By definition, we have
Laf(t)b?(t)
a L f(t) b L?(t)
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This is the desired property. In particular, for
ab1, we have L f(t) ?(t) L f(t)
L?(t) and for a -b 1, we have
L f(t) - ?(t) L f(t) - L?(t)
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2. Change of scale property If L f(t) F(s),
then Lf(at)
where a is a positive constant. Proof - By
definition, we have Lf(at)
(1)
Let us set at x. Then expression (1) becomes,

L f(at)
This is the desired property.
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3. Shifting property - Let a be any real
constant. Then L eat f(t)
F(s-a) Proof - By definition, we have
L e at f(t)
F(s-a)
This is the desired property. Here we note that
the Laplace transform of eat f(t) can be written
down directly by changing s to s-a in the Laplace
transform of f(t).
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TRANSFORMS OF SOME FUNCTIONS 1. Let a be a
constant. Then
L(eat)
s gt a
Thus, L(eat)
In particular, when a0, we get
s gt 0
L(1)
By inversion formula, we have
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Laplace Tranforms
2.L (cosh at)
Let s gt a . Then,
Thus, L (cosh at)
, s gt a
and so
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Laplace Tranforms
3. L (sinh at)
, s gt a
Thus, L (sinh at)
, s gt a
and so,
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LAPLACE TRANSFORMS
contd
Dr.A.T.Eswara Professor and Head Dept of
Mathematics PES College of Engineering Mandya -
571401
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Laplace Tranforms
dt
4. L (sin at)
Here we suppose that s gt 0 and then integrate
by using the formula

Thus, L (sin at)
,s gt 0
and so
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Laplace Tranforms
5. L (cos at)
Here we suppose that sgt0 and integrate by using
the formula
Thus, L (cos at)
, s gt 0
and so
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Laplace Tranforms
6. Let n be a constant, which is a non-negative
real number or a negative non-integer. Then
L(tn)
Let s gt 0 and set st x, then
is called gamma function of (n1) denoted by
The integral
Thus
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Laplace Tranforms
Thus
In particular, if n is a non-negative integer
then
n!. Hence
and so
or
as the case may be
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TABLE OF LAPLACE TRANSFORMS
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Application of shifting property - The shifting
property is If L f(t) F(s), then L eatf(t)
F(s-a) Application of this property leads to
the following results
1
Thus, L(eatcoshbt)
and
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Laplace Tranforms
and
and
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Laplace Tranforms
and
or
as the case may be
Hence
or
as the case may be
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Laplace Tranforms
Examples -
1. Find Lf(t) given t, 0 lt t lt 3
f(t)
4, t gt 3
Here
Lf(t)
Integrating the terms on the RHS, we get
Lf(t)
This is the desired result.
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2. Find Lf(t) given sin2t, 0 lt t
? ? 0, t gt ? Here
f(t)
Lf(t)
This is the desired result.
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3. Evaluate (i) L(sin3t sin4t)
(ii) L(cos2 4t) (iii)
L(sin32t)
(i) Here
L(sin3t sin4t) L
by using linearity property
(ii) Here
L(cos24t)
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(iii) We have
For ?2t, we get
so that
This is the desired result.
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4. Find L(cost cos2t cos3t) Here cos2t
cos3t
so that cost cos2t cos3t
Thus L(cost cos2t cos3t)
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5. Find L(cosh22t) We have
For ? 2t, we get
Thus,
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LAPLACE TRANSFORMS
contd
Dr.A.T.Eswara Professor and Head Dept of
Mathematics PES College of Engineering Mandya -
571401
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Laplace Tranforms
6. Evaluate (i) L
(ii)
(iii) L(t-3/2)
We have L(tn)
i) For n
we get
L(t1/2)
Since
, we have
Thus,
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(ii) For n -
, we get
(iii) For n -
, we get
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• Evaluate (i) L(t2) (ii) L(t3)
  
  We have,

L(tn)
(i) For n 2, we get
L(t2)
(ii) For n3, we get
L(t3)
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8. Find Le-3t (2cos5t 3sin5t)
Given 2L(e-3t cos5t) 3L(e-3t sin5t )
by using shifting property
, on simplification
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9. Find Lcoshat sin at Here
Lcoshat sinat
, on simplification
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Laplace Tranforms
10. Find L(cosht sin3 2t)
Given
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11. Find
We have
Put n -5/2. Hence
L(tn)
L(t-5/2)
Change s to s4. Therefore,
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Laplace Tranforms
Transform of tn f(t) Here we suppose that n is a
positive integer. By definition, we have
F(s)
Differentiating n times on both sides w.r.t. s,
we get
Performing differentiation under the integral
sign, we get
Multiplying on both sides by (-1)n , we get
by definition
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Laplace Tranforms
Thus, Ltnf(t)
This is the transform of tn f(t). Also,
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Laplace Tranforms
In particular, we have Lt f(t)
for n1
for n2, etc.
Lt2 f(t)
Also,
and
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Transform of
We have , F(s)
Therefore,
Thus,
This is the transform of
Also,
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Examples 1. Find Lte-t sin4t We have,
So that, Lte-t sin4t
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2. Find L(t2 sin3t) We have L(sin3t)
So that, L(t2 sin3t)
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3. Find
We have
Hence
cot 1 (s1)
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4. Find
Using this, evaluate L
We have
L(sint)
So that
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Laplace Tranforms
Consider
in view of the change of scale property
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5. Find L
We have
L cosat cosbt
So that
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Laplace Tranforms
6. Prove that
We have
Putting s 3 in this result, we get
This is the result as required.
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ASSIGNMENT
I Find L f(t) in each of the following cases
1. f(t) et , 0 lt t lt 2 0 , t gt 2
2. f(t) 1 , 0 ? t ? 3 t ,
t gt 3
, 0 ? t lt a
3. f(t)

• , t ? a

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II. Find the Laplace transforms of the following
functions
4. cos(3t 4) 5. sin3t sin5t 6. cos4t
cos7t 7. sin5t cos2t 8. sint sin2t sin3t 9.
sin2 5t 10. sin2(3t5) 11. cos3 2t 12. sinh2 5t
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13. t5/2
14
15. 3t 16. 5-t 17. e-2t cos2 2t 18. e2t sin3t
sin5t 19. e-t sin4t t cos2t 20. t2 e-3t cos2t
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21 22 23 24
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III. Evaluate the following integrals using
Laplace transforms
25 26 27 28
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Transforms of the derivatives of f(t)
Consider
L
by using integration by parts
0 - f (0) s Lf(t)
Thus
s Lf(t) f(0)
L
Similarly,
we get
s2 L f(t) s f(0) -
L
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In general, we have
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Transform of
Let ? (t)
?(0) 0 and
Then
(t) f(t)
Now,
L ?(t)
Thus,
Also,
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• Examples
• By using the Laplace transform of sinat, find the
  Laplace
• transform of cosat.

Let f(t) sin at, then Lf(t)
We note that
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Taking Laplace transforms, we get
or L(cosat)
Thus L(cosat)
This is the desired result
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2. Given
show that
Let f(t)
given Lf(t)
We note that,
Taking Laplace transforms, we get
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Hence
Thus
This is the result as required.
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3. Find
Here
Lf(t)
Using the result
We get,
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4. Find
Here



)
1
(
8
s
-

t
4
sin
t
te
L


2
2
)
17
2
(
s
s
Thus
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Transform of a periodic function A function f(t)
is said to be a periodic function of period T gt
0 if f(t) f(t nT) where n1,2,3,.. The
graph of the periodic function repeats itself in
equal intervals. For example, sint, cost are
periodic functions of period 2? since sin(t
2n?) sin t, cos(t 2n?) cos t.
The graph of f(t) sin t is shown below
Note that the graph of the function between 0 and
2? is the same as that between 2? and 4? and so
on.
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The graph of f(t) cos t is shown below
Note that the graph of the function between 0 and
2? is the same as that between 2? and 4? and so
on.
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Formula Let f(t) be a periodic function of
period T. Then
Proof By definition, we have
L f(t)
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Let us set u t nT, then
T

å
L f(t)
ò
-

)
(
nT
t
s
)
(
dt
nT
t
f
e


0
n

0
t
Here f(tnT) f(t), by periodic property
Hence
identifying the above series as a geometric
series.
Thus
L f(t)
This is the desired result.
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Examples-

• For the periodic function f(t) of period 4,
  defined by
• f(t) 3t, 0 lt t lt 2
• 6 , 2 lt t lt 4
• find L f(t)
• Here, period of f(t) T 4
• We have,

L f(t)
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Thus,
Lf(t)
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2. A periodic function of period
is defined by
Esin?t, 0 ? t lt
f(t)
0 ,
? t ?
where E and ? are positive constants. Show
that
L f(t)
Here
Therefore
T
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L f(t)
This is the desired result.
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3. A periodic function f(t) of period 2a, agt0 is
defined by
E , 0 ? t ? a f(t)
-E, a lt t ? 2a
show that L f(t)
Here T 2a. Therefore
L f(t)
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This is the result as desired.
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Step Function In many Engineering applications,
we deal with an important discontinuous function
H(t-a) defined as follows
0, t ? a
H (t-a)
1, t gt a
where a is a non-negative constant. This
function is known as the unit step function or
the Heaviside function. The function is named
after the British electrical engineer Oliver
Heaviside. The function is also denoted by
u(t-a). The graph of the function is shown below
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H(t-a)
1
t
0
a
Note that the value of the function suddenly
jumps from value zero to the value 1 as
from the left and retains the value 1 for all
tgta. Hence the function H(t-a) is called the
unit step function. For a0, We have
0, t ? 0
H (t)
1, t gt 0
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Transform of step function By definition, we have
LH(t-a)
In particular, we have
L H(t)
Also,
and
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Heaviside shift theorem Statement - L f(t-a)
H(t-a) e-as Lf(t) Proof - We have
L f(t-a) H(t-a)
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Setting t-a u, we get
Lf(t-a) H(t-a)
e-as L f(t)
This is the desired shift theorem. Also,
L-1 e-as L f(t) f(t-a) H(t-a)
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Examples 1. Find Let-2 sin(t-2) H(t-2)
Let f(t-2) et-2 sin(t-2) Then f(t)
et sint so that
L f(t)
By Heaviside shift theorem, we have Lf(t-2)
H(t-2) e-2s Lf(t) Thus,
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2. Find L(3t2 2t 3) H(t-1) Let f(t-1)
3t2 2t 3 so that f(t) 3(t1)2 2(t1)
3 3t2 8t 8 Hence
Thus L3t2 2t 3 H(t-1) Lf(t-1) H(t-1)
e-s L f(t)
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3. Find Le-t H(t-2) Let f(t-2) e-t ,
so that, f(t) e -(t2) Thus,
L f(t)
By shift theorem, we have
Thus
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4. Let f(t) f1 (t) , t ? a
f2 (t) , t gt a
Verify that f(t) f1(t) f2(t) f1(t)H(t-a)
Consider
f1(t) f2(t) f1(t)H(t-a) f1(t) f2
(t) f1(t), t gt a
0 , t ? a
f2 (t), t gt a f1(t), t ? a
f(t), given
Thus the required result is verified
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5. Express the following functions in terms of
unit step function and hence find their Laplace
transforms.
1. t2 , 1 lt t ? 2 f(t) 4t ,
t gt 2
2. cost, 0 lt t lt ? f(t)
sint, t gt ?
1. Here, f(t) t2 (4t-t2) H(t-2)
Hence,
(i)
L f(t)
Let ?(t-2) 4t t2 so that
?(t) 4(t2) (t2)2 -t2 4
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Now,
Expression (i) reads as
L f(t)
This is the desired result.
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2. Here f(t) cost (sint-cost)H(t-?) Hence,
L f(t)
(ii)
Let ? (t-?) sint cost Then ?(t) sin(t
?) cos(t ?) -sint cost so that
L ?(t)
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Expression (ii) reads as
L f(t)
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CONVOLUTION
The convolution of two functions f(t) and g(t)
denoted by f(t) g(t) is defined as
f(t) g(t)
Property f(t) g(t) g(t) f(t) Proof - By
definition, we have
f(t) g(t)
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Setting t-u x, we get
f(t) g(t)
This is the desired property. Note that the
operation is commutative.
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Convolution theorem - Lf(t) g(t) L f(t)
. L g(t) Proof - Let us denote
f(t) g(t) ?(t)
Consider
(1)
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We note that the region for this double integral
is the entire area lying between the lines u 0
and u t. On changing the order of
integration, we find that t varies from u to ?
and u varies from 0 to ?.
U
ut
tu
t ?
t
u0
0
91
Hence (1) becomes L?(t)
, where v t-u
L g(t) . L f(t)
Thus L f(t) . L g(t) Lf(t) g(t) This is
desired property.
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Examples
1. Verify Convolution theorem for the functions
f(t) and g(t) in the following cases (i) f(t)
t, g(t) sint (ii) f(t) t, g(t) et (i)
Here,
f g
Employing integration by parts, we get
f g t sint
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so that L f g
(1)
Next consider
(2)
L f(t) . L g(t)
From (1) and (2), we find that L f g
L f(t) . L g(t) Thus convolution theorem
is verified
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(ii) Here
f g
Employing integration by parts, we get f g
et t 1 so that
Lf g
(3)
Next
L f(t) . L g(t)
(4)
From (3) and (4) we find that Lf g L
f(t) . L g(t) Thus convolution theorem is
verified.
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2. By using the Convolution theorem, prove that
Let us define g(t) 1, so that g(t-u) 1
Then
L f(t) . L g(t) L f(t) .
Thus
This is the result as desired.
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3. Using Convolution theorem, prove that
Let us denote, f(t) e-t g(t) sin t, then
L f(t) . L g(t)
This is the result as desired
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ASSIGNMENT
1. By using the Laplace transform of coshat, find
the Laplace transform of sinhat.
2. Find (i)
(ii)
(iii)
(iv)
(v)
3. If f(t) t2, 0 lt t lt 2 and f(t2) f(t)
for t gt 2, find L f(t)
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4. Find L f(t) given t , 0 ? t ? a
f(2at) f(t) 2a
t, a lt t ? 2a
f(t)
5. Find L f(t) given 1, 0 lt t lt
f(at) f(t)
f(t)
-1, lt t lt a
6. Find the Laplace transform of the following
functions (i) et-1 H(t-2) (ii) t2 H(t-2)
(iii) (t2 t 1) H(t 2) (iv) (e-t sint)
H(t - ?)
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7. Express the following functions in terms of
unit step function and hence find their Laplace
transforms
(i) 2t, 0 lt t ? ? (ii) t2 ,
2 lt t ? 3 1 , t gt ? 3t ,
t gt 3 (iii) sin2t, 0 lt t ? ? (iv)
sint, 0 lt t ? ?/2 0
, t gt ? cost, t gt ?/2
f(t)
f(t)
f(t)
f(t)
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8. Let f1 (t) , t ? a f(t) f2 (t) , altt
? b f3 (t) , t gt b

Verify that f(t) f1(t) f2(t) f1(t)H(t-a)
f3(t) - f2(t) H(t-b)
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• 9. Express the following function in terms of
  unit step
• function and hence find its Laplace
  transform.

Sint, 0 lt t ? ? f(t)
Sin2t, ? lt t ? 2? Sin3t, t gt 2?
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10. Verify convolution theorem for the
following pair of functions (i) f(t)
cosat, g(t) cosbt (ii) f(t) t, g(t)
t e-t (iii) f(t) et, g(t) sint
11. Using the convolution theorem, prove the
following
i
ii
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INVERSE LAPLACE TRANSFORMS
Let L f(t) F(s). Then f(t) is defined as the
inverse Laplace transform of F(s) and is denoted
by L-1 F(s). Thus L-1 F(s) f(t). Linearity
Property Let L-1 F(s) f(t) and L-1 G(s) g(t)
and a and b be any two constants. Then L-1
a F(s) b G(s) a L-1 F(s) b L-1 G(s)
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Table of Inverse Laplace Transforms
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n 0, 1, 2, 3, . . .
n gt -1
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Examples 1. Find the inverse Laplace transforms
of the following
Here
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Evaluation of L-1 F(s a)
We have, if L f(t) F(s), then Leat f(t)
F(s a), and so L-1 F(s a) eat f(t)
e at L-1 F(s) Examples
Using the formula
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(1),
Then
2s25s-4 A(s2) (s-1) Bs (s-1) Cs (s2)
For s 0, we get A 2, for s 1, we get C 1
and for s -2, we get B -1. Using these
values in (1), we get
Hence
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Let us take
Then 4s 5 A(s 2) B(s 1) (s 2) C (s
1)2 For s -1, we get A 1, for s -2, we get
C -3
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Comparing the coefficients of s2, we get B C
0, so that B 3. Using these values in (1), we
get
Hence
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Hence
s3 A(s a) (s2 a2) B (s-a)(s2a2)(Cs D)
(s2 a2)
For s a, we get A ¼ for s -a, we get B
¼ comparing the constant terms, we get D
a(A-B) 0 comparing the coefficients of s3, we
get 1 A B C and so C ½. Using these
values in (1), we get
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Taking inverse transforms, we get
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Consider
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Evaluation of L-1e-as F(s) We have, if Lf(t)
F(s), then Lf(t-a) H(t-a) e-as F(s), and so
L-1e-as F(s) f(t-a) H(t-a) Examples
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Inverse transform of logarithmic and inverse
functions
Examples
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(2) Evaluate
or
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Since
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Examples
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,on integration by parts
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This expression is called the convolution theorem
for inverse Laplace transform
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Examples Employ convolution theorem to evaluate
the following
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ASSIGNMENT By employing convolution theorem,
evaluate the following
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LAPLACE TRANSFORM METHOD FOR DIFFERENTIAL
EQUATIONS
As noted earlier, Laplace transform technique is
employed to solve initial-value problems. The
solution of such a problem is obtained by using
the Laplace Transform of the derivatives of
function and then the inverse Laplace
Transform. The following are the expressions for
the derivatives derived earlier.

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Examples 1) Solve by using Laplace transform
method
Taking the Laplace transform of the given
equation, we get
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This is the solution of the given equation
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2.Solve by using Laplace transform method
Taking the Laplace transform of the given
equation, we get
Using the given conditions, we get
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The circuit is an LR circuit. The differential
equation with respect to the circuit is
Here L denotes the inductance, i denotes current
at any time t and E(t) denotes the E.M.F. It is
given that E(t) E e-at. With this, we have
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(6) Solve the simultaneous equations for
x and y in terms of t given
Taking Laplace transforms of the given equations,
we get
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(1) and (2)together represents the solution of
the given equations
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ASSIGNMENT
Employ Laplace transform method to solve the
following initial value problems
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