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Laplace Transforms
Laplace Transform
Inverse Laplace transform
Dirac delta and unit step function Solution of initial value problems
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Conditions for the Existence of a Laplace Transform of f(t)
1. f(t) is piecewise continuous on 0 t < .
2. f(t) is of exponential order as t →.
where s is a complex constant with real part greater than zero,
and L is the Laplace Transform operator.
If s is real then s > 0.
The Laplace Transform of a function defined on
is denoted by and is given by( )f t
0 t
( ) ( ) ( ) 0
stF s f t e dt L f t
−= =
( )F s
Definition
Note that conditions 1 and 2 are sufficient, but not necessary
( ) for all atf t Ke t T
That is there exist real constants K and T such that
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Example: Find the Laplace transform F(s) of f(t) = A t.
Solution: The Laplace transform F(s) of f(t) is given by
( ) ( ) ( ) stF s L f t f t e dt
−= = 0
Integrating by parts
( )st ste es s
F s A t dt− −
− −
= − =
0 0
1
( ) st stF s Ate dt A te dt
− − = = 0 0
( )st
stAs
A eA e dt
s s
−−− + =
−0 0
0 0
stA e
s s
−
= =−
0
A
s=
2
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Example: Find the Laplace transform F(s) of f(t) = eat.
Solution: The Laplace transform F(s) of f(t) is given by
( ) ( ) ( ) stF s L f t f t e dt
−= = 0
( ) ( )at st s a tF s e e dt e dt
− − − = = 0 0
( )
0( )
ts a t
t
e
s a
=− −
=
=− −
0 1
( ) ( )s a s a= −− − − −
1( ) atF s L es a
= =−
(Only if s > a)
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Laplace Transforms Table
( ) ( ) ( )f t F s L f t=
s
11
!nn
nt
s +1
ts2
1
ates a−
1
( )
attes a−
2
1
( )( ) Re( ) Re( )
aTa t T ee a s
s a
−−
−
sina
ats a+2 2
coss
ats a+2 2
sinha
ats a−2 2
coshs
ats a−2 2
( )
!n atn
nt e
s a+
−1
( ) ( ) ( )f t F s L f t=
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for ( )
for
delayed unit step function
sTt T eu t T
t T s
−− =
1
0
( ) unit impulse
( )
( ) for
t
t as t
t t
→ →
=
1
0
0 0
for ( )
for
unit step function
tu t
t s
=
1 0 1
0 0
( ) ( ) ( )f t F s L f t=
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Properties of Laplace Transforms
I. Linearity Property:
If a and b are constants and f(t), g(t) are two functions having
Laplace Transforms F(s), G(s) respectively, then
L{a f(t) ± b g(t)} = a L{f(t)} ± bL{g(t)}=a F(s) ± b G(s)
By definition, we have
{ ( ) ( )} [ ( ) ( )] stL a f t b g t a f t b g t e dt
−+ = +
0
( ) ( )st sta f t e dt b g t e dt
− −= +
0 0
{ ( )} { ( )}aL f t bL g t= +
( ) ( )aF s bG s= +
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III.Change of scale Property:
If L{f(t)} = F(s), then
By definition, we have
{ ( )} ( ) stL f at f at e dt
−=
0
{ ( )} ( )
su
a duL f at f u ea
−=
0
sF
a a
=
1
II. First Shifting Property:
If L{f(t)} = F(s), then L{eatf(t)} = F(s-a), a < s if real.
By definition, we have
{ ( )} ( )at at stL e f t e f t e dt
−=
0
( )( ) s a tf t e dt
− −=
0
( )F s a= −
( ){ ( )} sa aL f at F=1
Put at = u then a dt = du
( )
su
af u e dua
−=
0
1
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Proof: We have
( ) ( )
( ) ( ) ( ) ( )22 2(0)
0 0
d
dt
d
dt
L f t sF s f
L f t s F s sf f
= −
= − −
( ) ( )std ddt dtL f t e f t dt
−= 0
( ) ( )st ste f t se f t dt
− −= − −0
0
( ) ( ) f sL f t= − +0 0 ( ) ( )sL f t f= − 0
( ) 22 ( )ddtd d
L f t L f tdt dt
=
( ) (0)
dsL f t f
dt
= −
( )( ) (0) (0)s sL f t f f = − − 2 ( ) (0) (0)s L f t sf f = − −
Laplace Transform of Derivatives
If and is continuous, show that( ) ( )L f t F s= ( )f t
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( ) nnddtL f t = ( 1)1 2( ) (0) (0) ....... (0)nn n ns L f t s f s f f−− − − − − −
For nth derivative we have
( 1)1 2( ) (0) (0) ....... (0)nn n ns F s s f s f f−− − = − − − −
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Example: Find the Laplace transform F(s) of 2( ) cosf t t=
Solution:
First Approach: From definition
2
0 0
( ) ( ) cosst stF s f t e dt t e dt
− −= =
0
1 cos 2
2
stt e dt
−+= 2 2
1 1
2 4
s
s s
= +
+
Second Approach: Using known results
2( ) { ( )} {cos }F s L f t L t= =
1(1 cos 2 )
2L t
= +
We shall solve it by two ways.
(a)
2 2
1 1
2 4
s
s s
= +
+
( )
2 2
2 2
2
4
s
s s
+=
+
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2( ) cosf t t=(b) (0) 1f =
( ) 2 cos sinf t t t = − sin 2 t = −
We have
{ ( )} { ( )} (0)L f t sL f t f = −
{ sin 2 } { ( )} 1L t sL f t − = −
2 2
2{ ( )} 1
4sL f t
s
− = −
+
2 2
2{ ( )} 1
4sL f t
s
= −
+
2 2 2
2 2
4 2
4
s
s
+ −=
+
2 2
2 2
2
4
s
s
+=
+
( )
2 22
2 2
2{cos }
4
sL t
s s
+=
+
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Example: Find the Laplace transform F(s) of ( ) coshf t t=
Solution: We have
( ) coshf t t=2
t te e−+=
1{ ( )} ( )
2
t tL f t L e e−
= +
1( { } { })
2
t tL e L e−= +
1 1 1( )
2 1 1s s= +
− +
2
1 1 1
2 1
s s
s
+ + −=
− 2 1
s
s=
−
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Example: Find the Laplace transform F(s) of ( ) sin 2 cosf t t t=
Solution: Applying the identity
2sin cos sin( ) sin( )A B A B A B= + + −
( ) sin 2 cosf t t t=1
(sin 3 sin )2
t t= +
1( ) {sin 2 cos } ( {sin 3 } {sin })
2F s L t t L t L t = = +
2 2
1 3 1
2 9 1s s
= +
+ +
2 2
2 2
1 3( 1) 9
2 ( 9)( 1)
s s
s s
+ + +=
+ +
2
2 2
2 6
( 9)( 1)
s
s s
+=
+ +
we obtain
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Example: Find the Laplace transform F(s) of ( ) sinf t t at=
( ) sin (0) 0f t t at f= =Solution:
2( ) 2 cos ( )L f t L a at a f t = −
2( ) 2 cos ( )L f t aL at a L f t = −
2 22 2
( ) (0) (0) 2 ( )s
s L f t sf f a a L f ts a
− − = −+
( ) 2 2 2 22
( )as
s a L f ts a
+ =+
( )2
2 2
2( ) ( )
asF s L f t
s a
= =
+
( ) sin cos (0) 0f t at at at f = + =
2 2( ) 2 cos sin 2 cos ( )f t a at a t at a at a f t = − = −
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t
f(t)
O
1
1
Example: Find the Laplace transform graph of
Solution: From the graph
f(t) =1 for 0 ≤ t ≤ 1
f(t) = 0 for 1 < t
0
{ ( )} ( ) stL f t f t e dt
− =
1
0 1
(1) (0)st ste dt e dt
− −= +
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1
0
{ ( )} stL f t e dt−= 1
0
tst
t
e
s
=−
=
=−
1se
s s
−
= −− −
1 se
s
−−=
To draw the graph of
1( ) { ( )}
seF s L f t
s
−−= =
As 0, ( )s F s→ → 1 As , ( )s F s→ → 0
For 0 ,s 1 ( ) 0F s
1
F(s)
sO
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Proof: We know that
0
( ) { ( )} ( ) stF s L f t f t e dt
−= =
Differentiating both sides with respect to s
0
( ) ( ) std d
F s f t e dtds ds
−=
0
( ) std
f t e dtds
−=
0
( ) sttf t e dt
−= − { ( )}L tf t= −
Thus the result holds for n =1.
Differentiation theorem or Multiplication by
( ) ( ) ( ) ( )1 n nnL t f t F s= −nt
{ ( )} ( )d
L tf t F sds
= −
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( ) ( ) ( ) ( )1 m mmL t f t F s= −
Thus the result holds for n =2 also.
Now assume that the result holds for n = m i.e.
2{ ( )}L t f t { [ ( )]}L t tf t=
{ ( )}d
L tf tds
= − { ( )}d d
L f tds ds
= − − =
2
2( )
dF s
ds
( ) 1 ( )m mL t f t L t tf t+ =
( ) 1 { ( )}m
m
m
dL tf t
ds= −
( )1 ( )m
m
m
d dF s
dsds
= − −
( )
1 ( 1)1 ( )m mF s+ += −
As the result holds for n = m.
Hence the result holds for all n= 1, 2, 3, …….. ………...
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Integration theorem or Division by t
( )( )
s
f tL F s ds
t
=
Proof: We know that
0
( ) { ( )} ( ) stF s L f t f t e dt
−= =
Integrating both sides with respect to s from s to ∞.
0
( ) ( ) st
s s
F s ds f t e dt ds
−
=
0
( ) st
s
f t e ds dt
−
=
0
( ) st
s
f t e ds dt
−
=
0
( )st
s
ef t dt
t
−
=−
0
0( )
stef t dt
t t
− = −
− −
0
( ) stf t e dtt
−=
( )f tL
t
=
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Laplace Transform of Integrals
If , show that( ) ( )L f t F s= ( )0
( )t
F sL f u du
s
=
( )( )
dg tf t
dt=Let Then ( )
0
( ) (0)
t
f u du g t g= −
( ) 0
( ) (0)
t
L f u du L g t g
= −
( ) (0)L g t L g= − (0)
( ) (1)g
L g ts
= −
( )
( )dg t
L L f tdt
=
(0) ( )
( ) (2)g F s
L g ts s
− =
( ) (0) ( )sL g t g F s− =
Combining (1) and (2) we get ( )0
( )t
F sL f u du
s
=
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Inverse Laplace Transform
If the Laplace transform of a function f(t) is F(s) then f(t) is
Inverse Laplase Transform of F(s).
Symbolically, if ( ) 1then ( )f t L F s−={ ( )} ( )L f t F s=
The inverse Laplace transform of F(s) is given by the complex
integral
1( ) ( )
2
i
st
i
f t F s e dsi
+
−
=
However in most of cases we can find inverse Laplace transform
by simplifying and rearranging the function F(s) in terms of
functions whose inverse transforms are available in tables.
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Therefore ( )5 1 3 1 1 1
4 8 2 8 6F s
s s s= − +
+ +
Example 1: Find given that
Solution: Let
( ) 1( )f t L F s−=
( 5)( 3)( )
( 2)( 6)
s sF s
s s s
+ +=
+ +
31 2( )2 6
cc cF s
s s s= + +
+ +
We determine each constant by setting s equal to the root from
the corresponding denominator
( ) 2 65 3 14 8 8t t
f t e e− − = − +We have { }L
s=
11 atL e
s a=
−
1
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Example 2 Find given that ( ) 24 16
4 5
sF s
s s
+=
+ +( ) ( ) 1f t L F s−=
Solution: First Approach
( ) 24 16
4 5
sF s
s s
+=
+ +
4 16
( 2 )( 2 )
s
s i s i
+=
+ + + − 2 2
A B
s i s i= +
+ + + −
2
4 16
2 s i
sA
s i =− −
+=
+ −
4( 2 ) 16
2 2
i
i i
− − +=− − + −
8 4
2
i
i
−=
−4 2i= +
2
4 16
2 s i
sB
s i =− +
+=
+ +
4( 2 ) 16
2 2
i
i i
− + +=− + + +
8 4
2
i
i
+= 2 4i= −
2 4 2 4( )
2 2
i iF s
s i s i
+ − = +
+ + + −
(2 ) (2 )( ) (2 4 ) (2 4 )i t i tf t i e i e− + − − = + + −
atL es a
=−
1
2 [(2 4 ) (2 4 ) ]t it ite i e i e− −= + + −2 [2( ) 4 ( )]t it it it ite e e i e e− − −= + − − +
2 [4cos 8sin ]te t t−= +
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Since denominator is already in this form, we rearrange the
numerator in F(s) also and get
Second Approach
( ) 24 16
4 5
sF s
s s
+=
+ + 24 16
4 4 1
s
s s
+=
+ + +2
4 16
( 2) 1
s
s
+=
+ +
We know that and 2 2sin ( )atL e t
s a
− =+ +
2 2cos ( )at s aL e t
s a
− +=+ +
2
4 16( )
( 2) 1
sF s
s
+=
+ + 24( 2) 8
( 2) 1
s
s
+ +=
+ +
2 2
2 14 8
( 2) 1 ( 2) 1
s
s s
+= +
+ + + +
2( ) [4cos 8sin ]tf t e t t− = +
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Some important results used for inverse Laplace transforms
( )If { ( )} , thenL f t F s= ( )1 1{ } { ( )} ( )at atL F s a e L F s e f t− − − −+ = =
( )If { ( )} and (0) 0, thenL f t F s f= = ( )1{ } ( )d
L sF s f tdt
− =
( )If { ( )} , thenL f t F s=( )1
0
( )
tF sL f u du
s
−
=
( )If { ( )} , thenL f t F s= 1 ( ) ( )d
L F s tf tds
− − =
( ) ( )If { ( )} and { ( )} , thenL f t F s L g t G s= =
1
0
( ) ( ) ( ) ( )
t
L F s G s f u g t u du− = −
Convolution Theorem
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Example: Find the inverse Laplace transform of
1 1
2 2
0
1 1
( ) ( )
t
L L dts s a s a
− − = + +
1
2
0
1
( )
t
atL te dts s a
− − = +
21
( )
atL tes a
=−
2
1
( )s s a+
Solution: We have( )1
0
( )
tF sL f u du
s
−
=
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Example
Solve the following initial value problem using Laplace Transform
4 3 6 8, (0) 0, (0) 0y y y t y y − + = − = =
Solution: Taking the Laplace transform of DE
{ 4 3 } {6 8}L y y y L t − + = −
{ } 4 { } 3 { } 6 { } 8 {1}L y L y L y L t L − + = −
2
2
1 1{ } (0) (0) 4[ { } (0)] 3 { } 6 8s L y sy y sL y y L y
s s− − − − + = −
The Laplace transform is linear and we have
( 1)( ) 1 2( ) ( ) (0) (0) ....... (0)nn n n nL f t s F s s f s f f−− − = − − − −
1!
, 0,1,2,3,.....nn
nL t n
s += =
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2
2
1 1or { } 4 { } 3 { } 6 8s L y sL y L y
s s− + = −
2
2
1 1( 4 3) { } 6 8s s L y
s s− + = −
2 2 2
1 1{ } 6 8
( 4 3) ( 4 3)L y
s s s s s s= −
− + − +
1
2 2 2
1 16 8
( 4 3) ( 4 3)y L
s s s s s s
− = − − + − +
1 1
2 2 2
1 16 8 (1)
( 4 3) ( 4 3)y L L
s s s s s s
− − = − − + − +
2 2
1
( 4 3)s s s− + 21
( 3)( 1)s s s=
− − 2 3 1
A B C D
s s s s= + + +
− −
As (s-3) and (s-1) are linear factors, we have
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2
3
1
( 1)s
Cs s
=
=−
2
1
3 (3 1)=
−
1
18=
2
1
1
( 3)s
Ds s
=
=− 2
1
1
1 (1 3)s=
=−
1
2= −
To determine A and B, we have
2 2 2 21 ( 4 3) ( 4 3) ( 1) ( 3)As s s B s s Cs s Ds s= − + + − + + − + −
2 2 2 21 11 ( 4 3) ( 4 3) ( 1) ( 3)18 2
As s s B s s s s s s= − + + − + + − − −
Comparing the coefficients of s3 and constant term both sides,
we get 1 10 ,
18 2A= + − 1 3B=
1 1 9 1 4,
2 18 18 9A
− = − = =
1
3B =
2 2
1
( 4 3)s s s− + 24 1 1 1 1 1 1 1
9 3 18 3 2 1s s s s= + + −
− −
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1
2 2
1
( 4 3)L
s s s
−
− +
1 1 1 1
2
4 1 1 1 1 1 1 1
9 3 18 3 2 1L L L L
s s s s
− − − − = + + − − −
1!
, 0nn
nL t n
s += atL e
s a=
−
1Using the results
1
2 2
1
( 4 3)L
s s s
−
− +
34 1 1 1 (2)9 3 18 2
t tt e e= + + −
2
1 1
( 4 3) ( 3)( 1)s s s s s s=
− + − − 3 1
A B C
s s s= + +
− −
As s, (s-3) and (s-1) are linear factors, we have
2
0
1
4 3 sA
s s ==
− +
1
3=
3
1
( 1)s
Bs s
=
=−
1 1
3(3 1) 6= =
−
1
1
( 3)s
Cs s
=
=−
1 1
1(1 3) 2= = −
−
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2
1 1 1 1 1 1 1
( 4 3) 3 6 3 2 1s s s s s s= + −
− + − −
1 1 1 1
2
1 1 1 1 1 1 1
( 4 3) 3 6 3 2 1L L L L
s s s s s s
− − − − = + − − + − −
31 1 1 (3)3 6 2
t te e= + −
Using (2) and (3) in (1), we have
3 34 1 1 1 1 1 16 89 3 18 2 3 6 2
t t t ty t e e e e
= + + − − + −
3 38 1 8 42 3 43 3 3 3
t t t tt e e e e
= + + − − + −
( )31 4
2 3 43 3
t tt e e
= + − − −
32 t tt e e= − +
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Example: Solve the following problem using Laplace Transform
2 , ( ) , ( ) 2 24 2 4
y y t y y
+ = = = −
Solution: In this problem the initial conditions refer to some
later instant t = t0.
Such problems are called Shifted Data Problems.
As the values y and its derivatives occurring in Laplace transform
are not readily available, we may proceed to solve such problems in
two ways:
First method: Taking Laplace transform of DE we get
2
2
2{ } (0) (0) { }s L y sy y L y
s− − + =
2
2
2( 1) { } (0) (0)s L y sy y
s+ = + +
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2 2 2 2
(0) (0) 2{ }
1 1 ( 1)
sy yL y
s s s s
= + +
+ + +
1 1 1
2 2 2 2
1 2( ) (0) (0)
1 1 ( 1)
sy t y L y L L
s s s s
− − − = + + + + +
1 1 1
2 2 2 2
1 1 1(0) (0) 2
1 1 1
sy L y L L
s s s s
− − − = + + − + + +
( ) (0)cos (0)sin 2 2siny t y t y t t t = + + −
( ) (0)sin (0)cos 2 2cosy t y t y t t = − + + −
Applying the conditions ( ) , ( ) 2 24 2 4
y y
= = −
1 1 1(0) (0) 2 2
2 42 2 2y y
= + + −
1 1 12 2 (0) (0) 2 2
2 2 2y y− = − + + −
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Solving these equations we get (0) (0) 1y y= =
( ) 2 cos siny t t t t = + −
Second Method: Put 1
4t t
= + then
1 1 14( ) ( ) ( )y t y t y t= + =
1 1( )( ) dy tdy tydt dt
= = 1 1 1
1
( )dy t dt
dt dt= = 1 1 1
1
( )dy ty
dt= 1y y =
1for , 04
t t
= =
The IVP takes the form
( )1 1 1 142 , (0) , (0) 2 22
y y t y y
+ = + = = −
Taking the Laplace transform we get
2
1 1 1 1 2
2{ } (0) (0) { }
2s L y sy y L y
s s
− − + = −
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Example: Solve the following simultaneous differential equations
using Laplace Transform
, sintdx dy
y e x tdt dt
+ = + =
given that x = 1, y = 0 when t = 0.
Solution: Taking the Laplace Transform of both equations
2
1{ } (0) { }
1
1{ } { } (0)
1
sL x x L ys
L x sL y ys
− + =−
+ − =+
2
1or { } 1 { }
1
1{ } { }
1
sL x L ys
L x sL ys
− + =−
+ =+ 2
1or { } { } 1
1
1{ } { }
1
sL x L ys
L x sL ys
+ = +−
+ =+
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Solving these equations L{x} and L{y}, we get
2
1or { } { } 1
1
1{ } { }
1
sL x L ys
L x sL ys
+ = +−
+ =+
2 2
2 2
1 1{ }
1 1 1
1 1{ } 1
1 1 1
sL x s
s s s
sL y
s s s
= + −
− − +
= − −
− + −
2 2 2 2
2 2 2 2
1{ }
( 1)( 1) 1 ( 1)( 1)
1 1{ }
( 1)( 1) ( 1)( 1) 1
s sL x
s s s s s
sL y
s s s s s
= + −− − − + −
= − −− + − − −
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42
2 2 2 2
1{ }
( 1)( 1) 1 ( 1)( 1)
s sL x
s s s s s= + −
− − − + −
2 2
1
( 1) ( 1) ( 1)( 1) ( 1)( 1)( 1)
s s
s s s s s s s= + −
− + − + + − +
2
2
1 1 1
2( 1) 4( 1) 4( 1)
1 1 1 1 1
2( 1) 2( 1) 4( 1) 4( 1) 2( 1)
s s s
s s s s s
= + −− − +
+ + − + +− + − + +
2 2
1 1 1 1
2( 1) 2( 1) 2( 1) 2( 1)s s s s= + + +
− − + +
1
2 2
1 1 1 1
2( 1) 2( 1) 2( 1) 2( 1)x L
s s s s
− = + + + − − + +
1
sin2
t t tx te e e t−= + + +
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43
2 2 2 2
1 1{ }
( 1)( 1) ( 1)( 1) 1
sL y
s s s s s= − −
− + − − −
2 2
1 1
( 1)( 1)( 1) ( 1)( 1) ( 1)( 1)
s
s s s s s s s= − −
− + + + − − +
2
2
1 1
4( 1) 4( 1) 2( 1)
1 1 1 1 1
2( 1) 4( 1) 4( 1) 2( 1) 2( 1)
s
s s s
s s s s s
= + −− + +
− + − − −− − + − +
2 2
1 1
2( 1) 2( 1) 2( 1)
s
s s s= − − −
+ − +
1
2 2
1 1
2( 1) 2( 1) 2( 1)
sy L
s s s
− = − − − + − +
1
cos2
t ty t te e−= − + +
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Evaluation of Integrals by Laplace transform
Evaluate the integral
3
0
sintte tdt
−
0
( ) ( ) stL f t f t e dt
−= We have
Therefore 3
0
sintte tdt
−
3sin sL t t ==
3
( 1) sins
dL t
ds == −
2
3
1( 1)
1 s
d
ds s == −
+
( )2
2
3
2
1s
s
s=
=+
3
50=
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Unit Step Function or Heaviside’s unit function
Definition: The unit step function u(t - a) is defined as
0 for ( )
1 for
t au t a
t a
− =
where a ≥ 0.
Laplace transform of Unit Step Function
0
{ ( )} ( )stL u t a e u t a dt
−− = −0
.0 .1
a
st st
a
e dt e dt
− −= + 0st
a
e
s
−
= +−
{ ( )}ase
L u t as
−
− =
This function is used when force or input is turned on and of for
certain periods of time. When combined with a function it can
produce varied results.
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46
For example, the function
( ) ( ) ( )f t u t a u t b= − − − = 0 for 0
1 for
0 for
t a
a t b
b t
O t
f(t)
a
1
b
This function is referred to as Single Box function and represents
a Single Square Wave.
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47
( ) 3sin , 0 2f t t t =
Then the function
( ) ( 1)f t u t − = 0 0 1
( ) 1 2
for t
f t for t
and the function
( 1) ( 1)f t u t− − = 0 0 1
( ) 1 2 1
for t
f t for t
+
Now consider the function
-4
-3
-2
-1
0
1
2
3
4
0 2 4 6 8
-4
-3
-2
-1
0
1
2
3
4
0 2 4 6 8
-4
-3
-2
-1
0
1
2
3
4
0 2 4 6 8
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Second Shifting Property
If , then{ ( )} ( )L f t F s= { ( ) ( )} ( )asL f t a u t a e F s−− − =
0
{ ( ) ( )} ( ) ( ) stL f t a u t a f t a u t a e dt
−− − = − −
( ) st
a
f t a e dt
−= − Put t – a = u
( )
0
( ) s u af u e du
− += 0
( )as sue f u e du
− −= ( )ase F s−=
Applications:
1. Find the Laplace transform of
2 0
( ) 0 2
sin 2
t
f t t
t t
=
( ) 2 ( ) 2 ( ) ( 2 )sinf t u t u t u t t = − − + −
2
2
2 2 1( ) { ( )}
1
s sF s L f t e es s s
− −= = − ++
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49
2. Find the inverse Laplace transform of 2
2 2( )
ssse e
F ss
− −+=
+
22 2 2 2
( )s
ssF s e es s
− −= ++ +
From second Shifting Property
1 ( ) ( ) ( )asL e F s f t a u t a− − = − −
Since 1
2 2cos
sL t
s
− = +
1
2 2sinL t
s
− = +
therefore
1 1 122 2 2 2( ) ( )s
ssf t L F s L e L es s
−− − − − = = + + +
( ) ( )1 1
cos sin 1 12 2
t u t t u t
= − − + − −
( )1
sin sin 12
t u t t u t
= − − −
( )1
1 sin2
u t u t t
= − − −
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50
Unit Impulse and Dirac Delta function
Definition: Unit Impulse is defined as the integral of a large
force taken over a small time interval for which it acts..
A large force acting for a small time can be defined as
1, 0
( )
0 otherwisek
a t a k kf t a k
+ →
− =
0lim ( ) ( )kk
f t a t a→
− = −
The Dirac Delta function. It is not a function in true sense.
for ( )
0 otherwise.
t at a
=− =
1
( ) [ ( ) ( { })]kf t a u t a u t a kk
− = − − − +
0
( ) 1t a dt
− =
( )1{ ( )} [ ]as a k skL f t a e eks
− − +− = −1
[1 ]as kse eks
− −= −
{ ( )} asL t a e −− =Taking limit as k → 0, we get
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51
Example: Solve the following initial value problem using Laplace
Transform
3 2 ( ), (0) 0, (0) 0
( ) ( ) ( 1) ( 2)
( ) ( ) ( 1)
y y y r t y y
a r t u t u t Square wave
b r t t Unit impulse
+ + = = =
= − − −
= −
Solution: Laplace transform of the DE is
2 { } (0) (0) 3[ { } (0)] 2 { } { ( )}s L y sy y sL y y L y L r t− − + − + =
2( 3 2) { } { ( )}s s L y L r t+ + =
2
1{ } { ( )} (1)
3 2L y L r t
s s=
+ +
( ) ( ) ( 1) ( 2)a r t u t u t= − − −2
{ ( )}s se e
L r ts s
− −
= −
2 2
2
1{ } ( ) ( )( )
( 3 2)
s s s sL y e e F s e es s s
− − − −= − = −+ +
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52
2
1( )
( 3 2)F s
s s s=
+ +
1
( 2)( 1)s s s=
+ +
1 1 1
2 1 2( 2)s s s= − +
+ +
1 21 1( ) { ( )}2 2
t tf t L F s e e− − −= = − +
1 2( )( ) ( 1) ( 1) ( 2) ( 2)s sy L F s e e f t u t f t u t− − −= − = − − − − −
0 0 1
( 1) 1 2
( 1) ( 2) 2
t
f t t
f t f t t
= − − − −
( 1) 2( 1)
( 1) 2( 1) ( 2) 2( 2)
0 0 1
1 11 2
2 2
1 12
2 2
t t
t t t t
t
y e e t
e e e e t
− − − −
− − − − − − − −
= − + − + + −
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53
( ) ( ) ( 1)b r t t= −
{ ( )} sL r t e−=
2
1{ } ( )
3 2
s sL y e F s es s
− −= =+ +
2
1 1( )
3 2 ( 1)( 2)F s
s s s s= =
+ + + +
1 1
1 2s s= −
+ +
2( ) t tf t e e− −= −
1{ ( ) } ( 1) ( 1)sy L F s e f t u t− −= = − −
0 0 1
( 1) 1
t
f t t
=
−
1 2( 1)
0 0 1
1t t
t
e e t− + − −
=
−
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54
Example: Find the current i(t) in the circuit
if a single square wave with voltage V0 is applied. The circuit is
supposed to be quiescent before square wave is applied.
R
C
v
Solution: The equation of the circuit is
0
1( ) ( ) ( )
t
v t R i t i u dtC
= +
0( )
0 otherwise
V a t bv t
=
0( ) [ ( ) ( )]v t V u t a u t b= − − −
Taking the Laplace transform
( )( ) ( )
I sV s R I s
Cs= +
0( ) [ ]as bse e
V s Vs s
− −
= −
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0
1( ) [ ]
1
as bse eI s V
s sR
Cs
− −
= −
+0[ ]
1
as bse e CsV
s s RCs
− −
= −+
0 1( ) [ ]1
as bsVI s e eR
sRC
− −= −
+
0 [ ] ( )as bsV
e e F sR
− −= −
1 10( ) { ( )} [ ] ( )as bsV
i t L I s L e e F sR
− − − −= = −
1 10 ( ) ( )as bsV
L e F s L e F sR
− − − − = −
0 ( ) ( ) ( ) ( )V
u t a f t a u t b f t bR
= − − − − −
1( )
1F s
sRC
=
+
( )t
RCf t e−
=
0( ) ( ) ( )t a t b
RC RCV
i t u t a e u t b eR
− −− −
= − − −
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56
( ) 0 0i t for t a =
0( )t a
RCV
i t e for a t bR
−−
=
0( )t a t b
RC RCV
i t e eR
− −− −
= −
0 .a b t
RC RC RCV
e e e for t bR
− = −
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Calculus.ppt