Inverse laplace transform
The Inverse Laplace Transform
Def : if the Laplace transform of a function f(t) is F(s).
If
f(t)= F(s) , then
f(t) = 𝓛
Where
operator.
𝟏
F(s)
is called the inverse Laplace transformation
Ex. Since
=
we can write
=
*********************************************
Some Inverse Laplace Transforms
F(s)
f(t)=
1
t
cos at
sin at
sinhat
coshat
1
Inverse laplace transform
Properties of Inverse Laplace transform
1- Linearity property :
If c1and c2 are any constants while F1(s) and F2(s) are the
laplce transforms of f1(t) and f2(t) respectively, then
𝓛
𝟏
{c1F 1(s)+ c2F 2(s)} = c 1 𝓛
𝟏
F 1(t)+ c2 𝓛 𝟏 F 2(t)
= c 1f1(t)+ c2f2(t)
Ex. Find
{
Sol:
=4
-3
=4
+
-3 cos4t + sin2t
*********************************************
2- First shifting property :
If
F(s) = f(t) , then
𝓛
𝟏
F(s-a) = a f(t)
Ex. Find
Sol:
=
.
=t
****************
Ex. Find
Sol:
=
=
2
sin2t
Inverse laplace transform
3- Second shifting property:
If
F(s) = f(t) , then
𝓛
𝟏
𝐚𝐬
𝐞
𝑭 𝒔 =𝒇 𝒕
Ex. Find
𝒂𝒖 𝒕
𝒂
= sint
= sin (t- ) u(t-
*******************************
Ex. Find
Sol:
F(s)=
f(t)=
F(s) = sint
= sin (t-2) u(t-2)
***************************************************
Note :
𝓛
𝟏𝜞 𝒏 𝟏
𝒔𝒏+𝟏
= 𝒕𝒏
Ex. Find
=
=
=
=
*******************
4- Inverse Laplace Transform of derivatives:
If
F(s)= f(t) , then
3
Inverse laplace transform
𝓛 𝟏 { Fn (s)}= 𝓛
𝑛
𝟏 𝑑
𝑑𝑠𝑛
F(s) = (-1)n tn
f(t)
***************************************************
Ex. Find
{
}
Sol:
=
=
=(-1) t f(t)
-
= -t f(t)
2 cost -2 = -t f(t)
f(t)=
=
**********************************
tan-1s
Ex. Find
Sol:
F(s) = tan-1s
F'(s) =
= (-1) t f(t)
= -t f(t)
sin t = -t f(t)
f(t)=
*****************************************
5- Inverse Laplace of Integrals :
If
F(s) = f(t) , then
𝓛
𝟏
{
∞
𝒔
𝑭 𝒔 𝒅𝒔
=
𝒇 𝒕
𝒕
****************************
4
Inverse laplace transform
6- Multiplication by s :
If
F(s) = f(t) and f(0)=0 , then
𝓛
𝟏
S F(s) = f'(t)
{
Ex. Find
Sol:
}
= sint
=
(sin t )= cost
*********************************************
7- Division by s:
If
𝓛
F(s) = f(t)
𝟏𝐅𝐬
=
𝐬
𝓛
𝟏𝐅𝐬
𝓛
𝟏𝐅𝐬
𝐬𝟐
𝐬𝐧
𝐭
𝐟
𝟎
𝐭
, then
𝐝𝐭
=
𝐭 𝐭
𝟎 𝟎
=
𝐭 𝐭
………
𝟎 𝟎
𝐟 𝐭
𝐝𝐭
𝐭
𝐟
𝟎
𝐭
𝐝𝐭 𝐧
{
Ex. Find
=
=
= -cos t
=
= -cost+1=1-cost
=[ t- sint
=
=
=
=
+cost
+cost-1
5
= t-sint
= +cost-1
Inverse laplace transform
H.w
Find the Inverse Laplace transform of the
following:123-
4-
√
56-
√
6
Inverse laplace transform
Evaluation of Inverse Laplace Transform
The partial fractions method is one from the most method to
determine the inverse Laplace transform of any rational
function of the form
F(s)=
where p(s) and q(s) are
polynomials in s, with the degree of p(s) less than that of q(s),
can be written as the sum of rational functions having the
form
,
where A,B, a,b,c are constants.
Case1: if F(s)=
where q(s) is in the form of in repeated
factors i.e.:
F(s)=
=
***************************************************
Ex. Find f(t) if F(s)=
Sol:
First
is resolved into partial fractions, giving:
F(s)=
=
To find A1 multiply F(s) by (s+3) and setting s=-3
A1=
|
=
=
To find A2 multiply F(s) by (s-2) and setting s=2
7
Inverse laplace transform
|
A2=
=
=
Second step: using
of the result, we get
f(t)=
=
=
***************************************************
Ex. Find f(t) if F(s)=
Sol:
=
To find A1 multiply F(s) by s and setting s=0
|
A1=
= =
To find A2 multiply F(s) by (s+1) and setting s=-1
A2=
|
=
=
To find A3 multiply F(s) by (s+2) and setting s=-2
A2=
|
=
=
F(s)=
Second step: using
f(t)= 1-3
of the result, we get
+3
8
Inverse laplace transform
Case 2:- if q(s) is in the form of polynomial raised to same
power i.e. :(s-a)m
F(s)=
=
=
+
f(t)=
F(s)
=
To evaluate Am by the formula:
Am =
|
|
Ak=
k=1,2.3,…………m-1
Where
Q(s)=
***************************************************
Ex. Find f(t) if F(s)=
Sol:
F(s)=
=
f(t)=
F(s)
=
=
+
9
Inverse laplace transform
=
+
=
+
+ ]
Q(s) =F(s). (s-2)4
. (S-2)4
=
Q(s) =
A4=
|
=
|
Ak=
Where k=3
A 3=
|
=
|
=
|
=
Where k=2
A2=
Where k=1
A1=
f(t) =
+
=
-2 ]
]
***************************************************
10
Inverse laplace transform
Case 3: if f(t) =
, where p(s) and q(s) are polynomials
in s and the degree of p(s) is less than the degree of q(s), then
the terms in f(t) which corresponds to an unrepeated,
irreducible quadratic factor
of q(s).
=
F(s)=
The inverse Laplace transform is:f(t) = 𝓛 𝟏 F(s) =
𝒆 𝜶𝒕
𝜷
𝒑 𝐜𝐨𝐬 𝜷𝒕
𝒒 𝐬𝐢𝐧 𝜷𝒕
Where p and q are respectively, the real and imaginary parts of
R= q+j p = (
=
+B
=
q= 𝛼 𝐴
p= 𝛽 𝐴
************************************************
Ex. Find f(t) if F(s) =
Sol:
S2+2s+2=0
=
=
f(t) =
F(s) =
s=
√
=
=
11
Inverse laplace transform
q=
= -1
p=
=1
f(t) =
****************************
2nd method
=
F(s)= =
=
To find A1multiply F(s) by (s+1-j) and setting s=-1+j
|
A1=
=
=
=
To find A2multiply F(s) by (s+1+j) and setting s=-1-j
|
A2=
=
=
=
+
F(s)=
f(t)=
F(s)
+
=
=
Note:
𝒆𝒂
𝒋𝒃
= 𝒆𝒂𝒕 𝒄𝒐𝒔 𝒃𝒕
𝒋𝒔𝒊𝒏𝒃𝒕
𝒆𝒂
𝒋𝒃
= 𝒆𝒂𝒕 𝒄𝒐𝒔 𝒃𝒕
𝒋𝒔𝒊𝒏𝒃𝒕
f(t)=
=
f(t) =
[
]
12
Inverse laplace transform
f(t)=
***********************************************************
Inverse Laplace Transform by using Convolution property
If F(s) = G(s).H(s)
G(s)=g(t)
f(t)= 𝓛
𝟏
and
F(s) = 𝓛
H(s)=h(t)
𝟏
G(s). H(s) =
=
or
𝑡
𝑔 𝜏
𝑡
𝑡
𝜏 𝑑𝜏
𝜏 𝑔 𝑡
𝜏 𝑑𝜏
Ex. Find f(t) by using convolution if F(s)=
Sol:
G(s)=
, g(t)=
G(s)=
=
H(s)=
, h(t)=
H(s)=
=
f(t)=
F(s) =
H(s). G(s) =
=
=
]
=
=
[
***********************************************
13
Inverse laplace transform
Ex. Find the inverse Laplace transform of
F(s)=
Sol:
G(s)=
g(t)=
H(s)=
G(s)= t
h(t)=
f(t) =
H(s)= sin(at)
G(s). H(s) =
=
Using Integration by parts
Let u=
du=1 dt
dv=
v=
f(t)=
==
f(t)=
+ [0- sin(at)]
-
sin(at)
***************************************************
14
Inverse laplace transform
Solution of Differential Equations using Laplace Transform
The advantage of using the Laplace transform is replace the
operation of differential equations by an algebraic operation by
three steps:1- Taking the Laplace transform of each term in a D.E, it is
converted into algebraic equation in the variable S.
2- Setting the value initial value in an algebraic equation and
resolving the rational terms.
3- Taking the inverse laplace transforms of the result to get
on f(t).
***********************************************
Ex. Solve the D.E
y'+3 y=
,
y(0)=2
First step: taking the Laplace transforms
=
=
Second step: setting the value y(0)=2
=
(s+3) Y(s)=
+2
Y(s) =
y(t) =
y(t) =
=
15
Inverse laplace transform
A1=
|
=
A2=
|
=
y(t) =
=
y(t)=
***********************************************************
Ex.2/ solve the D.E y''+2y'+y=t
Sol:
=
s2 Y(s) –s y(0) –y'(0)+2[ sY(s)-y(0)]+Y(s)=
s2Y(s) –s +2+2sY(s) -2+Y(s)=
Y(s)[ s2+2s+1] =
Y(s) =
y(t)=
y(t)=
=
=
=
16
Inverse laplace transform
H.w
A- Find
of the following
1- F(S)=
2- F(s) =
3- F(s)=
4- F(s)=
5- F(s)=
B- Solve the D.E of the following:
1- y'+3y+2
=
for y(0)=1 if f(t) is the function whose
graph is shown below
2
1
2- y" +4y = u(t)
for y(0)=y'(0)=0
3- y"+4y'+13y=
for y(0)=1 and y'(0)=-2
17
2