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Title: Laplace transform mathes
Description: This is laplace transform class notes.if you learn this note you don't need to learn other things.
Description: This is laplace transform class notes.if you learn this note you don't need to learn other things.
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Unit -2
Syllabus:
Laplace transform, Existence theorem, Laplace transforms of derivatives and integrals, Initial and final value
theorems, Unit step function, Dirac- delta function, Laplace transform of periodic function, Inverse Laplace
transform, Convolution theorem, Application to solve simple linear and simultaneous differential equations
...
The basic knowledge of Laplace Transforms and its applications in solving differential equations
...
1
1
...
3
1
...
5
1
...
7
1
...
9
1
...
11
1
...
13
1
...
15
1
...
17
1
...
19
1
...
21
Inverse
Laplace 25
Transform
1
...
23
Linearity Property
27
1
...
25
Second shifting theorem
27-28
1
...
27
Inverse
32
Laplace
Transform
of
Derivatives
1
...
29
Multiplication by s:Heaviside
Formula
32
Expansion 33
for
Inverse
Laplace Transform
1
...
31
...
2
1
...
With the increasing complexity of engineering problems, Laplace transforms help in solving complex
problems with a very simple approach just like the applications of transfer functions to solve ordinary
differential equations
...
•
Analysis of Electrical Circuits
In electrical circuits, a Laplace transform is used for the analysis of linear time-invariant systems
...
•
Digital Signal Processing
One cannot imagine solving DSP (Digital Signal processing) problems without employing Laplace transform
...
It makes studying analytic part
of Nuclear Physics possible
...
It helps analyse the variables, which when altered,
produces desired manipulations in the result
...
This is an efficient and easier way to control processes that are
guided by differential equations
...
2 Introduction
The knowledge of Laplace Transformations has in recent years became an essential part of Mathematical background
required of engineers and scientists
...
The transform turns integral equations and differential equations to polynomial equations, which are much
easier to solve
...
This subject originated from the operational methods applied by the English engineer Oliver Heaviside (1850 – 1925)
to problems in electrical engineering
...
06
...
08
09
...
s n +1
1
s−a
at
1
s+a
e-at
(n -1)
at
(k -1)
at
t
e
(n − 1)!
t
e
Γ(k )
1
( s − a) n
n = 1,2,3,
...
a
s + a2
2
sin at
6
s
s + a2
2
11
...
13
...
...
cosh at
16
...
f(t – a)
...
6 First shifting theorem
Statement: If L ( F(t) ) = f(s) , then L ( eat F(t) ) = f(s-a)
∞
Proof: Definition of Laplace transformation
...
1
...
t
Proof:
a
−∞
∞
|------------------------ t < a ------------|0------|--- t > a -----|
∞
L { G(t) } =∫0 𝑒𝑒 −𝑠𝑠𝑠𝑠 𝐺𝐺(𝑡𝑡)𝑑𝑑𝑑𝑑
=
a
∞
0
a
−s t
∫ e
...
1
...
∫e
F ( x) dx
0
1
=
a
∞
1
=
a
∞
1
𝑎𝑎
x
−s
a
∫e
s
−x
a
F ( x) dx
0
∫e
s
−t
a
F (t ) dt
[since
0
b
∫
a
𝑠𝑠
𝑎𝑎
= 𝑓𝑓( )
By Leibnitz principle for differentiation under integral sign,
8
f ( x) dx =
b
∫ f (t ) dt
a
]
1
...
r
...
s,
∞
d
d −s t
{f(s)}
∫ e F (t ) dt =
ds 0
ds
∞
d
d −s t
∫0 ds (e ) F (t ) dt = ds { f(s) } ⇒
⇒
∞
∫ −t e
−s t
F (t ) dt =
0
∞
∫te
−s t
F (t ) dt = −
0
for n = m,
∞
∫e
−s t
[t
F (t ) ] dt = ( − 1)m
m
0
d
{ f(s) }
ds
d
{ f(s)}
ds
dm
{f(s) }
ds m
This completes the proof of multiplication by tn
...
10 Division by t
Statement: If L { F(t) } = f(s) then L {
∞
Proof: since = f(s) = ∫ e − s t F (t ) dt
1
∞
F(t) } = ∫𝑠𝑠 𝑓𝑓(𝑠𝑠)𝑑𝑑𝑑𝑑
t
0
On integrating both sides w
...
t
...
1
∞
F(t)} = ∫𝑠𝑠 𝑓𝑓(𝑠𝑠)𝑑𝑑𝑑𝑑
t
This completes the proof of theorem
...
Find L [3e2t sin t }
Solution:
1
L {sin t} = 𝑠𝑠2 +1
L [3e2t sin t }=
3
(𝑠𝑠−2)2 +1
( Use First shifting theorem)
Example 2
...
a
a
d
{
}
−
2
2
L { t sin at } =
= 2
ds s + a
s + a2
(
∴ L { t sin at } =
2𝑠𝑠
...
Example 3
...
a
2 =−
= ds 2
s + a 2
)
(
(
)
(
)
8
...
a s 2 + a 2 2 − 2as
...
2 s
4
s2 + a2
8
...
∴L { t2 sin at } = =
2
2 2
s + a
(
)
Example 4
...
− 2
(s + 3)3
d 2
3
= - ds
(s + 3)
2
...
Example 5
...
2
(1 − e t )
t
Example 6
...
11
(𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶−𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶)
�
𝑡𝑡
Example 7
...
= Log
∴L
(s 2 + b 2 )
t
∞
Example 8
...
−𝑠𝑠𝑠𝑠
∴ ∫0 𝑡𝑡𝑒𝑒 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆=(𝑠𝑠2 +1)2 and s = 2
4
∞
−2𝑡𝑡
∴ ∫0 𝑡𝑡𝑒𝑒 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆= 25
Example 9
...
Sin mt
dt
t
∫
0
m
Solution: L {Sin m t} =
Title: Laplace transform mathes
Description: This is laplace transform class notes.if you learn this note you don't need to learn other things.
Description: This is laplace transform class notes.if you learn this note you don't need to learn other things.