The document discusses the Laplace transform, which is defined as the integral of a function f(t) multiplied by e-st from 0 to infinity. The key points are: I. For a function to have a Laplace transform, it must be piecewise continuous and of exponential order. II. Important properties of the Laplace transform include linearity, shifting, and how it handles derivatives. III. The Laplace transform can be used to solve differential equations and analyze systems in fluid mechanics, electrical engineering, and other fields.

1. 5-May-18 1

2. LAPLACE
TRANSFORM
5-May-18 2presentation of fluid theory

3. LAPLACE TRANSFORM
We discuss in Laplace transform are as
follows
I. Definition Of Laplace transform
II. Conditions Of Laplace Transform
III. Properties Of Laplace Transform
IV. Applications Of Laplace Transform
V. Limitation Of Laplace Transform
5-May-18 3presentation of fluid theory

4. Definition Of Laplace Transform
• If ‘f’ is a real or complex valued function then the
laplace transform of ‘f’ is defined as
𝐿{𝑓} =
0
∞
𝑒−𝑠𝑡
𝑓 𝑡 𝑑𝑡
where 𝑓 𝑡 𝑖𝑠 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑛𝑑
𝐿 𝑓 = lim
𝒯→∞
0
∞
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡 → (1)
If this limit is exist then the function is laplace
transformable
5-May-18 4presentation of fluid theory

5. Example Of Laplace Transform
𝑓 𝑡 = 1
use definition of laplace transform
1 ⇒ 𝐿 1 = lim
𝒯→∞
0
𝒯
𝑒−𝑠𝑡(1)𝑑𝑡 → (1)
𝐿 1 = lim
𝒯→∞
𝑒−𝑠𝑡
−𝑠
𝒯
|
0
𝐿 1 =
−1
𝑠
lim
𝒯→∞
𝑒−𝑠𝑡 − 1
Now
I. If S>0 then 𝑒−𝑠𝑡 → 0
II. If S<0 then 𝑒−𝑠𝑡 → ∞
III. S=0 then
1
𝑠
→ ∞
So, 𝐿 1 =
1
𝑆
(𝑆 > 0)
5-May-18 5presentation of fluid theory

6. Suffificent Conditions For Existance
Of Laplace Transform
• We have two conditions for existance of
laplace transform .
I. f(t) piecewise continuity at 𝑡 ≥ 0
Piece wise continuity simply one can write
finitely many continuous function.
5-May-18 6presentation of fluid theory

7. Definition Of Piecewise Continuity
• A function f is called a piecewise continuous on
[a,b] if there are finite number of points
i.e;
𝑎 < 𝑡1 < 𝑡2 … … … . . < 𝑡 𝑛 < 𝑏
Such that ‘f’ is continuous on each open sub interval
𝑎, 𝑡1 , 𝑡1, 𝑡2 … … … (𝑡 𝑛, 𝑏) and all the limits exists
lim
𝑡→ 𝑎+
𝑓 𝑡 , lim
𝑡→ 𝑏−
f t , lim
𝑡→ 𝑡 𝑗
+
𝑓 𝑡 , lim
𝑡→ 𝑡 𝑗
−
𝑓 𝑡 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑗
5-May-18 7presentation of fluid theory

8. Example Of Piecewise Continuity
Q.Discuss piecewise continuity of f t =
1
𝑡−1
Solution:- 𝑓 𝑡 is not continous in any interval
containing 1.
Since,
lim
𝑡→1±
f(t) donot exist.
5-May-18 8presentation of fluid theory

9. Definition Of Exponential Order
• A function f is said to be of exponential order
𝛼 if there exist constant M and 𝛼 such that for
𝑡0 ≥ 0
𝑓 𝑡 ≤ 𝑀𝑒 𝛼𝑡
, 𝐹𝑜𝑟 𝐴𝑙𝑙 𝑡 ≥ 𝑡0
Or
A function is said to be of exponential order 𝛼 if
lim
𝑡→∞
𝑒−𝛼𝑡
𝑓 𝑡 = finite quantity
Then will be exponential order
5-May-18 9presentation of fluid theory

10. Examples of Exponential Order
Q. The function f(t)=𝑒 𝑡2
is not an exponential
order.
Solution:- By defintion:lim
𝑡→∞
𝑒−𝛼𝑡
𝑓 𝑡
= lim
𝑡→∞
𝑒−𝛼𝑡
𝑒 𝑡2
= lim
𝑡→∞
𝑒 𝑡(𝑡−α)
=∞, ∀ value of α
So the function is not an exponential order
5-May-18 10presentation of fluid theory

11. Examples of Exponential Order
Note:
If a function give a finite value or real then
will be exponential order
5-May-18 11presentation of fluid theory

12. Properties Of Laplace Transform
1.Linearity Property:
𝑳 𝒄 𝟏 𝒇 𝟏 𝒕 + 𝒄 𝟐 𝒇 𝟐(𝒕) = 𝒄 𝟏 𝑳{𝒇 𝟏 𝒕 } + 𝒄 𝟐 𝑳 𝒇 𝟏 𝒕
OR
𝑳
𝒌=𝟎
∞
𝒂 𝒌 𝒇 𝒌(𝒕) =
𝒌=𝟎
∞
𝒂 𝒌 𝑳{ 𝒇 𝒌(𝒕)}
For example:- L 𝑐𝑜𝑠𝜔𝑡 = 𝐿
𝑒 𝜔𝑡+𝑒−𝑖𝜔𝑡
2
=
1
2
𝐿{𝑒 𝑖𝜔𝑡} +
1
2
{𝑒−𝑖𝜔𝑡}
=
1
2
1
𝑠 − 𝑖𝜔
+
1
𝑠 + 𝑖𝜔
=
1
2
2𝑠
𝑠2 +𝜔2
𝐿 𝑐𝑜𝑠𝜔𝑡 =
𝑠
𝑠2 + 𝜔2
5-May-18 12presentation of fluid theory

13. 2. First Shifting roperty
If 𝐿 𝑓 𝑡 = 𝑓 𝑠 𝑡ℎ𝑒𝑛 𝐿 𝑒 𝑎𝑡 𝑓 𝑡 = F(s − a)
Proof:-
𝐿 𝑒 𝑎𝑡
𝑓 𝑡 =
0
∞
𝑒−𝑠𝑡
𝑒 𝑎𝑡
𝑓 𝑡 𝑑𝑡
=
0
∞
𝑒−(𝑠−𝑎)𝑡 𝑓 𝑡 𝑑𝑡
𝐿 𝑒 𝑎𝑡 𝑓 𝑡 = F(s-a)
. 𝐹 𝑆 = 0
∞
𝑒−𝑠𝑡
𝑒 𝑡
𝑓 𝑡 𝑑𝑡
5-May-18 13presentation of fluid theory

14. Example of First Shifting Property
Q. Simplify 𝐿{𝑒−𝑡
𝑠𝑖𝑛2
𝑡}
Solution:-𝐿 𝑠𝑖𝑛2 𝑡 = 𝐿
1−𝑐𝑜𝑠2𝑡
2
= 𝐿
1
2
−
𝑐𝑜𝑠2𝑡
2
=
1
2
𝐿 1 −
1
2
𝐿 𝑐𝑜𝑠2𝑡 =
1
2
.
1
𝑠
−
1
2
.
𝑠
𝑠2 + 4
=
1
2
1
𝑠
−
𝑠
𝑠2 + 4
=
1
2
4
𝑠(𝑠2 + 4)
=
2
𝑠(𝑠2 + 4)
= 𝐹 𝑠
So, 𝐿 𝑒−𝑡 𝑠𝑖𝑛2 𝑡 = 𝐹 𝑆 + 1 → . By Shift Theorem
2
(𝑆 + 1)( 𝑠 + 1 2 + 4)
𝐿 𝑒−𝑡 𝑠𝑖𝑛2 𝑡 =
2
(𝑠 + 1)(𝑠2 + 2𝑠 + 𝑠)
5-May-18 14presentation of fluid theory

15. 3.Second Shifting Property
If 𝐿 𝑓 𝑡 = 𝐹 𝑠 and
𝑔 𝑡 =
𝑓 𝑡 − 𝑎 ; 𝑡 > 𝑎
0 ; 0 < 𝑡 < 𝑎
then
𝐿 𝑔 𝑡 = 𝑒−𝑎𝑠 𝐹 𝑠
Proof:-
𝐿 𝑔 𝑡 =
0
∞
𝑒−𝑠𝑡 𝑔 𝑡 𝑑𝑡 =
0
∞
𝑒−𝑠𝑡 𝑓(𝑡 − 𝑎)𝑑𝑡
By Substitution: 𝑡 − 𝑎 = 𝑈 ⇒ 𝑡 = (𝑈 + 𝑎) and 𝑑𝑡 = 𝑑𝑢
So,
𝐿 𝑔 𝑡 =
0
∞
𝑒−𝑠 𝑢+𝑎 𝑓 𝑈 𝑑𝑡 = 𝑒−𝑠𝑎
0
∞
𝑒−𝑠𝑢 𝑓 𝑈 𝑑𝑈
𝐿 𝑔 𝑡 = 𝑒−𝑠𝑎 𝐹 𝑆 → (𝐴)
5-May-18 15presentation of fluid theory

16. Example of Second Shifting Property
𝑄. 𝐿 𝑔 𝑡 =? , 𝑔 𝑡 =
0,0 ≤ 𝑡 < 1
(𝑡 − 1)2
, 𝑡 ≥ 1
Sol:- We know that:𝐿 𝑡 𝑛
=
𝑛!
𝑠 𝑛+1
So, 𝐿 𝑡2
=
2
𝑠3
= 𝑒−𝑠(1)
.
2
𝑠3 = 𝑒−𝑠 2
𝑠3 . 𝑎 = 1
𝐿 𝑔 𝑡 = 𝑒−𝑠
2
𝑠3
5-May-18 16presentation of fluid theory

17. Change Of Scale Property
If L{f(t)}=F(S) 𝑡ℎ𝑒𝑛 L{f(at)}=1
𝑎
F( 𝑆
𝑎
)
Proof:
L{f(at)}=
0
∞
𝑒−𝑠𝑡
𝑓 𝑎𝑡 𝑑𝑡 → (1)
let 𝑢 = 𝑎𝑡 then 𝑎𝑑𝑡 = 𝑑𝑢 also 𝑑𝑡 =
𝑑𝑢
𝑎
put values in ‘ 1’ L{f(at)}
L{f(at)}=
0
∞
𝑒
−𝑠
𝑢
𝑎 𝑓 𝑢
𝑑𝑢
𝑎
→ (1)
L{f(at)}=
0
∞
𝑒
−
𝑠
𝑎
𝑢
𝑓 𝑢
𝑑𝑢
𝑎
→ (1)
L{f(at)}=1
𝑎
F
𝑠
𝑎
5-May-18 17presentation of fluid theory

18. Example Of Change Of Scale Property
Q.L{sin3t}= 1
3
f(s)
Solution: L{sin3t}= 1
3
1
𝑠
3
2
+1
L{sin3t}= 1
3
1
𝑠2+1
9
L{sin3t}=
1
3
9
𝑠2+9
5-May-18 18
L{sin3t}=
3
𝑠2+32
presentation of fluid theory

19. Laplace Transform Of Derivative
• Suppose f is continuous on [0,∞) and of exponential order
and that 𝑓′ is piecewise continuous on [0,∞) then 𝑓′ 𝑡 =
𝑠𝐿 𝑓 𝑡 = 𝑓(0)
• Proof:-
𝐿 𝑓′(𝑡) =
0
∞
𝑓′(𝑡)𝑒−𝑠𝑡 𝑑𝑡
So By Parts Integration
= 𝑓 𝑡 𝑒−𝑠𝑡
∞
|
0
−
0
∞
𝑓 𝑡 𝑒−𝑠𝑡 −𝑠 𝑑𝑡
= −𝑓 0 + 𝑠𝐿 𝑓 𝑡
So, 𝐿{𝑓′(𝑡)} = −𝑓 0 + 𝑠𝐿{𝑓(𝑡)} → (∗)
5-May-18
19presentation of fluid theory

20. Laplace Transform Of Derivative
• The above result also generalized that
𝐿 𝑓′′
(𝑡) = −𝑓′
(0) + 𝑠𝐿 𝑓′
(𝑡)
= − 𝑓′
(0) + 𝑠(−𝑓′
(0) + 𝑠𝐿 𝑓(𝑡) )
So,
𝐿 𝑓′′
(𝑡) = 𝑠2
𝐿𝑓 𝑡 − 𝑠𝑓 0 − 𝑓′
(0)
In general
𝐿 𝑓 𝑛
(𝑡) = 𝑠 𝑛
𝐿𝑓 𝑡 − 𝑠 𝑛−1
𝑓 0 − 𝑠 𝑛−2
𝑓′
(0)
… … … 𝑓 𝑛−1
(0)
5-May-18 20presentation of fluid theory

21. Example of Laplace Transform Of
Derivative
• Determine 𝐿{𝑠𝑖𝑛2
𝜔𝑡}
Solution:
𝑓(𝑡) = 𝑠𝑖𝑛2
𝜔𝑡
𝑓′
(𝑡) = 2𝑠𝑖𝑛𝜔𝑡.𝑐𝑜𝑠𝜔𝑡. 𝜔
𝑓′
𝑡 = 𝜔. 𝑠𝑖𝑛2𝜔𝑡 → (𝐴)
So from ∗
𝐿{𝑓′
(𝑡)} = 𝑠𝐿 𝑓 𝑡 − 𝑓 0
Put values of (A)
So,
𝐿 𝜔. 𝑠𝑖𝑛2𝜔𝑡 = 𝑠𝐿{𝑠𝑖𝑛2
𝜔𝑡} − 𝑠𝑖𝑛2
𝜔(0)
5-May-18 21presentation of fluid theory

22. Example of Laplace Transform Of
Derivative
𝐿 𝜔. 𝑠𝑖𝑛2𝜔𝑡 = 𝑠𝐿 𝑠𝑖𝑛2
𝜔𝑡 − 𝑠𝑖𝑛2
𝜔 0
= 𝑠𝐿 𝑠𝑖𝑛2
𝜔𝑡 − 0
𝐿 𝜔. 𝑠𝑖𝑛2𝜔𝑡 = 𝑠𝐿 𝑠𝑖𝑛2
𝜔𝑡
5-May-18 22
𝜔
𝑠
2𝜔
𝑠2 + 4𝜔2 = 𝐿{𝑠𝑖𝑛2 𝜔𝑡}
presentation of fluid theory

23. Multiplication By 𝑡 𝑛
• If 𝐿 𝑓 𝑡 = 𝐹 𝑠 then
𝐿 𝑡𝑓 𝑡 = −
𝑑
𝑑𝑠
𝐹 𝑠 → 1 . From general
In general will negative
𝐿{𝑡 𝑛 𝑓(𝑡)} = −1 𝑛
𝑑 𝑛
𝑑𝑠 𝑛
𝐹(𝑠) → (𝐵)
Proof:-
Given
𝐹 𝑠 =
0
∞
𝑒−𝑠𝑡 𝑓 𝑡 𝑑𝑡
5-May-18 23presentation of fluid theory

24. Multiplication By 𝑡 𝑛
𝑑𝐹(𝑠)
𝑑𝑠
=
0
∞
𝑒−𝑠𝑡
−𝑡 𝑓 𝑡 𝑑𝑡
= −
0
∞
𝑒−𝑠𝑡
𝑡𝑓 𝑡 𝑑𝑡 = −𝐿{𝑡𝑓(𝑡)}
5-May-18 24
−
𝑑
𝑑𝑠
𝐹 𝑠 = 𝐿{𝑡𝑓(𝑡)}
presentation of fluid theory

25. Example of Multiplication By 𝑡 𝑛
𝐿 𝑡𝑒 𝑡
= −1 1
𝑑
𝑑𝑠
𝐹 𝑠
𝐹 𝑠 = 𝐿 𝑒 𝑡
=
1
𝑠−1
. 𝑒 𝑎𝑡
=
1
𝑠−𝑎
So,
𝑑
𝑑𝑠
1
𝑠−1
=
𝑑
𝑑𝑠
(𝑠 − 1)−1
= −1(𝑠 − 1)−2
Note:
‘t’ just show derivative how many time will take
derivative.
5-May-18 25
𝐿 𝑡𝑒 𝑡 =
−1
(𝑠 − 1)2
presentation of fluid theory

26. Division By t
If ‘t’ is piecewise continuous on [0, ∞) and of
exponential order such that
lim
𝑡→0+
𝑓(𝑡)
𝑡
exists , then
𝐿
𝑓(𝑡)
𝑡
=
0
∞
𝐹 𝑢 𝑑𝑢 ( . 𝑠 > 𝑑)
Proof:- Let g(t) =
𝑓(𝑡)
𝑡
so that 𝑓 𝑡 = 𝑡𝑔 𝑡
then,
𝐹 𝑆 = 𝐿 𝑓 𝑡 = 𝐿 𝑡𝑔 𝑡 = −
𝑑
𝑑𝑠
𝐿 𝑔 𝑡 → (1)
5-May-18 26presentation of fluid theory

27. Division By t
Integrating 1 𝑤. 𝑟. 𝑡 ′
𝑆′
to infinity
−𝐿 𝑔(𝑡)
∞
|
𝑠
=
𝑠
∞
𝐹 𝑢 𝑑𝑢
5-May-18 27
𝐿{𝑔 𝑡 } =
𝑠
∞
𝐹 𝑠 𝑑𝑠
presentation of fluid theory

28. Example Of Division By t
Q.Find 𝐿
𝑠𝑖𝑛𝑎𝑡
𝑡
Solution:- 𝐿 𝑠𝑖𝑛𝑎𝑡 =
𝑎
𝑠2+𝑎2
=
𝑠
∞
𝑎
𝑠2 +𝑎2
𝑑𝑠 = 𝑡𝑎𝑛−1
𝑠
𝑎
∞
|
𝑠
𝐿
𝑠𝑖𝑛𝑎𝑡
𝑡
=
𝜋
2
− 𝑡𝑎𝑛−1 𝑠
𝑎
5-May-18 28presentation of fluid theory

29. Laplace Transform Of Integral
• Suppose f(t) is piecewise continuous on 0, ∞
and the function
𝑔 𝑡 =
0
𝑡
𝑓 𝑢 𝑑𝑢
Is an exponential order , then 𝐿 𝑔 𝑡 =
1
𝑠
𝐹 𝑠
Proof:- Clearly 𝑔 0 = 0 and 𝑔′
(𝑡) = 𝑓 𝑡
Using derivative theorem
⇒ 𝐿{𝑔′
(𝑡)} = 𝑠𝐿 𝑔 𝑡 − 𝑔(0)
5-May-18 29presentation of fluid theory

30. Laplace Transform Of Integral
⇒ 𝐿 𝑔 𝑡 =
1
𝑠
𝑓(𝑡)
. 𝑔′
(𝑡)=𝑓(𝑡)
So,
5-May-18 30
𝐿 𝑔 𝑡 =
1
𝑠
𝑓(𝑠)
presentation of fluid theory

31. APPLICATION OF LAPLACE
TRANSFORM
STEPS TO SOLVE L.T
I. Take Laplace transform on both
sides.
II. Use the formula.
III. Substitute initial value.
IV. Take inverse transform.
V. Resolve it in partial fraction.
VI. Divide the coefficient of L{Y}
presentation of fluid theory 315-May-18

32. EXAMPLES:
1. y′′+4y=0 ,y(0)=2, y′(0)=0
solution: y′′+4y=0
L{y+4y}=L{0}
L{y′′}+L{4y}=0
L{y′′}+4L{y}=0
Use formula of derivative
S2L{y}-SY(0)- y′′ (0)+4L{y}=0
presentation of fluid theory 325-May-18

33. S2L{y}-S(2)-0+4L{y}=0
S2L{y}+4L{y}-2S=0
(S2+4)L{y}=2S
L{y}=2S/(S2+4)
Take Laplace inverse
L-1 L{y}= L-1 {2S/S2+4}
y=2 L-1 {S/S2+4}
presentation of fluid theory 335-May-18

34. 2. Y′+3Y=13sin2t , y(0)=6
Solution:
Y′ +3Y=13sin2t
L{Y′+3Y}=L{13sin2t}
L {Y′}+3L{y}=13L{sin2t}
SY(S)-Y(0)+3Y(S)=13(2/S2 +4)
SY(S)-6+3Y(S)=26/S2 +4
Y(S)[S+3]=6+ 26/S2 +4
Y(S)=1/(S+3)[6+26/S2 +4]
presentation of fluid theory 345-May-18

35. Y(S)= 26
(𝑆+3)(𝑆2+4)
+ 6
𝑆+3
Take Laplace inverse
L-1(y(s))=L-1[ 26
(𝑆+3)(𝑆2+4)
+ 6
𝑆+3
]
y(t)=L-1[ 26
(𝑆+3)(𝑆2+4)
] +6L-1[ 1
𝑠+3
]→ (A)
BY Partial fraction
26
(𝑆+3)(𝑆2+4)
= 𝐴
𝑆+3
+ 𝐵𝑆+𝐶
𝑆2+4
→ (∗)
26=A(S2+4)+BS+C(S+3) →(1)
Put s=-3 in ‘1’
26 = 13𝐴 ⇒ 𝐴 = 2
So,
presentation of fluid theory 355-May-18

36. And A+B=0 so B=-2
Now compare coefficient of s,
26=AS2+4A+BS+CS+3C
26=4A+3C
26=4(2)+3C
26-8=3C
C=6 so ‘*’ implies that
26
(𝑆+3)(𝑆2+4)
=
2
𝑠+3
+ −2𝑠+6
𝑠2+4
Equation ‘A’ implies that
presentation of fluid theory 365-May-18

37. L-1[ 26
(𝑆+3)(𝑆2+4)
]=2 L-1[ 1
𝑠+3
]-2 L-1[ 𝑠
𝑠2+4
]+3 L-1[
2
𝑠2+4
]
L-1[ 26
(𝑆+3)(𝑆2+4)
]=2e-3t -2cos2t+3sin2t
‘A’ implies that,
Y(t)=2e-3t -2cos2t+3sin2t+6 L-1[
1
𝑠+3
]
presentation of fluid theory 375-May-18

38. 3. y′′-3y′+2y=e-4t , y(0)=1 , y′(0)=5
solution: Take Laplace transform
L{y′′}-3L{y′}+2L{y}=L{e-4t}
S2y(s)-sy(0)-y′(0)-3{sy(s)-y(0)}+2y(s)=
1
𝑠+4
S2y(s)-s(1)-5-3sy(s)+3+2y(s)=
1
𝑠+4
y(s){S2-3s+2}=
1
𝑠+4
+s+5-3=
1
𝑠+4
+s+2
y(s)=
1
(𝑠+4)(𝑠2−3𝑠+2)
+ 𝑠+2
(𝑠2−3𝑠+2)
→ (A)
𝑠+2
(𝑠2−3𝑠+2)
= 𝑠+2
(𝑠2−2𝑆−𝑆+2)
= 𝑠+2
(𝑆−2)(𝑆−1)
By partial fraction
presentation of fluid theory 385-May-18

39. 𝑠+2
(𝑠2−3𝑠+2)
= 𝐴
(𝑆−2)
+
𝐵
(𝑆−1)
→ (1)
S+2=A(S-1)+B(S-2) → (2)
put s=2 in 2
2+2=A(2-1)+0⇒A=4
now put s=1 in 2
1+2=A(1-1)+B(1-2)
3=-B so B=-3
Equation ‘1’ IMPLIES THAT
𝑠+2
(𝑠2−3𝑠+2)
= 4
(𝑆−2)
−
3
(𝑆−1)
→ (*1)
So, 1
(𝑠+4)(𝑠2−3𝑠+2)
= 𝐴
(𝑆+4)
+
𝐵
(𝑆−2)
+
𝐶
(𝑆−1)
→ (B)
presentation of fluid theory 395-May-18

40. 1=A(S-2)(S-1)+B(S+4)(S-1)+C(S+4)(S-2) →( 2)
put s=2 in ‘2’ then
1=0+B(6)(1)+0
6B=1 so B=1/6
put s=1 in ‘2’
1=0+0+C(5)(-1)⇒C=-1/5
Now put s=-4 in ‘2’
A=1/30 so ‘B’ IMPLIES THAT
1
(𝑠+4)(𝑠2−3𝑠+2)
=
1
30 𝑆+4
+
1
6(𝑆−2)
-
1
5(𝑆−1)
→ (∗2)
presentation of fluid theory 405-May-18

41. put Equation (*1) and (*2) in A
Y(S)=
1
30 𝑆+4
+
1
6(𝑆−2)
-
1
5(𝑆−1)
+
4
(𝑆−2)
-
3
(𝑆−1)
Take Laplace inverse
L-1{Y(s)}= 1
30
L−1 { 1
𝑠+4
} - 1
5
L−1{ 1
𝑠−1
}+ 1
6
L−1{ 1
𝑠−2
} + L−1{
4
𝑠−2
} - L−1 { 3
𝑆−1
}
Y(t)= 1
30
e-4t - 1
5
et +4e2t + 1
6
e2t -3et
presentation of fluid theory 415-May-18

42. 4. y′′-6y′+9y=t2e-3t , y(0)=2, y′-(0)=17
SOLUTION:
y′′-6y′+9y=t2e-3t
take Laplace transform
L{y′′}-6L{y′}+9L{Y}=L{t2e-3t}
S2y(s)-sy(0)-y′(0)-6{sy(s)-y(0)}+9y(s)=
2
𝑠−3 3
y(s){S2-6s+9}=-2s-17+12=
2
𝑠−3 3
y(s)=
2
(𝑠−3)(𝑠2−6𝑠+9)
+ 2𝑠+5
(𝑠2−6𝑠+9)
take Laplace inverse
presentation of fluid theory 425-May-18

43. L-1{y(s)}= L-1{ 2
𝑠−3 3 𝑆2−6𝑠+9
}+L-1{
2𝑠+5
𝑆−3 2}
BY PARTIAL FRACTION,
y(t)= L-1{ 2
𝑠−3 5}+ 2e3t +11e3t
presentation of fluid theory 435-May-18

44. LAPLACE INVERSE TRONSFORM
1. L{1}=1
𝑆
and L-1{1
𝑆
}=1
2. L{t}= 1
𝑆2 and L-1{ 1
𝑆2}=t
3. L{tn}={ 𝑛!
𝑠 𝑛+1
} 𝑎𝑛𝑑 L-1{ 𝑛!
𝑠 𝑛+1
}=tn
4. L{sinkt}= 𝑘
𝑆2+𝑘2 and L-1{ 𝑘
𝑆2+𝑘2 }=sinkt
5. L{eat}={
1
𝑠−𝑎
} and L-1{
1
𝑠−𝑎
}=eat
6. L{coskt}= 𝑠
𝑆2+𝑘2 and L-1{ 𝑠
𝑆2+𝑘2 }=coskt
And so on many application
presentation of fluid theory 445-May-18

45. 1.Find Laplace inverse of L-1{
−2𝑠+6
𝑠2+4
}
SOLUTION:
= L-1{
−2𝑠
𝑠2+4
}+3L-1{
6
𝑠2+4
}
= -2L-1{
𝑠
𝑠2+4
}+3L-1{
6
𝑠2+4
}
presentation of fluid theory 455-May-18
L-1[
−𝟐𝒔+𝟔
𝒔 𝟐+𝟒
]=-2cos2t+3sin2t

46. 2. Find Laplace inverse of L-1{
1
𝑠5}?
SOLUTION: We know that
L{tn}={ 𝑛!
𝑠 𝑛+1
}
1
𝑛!
L{tn}={ 1
𝑠 𝑛+1
}
L{𝑡 𝑛
𝑛!
}=
1
𝑠 𝑛+1
L-1{
1
𝑠 𝑛+1
}=𝑡 𝑛
𝑛!
n+1=5, Then n=4
presentation of fluid theory 465-May-18

47. Limitation of Laplace transform
• The inverse Laplace transform is in
general
Complicated(using matlab the
solution is available).
• There is no Laplace transform (or not
used),instead you will use Laplace
transform in discrete signal in the
same place as Laplace transform in
continuous signals.
presentation of fluid theory 47
5-May-18

48. • There is fast Laplace transform
corresponding to fast Fourier
transform.
• It some times make the problem
much worse.
• The Laplace transform method is
usually of no real use in non linear
problems, just because you do not get
a nice algebraic equation out of it.
• Transform function can be defined for
linear systems only.
presentation of fluid theory 515-May-18

49. • Initial condition lose their
importance since transform
functions does not take in to
account the initial condition.
• No interferences can be drawn
about the physical structure of a
system from its transform
function.
presentation of fluid theory
49
5-May-18

50. 5-May-18 50presentation of fluid theory