The Laplace Transform

In mathematics, the Laplace transform, named after its inventor Pierre-Simon Laplace, is an integral
transform that converts a function of a real variable 𝑡 (often time) to a function of a complex variable 𝑠
(complex frequency). The transform has many applications in science and engineering because it is a tool
for solving differential equations. In particular, it transforms linear differential equations into algebraic
equations and convolution into multiplication
∞
For suitable functions 𝑓, the Laplace transform is the integral F 𝑠 = ℒ 𝑓 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡.
Evaluate: (1) ℒ 𝑒 𝑎𝑡 (2) ℒ 𝑒 −𝑎𝑡 (3) ℒ 𝑡 (4) ℒ 𝑡 𝑛 (5) ℒ sin 𝑎𝑡 (6) ℒ cos 𝑎𝑡 (7) ℒ sinh 𝑎𝑡 (8)
ℒ cosh 𝑎𝑡
Solution of (1):
∞
We know that, ℒ 𝑓 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡
∞
∞ ∞ ∞ 𝑒 −𝑡(𝑠−𝑎) 1 1
⸫ ℒ 𝑒 𝑎𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝑒 𝑎𝑡 𝑑𝑡 = ‫׬‬0 𝑒 −𝑠𝑡+𝑎𝑡 𝑑𝑡 = ‫׬‬0 𝑒 −𝑡(𝑠−𝑎) 𝑑𝑡 = =0− = (since 𝑒 −∞ = 0
−(𝑠−𝑎) 0 − 𝑠−𝑎 𝑠−𝑎

and 𝑒 0 = 1)
Solution of (2):
∞
We know that, ℒ 𝑓 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡
∞ ∞
∞ 𝑒 −𝑠𝑡 1 ∞ −𝑠𝑡 1 𝑒 −𝑠𝑡 1
⸫ℒ 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝑡𝑑𝑡 = 𝑡
−𝑠 0
+ ‫𝑒 ׬‬
𝑠 0
𝑑𝑡 = 0 −
𝑠 −𝑠 0
=
𝑠2
 The Laplace Transform
Solution of (4):
∞
We know that, ℒ 𝑓 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡
∞
⸫ ℒ 𝑡 𝑛 = ‫׬‬0 𝑒 −𝑠𝑡 𝑡 𝑛 𝑑𝑡 (1)
1 𝑦𝑛
Putting 𝑠𝑡 = 𝑦 so that 𝑠𝑑𝑡 = 𝑑𝑦 ⸫𝑑𝑡 = 𝑑𝑦 . Also 𝑡𝑛 = , limit : when 𝑡 = 0 then 𝑦 = 0 and when 𝑡 = ∞
𝑠 𝑠𝑛
then 𝑦 = ∞
Thus from equation (1), we have
∞ −𝑦 𝑦 𝑛 1 1 ∞ −𝑦 1 ∞ −𝑦 1
ℒ 𝑡𝑛 = ‫׬‬0 𝑒 𝑠𝑛 𝑠 𝑑𝑦 = ‫ 𝑒 ׬‬. 𝑦 𝑛 𝑑𝑦 = ‫ 𝑒 ׬‬. 𝑦 𝑛+1 −1 𝑑𝑦 = Γ(𝑛 + 1) if 𝑛 > −1 and s>
𝑠 𝑛+1 0 𝑠 𝑛+1 0 𝑠 𝑛+1
0
𝑛!
=
𝑠 𝑛+1

𝑒 𝑖𝑎𝑡 −𝑒 −𝑖𝑎𝑡 𝑒 𝑖𝑎𝑡 +𝑒 −𝑖𝑎𝑡 𝑒 𝑎𝑡 −𝑒 −𝑎𝑡

We have (1) sin 𝑎𝑡 = (2) cos 𝑎𝑡 = (3) sinh 𝑎𝑡 =
2𝑖 2 2
𝑒 𝑎𝑡 +𝑒 −𝑎𝑡
(4) cosh 𝑎𝑡 =
2
 The Laplace Transform
Solution of (5):
∞ −𝑠𝑡
We know that, ℒ 𝑓 𝑡 = ‫׬‬0 𝑒 𝑓 𝑡 𝑑𝑡
∞ −𝑠𝑡 ∞ −𝑠𝑡 𝑒 𝑖𝑎𝑡 −𝑒 −𝑖𝑎𝑡
⸫ ℒ sin 𝑎𝑡 = ‫׬‬0 𝑒 sin 𝑎𝑡 𝑑𝑡 = ‫׬‬0 𝑒 . 𝑑𝑡
2𝑖
∞ ∞ ∞ ∞
1 −𝑠𝑡+𝑖𝑎𝑡 −𝑠𝑡−𝑖𝑎𝑡
1
= න𝑒 𝑑𝑡 − න 𝑒 𝑑𝑡 = න 𝑒 −𝑡(𝑠−𝑖𝑎) 𝑑𝑡 − න 𝑒 −𝑡(𝑠+𝑖𝑎) 𝑑𝑡
2𝑖 2𝑖
0 0 0 0
∞ ∞
1 𝑒 −𝑡(𝑠−𝑖𝑎) 𝑒 −𝑡(𝑠+𝑖𝑎) 1 1 1 1 𝑠+𝑖𝑎−𝑠+𝑖𝑎
= − = − =
2𝑖 −(𝑠−𝑖𝑎) 0 −(𝑠+𝑖𝑎) 0 2𝑖 𝑠−𝑖𝑎 𝑠+𝑖𝑎 2𝑖 (𝑠−𝑖𝑎)(𝑠+𝑖𝑎)

1 2𝑎𝑖 𝑎
= =
2𝑖 𝑠 2 +𝑎2 𝑠 2 +𝑎2
Similarly,
1
ℒ sin 𝑡 =
𝑠 2 +1
 The Laplace Transform
Solution of (8):
∞ −𝑠𝑡
We know that, ℒ 𝑓 𝑡 = ‫׬‬0 𝑒 𝑓 𝑡 𝑑𝑡
∞ ∞ −𝑠𝑡 𝑒 𝑎𝑡 +𝑒 −𝑎𝑡
⸫ ℒ cosh 𝑎𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 cosh 𝑎𝑡 𝑑𝑡 = ‫׬‬0 𝑒 . 𝑑𝑡
2
∞ ∞ ∞ ∞
1 −𝑠𝑡+𝑎𝑡 −𝑠𝑡−𝑎𝑡
1
= න𝑒 𝑑𝑡 + න 𝑒 𝑑𝑡 = න 𝑒 −𝑡(𝑠−𝑎) 𝑑𝑡 + න 𝑒 −𝑡(𝑠+𝑎) 𝑑𝑡
2 2𝑖
0 0 0 0
∞ ∞
1 𝑒 −𝑡(𝑠−𝑎) 𝑒 −𝑡(𝑠+𝑎) 1 1 1 1 𝑠+𝑎+𝑠−𝑎
= + = + =
2 −(𝑠−𝑎) 0 −(𝑠+𝑎) 0 2 𝑠−𝑎 𝑠+𝑎 2 (𝑠−𝑎)(𝑠+𝑎)

1 2𝑠 𝑠
= =
2 𝑠 2 −𝑎2 𝑠 2 −𝑎2
Similarly,
𝑠
ℒ cosh 𝑡 =
𝑠 2 −1
 The Laplace Transform
Formula
 The Laplace Transform
First Translation (or Shifting) Property:
Statement: If ℒ 𝐹 𝑡 = 𝑓(𝑠) then ℒ 𝑒 𝑎𝑡 𝐹 𝑡 = 𝑓(𝑠 − 𝑎)
Proof: By definition of Laplace transform, we have
∞

ℒ 𝐹 𝑡 = න 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡 = 𝑓(𝑠)
0
⸫ ℒ 𝑒 𝑎𝑡 𝐹 𝑡 = ‫׬‬0∞ 𝑒 −𝑠𝑡 . 𝑒 𝑎𝑡 𝑓 𝑡 𝑑𝑡 = ‫׬‬0∞ 𝑒 −𝑡(𝑠−𝑎) 𝑓 𝑡 𝑑𝑡 = 𝑓(𝑠 − 𝑎)
Second Translation (or Shifting) Property:

Statement: If ℒ 𝐹 𝑡 = 𝑓(𝑠) and G 𝑡 = ቊ𝐹 𝑡 − 𝑎 , 𝑡 > 𝑎 then ℒ 𝐺 𝑡 = 𝑒 −𝑎𝑠 𝑓(𝑠)

0, 𝑡 < 𝑎
Proof: By definition of Laplace transform, we have
∞

ℒ 𝐹 𝑡 = න 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡 = 𝑓(𝑠)
0
∞ 𝑎 ∞ 𝑎 ∞
⸫ℒ 𝐺 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 G 𝑡 𝑑𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 G 𝑡 𝑑𝑡 + ‫ 𝑒 𝑎׬‬−𝑠𝑡 G 𝑡 𝑑𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 . 0. 𝑑𝑡 + ‫ 𝑒 𝑎׬‬−𝑠𝑡 𝐹 𝑡 − 𝑎 𝑑𝑡
 The Laplace Transform
Proof:
∞
⸫ℒ 𝐺 𝑡 = 0 + ‫ 𝑒 𝑎׬‬−𝑠𝑡 𝐹 𝑡 − 𝑎 𝑑𝑡
Let 𝑡 − 𝑎 = 𝑢 ⇒ 𝑡 = 𝑎 + 𝑢 ⸫ d𝑡 = 𝑑𝑢
Limits: when 𝑡 = 𝑎 then 𝑢 = 0 and when 𝑡 = ∞ then 𝑢 = ∞
Hence from (1), we have
∞ ∞
ℒ 𝐺 𝑡 = ‫׬‬0 𝑒 −𝑠(𝑎+𝑢) 𝐹 𝑢 𝑑𝑢 = 𝑒 −𝑠𝑎 ‫׬‬0 𝑒 −𝑠𝑢 𝐹 𝑢 𝑑𝑢 = 𝑒 −𝑠𝑎 𝑓(𝑠) (Proved)
Find the Laplace transform of (1) 𝐹 𝑡 = 𝑒 −2𝑡 sin 4𝑡 (2) 𝐹 𝑡 = 𝑒 −𝑡 cos 2𝑡
Solution: Let 𝐹 𝑡 = sin 4𝑡
4 4
Now, ℒ 𝐹 𝑡 = ℒ sin 4𝑡 = = = 𝑓(𝑠)
𝑠 2 +4 2 𝑠 2 +16

We have, If ℒ 𝐹 𝑡 = 𝑓(𝑠) then ℒ 𝑒 𝑎𝑡 𝐹 𝑡 = 𝑓(𝑠 − 𝑎)

4 4
⸫ ℒ 𝑒 −2𝑡 sin 4𝑡 = =
(𝑠+2)2 +16 𝑠 2 +4𝑠+20
 The Laplace Transform
Change of scale Property
1 𝑠
Statement: If ℒ 𝐹 𝑡 = 𝑓(𝑠) then ℒ 𝐹(𝑎𝑡) = 𝑓
𝑎 𝑎
∞
Proof: We have ℒ 𝐹 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡 = 𝑓(𝑠)
∞
So ℒ 𝐹 𝑎𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝐹 𝑎𝑡 𝑑𝑡
1
Let, 𝑎𝑡 = 𝑢 ⇒ 𝑑𝑡 = 𝑑𝑢 Limits: when 𝑡 = 0 then 𝑢 = 0 and when 𝑡 = ∞ then 𝑢 = ∞
𝑎
∞ −𝑠𝑢 1 1 ∞ −𝑠𝑢 1 𝑠
⸫ ℒ 𝐹 𝑎𝑡 = ‫׬‬0 𝑒 𝑎 𝐹 𝑢 𝑑𝑢 = ‫𝐹 𝑎 𝑒 ׬‬ 𝑢 𝑑𝑢 = 𝑓 (Proved)
𝑎 𝑎 0 𝑎 𝑎
 The Laplace Transform
Laplace Transform for Derivatives
If ℒ 𝐹 𝑡 = 𝑓(𝑠) then (1) ℒ 𝐹 ′ (𝑡) = 𝑠𝑓 𝑠 − 𝐹 0 (2) ℒ 𝐹 ′′ (𝑡) = 𝑠 2 𝑓 𝑠 − 𝑠𝐹 0 − 𝐹 ′ 0
(3) ℒ 𝐹 ′′′ (𝑡) = 𝑠 3 𝑓 𝑠 − 𝑠 2 𝐹 0 − 𝑠𝐹 ′ 0 − 𝐹 ′′ 0
∞
Proof of (1): We have ℒ 𝐹 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡 = 𝑓(𝑠)
∞ ∞ ∞
So ℒ 𝐹 ′ (𝑡) = ‫׬‬0 𝑒 −𝑠𝑡 𝐹 ′ 𝑡 𝑑𝑡 = 𝑒 −𝑠𝑡 𝐹 𝑡 ∞
0 − ‫׬‬0 −𝑠 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡 = 0 − 𝐹 0 + 𝑠 ‫׬‬0 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡
⸫ ℒ 𝐹 ′ (𝑡) = −𝐹 0 + 𝑠𝑓(𝑠) (Proved)
∞
Proof of (2): We have ℒ 𝐹 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡 = 𝑓(𝑠)
∞ ∞ ∞
So ℒ 𝐹 ′′ (𝑡) = ‫׬‬0 𝑒 −𝑠𝑡 𝐹 ′′ 𝑡 𝑑𝑡 = 𝑒 −𝑠𝑡 𝐹 ′ 𝑡 ∞
0 − ‫׬‬0 −𝑠 𝑒 −𝑠𝑡 𝐹 ′ 𝑡 𝑑𝑡 = 0 − 𝐹 ′ 0 + 𝑠 ‫׬‬0 𝑒 −𝑠𝑡 𝐹 ′ 𝑡 𝑑𝑡
∞ ∞
∞
= −𝐹 ′ 0 + 𝑠 𝑒 −𝑠𝑡 𝐹 𝑡 0 − න −𝑠 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡 = −𝐹 ′ 0 + 𝑠{0 − 𝐹 0 } + 𝑠 2 න 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡
0 0

⸫ ℒ 𝐹 ′′ (𝑡) = −𝐹 ′ 0 − 𝑠𝐹 0 + 𝑠 2 𝑓(𝑠)
 The Laplace Transform
Laplace Transform for Derivatives
∞
Proof of (3): We have ℒ 𝐹 𝑡 = ‫׬‬0 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡 = 𝑓(𝑠)
∞ ∞ ∞
So ℒ 𝐹 ′′′ (𝑡) = ‫׬‬0 𝑒 −𝑠𝑡 𝐹 ′′′ 𝑡 𝑑𝑡 = 𝑒 −𝑠𝑡 𝐹 ′′ 𝑡 ∞
0 − ‫׬‬0 −𝑠 𝑒 −𝑠𝑡 𝐹 ′′ 𝑡 𝑑𝑡 = 0 − 𝐹 ′′ 0 + 𝑠 ‫׬‬0 𝑒 −𝑠𝑡 𝐹 ′′ 𝑡 𝑑𝑡
∞ ∞
∞
= −𝐹 ′′ 0 + 𝑠 𝑒 −𝑠𝑡 𝐹 ′ 𝑡 0 − 𝑠 න −𝑠 𝑒 −𝑠𝑡 𝐹 ′ 𝑡 𝑑𝑡 = −𝐹 ′′ 0 + 𝑠{0 − 𝐹 ′ 0 } + 𝑠 2 න 𝑒 −𝑠𝑡 𝐹 ′ 𝑡 𝑑𝑡
0 0
∞
∞
= −𝐹 ′′ 0 − 𝑠𝐹 ′ 0 + 𝑠 2 𝑒 −𝑠𝑡 𝐹 𝑡 0 − න −𝑠 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡
0
∞

= −𝐹 ′′ 0 − 𝑠𝐹 ′ 0 + 𝑠 2 {0 − 𝐹 0 } + 𝑠 3 න 𝑒 −𝑠𝑡 𝐹 𝑡 𝑑𝑡
0

⸫ ℒ 𝐹 ′′′ (𝑡) = −𝐹 ′′ 0 − 𝑠𝐹 ′ 0 − 𝑠 2 𝐹 0 + 𝑠 3 𝑓(𝑠)