Lecture 1 Sujin Khomrutai – 1 / 21
Method of Applied Math
Lecture 6: Laplace Transform
Sujin Khomrutai, Ph.D.
More examples
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 2 / 21
EX. Find the Laplace transform for each of the following functions f (t).
1. f (t) = 4
2. f ( t ) = 3 e − 2 t
3. f ( t ) = t 4 + t 2 + 1
4. f (t) = 3 sin t − 5 cos(2t)
5. f ′′ (t) = e 2 t , f (0) = f ′ (0) = 0
More examples
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 3 / 21
EX. Find the inverse Laplace transform for each of the following functions F (s).
1. F (s) = 1 s − 5 2. F (s) = 2
3 s + 6 3. F ( s ) = s − 3 4. F (s) = 1
s 2 + 9 5. F ( s ) = s + 5
s 2 + 1 6. F (s) = 2
s 2 + 4s + 3
Laplace Transform of Integrals
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 4 / 21
Theorem. Let F ( s ) = L [ f ( t )]. Then
L
Z t
0
f ( τ ) dτ
= F (s)
s = L[f (t)]
s Also,
L − 1
F ( s ) s
=
Z t
0
f (τ ) dτ =
Z t
0
L − 1 [F ](τ ) dτ
Laplace Transform of Integrals
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 5 / 21
Proof. Let g ( t ) = R t
0 f ( τ ) dτ . Then g (0) = 0 and g ′ ( t ) = f ( t ).
Then
L[g ′ (t)] = sG(s) − g(0) = sG(s).
On the other hand, L [ g ′ ( t )] = L [ f ( t )] = F ( s ), hence
G ( s ) = F (s)
s .
Example 4
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 6 / 21
EX. Find the Laplace transforms
L
Z t
0
e 2 τ dτ
, L
Z t
0
( τ 2 + 3 sin τ ) dτ
.
Example 5
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 7 / 21
EX. Find the inverse Laplace transforms
L − 1
1
s(s 2 + 1)
, L − 1
1
s(s − 3)
.
Shifting in s
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 8 / 21
Theorem. Let F ( s ) = L [ f ( t )]. Then L
e at f (t)
= F (s − a) = L [f ](s − a), thus
L − 1 [F (s − a)] = e at f (t) = e at L − 1 [F (s)].
Shifting in s
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 9 / 21
Proof. By definition,
L [e at f (t)] =
Z ∞
0
e −st e at f (t) dt
=
Z ∞
0
e − ( s−a ) t f (t) dt
=
Z ∞
0
e −st f ( t ) dt
s→s−a
= F ( s − a )
Example 6
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 10 / 21
EX. Find the Laplace transforms L
e 3 t sin t
, L [e − 2 t cos 5t], L [t 4 e t ].
Example 7
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 11 / 21
EX. Find the inverse Laplace transforms
L − 1
1 (s − 3) 2
, L − 1
s + 1
(s + 1) 2 + 4
, L − 1
1
s 2 + 4s + 5
.
Example 8
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 12 / 21
EX. Solve the IVP
y ′′ + 2y ′ + 5y = 0, y(0) = 0, y ′ (0) = 3.
ANS. y(t) = 3 2 e −t sin 2t.
Heaviside Functions
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 13 / 21
Definition. The Heaviside function (or unit step function ) is
H ( t ) =
( 0, t < 0 1, t ≥ 0
For a number a ≥ 0, we have
H ( t − a ) =
( 0 t < a
1 t ≥ a
Heaviside Functions
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 14 / 21
The Laplace Transform of H ( t − a )
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 15 / 21
Theorem. For a number a ≥ 0, we have
L [H (t − a)] = e −as s . In particular,
L [ H ( t )] = 1 s .
Proof. By the definition of H (t − a),
L [ H ( t − a )] =
Z ∞
0
e −st H ( t − a ) dt =
Z ∞
a
e −st dt = e −as
s .
Example 9
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 16 / 21
EX. Find the Laplace transform of the following functions:
f (t) =
( 0 if 0 < t < 3
5 if t ≥ 3 , g(t) = 3H (t) − H (t − π).
Example 9
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 17 / 21
EX. Find the inverse Laplace transforms
L − 1
e − 2 s s
, L − 1
2e −s − e − 2 s s
Pulses
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 18 / 21
Def. A pulse is a function defined by
k [ H ( t − a ) − H ( t − b )] =
0 if t < a
k if a ≤ t < b, 0 if t ≥ b,
where k 6= 0 and 0 ≤ a < b are constants.
Laplace Transform of Pulses
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 19 / 21
Theorem.
L [ k ( H ( t − a ) − H ( t − b ))] = k e −as − e −bs
s .
Example 10
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 20 / 21
EX. Find the Laplace transform of the functions
f ( t ) =
0 0 < t < 4 2 4 ≤ t < 6 0 t ≥ 6
, g ( t ) = 10( H ( t − 2) − H ( t − 5)) .
Example 11
EX 1.
EX 2.
Prop 3: Integration EX 4.
EX 5.
Prop 4: s-shifting EX 6.
EX 7.
EX 8.
Heaviside EX 9.
Def: Pulse Pulse EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 21 / 21
EX. Find the inverse Laplace transform of the function
F (s) = 4 e − 3 s − e − 4 s
s .
