Laplace Transforms

Laplace Transforms
Nirav B. Vyas
Department of Mathematics
Atmiya Institute of Technology and Science
Yogidham, Kalavad road
Rajkot - 360005 . Gujarat
N. B. Vyas Laplace Transforms

nition:
Let f(t) be a function of t de

ned for all t 0 then Laplace
transform of f(t) is denoted by Lff(t)g or f(s) and is
de

ned as
L ff (t)g = f (s) =
Z1
0
estf (t) dt
provided the integral exists where s is a parameter ( real or
complex).

Laplace Transforms
NOTATIONS:
The original functions are denoted by lowercase letters such
as f(t); g(t); :::
Laplace transforms by the same letters with bars such as
f(s)g(s); :::

Linearity of the Laplace Transforms
Theorem 1:
If Lff (t)g = f (s) and Lfg (t)g = g (s) then for any constants a and b
Lfaf (t) + bg (t)g = aL ff (t)g + bL fg (t)g

Theorem 1:
Corollary 1:
Putting a = 0 and b = 0, we get L[0] = 0

Theorem 1:
Corollary 1:
Corollary 2:
Putting b = 0, we get L[af(t)] = aL[f(t)]

Theorem 1:
Corollary 1:
Corollary 2:
Putting b = 0, we get L[af(t)] = aL[f(t)]
Corollary 3:
L[a1f1 (t) + a2f2 (t) + ::: + anfn (t)]
= a1L[f1(t)] + a2L[f2(t)] + ::: + anL[fn(t)]

Laplace Transforms of some elementary functions
1 L(1) =
1
s

1 L(1) =
1
s
2 L(eat) =
1
s a

1 L(1) =
1
s
2 L(eat) =
1
s a
cor.1 If a = 0 ) L(1) =
1
s

1 L(1) =
1
s
2 L(eat) =
1
s a
cor.1 If a = 0 ) L(1) =
1
s
cor.2 L[eat] =
1
s + a
if s a

1 L(1) =
1
s
2 L(eat) =
1
s a
cor.1 If a = 0 ) L(1) =
1
s
cor.2 L[eat] =
1
s + a
if s a
cor.3 L[cat] = L[eat log c] =
1
s a logc
if s a log c and c 0

3 L[sinh at] =
a
s2 a2

3 L[sinh at] =
a
s2 a2
4 L[cosh at] =
s
s2 a2

3 L[sinh at] =
a
s2 a2
4 L[cosh at] =
s
s2 a2
5 L[sin at] =
a
s2 + a2

3 L[sinh at] =
a
s2 a2
4 L[cosh at] =
s
s2 a2
5 L[sin at] =
a
s2 + a2
6 L[cos at] =
s
s2 + a2 ; s 0

3 L[sinh at] =
a
s2 a2
4 L[cosh at] =
s
s2 a2
5 L[sin at] =
a
s2 + a2
6 L[cos at] =
s
s2 + a2 ; s 0
cor.1 L[sin t] =
1
s2 + 1
; s 0

3 L[sinh at] =
a
s2 a2
4 L[cosh at] =
s
s2 a2
5 L[sin at] =
a
s2 + a2
6 L[cos at] =
s
s2 + a2 ; s 0
cor.1 L[sin t] =
1
s2 + 1
; s 0
cor.2 L[cos t] =
s
s2 + 1
; s 0

7 L[tn] =
(n + 1)
sn+1

7 L[tn] =
(n + 1)
sn+1
=
n!
sn+1 ; n = 0; 1; 2; :::

7 L[tn] =
(n + 1)
sn+1
=
n!
sn+1 ; n = 0; 1; 2; :::
cor.1 If n = 0, L[1] =
1
s

7 L[tn] =
(n + 1)
sn+1
=
n!
sn+1 ; n = 0; 1; 2; :::
cor.1 If n = 0, L[1] =
1
s
cor.2 If n =
1
2

7 L[tn] =
(n + 1)
sn+1
=
n!
sn+1 ; n = 0; 1; 2; :::
cor.1 If n = 0, L[1] =
1
s
cor.2 If n =
1
2
L

t1
2

=
(1
)
2 1
s
2
=
r

s

Examples of Laplace Transform
1 L

2t3 + e2t + t
4
3


1 L

2t3 + e2t + t
4
3

2 L

A + B t
1
2 + C t
1
2


1 L

2t3 + e2t + t
4
3

2 L

A + B t
1
2 + C t
1
2

3 L

eat 1
a


1 L

2t3 + e2t + t
4
3

2 L

A + B t
1
2 + C t
1
2

3 L

eat 1
a

4 Lfsin(at + b)g

1 L

2t3 + e2t + t
4
3

2 L

A + B t
1
2 + C t
1
2

3 L

eat 1
a

4 Lfsin(at + b)g
5 Lfsin 2t cos 3tg

1 L

2t3 + e2t + t
4
3

2 L

A + B t
1
2 + C t
1
2

3 L

eat 1
a

4 Lfsin(at + b)g
5 Lfsin 2t cos 3tg
6 L

cos24t


1 L

2t3 + e2t + t
4
3

2 L

A + B t
1
2 + C t
1
2

3 L

eat 1
a

4 Lfsin(at + b)g
5 Lfsin 2t cos 3tg
6 L

cos24t

7 L

cos32t


First Shifting Theorem
If Lff (t)g = f (s) then L

eatf (t)

= f (s a)


eatf (t)

= f (s a)
Proof: By the def. of Laplace


eatf (t)

= f (s a)

1R
L
eatf (t)
=
0
esteatf (t) dt


eatf (t)

= f (s a)

1R
L
eatf (t)
=
0
esteatf (t) dt
=
1R
0
e(sa)tf (t) dt


eatf (t)

= f (s a)

1R
L
eatf (t)
=
0
esteatf (t) dt
=
1R
0
e(sa)tf (t) dt
=
1R
0
ertf (t) dt


eatf (t)

= f (s a)

1R
L
eatf (t)
=
0
esteatf (t) dt
=
1R
0
e(sa)tf (t) dt
=
1R
0
ertf (t) dt
= f(r)


eatf (t)

= f (s a)

1R
L
eatf (t)
=
0
esteatf (t) dt
=
1R
0
e(sa)tf (t) dt
=
1R
0
ertf (t) dt
= f(r)
= f(s a)


eatf (t)

= f (s a)

1R
L
eatf (t)
=
0
esteatf (t) dt
=
1R
0
e(sa)tf (t) dt
=
1R
0
ertf (t) dt
= f(r)
= f(s a)
Thus if we know the transformation f(s) of f(t) then we can
write the transformation of eatf(t) simply replacing s by s a to
get F(s a)

Note:
1 L(eat) =
1
s a

Note:
1 L(eat) =
1
s a
2 L[eattn] =
(n + 1)
(s a)n+1

Note:
1 L(eat) =
1
s a
2 L[eattn] =
(n + 1)
(s a)n+1
3 L[eatsinh bt] =
b
(s a)2 b2

Note:
1 L(eat) =
1
s a
2 L[eattn] =
(n + 1)
(s a)n+1
3 L[eatsinh bt] =
b
(s a)2 b2
4 L[eatcosh bt] =
s
(s a)2 b2

Note:
1 L(eat) =
1
s a
2 L[eattn] =
(n + 1)
(s a)n+1
3 L[eatsinh bt] =
b
(s a)2 b2
4 L[eatcosh bt] =
s
(s a)2 b2
5 L[eatsin bt] =
b
(s a)2 + b2

Note:
1 L(eat) =
1
s a
2 L[eattn] =
(n + 1)
(s a)n+1
3 L[eatsinh bt] =
b
(s a)2 b2
4 L[eatcosh bt] =
s
(s a)2 b2
5 L[eatsin bt] =
b
(s a)2 + b2
6 L[eatcos bt] =
s a
(s a)2 + b2 ; s 0

Examples
1 Find out the Laplace transform of e3t (2 cos 5t 3 sin 5t)

Examples
2 L[eatsinhbt]

Examples
2 L[eatsinhbt]
3 L[t3e3t]

Examples
2 L[eatsinhbt]
3 L[t3e3t]
4 L[(t + 2)2et]

Examples
2 L[eatsinhbt]
3 L[t3e3t]
4 L[(t + 2)2et]
5 L[etsin2t]

Examples
2 L[eatsinhbt]
3 L[t3e3t]
4 L[(t + 2)2et]
5 L[etsin2t]
6 L[cosh at sin at]

Examples
Ex. Find the Laplace transform of the function which is de

ned as
f(t) =

t/T 0 t T
1 when t T

Examples
Ex. Find Laplace transform of f(t) =

sin t 0 t
0 when t

Examples
Ex. Find Laplace transform of f(t) where f(t) =

t 0 t 4
5 when t 4

Change of Scale property

eatf (bt)

=
1
b
f

s a
b

; b 0


eatf (bt)

=
1
b
f

s a
b

; b 0


eatf (bt)

=
1
b
f

s a
b

; b 0
Lff (t)g =
1 Z
0
estf (t) dt


eatf (bt)

=
1
b
f

s a
b

; b 0
Lff (t)g =
1 Z
0
estf (t) dt
L

eatf (bt)

=
1 Z
0
esteatf (bt) dt


eatf (bt)

=
1
b
f

s a
b

; b 0
Lff (t)g =
1 Z
0
estf (t) dt
L

eatf (bt)

=
1 Z
0
esteatf (bt) dt
=
1R
0
e(sa)tf (bt) dt


eatf (bt)

=
1
b
f

s a
b

; b 0
Lff (t)g =
1 Z
0
estf (t) dt
L

eatf (bt)

=
1 Z
0
esteatf (bt) dt
=
1R
0
e(sa)tf (bt) dt
=
1R
0
e( sa
b )btf (bt) dt


eatf (bt)

=
1
b
f

s a
b

; b 0
Lff (t)g =
1 Z
0
estf (t) dt
L

eatf (bt)

=
1 Z
0
esteatf (bt) dt
=
1R
0
e(sa)tf (bt) dt
=
1R
0
e( sa
b )btf (bt) dt
Let bt = u ) b dt = du


eatf (bt)

=
1
b
f

s a
b

; b 0
Lff (t)g =
1 Z
0
estf (t) dt
L

eatf (bt)

=
1 Z
0
esteatf (bt) dt
=
1R
0
e(sa)tf (bt) dt
=
1R
0
e( sa
b )btf (bt) dt

1R
L
eatf (bt)
=
0
e( sa
b )uf (u)
du
b


eatf (bt)

=
1
b
f

s a
b

; b 0
Lff (t)g =
1 Z
0
estf (t) dt
L

eatf (bt)

=
1 Z
0
esteatf (bt) dt
=
1R
0
e(sa)tf (bt) dt
=
1R
0
e( sa
b )btf (bt) dt

1R
L
eatf (bt)
=
0
e( sa
b )uf (u)
du
b
=
1
b
f

s a
b


Inverse Laplace Transform
If Lff (t)g = f (s) then f(t) is called the inverse
Laplace
transform of f(s) and it is denoted by L1
f(s)

= f (t)

Laplace
f(s)

= f (t)
1 L1

1
s

= 1

Laplace
f(s)

= f (t)
1 L1

1
s

= 1
2 L1

1
s a

= eat

Laplace
f(s)

= f (t)
1 L1

1
s

= 1
2 L1

1
s a

= eat
3 L1

1
s2 + a2

=
1
a
sin at

Laplace
f(s)

= f (t)
1 L1

1
s

= 1
2 L1

1
s a

= eat
3 L1

1
s2 + a2

=
1
a
sin at
4 L1

s
s2 + a2

= cos at

Laplace
f(s)

= f (t)
1 L1

1
s

= 1
2 L1

1
s a

= eat
3 L1

1
s2 + a2

=
1
a
sin at
4 L1

s
s2 + a2

= cos at
5 L1

1
s2 a2

=
1
a
sinh at

Laplace
f(s)

= f (t)
1 L1

1
s

= 1
2 L1

1
s a

= eat
3 L1

1
s2 + a2

=
1
a
sin at
4 L1

s
s2 + a2

= cos at
5 L1

1
s2 a2

=
1
a
sinh at
6 L1

s
s2 a2

= cosh at

Laplace
f(s)

= f (t)
1 L1

1
s

= 1
2 L1

1
s a

= eat
3 L1

1
s2 + a2

=
1
a
sin at
4 L1

s
s2 + a2

= cos at
5 L1

1
s2 a2

=
1
a
sinh at
6 L1

s
s2 a2

= cosh at
7 L1

1
sn

=
tn1
(n 1)!

Partial Fractions

Examples of Inverse Laplace Transform - I
1 L1

s2 3s + 4
s3


1 L1

s2 3s + 4
s3

2 L1

3
2

s4 2s2 + 1
s5


1 L1

s2 3s + 4
s3

2 L1

3
2

s4 2s2 + 1
s5

3 L1

s + 7
(s + 1)2 + 1


1 L1

s2 3s + 4
s3

2 L1

3
2

s4 2s2 + 1
s5

3 L1

s + 7
(s + 1)2 + 1

4 L1

3s + 5
(s + 1)4


1 L1

s2 3s + 4
s3

2 L1

3
2

s4 2s2 + 1
s5

3 L1

s + 7
(s + 1)2 + 1

4 L1

3s + 5
(s + 1)4

5 L1

3s
s2 + 2s 8


1 L1

s2 3s + 4
s3

2 L1

3
2

s4 2s2 + 1
s5

3 L1

s + 7
(s + 1)2 + 1

4 L1

3s + 5
(s + 1)4

5 L1

3s
s2 + 2s 8

6 L1

3s + 7
s2 2s 3


1 L1

s2 3s + 4
s3

2 L1

3
2

s4 2s2 + 1
s5

3 L1

s + 7
(s + 1)2 + 1

4 L1

3s + 5
(s + 1)4

5 L1

3s
s2 + 2s 8

6 L1

3s + 7
s2 2s 3

7 L1

2s2 6s + 5
s3 6s2 + 11s 6


Examples of Inverse Laplace Transform - II
1 L1

s + 29
(s + 4)(s2 + 9)


1 L1

s + 29
(s + 4)(s2 + 9)

2 L1

s
(s2 1)


1 L1

s + 29
(s + 4)(s2 + 9)

2 L1

s
(s2 1)

3 L1

4s + 5
(s 1)2(s + 2)


1 L1

s + 29
(s + 4)(s2 + 9)

2 L1

s
(s2 1)

3 L1

4s + 5
(s 1)2(s + 2)

4 L1

2s2 1
(s2 + 1)(s2 + 4)


1 L1

s + 29
(s + 4)(s2 + 9)

2 L1

s
(s2 1)

3 L1

4s + 5
(s 1)2(s + 2)

4 L1

2s2 1
(s2 + 1)(s2 + 4)

5 L1

s
s4 + s2 + 1


1 L1

s + 29
(s + 4)(s2 + 9)

2 L1

s
(s2 1)

3 L1

4s + 5
(s 1)2(s + 2)

4 L1

2s2 1
(s2 + 1)(s2 + 4)

5 L1

s
s4 + s2 + 1

6 L1

s
s4 + 4a4


1 L1

s + 29
(s + 4)(s2 + 9)

2 L1

s
(s2 1)

3 L1

4s + 5
(s 1)2(s + 2)

4 L1

2s2 1
(s2 + 1)(s2 + 4)

5 L1

s
s4 + s2 + 1

6 L1

s
s4 + 4a4

7 L1

s + 3
s2 + 6s + 13


Transformation of Derivatives
Thm: If f0(t) be continuous and L[f(t)] = f(s) then
Lff0(t)g = s f(s) f(0) provided lim
t!1
estf(t) = 0

t!1
estf(t) = 0
i.e. Lff0(t)g = sL ff(t)g f(0)

t!1
estf(t) = 0
Simillarly Lff00(t)g = sL ff0(t)g f0(0)

t!1
estf(t) = 0
= s [sL ff(t)g f(0)] f0(0)

t!1
estf(t) = 0
= s [sL ff(t)g f(0)] f0(0)
= s2Lff(t)g sf(0) f0(0)

t!1
estf(t) = 0
= s [sL ff(t)g f(0)] f0(0)
= s2Lff(t)g sf(0) f0(0)
= s2 f(s) sf(0) f0(0)

t!1
estf(t) = 0
= s [sL ff(t)g f(0)] f0(0)
= s2Lff(t)g sf(0) f0(0)
= s2 f(s) sf(0) f0(0)
In general
Lffn(t)g = sn f(s) sn1f(0) sn2f0(0) : : : fn1(0)

Examples of Transformation of Derivatives
Ex. Derive the Laplace transform of sin at and cos at

Ex. Obtain Lftng from L(1) =
1
s

1
s
Ex. Find L(t sin at)

1
s
Ex. Find L(t sin at)
Ex. Find L(t cos at)

Transformation of Integrals
Thm: If L[f(t)] = f(s) then L
Z t
0

f(u)du
=
1
s
f(s)

Z t
0

f(u)du
=
1
s
f(s)
Proof: Let I(t) =
Z t
0
f(u)du

Z t
0

f(u)du
=
1
s
f(s)
Proof: Let I(t) =
Z t
0
f(u)du
) I0(t) =
d
dt
Z t
0

= f(t)
f(u)du

Z t
0

f(u)du
=
1
s
f(s)
Proof: Let I(t) =
Z t
0
f(u)du
) I0(t) =
d
dt
Z t
0

= f(t) and I(0) = 0
f(u)du

Z t
0

f(u)du
=
1
s
f(s)
Proof: Let I(t) =
Z t
0
f(u)du
) I0(t) =
d
dt
Z t
0

= f(t) and I(0) = 0
f(u)du
) Lff(t)g = LfI0(t)g = sI(s) I(0) = sI(s)

Z t
0

f(u)du
=
1
s
f(s)
Proof: Let I(t) =
Z t
0
f(u)du
) I0(t) =
d
dt
Z t
0

= f(t) and I(0) = 0
f(u)du
) Lff(t)g = sI(s)

Z t
0

f(u)du
=
1
s
f(s)
Proof: Let I(t) =
Z t
0
f(u)du
) I0(t) =
d
dt
Z t
0

= f(t) and I(0) = 0
f(u)du
) Lff(t)g = sI(s)
) Lff(t)g = sL fI(t)g

Z t
0

f(u)du
=
1
s
f(s)
Proof: Let I(t) =
Z t
0
f(u)du
) I0(t) =
d
dt
Z t
0

= f(t) and I(0) = 0
f(u)du
) Lff(t)g = sI(s)
) Lff(t)g = Z sL fI(t)g
t
) f(s) = sL
0

f(u)du

Z t
0

f(u)du
=
1
s
f(s)
Proof: Let I(t) =
Z t
0
f(u)du
) I0(t) =
d
dt
Z t
0

= f(t) and I(0) = 0
f(u)du
) Lff(t)g = sI(s)
t
) f(s) = sL
0

f(u)du
) 1
s
f(s) = L
Z t
0

f(u)du

Z t
0

f(u)du
=
1
s
f(s)
Proof: Let I(t) =
Z t
0
f(u)du
) I0(t) =
d
dt
Z t
0

= f(t) and I(0) = 0
f(u)du
) Lff(t)g = sI(s)
t
) f(s) = sL
0

f(u)du
) 1
s
f(s) = L
Z t
0

f(u)du
) L1

1
s
f(s)

=
Z t
0
f(u)du

Examples of Transformation of Integrals
Ex. Prove that: L1

1
s2 + 1

= sin t
Ex. Prove that: L1

1
s(s2 + 1)

= 1 cos t
Ex. Find inverse Laplace transform of
1
s3(s2 + a2)

Multiplication by tn
Thm: If L[f(t)] = f(s) then Lftnf(t)g = (1)n dn
dsn

f(s)


Multiplication by tn
Thm: If L[f(t)] = f(s) then Lftnf(t)g = (1)n dn
dsn

f(s)

if Lftf (t)g = (1)1 d
ds

f(s)

then L1

f0(s)

= tf (t)

Examples of Laplace transform when tn is in
multiplication
1 L

t2eat


multiplication
1 L

t2eat

2 L

t3e3t


multiplication
1 L

t2eat

2 L

t3e3t

3 Lftcos atg

multiplication
1 L

t2eat

2 L

t3e3t

3 Lftcos atg
4 L

tsin2t


multiplication
1 L

t2eat

2 L

t3e3t

3 Lftcos atg
4 L

tsin2t

5 L

te2tcos 3t


multiplication
1 L

t2eat

2 L

t3e3t

3 Lftcos atg
4 L

tsin2t

5 L

te2tcos 3t

6 Lftcos(4t + 3)g

Division by t

1
t

f(t)
=
Z 1
s
f(s) provided the
integral exists.

Examples of Laplace Transform when t is in division
1 L

sin t
t


1 L

sin t
t

2 L

1 cos 2t
t


1 L

sin t
t

2 L

1 cos 2t
t

3 L

eat ebt
t


1 L

sin t
t

2 L

1 cos 2t
t

3 L

eat ebt
t

4 L

cos 2t cos 3t
t


1 L

sin t
t

2 L

1 cos 2t
t

3 L

eat ebt
t

4 L

cos 2t cos 3t
t

5 L

1 et
t


1 L

sin t
t

2 L

1 cos 2t
t

3 L

eat ebt
t

4 L

cos 2t cos 3t
t

5 L

1 et
t

6 L

cos at cos bt
t


1 L

sin t
t

2 L

1 cos 2t
t

3 L

eat ebt
t

4 L

cos 2t cos 3t
t

5 L

1 et
t

6 L

cos at cos bt
t

7 L

etsin t
t


nite integral using Laplace Transform
1 Find
Z 1
0
te2tsin t dt

1 Find
Z 1
0
te2tsin t dt
2 Find
Z 1
0
sin mt
t
dt

1 Find
Z 1
0
te2tsin t dt
2 Find
Z 1
0
sin mt
t
dt
3 Find
Z 1
0
et e3t
t
dt

1 Find
Z 1
0
te2tsin t dt
2 Find
Z 1
0
sin mt
t
dt
3 Find
Z 1
0
et e3t
t
dt
4 Find
Z 1
0
etsin2 t
t
dt

Examples of Inverse Laplace Transform
1 L1

s
(s2 + a2)2


1 L1

s
(s2 + a2)2

2 L1

cot1 s
a


1 L1

s
(s2 + a2)2

2 L1

cot1 s
a

3 L1

log

s + 1
s 1


Convolution
Defn:
Convolution of function Z f(t) and g(t) is denoted f(t) g(t) and
t
de

ned as f(t) g(t) =
0
f(u)g(t u) du

ned as f(t) g(t) =
0
f(u)g(t u) du
Theorem:
Convolution theorem
If L1
f(s)

= f(t) and L1 fg(s)g = g(t) then
L1

f(s)g(s)

=
Z t
0
f(u)g(t u) du

Convolution
Proof: Let (t) =
Z t
0
f(u)g(t u) du

Convolution
Proof: Let (t) =
Z t
0
f(u)g(t u) du
then L((t)) =
Z 1
0
est
Z t
0

dt
f(u)g(t u) du

Convolution
Proof: Let (t) =
Z t
0
f(u)g(t u) du
then L((t)) =
Z 1
0
est
Z t
0

dt
f(u)g(t u) du
=
Z 1
0
Z t
0
estf(u)g(t u) du dt

Convolution
Proof: Let (t) =
Z t
0
f(u)g(t u) du
then L((t)) =
Z 1
0
est
Z t
0

dt
f(u)g(t u) du
=
Z 1
0
Z t
0
estf(u)g(t u) du dt
The region integration for this double integration is entire area
lying between the lines u = 0 and u = t. On changing the order
of integration, we get

Convolution
Proof: Let (t) =
Z t
0
f(u)g(t u) du
then L((t)) =
Z 1
0
est
Z t
0

dt
f(u)g(t u) du
=
Z 1
0
Z t
0
estf(u)g(t u) du dt
of integration, Z we get
1
L((t)) =
0
Z 1
u
estf(u)g(t u) dt du

Convolution
Proof: Let (t) =
Z t
0
f(u)g(t u) du
then L((t)) =
Z 1
0
est
Z t
0

dt
f(u)g(t u) du
=
Z 1
0
Z t
0
estf(u)g(t u) du dt
1
L((t)) =
0
Z 1
u
estf(u)g(t u) dt du
=
Z 1
0
esuf(u)
Z 1
u

du
est+sug(t u) dt

Convolution
Proof: Let (t) =
Z t
0
f(u)g(t u) du
then L((t)) =
Z 1
0
est
Z t
0

dt
f(u)g(t u) du
=
Z 1
0
Z t
0
estf(u)g(t u) du dt
1
L((t)) =
0
Z 1
u
estf(u)g(t u) dt du
=
Z 1
0
esuf(u)
Z 1
u

du
est+sug(t u) dt
=
Z 1
0
esuf(u)
Z 1
u

du
es(tu)g(t u) dt

Convolution
Proof: Let (t) =
Z t
0
f(u)g(t u) du
then L((t)) =
Z 1
0
est
Z t
0

dt
f(u)g(t u) du
=
Z 1
0
Z t
0
estf(u)g(t u) du dt
1
L((t)) =
0
Z 1
u
estf(u)g(t u) dt du
=
Z 1
0
esuf(u)
Z 1
u

du
est+sug(t u) dt
=
Z 1
0
esuf(u)
Z 1
u

du
es(tu)g(t u) dt
=
Z 1
0
esuf(u)
Z 1
u
esvg(v) dv

du, Putting t u = v

Convolution
=
Z 1
0
esuf(u)g(s)du

Convolution
=
Z 1
0
esuf(u)g(s)du
= g(s)
Z 1
0
esuf(u)du

Convolution
=
Z 1
0
esuf(u)g(s)du
= g(s)
Z 1
0
esuf(u)du
) L((t)) = g(s) f(s)

Convolution
=
Z 1
0
esuf(u)g(s)du
= g(s)
Z 1
0
esuf(u)du
) L((t)) = g(s) f(s)

L1
g(s) f(s)

= (t) =
Z t
0
f(u)g(t u) du

Examples of Convolution theorem
Apply convolution theorem to evaluate:
Ex. L1

1
s2(s 1)


Ex. L1

1
s2(s 1)

Ex. L1

s
(s2 + 4)2


Ex. L1

1
s2(s 1)

Ex. L1

s
(s2 + 4)2

Ex. L1

1
(s + a)(s + b)


Ex. L1

1
s2(s 1)

Ex. L1

s
(s2 + 4)2

Ex. L1

1
(s + a)(s + b)

Ex. L1

1
s(s2 + 4)


Application to Dierential Equations
Ex. Use transform method to solve y00 + 3y0 + 2y = et, y(0) = 1 ,
y0(0) = 0

Application to Dierential Equations
Ex. Use transform method to solve y00 + 3y0 + 2y = et, y(0) = 1 ,
y0(0) = 0
Ex. Solve the equation x00 + 2x0 + 5x = et sin t, x(0) = 0 , x0(0) = 1

Laplace transform of Periodic function
If f(t) is sectionally continuous function over an
interval of length p (0 t p) and f(t) is a
periodic function with period p (p 0), that is
f(t + p) = f(t), then its Laplace transform exists
and
1
Lff(t)g =
1 eps
Z p
0
estf(t)dt, (s 0)

Periodic Square Wave
Ex. Find the Laplace transform of the square wave
function of period 2a de

ned as
f(t) =

k if 0 t a
k if a t 2a

Periodic Triangular Wave
Ex. Find the Laplace transform of periodic function
f(t) =

t if 0 t a
2a t if a t 2a
with period 2a

Unit Step function or Heaviside's unit function
The Heaviside step function, or the unit step
function, usually denoted by H (but sometimes u
or ), is a discontinuous function whose value is
zero for negative argument and one for positive
argument.

The Heaviside step function, or the unit step
function, usually denoted by H (but sometimes u
or ), is a discontinuous function whose value is
zero for negative argument and one for positive
argument.
The function is used in the mathematics of control
theory, signal processing, structural mechanics,
etc..

It is denoted by ua(t) or
u(t a) or H(t a) and
is de

ned as H(t a) =
0 t a
1 t a

Laplace Transforms

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