TOPIC 2
LAPLACE TRANSFORMS
2.1 Definition of the Laplace transform
L { f (t )} F (s ) f (t )e st dt
s0
Find the Laplace transform of f (t ) 1
Example 1
Solution
e st
1
1
L {1} 1e dt
(0 1)
s
s
s 0
0
st
Example 2
Find the Laplace transform of t.
Solution
e st
L {t} te dt td
s
0
0
st
1 st
te
s
1 st
e dt
s 0
1
1 e st
[0 0]
s
s s 0
1
1
(0 1) 2
2
s
s
Find the Laplace transform of e at .
Example 3
Solution
L {e } e e dt e ( a s )t dt
at
at
st
e ( s a ) t
1
1
(0 1)
sa
sa
( s a) 0
2.2 Properties of the Laplace transform
2.2.1 Linearity
L { f (t ) g (t )} L { f (t )} L ( g (t )} F ( s) G ( s)
Example 1
Find the Laplace transform of
3 7e 3t 6 sin 2t
(a)
(b)
2t 5 cos 3t
Solution
L {3 7e 3t 6 sin 2t} 3 L {1} 7 L {e 3t } 6 L {sin 2t}
(a)
1
2
1
3 7
6 2
s3
s 4
s
3
7
12
2
s s3 s 4
L {2t 5 cos 3t} 2L {t} 5 L {cos 3t}
(b)
2
5s
2
2
s
s 9
2.2.2 First shift theorem
If L { f (t )} F ( s), then L {e at f (t )} F (s a )
Example 1
Use the first shift theorem to find the Laplace transform of
(a)
e 4t t 2
(b)
e 2t t 3
(c)
e 5t sin 3t
(d)
e t cosh 4t
Solution
(a)
Here
a 4 and
f (t ) t 2
Therefore
F ( s)
2
s3
Hence
L {e 3t t 2 } F ( s 4)
2
( s 4) 3
(b)
Here
and f (t ) t 3
a 2
Therefore
F ( s)
6
s4
Hence
L {e 2t t} F ( s 2)
(c)
6
( s 2) 4
Here
and f (t ) sin 3t
a5
Therefore
F ( s)
3
s 9
2
Hence L {e 5t sin 2t} F ( s 5)
(d)
3
( s 5) 2 9
Here
and f (t ) cosh 4t
a 1
Therefore
F ( s)
s
s 16
2
Hence L {e t cosh 4t} F ( s 1)
s 1
( s 1) 2 16
2.3
Laplace transform of derivatives
Let f (t ) be a function of t, and f ' and f " the first and second derivatives of f. The
Laplace trans form of f (t ) is F (s) . Then
L f ' sF ( s ) f (0)
L f " s 2 F (s ) sf (0) f ' (0)
Generally,
L f
Example 1
(n)
F ( s ) s n1 f (0) s n 2 f ' (0) ... f
( 0)
If L x (t ) X (s ) and x(0) 2, x' (0) 1, write the expressions for
(a)
2 x"3x ' x
(b)
x"2 x ' x
Solution
( n 1)
L x ' sX ( s ) x (0) sX ( s ) 2
L x" s 2 X ( s ) sx (0) x' (0) s 2 X (s ) 2s 1
(a)
(b)
2 s
L {2 x"3x ' x} 2 s 2 X (s ) 2 s 1 3sX (s ) 2 X (s )
2
3s 1 X ( s ) 4 s 8
L x"2 x' x ( s 2 X ( s ) 2s 1) 2( sX ( s ) 2) X ( s )
( s 2 2 s 1) X ( s ) 2 s 5
2.4
Inverse Laplace transform
Since L { f (t )} F ( s), we use the symbol L 1 to denote the inverse Laplace transfrom
and we write
f (t ) L 1 F (s )
Therefore, from the Table 3.1, we have
1
n!
s
L 1 1, L 1 n1 t n , L 1 2
cos at ,
2
s
s
s a
and so on. Like L, the operator L 1 can be shown to be a linear operator.
Example 1
Find the inverse Laplace transform of
2
s3
(a)
(b)
16
s3
(c)
7
s4
Solution
(a)
Since
2
2!
21 ,
3
s
s
we have
2
L 1 3 L 1
s
(b)
2! 2
21 t
s
Since
16
2!
8 21 ,
3
s
s
we have
16
2!
L 1 3 8L 1 21 8t 2
s
s
(c)
Since
7
7 3!
31
4
3! s
s
we have
7 1 3! 7 3
7
L 1 4
L 31 t
3!
s
s 6
Example 2
Find the inverse Laplace transform of
(a)
3s
s 4
(b)
4
s 9
(c)
s 3
s2 9
(d)
2s 7
s 2 16
Solution
(a)
3s
s
3 2
s 4
s 22
Since
we have
s
3s
1
L 1 2
3 cos 2t
3L 2
2
s 4
s 2
(b)
4
4
3
2
s 9 3 s 32
Since
we have
4 4 1 3 4
L 1 2
sin 3t
L 2
2
s 9 3
s 3 3
(c)
Since
s3
s
3
2
2
2
2
s 9 s 3
s 32
we have
s
3
s3
1
L 1 2
2
L 2
cos 3t sin 3t
2
s 32
s 9
s 3
(d)
Since
2s 7
s
7
4
2 2
2
2
2
4 s 43
s 16
s 4
we have
2s 7
1
L 1 2
2 L
s
16
s 7 1
L
2
2
s 4 4
4
2
2
s 4
7
2 cos 2t sin 2t
4
Example 3
Use the first shift theorem to find the inverse Laplace transform of the
following functions.
(a)
10
( s 2) 4
(b)
7
( s 3) 5
(c)
s 1
( s 1) 2 4
(d)
s2
( s 1) 2 9
(e)
s3
s 6s 13
(f)
2s 3
s 6s 13
Solution
From the first shift theorem, since L {e at f (t )} F (s a ) , then
L 1 {F ( s a)} e at f (t ).
(a)
Note that
10
10
3!
5
3!
5
F (s 2) .
4
31
31
( s 2)
3! ( s 2)
3 (s 2)
3
This shows that
a 2 and
F ( s)
3!
s 31
7
Then
3!
f (t ) L 1{F ( s )} =L 1 31 t 3
s
Therefore
10 5 1
L 1
L
4
( s 2) 3
3!
31
( s 2)
5
e 2t t 3
3
(b)
Note that
7
7
4!
7
F (s 3)
5
4 1
( s 3)
4! (s 3)
24
This shows that
a 3 and
F ( s)
4!
s 41
Then
4!
f (t ) L 1{F ( s )} =L 1 41 t 4
s
Therefore
7 7
4!
L 1
L 1
5
4 1
( s 3) 24
( s 3)
(c)
7 3t 4
e t
24
Note that
s 1
s 1
F ( s 1)
2
( s 1) 4 (s 1) 2 2 2
This shows that
a 1
and
F ( s)
s
s 22
2
Then
s
f (t ) L 1{F ( s )} =L 1 2
cos 2t
2
s 2
Therefore
s 1
1
L 1
L
2
( s 1) 4
(d)
s 1
e t cos 2t
2
2
(
s
1
)
Note that
s2
(s 1) 3
s 1
3
2
2
2
2
2
( s 1) 9 ( s 1) 3
( s 1) 3
( s 1) 2 32
Since
s2
s 1
3
1
L 1
L 1
L
2
2
2
2
2
( s 1) 9
( s 1) 3
( s 1) 3
we need to find the two inverses separately.
For the first inverse,
s 1
F1 ( s 1)
( s 1) 2 3 2
a 1 ,
and hence
F1 ( s )
s
s 32
2
Then,
s
f1 (t ) L 1{F1 ( s )} =L 1 2
cos 3t
2
s 3
Therefore
s 1
L 1
e t cos 3t
2
2
( s 1) 3
(1)
For the second inverse,
a 1 ,
3
F2 ( s 1)
( s 1) 2 3 2
and hence
F2 ( s )
3
s 32
2
Then
3
f 2 (t ) L 1{F2 ( s )} =L 1 2
sin 3t
2
s 3
Therefore
3
L 1
e t sin 3t
2
2
( s 1) 3
(2)
Finally, combining the two results (1) and (2), we obtain
s2
1
L 1
L
2
(
s
1
)
s 1
L 1
2
2
(
s
1
)
2
2
( s 1) 3
e t cos 3t e t sin 3t
(e)
Completing the square of the denominator,
s3
s3
F ( s 3)
s 6s 13 (s 3) 2 2 2
2
This shows that
a 3 and
F ( s)
s
s 22
2
and hence
s
f (t ) L 1{F ( s )} =L 1 2
cos 2t
2
s 2
s 3
s3
1
L 1 2
e 3t cos 2t
L
2
2
s 6s 13
( s 3) 2
(f)
Since
2s 3
2s 6 3
s3
3
2
2
2
2
2
2
s 6s 13 (s 3) 2
( s 3) 2
2 ( s 3) 2 2 2
2
we have
10
2s 3
1
L 1 2
2 L
s
6
s
13
3 1
s 3
L
2
2
( s 3) 2 2
2
2
( s 3) 2
3
2e 3t cos 2t e 3t sin 2t
2
2.5 Using partial fractions to find the inverse Laplace transform
Example 1
(a)
Find the inverse Laplace transform of the following expressions.
4s 1
s2 s
(b)
3s 4
s 3s 2
2
Solution
(a)
We express the given expression as partial fractions:
4s 1
4s 1
A
B
2
s s s ( s 1) s s 1
4s 1 A( s 1) Bs
Taking s 0 : A 1
Taking s 1 : B 3
Hence
4s 1 1
3
2
s s s s 1
and therefore
4s 1
t
L 1 2
1 3e
s s
(b)
We express the given expression as partial fractions:
3s 4
3s 4
A
B
s 3s 2 (s 1)(s 2) s 1 s 2
2
3s 4 A( s 2) B( s 1)
Taking s 1 :
A 1
11
B2
Taking s 2 :
Hence
3s 4
1
2
( s 1)(s 2) s 1 s 2
and therefore
3s 4
t
2 t
L 1
e 2e
( s 1)(s 2)
2.6 Solving linear constant coefficient differential equations using the Laplace
transform
Example 1
Solve the differential equation
dx
3x 0
dt
x(0) 2
Solution
Taking the Laplace transform of the equation, we have
dx
L 3 x L 0
dt
dx
L 3 L x 0
dt
sX ( s) 2 3 X ( s) 0
X (s )
2
s3
Taking the inverse Laplace transform, we obtain
2
L 1 X ( s ) L 1
s 3
x ( t ) 2e 3 t
12
Example 2
Solve the differential equation
dx
2 x 4e 2 t
dt
x(0) 1
Solution
Taking Laplace transforms and incorporating the initial condition x(0) 1
leads to
sX ( s ) 1 2 X ( s )
X (s)
4
s2
s2
( s 2) 2
Resolving this rational term into partial fractions gives
s2
1
4
2
( s 2)
s 2 (s 2) 2
That is
X (s)
1
4
s 2 (s 2) 2
Taking the inverse Laplace transform, we obtain
4
1
1
L 1 X ( s ) L 1
L
2
s 2
( s 2)
The first inverse Laplace transform on the right-hand side,
L
1
2t
e
s 2
(1)
We use the first shift theorem to obtain the inverse Laplace transform of
the second expression:
L
4
4te 2t
2
( s 2)
(2)
Combining the inverses (1) and (2), the desired solution is
x(t ) (1 4t )e 2t
13
Example 3
Solve
y" y t 2
y (0) 2
y ' (0) 0
Using the Laplace transform.
Solution
Taking Laplace transforms and incorporating the initial conditions leads to
s 2 Y ( s ) 2s Y (s )
2
s3
Rearranging for Y (s),
( s 2 1)Y ( s )
Y ( s)
2
2s
s3
2( s 4 1) 2( s 2 1)(s 2 1)
s3
s3
2( s 2 1)
s3
2 2
s s3
Taking the inverse Laplace transform gives
y (t ) 2 t 2
Example 4
Solve the differential equation
d 2x
dx
5 6 x 2e t
2
dt
dt
subject to the initial conditions x 1 and x' 0 at t 0.
Solution
Taking Laplace transforms
[s 2 X (s ) sx (0) x' (0)] 5[ sX (s ) x(0)] 6 X ( s )
2
s 1
which on incorporating initial conditions and rearrangement gives
( s 2 5 s 6) X ( s )
2
s5
s 1
That is
14
X (s)
2
s5
( s 1)( s 2)( s 3) (s 2)(s 3)
Resolving the rational terms into partial fractions gives
X (s)
1
1
1
s 1 s 2 s 3
Taking inverse transform gives the desired solution
x ( t ) e t e 2 t e 3 t
2.7 Sample problems related to engineering applications using Laplace transforms:
Problem 1
A first order differential equation involving current i in a series R-L circuit is given by:
di
E
5i
and i 0 at time t 0 . Use Laplace transforms to solve for i(t ) when:
dt
2
a) E = 20
b) E = 40e 3t
c) E 50 sin 5t
Answers:
a) i (t ) 2 2e 5t
b) i (t ) 10e 3t 10e 5t
c) i (t )
5 5t 5
5
e cos 5t sin 5t
2
2
2
Problem 2
The equation
d 2 i R di
1
i 0 represents a current i flowing in an electrical circuit
dt 2 L dt LC
containing resistance R, inductance L and capacitance C connected in series. Let R = 200
ohms, L = 0.20 henry and C = 20x10-6 farads. Use Laplace transforms to solve the
equation for i given the boundary conditions that when t = 0, i = 0 and
di
100 .
dt
Answer: 100te 500 t
15
