Fourier and Laplace
363
ANALYTICAL Solution
Taking the LT of the given equation
2
0
3
30
sY s
y
Y s
s
( )
( )
( )
Ϫ
 
  ϩ
ϭ
Y s s
s
( )[
]
2
3
30
ϩ ϭ
then
Y s s
s
( )(
)
2
3
30
ϩ ϭ
Y s
s s
s s
A
s
B
s
( )
(
)
(
. )
.
ϭ
ϩ
ϭ
ϩ
ϭ ϩ ϩ
30
2
3
15
1 5
1 5
by partial fraction expansion
Y s
s
s
( )
.
ϭ
ϩ
10
10
1 5
Ϫ
then
y t
Y s
( )
[ ( )]
ϭ £
Ϫ1
and
y t
u t
e
u t
t
( )
( )
( )
.
ϭ 10
10
1 5
Ϫ
Ϫ
e. Example (#5)
Let
dy t
dt
y t
( )
( )
ϩ
ϭ
4
0
with y(0) = 10.
ANALYTICAL Solution
Taking the LT of the preceding equation
sY s
Y s
( )
( )
Ϫ ϩ
10 4
0
ϭ
Y s s
Y s
s
( )(
)
,
( )
ϩ ϭ
ϭ ϩ
4
10
10
4
and
y t
Y s
( )
[ ( )]
ϭ £
Ϫ1
then
y t
e u t
t
( )
( )
ϭ 10
4
Ϫ
R.4.114 The MATLAB function F = laplace(f) returns the LT of ƒ, denoted by f(t) → F(s),
where f is a symbolic object with independent variable t.
CRC_47760_CH004.indd 363
CRC_47760_CH004.indd 363
7/28/2008 12:26:03 PM
7/28/2008 12:26:03 PM