648 Higher Engineering Mathematics
Hence,cn=
∫ 1
0
te−j^2 πnt=uv−
∫
vdu
=
[
t
e−j^2 πnt
−j 2 πn
] 1
0
−
∫ 1
0
e−j^2 πnt
−j 2 πn
dt
=
[
t
e−j^2 πnt
−j 2 πn
−
e−j^2 πnt
(−j 2 πn)^2
] 1
0
=
(
e−j^2 πn
−j 2 πn
−
e−j^2 πn
(−j 2 πn)^2
)
−
(
0 −
e^0
(−j 2 πn)^2
)
From equation (2),
cn=
(
cos2πn−jsin2πn
−j 2 πn
−
cos2πn−jsin2πn
(−j 2 πn)^2
)
+
1
(−j 2 πn)^2
However, cos2πn=1andsin2πn=0 for all positive
and negative integer values ofn.
Thus,cn=
1
−j 2 πn
−
1
(−j 2 πn)^2
+
1
(−j 2 πn)^2
=
1
−j 2 πn
=
1 (j)
−j 2 πn(j)
i.e. cn=
j
2 πn
From equation (13),
c 0 =a 0 =
1
L
∫ L
2
−L 2
f(t)dt
=
1
L
∫L
0
f(t)dt=
1
1
∫ 1
0
tdt
=
[
t^2
2
] 1
0
=
[
1
2
− 0
]
=
1
2
Hence, the complex Fourier series is given by:
f(t)=
∑∞
n=−∞
cnej
2 πnt
L from equation (11)
i.e. f(t)=
1
2
+
∑∞
n=−∞
j
2 πn
ej^2 πnt
=
1
2
+
j
2 π
∑∞
n=−∞
ej^2 πnt
n
Problem 3. Show that the exponential form of the
Fourier series for the waveform described by:
f(x)=
{
0when− 4 ≤x≤ 0
10 when 0≤x≤ 4
and has a period of 8, is given by:
f(x)=
∑∞
n=−∞
5 j
nπ
(cosnπ− 1 )ej
nπx
(^4).
From equation (12),
cn=
1
L
∫ L
2
−L 2
f(x)e−j
2 πLnx
dx
1
8
[∫ 0
− 4
0e−j
πnx 4
dx+
∫ 4
0
10e−j
πnx 4
dx
]
10
8
[
e−j
πnt
4
−jπ 4 n
] 4
0
10
8
(
4
−jπn
)[
e−jπn− 1
]
5 j
−j^2 πn
(
e−jπn− 1
)
5 j
πn
(
e−jπn− 1
)
From equation (2), e−jθ=cosθ−jsinθ, thus
e−jπn=cosπn−jsinπn=cosπn for all integer
values ofn. Hence,
cn=
5 j
πn
(
e−jπn− 1
)
5 j
πn
(cosnπ− 1 )
From equation (11), the exponential Fourier series is
given by:
f(x)=
∑∞
n=−∞
cnej
2 πnx
L
∑∞
n=−∞
5 j
nπ
(cosnπ−1)ej
nπx
4