648 Higher Engineering Mathematics
Hence, c n =
1
0
t e
− j 2πnt
= uv −
v du
=
t
e − j 2πnt
− j 2πn
1
0
−
1
0
e − j 2πnt
− j 2πn
dt
=
t
e
− j 2πnt
− j 2πn
−
e
− j 2πnt
(− j 2πn)
2
1
0
=
e − j 2πn
− j 2πn
−
e − j 2πn
(− j 2πn)
2
−
0 −
e 0
(− j 2πn)
2
From equation (2),
c n =
cos 2πn − j sin 2πn
− j 2πn
−
cos 2πn − j sin 2πn
(− j 2πn)
2
+
1
(− j 2πn) 2
However, cos 2πn = 1 and sin2πn =0 for all positive
and negative integer values of n.
Thus, c n =
1
− j 2πn
−
1
(− j 2πn)
2
+
1
(− j 2πn)
2
=
1
− j 2πn
=
1( j )
− j 2πn( j )
i.e. c n =
j
2π n
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
t dt
=
t 2
2
1
0
=
1
2
− 0
=
1
2
Hence, the complex Fourier series is given by:
f (t) =
∞
n=−∞
c n e
j
2πnt
L from equation (11)
i.e. f (t) =
1
2
+
∞
n=−∞
j
2πn
e
j 2πnt
=
1
2
+
j
2π
∞
n=−∞
e
j2πnt
n
Problem 3. Show that the exponential form of the
Fourier series for the waveform described by:
f (x) =
0 when −4 ≤ x ≤ 0
10 when 0 ≤ x ≤ 4
and has a period of 8, is given by:
f (x) =
∞
n=−∞
5 j
nπ
(cos nπ − 1) e
j
nπ x
4 .
From equation (12),
c n =
1
L
L
2
−
L
2
f (x) e
− j 2πnx
L dx
=
1
8
0
−4
0 e
− j πnx
4 dx +
4
0
10 e
− j πnx
4 dx
=
10
8
e
− j
πnt
4
− j
πn
4
4
0
=
10
8
4
− j πn
e
− j πn
− 1
=
5 j
− j 2 πn
e
− j πn
− 1
=
5 j
πn
e
− j πn
− 1
From equation (2), e − j θ = cos θ − j sin θ, thus
e − j πn = cos πn − j sin πn = cos πn for all integer
values of n. Hence,
c n =
5 j
πn
e
− j πn
− 1
=
5 j
πn
(cos nπ − 1)
From equation (11), the exponential Fourier series is
given by:
f (x) =
∞
n=−∞
c n e
j
2πnx
L
=
∞
n=−∞
5j
nπ
(cos nπ − 1) e
j nπx
4