The complex or exponential form of a Fourier series 647
Hence, the extended complex form of the Fourier series
shown in equation (14) becomes:
f (x) =
5
2
+
5
π
e
j
π x
2 −
5
3π
e
j
3π x
2 +
5
5π
e
j
5π x
2
−
5
7π
e
j 7π x
2 + · · · +
5
π
e
− j π x
2
−
5
3π
e
− j
3π x
2 +
5
5π
e
− j
5π x
2
−
5
7π
e
− j
7π x
2 + · · ·
=
5
2
+
5
π
e
j
π x
2 + e
− j
π x
2
−
5
3π
e
j 3π x
2 + e
− j 3π x
2
+
5
5π
e
5π x
2 + e
− j
5π x
2
− · · ·
=
5
2
+
5
π
(2)
e
j π x
2 + e
− j π x
2
2
−
5
3π
(2)
e
j
3π x
2 + e
− j
3π x
2
2
+
5
5π
(2)
e
j
5π x
2 + e
− j
5π x
2
2
− · · ·
=
5
2
+
10
π
cos
π x
2
−
10
3π
cos
3π x
2
+
10
5π
cos
5π x
2
− · · ·
(from equation (3))
i.e. f (x) =
5
2
+
10
π
cos
πx
2
−
1
3
cos
3πx
2
+
1
5
cos
5πx
2
−· · ·
which is the same as obtained on page 632.
Hence,
∞
n=−∞
5
πn
sin
nπ
2
e
j πnx
2 is equivalent to
5
2
+
10
π
cos
π x
2
−
1
3
cos
3πx
2
+
1
5
cos
5πx
2
−· · ·
Problem 2. Show that the complex Fourier series
for the function f (t ) =t in the range t = 0 to t = 1,
and of period 1, may be expressed as:
f (t ) =
1
2
+
j
2π
∞
n=−∞
e j 2πnt
n
The saw tooth waveform is shown in Figure 71.2.
Period L 5 1
f (t)
f (t) 5 t
21
0
1
2
t
Figure 71.2
From equation (11), the complex Fourier series is
given by:
f (t ) =
∞
n=−∞
c n e
j
2πnt
L
and when the period, L =1, then:
f (t ) =
∞
n=−∞
c n e
j 2πnt
where, from equation (12),
c n =
1
L
L
2
− L
2
f (t ) e
− j
2πnt
L dt =
1
L
L
0
f (t ) e
− j
2πnt
L dt
and when L =1 and f (t ) =t , then:
c n =
1
1
1
0
t e
− j 2πnt
1 dt =
1
0
t e
− j 2πnt dt
Using integration by parts (see Chapter 43), let u = t ,
from which,
du
dt
= 1, and dt = du, and
let dv = e − j 2πnt , from which,
v =
e
− j 2πnt dt =
e − j 2πnt
− j 2πn