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
ej
πx
(^2) −
5
3 π
ej
3 πx
(^2) +
5
5 π
ej
5 πx
2
−
5
7 π
ej
7 πx
(^2) +···+
5
π
e−j
πx
2
−
5
3 π
e−j
3 πx
(^2) +
5
5 π
e−j
5 πx
2
−
5
7 π
e−j
7 πx
(^2) + ···
5
2
- 5
π
(
ej
πx
(^2) +e−j
πx
2
)
−
5
3 π
(
ej
3 πx
(^2) +e−j
3 πx
2
)
5
5 π
(
e
5 πx
(^2) +e−j
5 πx
2
)
− ···
5
2
5
π
( 2 )
(
ej
πx
(^2) +e−j
πx
2
2
)
−
5
3 π
( 2 )
(
ej
3 πx
(^2) +e−j
3 πx
2
2
)
5
5 π
( 2 )
(
ej
5 π 2 x
+e−j
5 π 2 x
2
)
− ···
5
2
10
π
cos
(πx
2
)
−
10
3 π
cos
(
3 πx
2
)
10
5 π
cos
(
5 πx
2
)
− ···
(from equation (3))
i.e.f(x)=
5
2
10
π
[
cos
(πx
2
)
−
1
3
cos
(
3 πx
2
)
1
5
cos
(
5 πx
2
)
−···
]
which is the same as obtained on page 632.
Hence,
∑∞
n=−∞
5
πn
sin
nπ
2
ej
π 2 nx
is equivalent to
5
2
10
π
[
cos
(πx
2
)
−
1
3
cos
(
3 πx
2
)
1
5
cos
(
5 πx
2
)
−···
]
Problem 2. Show that the complex Fourier series
for the functionf(t)=tin the ranget=0tot=1,
and of period 1, may be expressed as:
f(t)=
1
2
j
2 π
∑∞
n=−∞
ej^2 πnt
n
The saw tooth waveform is shown in Figure 71.2.
Period L 51
f(t)
f(t) 5 t
2 10 12t
Figure 71.2
From equation (11), the complex Fourier series is
given by:
f(t)=
∑∞
n=−∞
cnej
2 πnt
L
and when the period,L=1, then:
f(t)=
∑∞
n=−∞
cnej^2 πnt
where, from equation (12),
cn=
1
L
∫ L
2
−L 2
f(t)e−j
2 πnt
L dt=^1
L
∫ L
0
f(t)e−j
2 πnt
L dt
and whenL=1andf(t)=t, then:
cn=
1
1
∫ 1
0
te−j
2 π 1 nt
dt=
∫ 1
0
te−j^2 πntdt
Using integration by parts (see Chapter 43), letu=t,
from which,
du
dt
= 1 ,and dt=du,and
let dv=e−j^2 πnt, from which,
v=
∫
e−j^2 πntdt=
e−j^2 πnt
−j 2 πn