THE COMPLEX OR EXPONENTIAL FORM OF A FOURIER SERIES 693L
=5
2+5
π(2)(
ejπx(^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 πx
(^2) +e−j
5 πx
2
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 678.
Hence,
∑∞
n=−∞
5
πn
sin
nπ
2
ej
πnx
(^2) 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= 0
tot=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 74.2.
From equation (11), the complex Fourier series is
given by:
f(t)=
∑∞
n=−∞
cnej
2 πnt
L
f(t) f(t) =t
−10 1 2t
Period L = 1
Figure 74.2
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=1 andf(t)=t, then:
cn=
1
1
∫ 1
0
te−j
2 πnt
(^1) 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
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
)