694 FOURIER SERIES

From equation (2),

cn=

(

cos 2πn−jsin 2πn

−j 2 πn

−

cos 2πn−jsin 2πn

(−j 2 πn)^2

)

+

1

(−j 2 πn)^2

However, cos 2πn=1 and sin 2πn=0 for all posi-

tive 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)=

{

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

(cosnπ− 1 )ej

nπx

4

From equation (12),

cn=

1

L

∫L

2

−L 2

f(x)e−j

2 πnx

L dx

=

1

8

[∫ 0

− 4

0e−j

πnx

(^4) dx+
∫ 4
0
10 e−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
Now try the following exercise.
Exercise 248 Further problems on the com-
plex form of a Fourier series

1. Determine the complex Fourier series for the
   function defined by:

f(t)=

{

0, when −π≤t≤ 0

2, when 0 ≤t≤π

The function is periodic outside of this range

of period 2π.

[

f(t)=

∑∞

n=−∞

j

nπ

(cosnπ− 1 )ejnt

= 1 −j

2

π

(

ejt+

1

3

ej^3 t+

1

5

ej^5 t+...

)

+j

2

π

(

e−jt+

1

3

e−j^3 t+

1

5

e−j^5 t+···

)]