648 Higher Engineering Mathematics

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

)

From equation (2),

cn=

(

cos2πn−jsin2πn

−j 2 πn

−

cos2πn−jsin2πn

(−j 2 πn)^2

)

+

1

(−j 2 πn)^2

However, cos2πn=1andsin2πn=0 for all positive

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)=

{

0when− 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 πLnx
dx

1
8
[∫ 0
− 4
0e−j
πnx 4
dx+
∫ 4
0
10e−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