Higher Engineering Mathematics

THE COMPLEX OR EXPONENTIAL FORM OF A FOURIER SERIES 695

L

2. Show that the complex Fourier series for the
   waveform shown in Figure 74.3, that has
   period 2, may be represented by:

f(t)= 2 +

∑∞

n=−∞

(n=0)

j 2

πn

(cosnπ− 1 )ejπnt

f(t)

4

− 10 1 2t

Period L = 2

Figure 74.3

3. Show that the complex Fourier series of
   Problem 2 is equivalent to:

f(t)= 2 +

8

π

(

sinπt+

1

3

sin 3πt

+

1

5

sin 5πt+...

)

4. Determine the exponential form of the
   Fourier series for the function defined by:
   f(t)=e^2 twhen− 1 <t<1 and has period 2.
   [

f(t)=

1

2

∑∞

n=−∞

(

e(^2 −jπn)−e−(2−jπn)

2 −jπn

)

ejπnt

]

74.4 Symmetry relationships

If even or odd symmetry is noted in a function, then
time can be saved in determining coefficients.
The Fourier coefficients present in the complex
Fourier series form are affected by symmetry. Sum-
marising from previous chapters:
Aneven functionis symmetrical about the verti-
cal axis and contains no sine terms, i.e.bn=0.

For even symmetry,

a 0 =

1

L

∫L

0

f(x)dx and

an=

2

L

∫L

0

f(x) cos

(

2 πnx

L

)

dx

=

4

L

∫ L

2

0

f(x) cos

(

2 πnx

L

)

dx

Anodd functionis symmetrical about the origin and

contains no cosine terms,a 0 =an=0.

For odd symmetry,

bn=

2

L

∫L

0

f(x) sin

(

2 πnx

L

)

dx

=

4

L

∫ L

2

0

f(x) sin

(

2 πnx

L

)

dx

From equation (7), page 690,cn=

an−jbn

2

Thus, foreven symmetry,bn=0 and

cn=

an

2

=

2

L

∫ L

2

0

f(x)cos

(

2 πnx

L

)

dx (15)

Forodd symmetry,an=0 and

cn=

−jbn

2

=−j

2

L

∫L 2

0

f(x)sin

(

2 πnx

L

)

dx (16)

For example, in Problem 1 on page 691, the func-

tionf(x) is even, since the waveform is symmetrical

about thef(x) axis. Thus equation (15) could have

been used, giving:

cn=

2

L

∫L

2

0

f(x) cos

(

2 πnx

L

)

dx

=

2

4

∫ 2

0

f(x) cos

(

2 πnx

4

)

dx

=

1

2

{∫ 1

0

5 cos

(πnx

2

)

dx+

∫ 2

1

0dx

}

=

5

2

⎡

⎢

⎣

sin

(πnx

2

)

πn

2

⎤

⎥

⎦

1

0

=

5

2

(

2

πn

)(

sin

nπ

2

− 0

)

=

5

πn

sin

nπ

2

which is the same answer as in Problem 1; how-

ever, a knowledge of even functions has produced

the coefficient more quickly.