Higher Engineering Mathematics, Sixth Edition

The complex or exponential form of a Fourier series 645

Rearranging gives:

f(x)=a 0 +

∑∞

n= 1

[(

an−jbn

2

)

ej

2 πLnx

+

(

an+jbn

2

)

e−j

2 πLnx

]

(5)

The Fourier coefficientsa 0 ,anandbnmay be replaced
by complex coefficientsc 0 ,cnandc−nsuch that

c 0 =a 0 (6)

cn=

an−jbn

2

(7)

and c−n=

an+jbn

2

(8)

wherec−nrepresents the complex conjugate ofcn(see
page 216).
Thus, equation (5) may be rewritten as:

f(x)=c 0 +

∑∞

n= 1

cnej

2 πnx

L +

∑∞

n= 1

c−ne−j

2 πnx

L (9)

Since e^0 =1, thec 0 term can be absorbed into the sum-
mation since it is just another term to be added to the
summation of thecnterm whenn=0. Thus,

f(x)=

∑∞

n= 0

cnej

2 πnx

L +

∑∞

n= 1

c−ne−j

2 πnx

L (10)

Thec−nterm may be rewritten by changing the limits
n=1ton=∞ton=−1ton=−∞.Sincenhas been
made negative, the exponential term becomes ej

2 πnx
L
andc−nbecomescn. Thus,

f(x)=

∑∞

n= 0

cnej

2 πLnx

+

−∞∑

n=− 1

cnej

2 πLnx

Since the summations now extend from−∞to−1and
from 0 to+∞, equation (10) may be written as:

f(x)=

∑∞

n=−∞

cnej

2 πnx

L (11)

Equation (11) is thecomplexorexponential formof
the Fourier series.

71.3 The complex coefficients

From equation (7), the complex coefficientcnwas

defined as:cn=

an−jbn

2

However,anandbnare defined (from page 630) by:

an=

2

L

∫ L

2

−L 2

f(x)cos

(

2 πnx

L

)

dx and

bn=

2

L

∫ L 2

−L 2

f(x)sin

(

2 πnx

L

)

dx

Thus, cn=

⎛

⎜

⎝

2

L

∫L 2

−L 2 f(x)cos

( 2 πnx

L

)

dx

−j^2 L

∫L 2

−L 2

f(x)sin

( 2 πnx

L

)

dx

⎞

⎟

⎠

2

=

1

L

∫ L 2

−L 2

f(x)cos

(

2 πnx

L

)

dx

−j

1

L

∫ L

2

−L 2

f(x)sin

(

2 πnx

L

)

dx

From equations (3) and (4),

cn=

1

L

∫ L

2

−L 2

f(x)

(

ej

2 πLnx

+e−j

2 πLnx

2

)

dx

−j

1

L

∫ L

2

−L 2

f(x)

(

ej

2 πnx

L −e−j

2 πnx

L

2 j

)

dx

from which,

cn=

1

L

∫ L

2

−L 2

f(x)

(

ej

2 πnx

L +e−j

2 πnx

L

2

)

dx

−

1

L

∫ L

2

−L 2

f(x)

(

ej

2 πnx

L −e−j

2 πnx

L

2

)

dx

i.e. cn=

1

L

∫ L

2

−L 2

f(x)e−j

2 πLnx

dx (12)

Careneeds to betaken when determiningc 0 .Ifnappears

in the denominator of an expression the expansion can

beinvalidwhenn=0.In such circumstances it is usually

simpler to evaluatec 0 by using the relationship:

c 0 =a 0 =

1

L

∫ L 2

−L 2

f(x)dx (from page 630). (13)