The complex or exponential form of a Fourier series 645
Rearranging gives:
f (x) = a 0 +
∞
n=1
a n − j b n
2
e
j
2πnx
L
+
a n + j b n
2
e
− j
2πnx
L
(5)
The Fourier coefficients a 0 , a n and b n may be replaced
by complex coefficients c 0 , c n and c −n such that
c 0 = a 0
(6)
c n =
a n − j b n
2
(7)
and c −n =
a n + j b n
2
(8)
where c −n represents the complex conjugate of c n (see
page 216).
Thus, equation (5) may be rewritten as:
f (x) = c 0 +
∞
n=1
c n e
j
2πnx
L +
∞
n=1
c −n e
− j
2πnx
L
(9)
Since e
0
= 1, the c 0 term can be absorbed into the summation since it is just another term to be added to the
summation of the c n term when n =0. Thus,
f (x) =
∞
n=0
c n e
j
2πnx
L +
∞
n=1
c −n e
− j
2πnx
L
(10)
The c −n term may be rewritten by changing the limits
n =1 to n =∞ to n =−1 to n =−∞. Since n has been
made negative, the exponential term becomes e
j
2πnx
L
and c −n becomes c n . Thus,
f (x) =
∞
n=0
c n e
j
2πnx
L +
−∞
n=−1
c n e
j
2πnx
L
Since the summations now extend from −∞ to −1 and
from 0 to +∞, equation (10) may be written as:
f (x) =
∞
n=−∞
c n e
j
2π nx
L
(11)
Equation (11) is the complex or exponential form of
the Fourier series.
71.3 The complex coefficients
From equation (7), the complex coefficient c n was
defined as: c n =
a n − j b n
2
However, a n and b n are defined (from page 630) by:
a n =
2
L
L
2
−
L
2
f (x) cos
2πnx
L
dx and
b n =
2
L
L
2
−
L
2
f (x) sin
2πnx
L
dx
Thus, c n =
⎛
⎜
⎝
2
L
L
2
−
L
2
f (x) cos
2πnx
L
dx
− j
2
L
L
2
−
L
2
f (x) sin
2πnx
L
dx
⎞
⎟
⎠
2
=
1
L
L
2
−
L
2
f (x) cos
2πnx
L
dx
− j
1
L
L
2
−
L
2
f (x) sin
2πnx
L
dx
From equations (3) and (4),
c n =
1
L
L
2
− L
2
f (x)
e
j
2πnx
L + e
− j
2πnx
L
2
dx
− j
1
L
L
2
−
L
2
f (x)
e
j
2πnx
L − e
− j
2πnx
L
2 j
dx
from which,
c n =
1
L
L
2
−
L
2
f (x)
e
j 2πnx
L + e
− j 2πnx
L
2
dx
−
1
L
L
2
−
L
2
f (x)
e
j
2πnx
L − e
− j
2πnx
L
2
dx
i.e.
c n =
1
L
L
2
−
L
2
f (x)e
−j 2π nx
L dx
(12)
Care needs to be taken when determining c 0 . If n appears
in the denominator of an expression the expansion can
be invalid when n =0. In such circumstances it is usually
simpler to evaluate c 0 by using the relationship:
c 0 = a 0 =
1
L
L
2
−
L
2
f (x)dx (from page 630). (13)
Rearranging gives:
f (x) = a 0 +
∞
n=1
a n − j b n
2
e
j
2πnx
L
+
a n + j b n
2
e
− j
2πnx
L
(5)
The Fourier coefficients a 0 , a n and b n may be replaced
by complex coefficients c 0 , c n and c −n such that
c 0 = a 0
(6)
c n =
a n − j b n
2
(7)
and c −n =
a n + j b n
2
(8)
where c −n represents the complex conjugate of c n (see
page 216).
Thus, equation (5) may be rewritten as:
f (x) = c 0 +
∞
n=1
c n e
j
2πnx
L +
∞
n=1
c −n e
− j
2πnx
L
(9)
Since e
0
= 1, the c 0 term can be absorbed into the summation since it is just another term to be added to the
summation of the c n term when n =0. Thus,
f (x) =
∞
n=0
c n e
j
2πnx
L +
∞
n=1
c −n e
− j
2πnx
L
(10)
The c −n term may be rewritten by changing the limits
n =1 to n =∞ to n =−1 to n =−∞. Since n has been
made negative, the exponential term becomes e
j
2πnx
L
and c −n becomes c n . Thus,
f (x) =
∞
n=0
c n e
j
2πnx
L +
−∞
n=−1
c n e
j
2πnx
L
Since the summations now extend from −∞ to −1 and
from 0 to +∞, equation (10) may be written as:
f (x) =
∞
n=−∞
c n e
j
2π nx
L
(11)
Equation (11) is the complex or exponential form of
the Fourier series.
71.3 The complex coefficients
From equation (7), the complex coefficient c n was
defined as: c n =
a n − j b n
2
However, a n and b n are defined (from page 630) by:
a n =
2
L
L
2
−
L
2
f (x) cos
2πnx
L
dx and
b n =
2
L
L
2
−
L
2
f (x) sin
2πnx
L
dx
Thus, c n =
⎛
⎜
⎝
2
L
L
2
−
L
2
f (x) cos
2πnx
L
dx
− j
2
L
L
2
−
L
2
f (x) sin
2πnx
L
dx
⎞
⎟
⎠
2
=
1
L
L
2
−
L
2
f (x) cos
2πnx
L
dx
− j
1
L
L
2
−
L
2
f (x) sin
2πnx
L
dx
From equations (3) and (4),
c n =
1
L
L
2
− L
2
f (x)
e
j
2πnx
L + e
− j
2πnx
L
2
dx
− j
1
L
L
2
−
L
2
f (x)
e
j
2πnx
L − e
− j
2πnx
L
2 j
dx
from which,
c n =
1
L
L
2
−
L
2
f (x)
e
j 2πnx
L + e
− j 2πnx
L
2
dx
−
1
L
L
2
−
L
2
f (x)
e
j
2πnx
L − e
− j
2πnx
L
2
dx
i.e.
c n =
1
L
L
2
−
L
2
f (x)e
−j 2π nx
L dx
(12)
Care needs to be taken when determining c 0 . If n appears
in the denominator of an expression the expansion can
be invalid when n =0. In such circumstances it is usually
simpler to evaluate c 0 by using the relationship:
c 0 = a 0 =
1
L
L
2
−
L
2
f (x)dx (from page 630). (13)