THE COMPLEX OR EXPONENTIAL FORM OF A FOURIER SERIES 691L
Since e^0 =1, thec 0 term can be absorbed into the
summation since it is just another term to be added
to the summation of thecnterm whenn=0. Thus,
f(x)=
∑∞n= 0cnej2 πnx
L +∑∞n= 1c−ne−j2 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= 0cnej2 πnx
L +−∞∑n=− 1cnej2 πnx
LSince the summations now extend from−∞to− 1
and from 0 to+∞, equation (10) may be written as:
f(x)=∑∞n=−∞cnej2 πnx
L (11)Equation (11) is thecomplexorexponential form
of the Fourier series.
74.3 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 2f(x) cos(
2 πnx
L)
dx andbn=
2
L∫L
2−L 2f(x) sin(
2 πnx
L)
dxThus, cn=
⎛⎜
⎝2
L∫L 2
−L 2f(x) cos( 2 πnx
L)
dx−j^2 L∫L 2
−L 2f(x) sin( 2 πnx
L)
dx⎞⎟
⎠2=1
L∫L
2−L 2f(x) cos(
2 πnx
L)
dx−j1
L∫L
2−L 2f(x) sin(
2 πnx
L)
dxFrom equations (3) and (4),cn=1
L∫L
2−L 2f(x)(
ej2 πnx
L +e−j2 πnx
L
2)dx−j1
L∫L
2−L 2f(x)(
ej2 πnx
L −e−j2 πnx
L
2 j)dxfrom which,cn=1
L∫L
2−L 2f(x)(
ej2 πnx
L + e−j2 πnx
L
2)dx−1
L∫ L 2−L 2f(x)(
ej2 πnx
L −e−j2 πnx
L
2)dxi.e. cn=1
L∫L
2−L 2f(x)e−j2 πnx
L dx (12)Care needs to be taken when determiningc 0 .Ifn
appears in the denominator of an expression the
expansion can be invalid whenn=0. In such cir-
cumstances it is usually simpler to evaluatec 0 by
using the relationship:c 0 =a 0 =1
L∫ L
2−L 2f(x)dx (from page 676). (13)Problem 1. Determine the complex Fourier
series for the function defined by:f(x)={
0, when− 2 ≤x≤− 1
5, when− 1 ≤x≤ 1
0, when 1 ≤x≤ 2The function is periodic outside this range of
period 4.This is the same Problem as Problem 2 on page 677
and we can use this to demonstrate that the two forms
of Fourier series are equivalent.
The functionf(x) is shown in Figure 74.1, where
the period,L=4.
From equation (11), the complex Fourier series is
given by:f(x)=∑∞n=−∞cnej2 πnx
L