The complex or exponential form of a Fourier series 649
Now try the following exercise
Exercise 236 Further problems on the
complexform of a Fourier series- Determine the complex Fourier series for the
function defined by:
f(t)={
0 , when−π≤t≤ 0
2 , when 0≤t≤π
The function is periodic outside of this range
of period 2π.
[
f(t)=∑∞n=−∞j
nπ(cosnπ− 1 )ejnt= 1 −j2
π(
ejt+1
3ej^3 t+1
5ej^5 t+···)+j2
π(
e−jt+1
3e−j^3 t+1
5e−j^5 t+···)]- Show that the complex Fourier series for the
waveform shown in Figure 71.3, that has
period 2, may be represented by:
f(t)= 2 +∑∞n=−∞
(n= 0 )j 2
πn(cosnπ− 1 )ejπnt2104(^12) t
Period L 52
f(t)
Figure 71.3
- Show that the complex Fourier series of
Problem 2 is equivalent to:
f(t)= 2 +8
π(
sinπt+1
3sin3πt+1
5sin5πt+···)- Determine the exponential form of the Fourier
series for the function defined by: f(t)=e^2 t
when− 1 <t<1 and has period 2.
[
f(t)=1
2∑∞n=−∞(
e(^2 −jπn)−e−(^2 −jπn)
2 −jπn)
ejπnt]71.4 Symmetry relationships
If even or oddsymmetry is noted ina function,then time
can be saved in determining coefficients.
The Fourier coefficients present in the complex Fourier
series form are affected by symmetry. Summarising
from previous chapters:
Aneven functionis symmetrical about the vertical axis
and contains no sine terms, i.e.bn=0.
For even symmetry,a 0 =1
L∫L0f(x)dx andan=
2
L∫L0f(x)cos(
2 πnx
L)
dx=4
L∫ L
2
0f(x)cos(
2 πnx
L)
dxAnodd functionis symmetrical about the origin and
contains no cosine terms,a 0 =an=0.
For odd symmetry,bn=2
L∫ L0f(x)sin(
2 πnx
L)
dx=4
L∫ L
2
0f(x)sin(
2 πnx
L)
dxFrom equation (7), page 645,cn=an−jbn
2
Thus, foreven symmetry,bn=0andcn=an
2=2
L∫ L
2
0f(x)cos(
2 πnx
L)
dx (15)Forodd symmetry,an=0andcn=−jbn
2=−j2
L∫ L
2
0f(x)sin(
2 πnx
L)
dx (16)For example, in Problem 1 on page 646, the function
f(x) is even, since the waveform is symmetrical about