Periodic Functions and
Fourier Series
Periodic Functions
A function
f  
is periodic
if it is defined for all real

and if there is some positive number,
T
such that
f   T   f   .
f  
0

T
f  
0

T
f  
0

T
Fourier Series
f  
be a periodic function with period
2
The function can be represented by
a trigonometric series as:


n 1
n 1
f    a0   an cos n   bn sin n


n 1
n 1
f    a0   an cos n   bn sin n
What kind of trigonometric (series) functions
are we talking about?
cos  , cos 2 , cos 3  and
sin , sin 2 , sin 3 
0

0
cos 
cos 2
2
cos 3
0

0
sin 
sin 2
2
sin 3
We want to determine the coefficients,
an
and
bn
.
Let us first remember some useful
integrations.

  cos n cos m d

1 
1 
  cosn  m  d   cosn  m  d
2 
2 


cos n cos m d  0
nm



cos n cos m d  

nm

  sin n cos m d

1 
1 
  sinn  m  d   sinn  m  d
2 
2 


sin n cos m d  0

for all values of m.

  sin n sin m d

1 
1 
  cosn  m  d   cosn  m  d
2 
2 


sin n sin m d  0



sin n sin m d  

nm
nm
Determine
a0
Integrate both sides of (1) from

to


  f  d





   a0   an cos n   bn sin n  d

n 1
n 1




  f   d



  a0d     an cos n  d


 n 1






    bn sin n  d

 n 1





  f  d    a d  0  0


0


f d  2a0  0  0

1
a0 
2


f d

You may integrate both sides of (1) from
0

to
2
instead.
f   d
0

2
2
0


 a0   an cos n   bn sin n  d
n 1
n 1




It is alright as long as the integration is
performed over one period.

2
f   d
0
2
2
0
0
  a0d  

2
0

2
0


  an cos n  d
 n 1




  bn sin n  d
 n 1


2
f   d   a0d  0  0
0

2
0
f   d  2a0  0  0
1
a0 
2

2
0
f   d
Determine
Multiply (1) by
an
cos m
and then Integrate both sides from

to


  f   cos m d





   a0   an cos n   bn sin n  cos m d

n 1
n 1



Let us do the integration on the right-hand-side
one term at a time.
First term,


a0 cos m  d  0

Second term,



  n 1
an cos n cos m  d
Second term,


  a

n
cos n cos m d  am
n 1
Third term,


b



n 1
n
sin n cos m  d  0
Therefore,



f   cos m d  am
am 
1


 

f   cos m d
m  1, 2, 
Determine
Multiply (1) by
bn
sin m
and then Integrate both sides from

to


  f   sin m d





  a0   a n cos n   bn sin n  sin m  d

n 1
n 1



Let us do the integration on the right-hand-side
one term at a time.
First term,



a0 sin m  d  0
Second term,


  a

n 1
n
cos n sin m  d
Second term,


  a

n 1
n
cos n sin m  d  0
Third term,


b



n
n 1
sin n sin m d  bm
Therefore,



f   sin m d  bm
bm 
1


 

f   sin md
m  1, 2 ,
The coefficients are:
1
a0 
2


f d


am 
f   cos m d

 
bm 
1
1



 

f   sin m d
m  1, 2, 
m  1, 2, 
We can write n in place of m:
1
a0 
2


f d


an 
f   cos n d

 
bn 
1
1



 

f   sin n d
n  1, 2 ,
n  1, 2 ,
The integrations can be performed from
0
to
2
1
a0 
2

2
0
an 

2
bn 
1
2
1
0


0
instead.
f   d
f   cos n d
f   sin n d
n  1, 2 ,
n  1, 2 ,
Example 1. Find the Fourier series of
the following periodic function.
f  
A
0
-A


2
3
4
5
f   A when 0    
  A when     2
f   2   f  
1 2


a0 
f

d


2 0
2
1  






f

d


f

d





0


2
2
1  


A
d



A
d





0


2
0
an 
1


2
0
f   cos n d
2
1 




A
cos
n

d



A
cos
n

d





0




2
1  sin n 
1
sin n 
 A
  A
0



n 0  
n 
bn 
1

2
0
f   sin n d
2
1 




A
sin
n

d



A
sin
n

d



  0

2
1
cos n 
1  cos n 
  A

A

n  0  
n  
A
 cos n  cos 0  cos 2n  cos n 

n
A
 cos n  cos 0  cos 2n  cos n 
bn 
n
A
1  1  1  1

n
4A

when n is odd
n
A
 cos n  cos 0  cos 2n  cos n 
bn 
n
A
 1  1  1  1

n
 0 when n is even
Therefore, the corresponding Fourier series is
4A 
1
1
1

 sin  sin 3  sin 5  sin 7  
 
3
5
7

the function looks like the following.
1.5
1
0.5
f()
0
0.5
1
1.5

Types of periodic waveforms
• Amplitude varies in a repeating manner amplitude modulation
• Frequency varies in a repeating manner frequency modulation
• The shape of the waveform varies in a
repeating manner - nonsinusoidal periodic
wave
Amplitude modulation (AM)
Carrier frequency
plus side bands
Frequency modulation (FM)
Modulation
determines the
number of side
bands
Periodic nonsinusoidal signals
Harmonic series
Harmonic series
• Harmonic frequencies
are integer multiples of
the fundamental frequency,
i.e. w, 2w, 3w, 4w …
• Dirichlet’s rule states that the energy in
higher harmonics falls off exponentially
with the frequency of the harmonic
• Note, however, that some animals alter the
amplitude of harmonics by selective
filtering during sound production
Compound signals
• Nonsinusoidal modulation of a sine wave
• Sinusoidal modulation of a nonsinusoidal
carrier wave
• Nonsinusoidal modulation of a
nonsinusoidal carrier wave
Pulsed sine wave (frog or insect)
Fourier analysis of aperiodic signals
• Most natural signals have a short, not
infinite, duration
• The more aperiodic a signal is, the more
frequency components are needed to
describe the signal with a Fourier series
• In the limit, an infinitely short signal has
constant amplitude at all frequencies, a delta
pulse
Finite sounds and Fourier ‘lobes’