f ( x)  a0  a1 cos x  a2 cos 2x  a3 cos 3x  ......  b1 sin x  b2 sin 2x  b3 sin 3x  .....

f ( x)  a0  (an cos nx  bn sin nx)
n1
Fourier Series
2.06715
2.5
0.900316 1
2
0.8
1.5
c( k)
0.6
0
f( x)
1
0.4
0.5
0.2
0 0
5
0638351
0
5
10
15
0
1
1
2
3
4
5
6
k
7
8
9
10
11 12
11
0.5
-3.14159
x
12.5584
1
INFINITE SERIES
A series is a sum of a sequence of terms. An infinite series is an
ordered set of terms, combined together by the addition operators (+
or -). The term infinite series is used to emphasize the fact that series
contains an infinite number of terms. The series is generally written as

a
n1
n
 a1  a2  a3  ......
The term an is called the general term. It is an expression that contains n.
For example if
an 
1
2n 1
A series can be expressed in summation form using general term in a
summation symbol  or in the expanded form using individual terms:


1
1 1 1 1
an  
     ......

3 5 7 9
n1
n1 2n  1
2
There are different types of series. For instance, if the difference an+1 - an
between successive terms of a series is a constant, the series is called an
arithmetic series. An example of an arithmetic series is:

(4n 1)  3  7 1115  ......
n1
an1
A series for which the ratio of each two consecutive termsa
n
is a
constant is called a geometric series. An example of a geometric series is

1
1 1 1

1

   ......

n
2 4 8
n 0 2
A series may converge to a definite value in which case it is called a
convergent series. If it does not converge, it is called a divergent series.
3
The terms of series do not have to be constants. They can be functions of
a variable, usually denoted by x. For example, there are power series that
are expressed as:
n
n
2
3
4
a
x

a

a
x

a
x

a
x

a
x
 ......
n
0
1
2
3
4
n 0
In the above expression, a0, a1,….. are called coefficients. Many functions
can be approximated by power series. For example
x3 x5
sin x  x    .....
3! 5!
x2 x4
cos x  x    .....
2! 4!
2
3
4
x
x
x
ex  1  x     .......
2! 3! 4!
4
One type of series contains terms that are sinusoidal trigonometric
functions. This type is called Fourier series
f ( x)  a0  c1 sin ( x  1)  c2 sin (2x  2 )  c3 sin (3x  3 )  .....
Fourier series are often shown decomposed into sine and cosine parts:
f ( x)  a0  a1 cos x  a2 cos 2x  a3 cos 3x  ......  b1 sin x  b2 sin 2x  b3 sin 3x  .....

f ( x)  a0  (an cos nx  bn sin nx)
n1
5
Fourier Series Analysis
The name of Fourier series comes from the French mathematician Joseph
Fourier (1768-1830) who used the series to solve problems in
thermodynamics.
Later the series were used by various researchers to solve mechanical
vibrations and problems with electrical signals and electromagnetic
radiation.
The Fourier series analysis is a mathematical tool used for analyzing
periodic functions.
A periodic function is a function for which f(t) = f(t+T)
f(t)
110
100
g( t )
50
T
0
0.001 0
-0.000333
0.001
0.002
0.003
0.004
t
0.005
0.006
0.007
0.008 0.009
0.008997
t
6
Fourier Series Analysis
Fourier series analysis decomposes any periodic function into an infinite
weighted sum of much simpler sine and cosine component functions.
The problems then can be solved for the individual simple sine and cosine
functions, and then the individual solutions can be recombined to obtain the
solution to the original problem or at least an approximation of the solution to
a desired accuracy.
The study of Fourier series is known as harmonic analysis.
For example a square wave can be decomposed into in infinite number of
sinusoidal functions that, when all added together, approximate the square
wave.
Let us consider a function given by
1
1
1


f (t )  10.00 sin t  sin 3t  sin 5t  sin 7t  .....
3
5
7


or

1
f (t )  10.00 
sin nt
n1,3,5... n
7
Adding first two terms (pink and blue) gives a graph shown in
red.
10
5
A1( t )
A3( t ) 0
C3( t )
5
10
0
2
4
6
8
10
t
8
Adding the third term gives:
10
5
A1( t )
A3( t )
0
C5( t )
A5( t )
5
10
0
2
4
6
8
10
t
f (t )  10.00sin t  3.33sin 3t  2.00sin 5t
9
Adding first four term results in a waveform that begins to resemble
a square wave. Adding large number of terms will give a square
wave. 10
A1( t )
5
A3( t )
C( t )
0
A5( t )
A7( t )
5
10
0
2
4
6
8
10
t
10
Finding the Terms of Fourier Series
We will start by writing a periodic function f(x) with a period of 2π as
a sum of sinusoidal components:
f ( x)  a0  a1 cos x  a2 cos 2x  a3 cos 3x  ......  b1 sin x  b2 sin 2x  b3 sin 3x  .....

f ( x)  a0  (an cos nx  bn sin nx)
(1)
n1
In the above equation, the non-sinusoidal term a0 is a constant that
will be equal to zero if the function f(x) oscillates around the x axis.
If f(x) oscillates around any other value, a0 will be different from
zero. The constant a0 is called an offset or average value.
Terms a1cos x and b1sin x are called the fundamental components.
The remaining terms are called harmonic components or just
harmonics; terms with argument 2x are second harmonics, terms
with argument 3x are third harmonics and so on.
11
2nd harmonic
components
Average
value
f ( x)  a0  a1 cos x  a2 cos 2x  a3 cos 3x  .....
 b1 sin x  b2 sin 2x  b3 sin 3x  .....
1st harmonic
components
3rd harmonic
components
12
Example 1
4
𝑎) 𝑡ℎ𝑖𝑟𝑑 ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 =
𝑠𝑖𝑛3𝜃
3𝜋
𝑐𝑜𝑠2𝑥 𝑠𝑖𝑛2𝑥
𝑏) 𝑠𝑒𝑐𝑜𝑛𝑑 ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 =
+
2
3
13
Example 2
4
1
1
𝑎) 𝑓 𝑥 =
𝑠𝑖𝑛𝑥 + 𝑠𝑖𝑛3𝑥 + 𝑠𝑖𝑛5𝑥 + … .
𝜋
3
5
4
=
𝜋
∞
𝑛=1,3,5,…
1
sin(𝑛𝑥)
𝑛
1
−2
1
1
𝑑) 𝑓 𝑦 = + ( )(𝑐𝑜𝑠𝑦 + 2 𝑐𝑜𝑠3𝑦 + 2 𝑐𝑜𝑠5𝑦 + ⋯ )
2
𝜋
3
5
1 −2
= +
2
𝜋
∞
𝑛=1,3,5,…
1
𝑐𝑜𝑠𝑛𝑦
2
𝑛
14
Finding the Coefficients
Since we know the function f(x), but we don’t know the offset and the
constants (coefficients) ai and bi , our aim is to find formulas for the
coefficients ai and bi in terms of f(x).
If we integrate both sides of equation (1) inside the limits of –π and
+π, we get:

f ( x)  a0   (an cos nx  bn sin nx)
(1)
n1




 f ( x)dx   a dx   (a cos nx  b sin nx)dx
0



 f ( x)dx  a x
0



n
  n1
n




n1

n1

  an  cos nx dx   bn  sin nx dx
15
Now we will find the value of the integrals inside the summation:

1
1

 cos nxdx  n sin nx   n [sin n  sin(n)]  0
because n is
an integer.

1
1

sin
nxdx


cos
nx


[cos n  cos(n)]  0


n
n

Therefore,
 f ( x)dx  2a
0


Solving for a0 gives
1
a0 
f ( x)dx

2  
16
SUMMARY
A periodic function f(x) can be replaced by a sum of sine and
cosine components called Fourier series
f ( x)  a0  a1 cos x  a2 cos 2x  a3 cos 3x  ......  b1 sin x  b2 sin 2x  b2 sin 2x  .....

f ( x)  a0  (an cos nx  bn sin nx)
n1
The coefficients of the series are given by

1
a0 
f ( x)dx

2  
1
an   f ( x)cos nx dx
 
1
bn   f ( x)sin nx dx
 
17
Example 3
Find the average value for the following waveforms:
𝑎) 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒, 𝑎0 = 0
𝑏) 𝑓 𝜃 = 0
𝑓 𝜃 = 𝑠𝑖𝑛𝜃
−𝜋 <𝜃 <0
0<𝜃<𝜋
18
1
𝑎0 =
2𝜋
𝜋
−𝜋
1
𝑓 𝜃 𝑑𝜃 =
2𝜋
𝜋
0
1
𝑓 𝜃 𝑑𝜃 =
[−𝑐𝑜𝑠𝜃]𝜋0
2𝜋
−1
−1
1
𝑎0 =
𝑐𝑜𝑠𝜋 − 𝑐𝑜𝑠0 =
−1 − 1 =
2𝜋
2𝜋
𝜋
19
Example 4
Find coefficients a1 and b1 for the above waveform:
1
𝑓 𝜃 = 𝜃−1
𝜋
1
𝑎1 =
𝜋
2𝜋
0
0 < 𝜃 < 2𝜋
1
1
𝜃 − 1 𝑐𝑜𝑠𝜃𝑑𝜃 = 2
𝜋
𝜋
2𝜋
0
1
1
2𝜋
𝑎1 = 2 [𝑐𝑜𝑠𝜃 + 𝜃𝑠𝑖𝑛𝜃]0 − 𝑠𝑖𝑛𝜃
𝜋
𝜋
1
𝜃𝑐𝑜𝑠𝜃𝑑𝜃 −
𝜋
2𝜋
𝑐𝑜𝑠𝜃𝑑𝜃
0
2𝜋
0
𝑎1
1
= 2 𝑐𝑜𝑠2𝜋 + 2𝜋𝑠𝑖𝑛2𝜋 − 𝑐𝑜𝑠0 − 0𝑠𝑖𝑛0
𝜋
1
− 𝑠𝑖𝑛2𝜋 − 𝑠𝑖𝑛0 = 0
𝜋
20
Example 4
Find coefficients a1 and b1 for the first waveform:
1
1) 𝑓 𝜃 = 𝜃 − 1
𝜋
1
𝑏1 = 2
𝜋
2𝜋
0
0 < 𝜃 < 2𝜋
1
1
𝜃 − 1 𝑠𝑖𝑛𝜃 = 2
𝜋
𝜋
2𝜋
0
1
𝜃𝑠𝑖𝑛𝜃𝑑𝜃 −
𝜋
1
1
2𝜋
𝑏1 = 2 [−𝜃𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃]0 − 𝑐𝑜𝑠𝜃
𝜋
𝜋
2𝜋
𝑠𝑖𝑛𝜃𝑑𝜃
0
2𝜋
0
1
1
𝑏1 = 2 −2𝜋𝑐𝑜𝑠2𝜋 + 𝑠𝑖𝑛2𝜋 − 0 − 𝑠𝑖𝑛0 − 𝑐𝑜𝑠2𝜋 − 𝑐𝑜𝑠0
𝜋
𝜋
−2
=
𝜋
21
Polar Form of Fourier Series
In order to easily compare relative magnitudes and phase shifts of
the individual harmonics, we want to get Fourier series in a polar
form.
f ( x)  a0  c1 sin ( x  1)  c2 sin (2x  2 )  c3 sin (3x  3 )  .....

f ( x)  a0   cn sin(nx  n )
n1
22
Polar Form of Fourier Series
cn sin (nx  n )  cn sin n cos nx  cn cosn sin nx
an
an  cn sinn
bn  cn cosn
bn
an
an +
bn -
an +
bn +
cn  an2  bn2
OR use
rectangular to
polar conversion
on your calculator
bn
an bn -
an bn +
23
Example:
Find the polar form of function
Will be solved in class.
24
Harmonic Spectrum
Harmonic spectrum is a graph showing the amplitude cn of the polar
representation of Fourier series as a function of the harmonic order n.
25
If the horizontal axis is labelled in Hz, the graph is called
frequency spectrum.
26
Appendix- Coefficients Calculation
To determine the coefficients an for n  1 we’ll start again with
equation (1). First we will multiply it by cos mx (where m is an
integer and m  1 and then we’ll integrate it from –π to +π.

f ( x)  a0   (an cos nx  bn sin nx)
(1)
n1

f ( x)cos mx  a0 cos mx   (an cos nx cos mx  bn sin nx cos mx)
n1




 f ( x)cos mx dx   a cos mx dx   
0




n1  

an cos nx cos mx dx    bn sin nx cos mx dx
n1  


 f ( x)cos mx dx  a  cos mx dx   a 
0


n1

n



n1

cos nx cos mx dx   bn  sin nx cos mx dx
27




 f ( x)cos mxdx  a  cos mxdx  a 
0


 cos mx dx 


n1
n



n1

cos nx cos mxdx  bn  sin nx cos mxdx
1
1

sin mx    [sin m  sin(m)]  0
m
m
1
cos nx cos mx  [cos(n  m) x  cos(n  m)x]
2
For all integers n and m, the product of two cosine functions is
the sum of two cosine functions if n  m and the integral of any
cosine function integrated from –π to π is zero.
28
In case n = m, we have



1
1  sin 2nx 
 cos nx dx   2 (1 cos 2nx) dx  2  x  2   
2
29




 f ( x)cos mxdx  a  cos mxdx  a 
0



 cos mx dx 

n1
n



n1

cos nx cos mxdx  bn  sin nx cos mxdx
1
1

sin mx    [sin m  sin(m)]  0
m
m
1
cos nx cos mx  [cos(n  m) x  cos(n  m)x]
2
1
sin nx cos mx  [sin(n  m)x  sin(n  m)x]
2
Each sine term integrated from –π to π is zero.
30




 f ( x)cos mxdx  a  cos mxdx  a 
0

n1

n



n1

cos nx cos mxdx  bn  sin nx cos mxdx
0

0
an  cos2 nx dx  an 

only if n = m




f ( x)cos nx dx  an  cos2 nx dx  an 

1
an   f ( x)cos nx dx
 
31
To find the coefficients bn of the sine terms we will employ a similar
method. We will multiply equation (1) by sin mx and then integrate
from –π to π is zero.

f ( x)sin mx  a0 sin mx   (an cos nx sin mx  bn sin nx sin mx)
n1




 f ( x)sin mx dx   a sin mx dx   
0




n1  

an cos nx sin mx dx    bn sin nx sin mx dx
n1  


 f ( x)sin mx dx  a  sin mx dx   a 
0


n1

n



n1

cos nx sin mx dx   bn  sin nx sin mx dx
32




 f ( x)sin mxdx  a  sin mxdx  a 
0

n1

n



n1

cos nx sin mxdx  bn  sin nx sin mxdx
0
0
1
sin nx sin mx  [cos(n  m)x  cos(n  m)x]
2
0 for n

m

 sin nx sin mx dx 




1
1  sin 2nx 
2
sin
nx
dx

(
1

cos
2
nx
)
dx

x


 2
2 
2   
33




 f ( x)sin mxdx  a  sin mxdx  a 
0

n1

n



n1

cos nx sin mxdx  bn  sin nx sin mxdx

bn  sin2 nxdx  bn
0
0

only if n = m




f ( x)sin mx dx  bn  sin2 nx dx  bn 

1
bn   f ( x)sin nx dx
 
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