Lecture 8
Fourier Series applied to pulses
Remember homework 1 for submission 31/10/08
Today
• How we define a pulse using Fourier series
• Complex number Fourier series
• Parseval’s Theorem
Hello!!!
I’m a
pulse
Remember Phils Problems for loads of stuff
http://www.hep.shef.ac.uk/Phil/PHY226.htm
Comment on homework
In question 1 you need to find an expression for q
You may end up with an answer that contains a complex
number as the denominator
Typically we don’t leave answers with complex numbers on
the bottom of a fraction so…
A
A(a  ib )
A(a  ib )

 2
2
(a  ib ) (a  ib )( a  ib ) (a  b )
Fourier Series
Summary
The Fourier series can be
written with period L as

1
2nx
2nx
f ( x)  a0   an cos
 bn sin
2
L
L
n 1
The Fourier series coefficients can be found by:2 L
2 L
2nx
2 L
2nx
a0   f ( x)dx an   f ( x) cos
dx
dx bn   f ( x) sin
0
0
0
L
L
L
L
L
We have seen in the last couple of
lectures how a periodically
repeating function can be
represented by a Fourier series
y axis
100
step periods
80
60
40
20
0
-30
-20
-10
0
-20
-40
10
x axis
20
30
Fourier Series applied to pulses
Can Fourier series be used to represent a single pulse event rather
than an infinitely repeating pattern ?
Yes. It’s fine if you have a single pulse occurring between 0 and d to
pretend that it is part of an infinitely repeating pattern. All you have to
do then is work out the Fourier series for the infinite pattern and say
that the pulse is represented by that function just between 0 and d
i.e.
becomes
but only look between 0 and d
What is period of repeating pattern?
Can we neglect sine or cosine terms?
This approach is fine but it leads to a lot of work in the integration stage.
Fourier Series applied to pulses
There is a better way….
If the only condition is that the pretend function be periodic, and since
we know that even functions contain only cosine terms and odd
functions only sine terms, why don’t we draw it either like this or this?
Odd function (only sine terms)
Even function (only cosine terms)
What is period of repeating pattern now?
Fourier Series applied to pulses
Half-range sine series
We saw earlier that for a function with period L the Fourier series is:
1
2nx
2nx
f ( x)  a0   an cos
 bn sin
2
L
L
n 1
where
an 
2 L
2nx
f
(
x
)
cos
dx
L 0
L
bn 
2 L
2nx
f
(
x
)
sin
dx
L 0
L
In this case we have a function of period 2d which is odd and so contains
only sine terms, so the formulae become:-
2nx 
nx
f ( x)   bn sin
  bn sin
where
(2d ) n 1
d
n 1

2
bn 
2d

d
d
2nx
1 d
nx
2 d
nx
f ( x) sin
dx   f ( x) sin
dx   f ( x) sin
dx

d
0
( 2d )
d
d
d
d
Remember, this is all to simplify the
Fourier series. We’re still only
allowed to look at the function
between x = 0 and x = d
I’m looking at
top diagram
Fourier Series applied to pulses
Half-range cosine series
Again, for a function with period L the Fourier series is:
1
2nx
2nx
f ( x)  a0   a n cos
 bn sin
2
L
L
n 1
where
an 
2 L
2nx
f
(
x
)
cos
dx
L 0
L
bn 
2 L
2nx
f
(
x
)
sin
dx
L 0
L
Again we have a function of period 2d but this time it is even and so
contains only cosine terms, so the formulae become:

1
2nx 1
nx
f ( x)  a0   an cos
 a0   an cos
2
( 2d ) 2
d
n 1
n 1
2
an 
2d

d
d
where
2nx
1 d
nx
2 d
nx
f ( x) cos
dx   f ( x) cos
dx   f ( x) cos
dx

d
0
( 2d )
d
d
d
d
Remember, this is all to simplify the
Fourier series. We’re still only
allowed to look at the function
between x = 0 and x = d
I’m looking at
top diagram
Fourier Series applied to pulses
Summary of half-range sine and cosine series
The Fourier series for a pulse such as
can be written as either a half range sine or cosine series. However the
series is only valid between 0 and d
Half range
sine series
nx
f ( x)   bn sin
d
n 1

where

1
nx
Half range
f ( x)  a0   an cos
cosine series
2
d
n 1
where
Let’s do an example to demonstrate this………
2 d
nx
bn   f ( x) sin
dx
0
d
d
2 d
nx
an   f ( x) cos
dx
0
d
d
Half Range Fourier Series
Find Half Range Sine Series which represents
the displacement f(x), between x = 0 and 6, of
the pulse shown to the right
The pulse is defined as
So bn 
f ( x)  x for 0  x  6 with a length d = 6
2 d
nx
2 6
nx
f
(
x
)
sin
dx

x
sin
dx Integrate by parts


0
0
d
d
6
6
so set u  x and sin
nx
dx  dv
6
v   sin
nx
6
nx
dx  
cos
as
6
n
6
 udv  uv   vdu
du  dx
6
6
6
6
1   6 x
nx 
6
nx  1   6 x
nx   36
nx  
bn  
cos

cos


cos

sin
3   n
6  0 0 n
6  3   n
6  0  n 2 2
6  0 
1   36
12
12
  36

bn   
cos n    2 2 sin n    2 2 sin n 
cos n
3   n
n

n

n

 

n=1
b1 
12

n=2
b2  
12
6

2

n=3
b3 
12 4

3 
n=4
b4  
12
3

4

n=5
b5 
12
5
Half Range Fourier Series
Find Half Range Sine Series which represents
the displacement f(x), between x = 0 and 6, of
the pulse shown to the right
nx
f ( x)   bn sin
d
n 1
f ( x) 
2 d
nx
bn   f ( x) sin
dx
0
d
d

Half range
sine series
12

sin
x
6

6

sin
x
3

where
4

sin
x
2

3

sin
2x 12
5x

sin
 .....
3
5
6
Can we check this on Fourier_checker.xls at
Phils Problems website???
n=1
b1 
12

n=2
b2  
12
6

2

n=3
b3 
12 4

3 
n=4
b4  
12
3

4

n=5
b5 
12
5
Half Range Fourier Series
f ( x) 
12

sin
n=1
b1 
12

x
6

6

sin
x
3

4

sin
n=2
b2  
12
6

2

x
2

3

sin
2x 12
5x

sin
 .....
3
5
6
n=3
b3 
12 4

3 
n=4
b4  
12
3

4

n=5
b5 
12
5
Fourier Series applied to pulses
Why is this useful????????????????
In Quantum you have seen that there exist specific solutions to the wave
equation within a potential well subject to the given boundary conditions.
nx
n ( x)  Bn sin
d
n  0 when x  0 and x  d
Typically a particle will be in a superposition of eigenfunctions so …


n 1
n 1
 ( x)   n ( x)   Bn sin
nx
d
Compare this with the Half range sine series we deduced earlier
f ( x) 
12

sin
x
d

6

sin
2x 4
3x 3
4x 12
5x
 sin
 sin

sin
 .....
d

d

d
5
d
Given a complicated general solution we can deconvolve into harmonic terms
Complex Fourier Series
For waves on strings or at the beach we need real standing waves.
But in some other areas of physics, especially Quantum, it is more convenient
to consider complex or running waves.
Let’s derive the complex form of the Fourier series assuming for simplicity
that the period is 2
Remember:-
cos
2nx
1
 cos nx  (einx  e inx )
L
2
2nx
1 inx inx
sin
 sin nx  (e  e )
L
2i
The complex form of the Fourier series can be derived by assuming a
solution of the form f ( x) 

 imx
inx
and
then
multiplying
both
sides
by
c
e
e
n
n  
and integrating over a period:

2
0
f ( x)e
imx
dx 

2
 cn  e e
n  
0
inx imx
dx 

2
i ( nm ) x
c
e
dx
 n
n  
0
For n  m integral vanishes. For n=m integral = 2. So c n 
1
2

2
0
f ( x)e inx dx
Complex Fourier Series
Therefore for a period of 2 the complex Fourier series is given as:-
f ( x) 

c e
n  
inx
n
where
1
cn 
2

2
0
f ( x)e inx dx
The more general expression for a function f(x) with period L can be
expressed as:-
f ( x) 

c e
n  
n
2inx
L
where

1 L
cn   f ( x)e
L 0
2inx
L
dx
Let’s try an example from the notes …..
Complex Fourier Series example
Example 6
Find the complex Fourier series for f(x) = x in the range -2 < x < 2 if the
repeat period is 4.
cn 
1
f ( x)e

0
L
L
2inx
L
Integration by parts
dx and period is 4, so we write
 u dv  uv   v du
2
So du  dx and v 
e
in
1   2x
cn  
e
4  in
inx
2
2
 e
in
inx
2
1 2
cn   xe
4 2
with u  x and dv  e
inx
2
inx
2
dx
dx
inx
2
2

dx 
 2
1   2x
e

4  in
inx
2
4
 22 2e
in
inx 2
2

 x
e
 
 2  2in
inx
2
1
 2 2e
 n
inx 2
2


 2
1 in   1 in
1 in   1 in in
1
  1 in
Cn  
e  2 2e 
e  2 2e 
e  e  2 2 e in  ein
 n
 n
 n
 in
  in
 in




Complex Fourier Series example
Example 6
Find the complex Fourier series for f(x) = x in the range -2 < x < 2 if the
repeat period is 4.
Since



i in in
1
1 i
  i then Cn 
e  e  2 2 e in  ein
i i
n
 n

We want to find actual values for Cn so it would be helpful to convert expression
for Cn into sine and cosine terms using the standard expressions:-
cos n 

1 in
e  ein
2

 1 in in
sin n 
e e
2i


So we can write…
Cn 
2i
2i
2i
cos n  2 2 sin n  cos n
n
 n
n
f ( x) 

c e
n  
n
2inx
L
so Cn 
2i
2i
 1 n and since
cos n 
n
n

2i
 1 ne
so f ( x)  
n   n
inx
2
Parseval’s Theorem applied to Fourier Series
The energy in a vibrating string or an electrical signal is proportional to the
square of the amplitude of the wave
Parseval’s Theorem applied to Fourier Series
Consider again the standard Fourier series with a period taken for
simplicity as 2

1
…. f ( x)  a0   an cos nx  bn sin nx
2
n 1
Square both sides then integrate over a period:
  f ( x) dx  
2
2
0
2
0
2


1

a

a
cos
nx

b
sin
nx

n
2 0  n
 dx
n 1
n 1


The RHS will give both squared terms and cross term. When we integrate, all
the cross terms will vanish. All the squares of the cosines and sines integrate
to give  (half the period). Hence:-

2
0

a0
 f ( x) dx  2   [an 2  bn 2 ]
4
n 1
2
2
Hence Parseval’s theorem tells us that the total energy in a vibrating
system is equal to the sum of the energies in the individual modes.
Fourier Series in movies!!!!
In music if the fundamental frequency of a note is 100 Hz
Then the harmonics are at 200 Hz, 300 Hz, 400 Hz, 500 Hz, 600 Hz
But the octaves are at 200 Hz, 400 Hz, 800 Hz, 1600 Hz, 3200 Hz
If you wanted to explain the Fourier series to an alien you’d probably pick
notes that showed you understood this ……
So as a greeting why not try
392Hz
440Hz
G, A, F, Foctave lower , C
349Hz
174Hz
261Hz
Why are these notes special ....We’ll take the 1st harmonic as Fbottom=21.8Hz
18th harmonic
20th harmonic
4th octave
(16th harmonic)
8th harmonic
12th harmonic
f (t )total  B18 sin( 2f18t )  B20 sin( 2f 20t )  B16 sin( 2f16t )  B8 sin( 2f8t )  B12 sin( 2f12t )
http://uk.youtube.com/watch?v=tUcOaGawIW0