Announcements 10/17/12
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Prayer
TA: on Fridays Clement can now only have office hours 3-4
pm. Available 12-1 pm by request.
Term project proposals due Sat night (emailed to me)
“Learn Smart” trial – I’ll send info by email
Exam 2 starts a week from tomorrow!
a. Review session: either Monday, Tues, or Wed. Please vote
by tomorrow night so I can schedule the room on Friday.
Anyone need my “Fourier series summary” handout?
Pearls
Before
Swine
From warmup
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Extra time on?
a. (nothing in particular)
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Other comments?
a. this stuff is so awesome!
b. This is very interesting stuff to me, especially how Dr.
Durfee related it to the real world. I remember doing
Fourier Transforms in Physics 145 and I didn't understand
them and thus absolutely hated them!
c. I work in the math lab and I was able to answer a question
from someone not in a math class about fourier transforms.
It was why the integral is zero if they have different
frequencies. My mom doesn't know anything about fourier
transforms but I still wanted someone to be proud of me.
d. In case you haven't already seen it: http://xkcd.com/26/
From warmup

The most interesting functions that you created
Warmup, cont.
Warmup, cont.
Warmup, cont.
Transforms
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Transform: one-to-one mapping from function to list of coefficients (or to
a function if “spacing” between coefficients becomes infinitely small)
Example: Taylor’s series for ex = 1 + x + x2/2! + x3/3! + …
ex  (1, 1, 1/2!, 1/3!, …)
The coefficients tell you the amplitudes of the polynomials that are
present.
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Fourier Transform: The coefficients tell you the amplitudes of the
frequencies that are present. Example:

Fourier Transform
20
10
600
400
200
200
400
600
10
20
Cos 0.9 x
Cos 0.91 x
Cos 0.92 x
Cos 0.93 x
Cos 0.94 x
Cos 0.95 x
Cos 0.96 x
Cos 0.97 x
Cos 0.98 x
Cos 0.99 x
Cos 1. x
Cos 1.03 x
Cos 1.04 x
Cos 1.05 x
Cos 1.06 x
Cos 1.07 x
Cos 1.08 x
Cos 1.09 x
Cos 1.1 x
Cos 1.01 x
Cos 1.02 x
Do the transform (or
have a computer do it)
 How did the computer
know the k values are all
multiples of k = 0.01?
Answer from computer: “There
are several components at
different values of k; all are
multiples of k = 0.01.
k = 0.01: amplitude = 0
k = 0.02: amplitude = 0
…
…
k = 0.90: amplitude = 1
k = 0.91: amplitude = 1
k = 0.92: amplitude = 1
…”
From warmup
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Fig. 6.2--what the two plots are showing?
a. The top plot is a Fourier transform of Dr. Durfee saying
the word "hello." [It can be represented by a sum of
sine waves with different frequencies.] The bottom plot is
the sine functions' amplitudes plotted as a function of
frequency.
Fourier Theorem

Any function periodic on a distance L can be written as a
sum of sines and cosines like this:

 2p nx 
f ( x)  a0 
an cos 


 L 
n1


 2p nx 
bn sin 

 L 
n1


Notation issues:

n1
a. a0, an, bn = how “much”
at that frequency
b. Time vs distance
c. a0 vs a0/2
d. 2p/L = k (or k0)
2p/T = w (or w0 )
e. 2pn/L = nfundamental
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The trick: finding the “Fourier coefficients”, an and bn
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compare to: f ( x)  a0 
an x n
Applications (a short list)

“What are some applications of Fourier transforms?”
a. Electronics: circuit response to non-sinusoidal signals
b. Data compression (as mentioned in PpP)
c. Acoustics: guitar string vibrations (PpP, next lecture)
d. Acoustics: sound wave propagation through dispersive
medium
e. Medical imaging: constructing 3D image from 2D pictures
f. Optics: spreading out of pulsed laser in dispersive medium
g. Optics: frequency components of pulsed laser can excite
electrons into otherwise forbidden energy levels
h. Quantum: wavefunction of an electron in a periodic
structure (e.g. atoms in a solid)
How to find the coefficients

 2p nx 
f ( x)  a0 
an cos 


 L 
n1

a0 
1
L
L

an 
f ( x )dx
0
bn 


What does
What does
1
a0 
L
2
a1 
L
L

0
2
L
2
L

0

0
 2p nx 
bn sin 

L


n1

L
L

 2p nx 
f ( x) cos 
dx

 L 
 2p nx 
f ( x)sin 
dx

 L 
L
 f ( x)dx
mean?
0
 2p x 
f ( x) cos 
dx

 L 
mean?
Let’s wait a
minute for
derivation.
Example: square wave

 2p nx 
f ( x)  a0 
an cos 


 L 
n1

a0 

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1
L
L

0
f ( x )dx
an 
2
L
L

0

 2p nx 
bn sin 

L


n1

 2p nx 
f ( x) cos 
 dx
L


bn 
2
L
L

0
 2p nx 
f ( x)sin 
dx

 L 
f(x) = 1, from 0 to L/2
f(x) = -1, from L/2 to L
(then repeats)
a0 = ? 0
an = ? 0
b1 = ? 4/p
b2 = ? Could work out each bn individually, but why?
bn = ? 4/(np), only odd terms
Square wave, cont.

f ( x) 

n 1
(odd only)
 4
 np

  2p nx 
 sin  L 
 

 4   2p x   4   6p x   4   10p x 
f ( x)    sin 
   sin 
 
sin 
 ...




 p   L   3p   L   5p   L 
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Plots with Mathematica:
Square
wave,
cont.
Square
wave,
cont.
nmax = 500
Square
wave,
cont.
1.0
0.5
4
2
2
0.5
1.0
Why is it still not quite getting
the corners right?
4
Deriving the coefficient equations

 2p nx 
f ( x)  a0 
an cos 


 L 
n1

a0 

1
L
L

0
f ( x )dx
an 
2
L
L

0

 2p nx 
bn sin 

L


n1

 2p nx 
f ( x) cos 
 dx
L


bn 
2
L
L

0
 2p nx 
f ( x)sin 
dx

 L 
From Warmup: There are really only two main steps in
PpP 6.4: 1) multiply by another cosine (but of a
different frequency), and 2) integrate the product over a
period. And the important thing to realize is that the
product of two cosines of different frequecies,
integrated over an integer number of periods, will give
you zero. So all of the terms in the summation except
one just integrate to zero.
Deriving the coefficient equations

 2p nx 
f ( x)  a0 
an cos 


 L 
n1

a0 




1
L
L

0
f ( x )dx
an 
2
L
L

0

 2p nx 
bn sin 

L


n1

 2p nx 
f ( x) cos 
 dx
L


bn 
2
L
L

0
 2p nx 
f ( x)sin 
dx

 L 
a0: just integrate LHS and RHS from 0 to L.
an: multiply LHS and RHS by cos(2pmx/L), then integrate
bn: multiply LHS and RHS by sin(2pmx/L), then integrate
Details:
a. If n  m, then cos(2pmx/L)cos(2pnx/L) integrates to 0.
(Same for sines.)
b. If n = m, then it integrates to (1/2)L (Same for sines.)
c. Either way, sin(2pmx/L)cos(2pnx/L) always integrates to 0.
Graphical “proof” with Mathematica