Announcements 10/20/10
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Prayer
Term project proposals due on Saturday night!
Email to me: proposal in body of email, 650 word
max. See website for guidelines, grading, ideas,
and examples of past projects.
a. If in a partnership, just one email from the two
of you
Exam 2 starts a week from tomorrow.
a. Exam 2 optional review session: vote on times
by tomorrow evening. Survey link sent out this
morning.
Anyone not get the Fourier transform handout last
lecture?
Quick Writing

Ralph doesn’t understand what a transform is. As
discussed last lecture and in today’s reading, how
would you describe the “transform” of a function
to him?
Reading Quiz

In the Fourier transform of a periodic function,
which frequency components will be present?
a. Just the fundamental frequency, f0 = 1/period
b. f0 and potentially all integer multiples of f0
c. A finite number of discrete frequencies centered
on f0
d. An infinite number of frequencies near f0,
spaced infinitely close together
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:
compare to: f ( x)  a0   an x n
a. a0, an, bn = how “much”
n1
at that frequency
a. Time vs distance
b. a0 vs a0/2
c. 2p/L = k (or k0)… compare 2p/T = w (or w0 )
d. Durfee:
– an and bn reversed
– Uses l0 instead of L

The trick: finding the “Fourier coefficients”, an and bn

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 







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   2p x   4
f ( x)    sin 


 p   L   3p

 4
 np

  2p nx 
 sin  L 
 

  6p x   4
 sin  L    5p
 
 
  10p x 
 sin  L   ...
 

Plots with Mathematica:
http://www.physics.byu.edu/faculty/colton/courses/phy123-fall10/lectures/lecture 22 - square
wave Fourier.nb
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 
To derive equation for a0, just integrate LHS and RHS from 0 to L.
To derive equation for an, multiply LHS and RHS by cos(2pmx/L),
then integrate from 0 to L.
(To derive equation for bn, multiply LHS and RHS by sin(2pmx/L),
then integrate from 0 to L.)
Recognize that when n and m are different,
cos(2pmx/L)cos(2pnx/L) integrates to 0. (Same for sines.)
Graphical “proof” with Mathematica
Otherwise integrates to (1/2)L (and m=n). (Same for sines.)
Recognize that sin(2pmx/L)cos(2pnx/L) always integrates to 0.
Sawtooth Wave, like HW 22-1
1

2

1
 2p nx 
sin 

np
 L 
(The next few slides from Dr. Durfee)
N 0
N 1
N 2
N 3
N  10
N  500
The Spectrum of a Saw-tooth Wave
0.6
Amplitude [m]
0.4
0.2
0
-0.2
-0.4
0
10
20
30
k [rad/m]
40
50
60
The Spectrum of a Saw-tooth Wave
0.6
0
0.4
-pi/4
0.3
0.2
-pi/2
0.1
0
0
10
20
30
k [rad/m]
40
50
60
Phase [rad]
Amplitude [m]
0.5
Electronic “Low-pass filter”

“Low pass filter” = circuit which preferentially lets
lower frequencies through.
What comes out?
Circuit
?
How to solve:
(1) Decompose wave into Fourier series
(2) Apply filter to each freq. individually
(3) Add up results in infinite series again
Low-Pass Filter – before filter
0.6
0
0.5
0.4
0.3
-pi/2
0.2
-3 pi/4
0.1
0
0
10
20
30
k [rad/m]
40
50
-pi
60
Phase [rad]
Amplitude [m]
-pi/4
Low-Pass Filter – after filter
0.6
0
0.5
0.4
0.3
-pi/2
0.2
-3 pi/4
0.1
0
0
10
20
30
k [rad/m]
40
50
-pi
60
Phase [rad]
Amplitude [m]
-pi/4
Low Pass Filter
1
y and y
filtered
[m]
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
x [m]
2
2.5
3
Actual Data from Oscilloscope