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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:
Notation issues: a. a0, an, bn = how “much”
at that frequencya. Time vs distanceb. a0 vs a0/2c. 2/L = k (or k0)… compare 2/T = (or 0 )d. Durfee:
– an and bn reversed– Uses 0 instead of L
The trick: finding the “Fourier coefficients”, an and bn
01 1
2 2( ) cos sinn n
n n
nx nxf x a a b
L L
01
compare to: ( ) nn
n
f x a a x
How to find the coefficients
What does mean?
What does mean?
0
0
1( )
L
a f x dxL
0
2 2( )cos
L
nnx
a f x dxL L
0
2 2( )sin
L
nnx
b f x dxL L
01 1
2 2( ) cos sinn n
n n
nx nxf x a a b
L L
0
0
1( )
L
a f x dxL
1
0
2 2( )cos
Lx
a f x dxL L
Let’s wait a minute for derivation.
Example: square wave
f(x) = 1, from 0 to L/2 f(x) = -1, from L/2 to L
(then repeats) a0 = ? an = ? b1 = ? b2 = ? bn = ?
0
0
1( )
L
a f x dxL
0
2 2( )cos
L
nnx
a f x dxL L
0
2 2( )sin
L
nnx
b f x dxL L
01 1
2 2( ) cos sinn n
n n
nx nxf x a a b
L L
004/Could work out each bn individually, but why?
4/(n), only odd terms
Square wave, cont.
Plots with Mathematica:http://www.physics.byu.edu/faculty/colton/courses/phy123-fall10/lectures/lecture 22 - square wave
Fourier.nb
1(odd only)
4 2( ) sin
n
nxf x
n L
4 2 4 6 4 10( ) sin sin sin ...
3 5
x x xf x
L L L
Deriving the coefficient equations
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(2mx/L),
then integrate from 0 to L.(To derive equation for bn, multiply LHS and RHS by sin(2mx/L), then integrate from 0 to L.)
Recognize that when n and m are different, cos(2mx/L)cos(2nx/L) integrates to 0. (Same for sines.)
Graphical “proof” with MathematicaOtherwise integrates to (1/2)L (and m=n). (Same for sines.)
Recognize that sin(2mx/L)cos(2nx/L) always integrates to 0.
0
0
1( )
L
a f x dxL
0
2 2( )cos
L
nnx
a f x dxL L
0
2 2( )sin
L
nnx
b f x dxL L
01 1
2 2( ) cos sinn n
n n
nx nxf x a a b
L L
0N 1N 2N
3N 10N 500N
1 1 2sin
2
nx
n L
Sawtooth Wave, like HW 22-1
(The next few slides from Dr. Durfee)
The Spectrum of a Saw-tooth WaveThe Spectrum of a Saw-tooth Wave
0 10 20 30 40 50 60-0.4
-0.2
0
0.2
0.4
0.6
Am
plitu
de
[m]
k [rad/m]
The Spectrum of a Saw-tooth WaveThe Spectrum of a Saw-tooth Wave
0 10 20 30 40 50 600
0.1
0.2
0.3
0.4
0.5
0.6
Am
plitu
de [
m]
k [rad/m]0 10 20 30 40 50 60
-pi/2
-pi/4
0
Pha
se [
rad]
Electronic “Low-pass filter” “Low pass filter” = circuit which
preferentially lets lower frequencies through.
?Circuit
What comes out?
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 filterLow-Pass Filter – before filter
0 10 20 30 40 50 600
0.1
0.2
0.3
0.4
0.5
0.6
Am
plitu
de [
m]
k [rad/m]0 10 20 30 40 50 60
0
0.1
0.2
0.3
0.4
0.5
0.6
Am
plitu
de [
m]
k [rad/m]0 10 20 30 40 50 60
0
0.1
0.2
0.3
0.4
0.5
0.6
Am
plitu
de [
m]
k [rad/m]0 10 20 30 40 50 60
-pi
-3 pi/4
-pi/2
-pi/4
0
Pha
se [
rad]
Low-Pass Filter – after filterLow-Pass Filter – after filter
0 10 20 30 40 50 600
0.1
0.2
0.3
0.4
0.5
0.6
Am
plitu
de [
m]
k [rad/m]0 10 20 30 40 50 60
0
0.1
0.2
0.3
0.4
0.5
0.6
Am
plitu
de [
m]
k [rad/m]0 10 20 30 40 50 60
0
0.1
0.2
0.3
0.4
0.5
0.6
Am
plitu
de [
m]
k [rad/m]0 10 20 30 40 50 60
-pi
-3 pi/4
-pi/2
-pi/4
0
Pha
se [
rad]
Low Pass FilterLow Pass Filter
0 0.5 1 1.5 2 2.5 30
0.2
0.4
0.6
0.8
1
y an
d y fil
tere
d
[m]
x [m]
Actual Data from OscilloscopeActual Data from Oscilloscope