Announcements 10/17/12 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?

PearlsBefore Swine

From warmup Extra time on?

a. (nothing in particular)

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 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.

Fourier Transform: The coefficients tell you the amplitudes of the frequencies that are present. Example:

Fourier Transform

Do the transform (or have a computer do it)

Answer from computer: “There are several components at different values of k; all are multiples of k = 0.01.

k = 0.01: amplitude = 0k = 0.02: amplitude = 0……k = 0.90: amplitude = 1k = 0.91: amplitude = 1k = 0.92: amplitude = 1…”

Cos0.9 x Cos0.91 x Cos0.92 x Cos0.93 x Cos0.94 x Cos0.95 x Cos0.96 x Cos0.97 x Cos0.98 x Cos0.99 x Cos1. x Cos1.01 x Cos1.02 x Cos1.03 x Cos1.04 x Cos1.05 x Cos1.06 x Cos1.07 x Cos1.08 x Cos1.09 x Cos1.1 x

600 400 200 200 400 600

20

10

10

20

How did the computer know the k values are all multiples of k = 0.01?

From warmup

 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:

Notation issues: a. a0, an, bn = how “much”

at that frequencyb. Time vs distancec. a0 vs a0/2d. 2/L = k (or k0) 2/T = (or 0 )e. 2n/L = nfundamental

The trick: finding the “Fourier coefficients”, an and bn

01 1

2 2( ) cos sinn nn n

nx nxf x a a bL L

01

compare to: ( ) nn

n

f x a a x

Applications (a short list) “What are some applications of Fourier transforms?”

a. Electronics: circuit response to non-sinusoidal signalsb. Data compression (as mentioned in PpP)c. Acoustics: guitar string vibrations (PpP, next lecture)d. Acoustics: sound wave propagation through dispersive

mediume. Medical imaging: constructing 3D image from 2D picturesf. Optics: spreading out of pulsed laser in dispersive

mediumg. Optics: frequency components of pulsed laser can excite

electrons into otherwise forbidden energy levelsh. Quantum: wavefunction of an electron in a periodic

structure (e.g. atoms in a solid)

How to find the coefficients

What does mean?

What does mean?

00

1 ( )L

a f x dxL

0

2 2( )cosL

nnxa f x dx

L L

0

2 2( )sinL

nnxb f x dx

L L

01 1

2 2( ) cos sinn nn n

nx nxf x a a bL L

00

1 ( )L

a f x dxL

10

2 2( )cosL

xa 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 = ?

00

1 ( )L

a f x dxL

0

2 2( )cosL

nnxa f x dx

L L

0

2 2( )sinL

nnxb f x dx

L L

01 1

2 2( ) cos sinn nn n

nx nxf x a a bL L

004/Could work out each bn individually, but why?4/(n), only odd terms

Square wave, cont.

Plots with Mathematica:

1(odd only)

4 2( ) sinn

nxf xn L

4 2 4 6 4 10( ) sin sin sin ...3 5

x x xf xL L L

Square wave, cont.

Square wave, cont.

Square wave, cont.

4 2 2 4

1 .0

0 .5

0 .5

1 .0

nmax = 500

Why is it still not quite getting the corners right?

Deriving the coefficient equations

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.

00

1 ( )L

a f x dxL

0

2 2( )cosL

nnxa f x dx

L L

0

2 2( )sinL

nnxb f x dx

L L

01 1

2 2( ) cos sinn nn n

nx nxf x a a bL L

Deriving the coefficient equations

a0: just integrate LHS and RHS from 0 to L. an: multiply LHS and RHS by cos(2mx/L), then integrate bn: multiply LHS and RHS by sin(2mx/L), then integrate Details:

a. If n m, then cos(2mx/L)cos(2nx/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(2mx/L)cos(2nx/L) always integrates to

0.

Graphical “proof” with Mathematica

00

1 ( )L

a f x dxL

0

2 2( )cosL

nnxa f x dx

L L

0

2 2( )sinL

nnxb f x dx

L L

01 1

2 2( ) cos sinn nn n

nx nxf x a a bL L