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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…

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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? Pearls Before 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 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 Answer from computer: “There Cos 0.96 x Cos 0.97 x Cos 0.98 x are several components at Cos 0.99 x Cos 1. x Cos 1.01 x Cos 1.02 x different values of k; all are 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 multiples of k = 0.01. k = 0.01: amplitude = 0 Do the transform (or k = 0.02: amplitude = 0 have a computer do it) … … How did the computer k = 0.90: amplitude = 1 know the k values are all k = 0.91: amplitude = 1 k = 0.92: amplitude = 1 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: 2p nx 2p nx f ( x) a0 n1 an cos L n1 bn sin L Notation issues: compare to: f ( x) a0 n1 an x n 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 = nfundamental The trick: finding the “Fourier coefficients”, an and bn 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 2p nx f ( x) a0 n1 an cos L n1 bn sin L L L 2p nx 1 2 a0 f ( x )dx an f ( x) cos dx Let’s wait a L 0 L 0 L minute for L 2p nx derivation. 2 bn f ( x)sin dx L 0 L L f ( x)dx 1 a0 What does L mean? 0 L 2p x 2 What does a1 L f ( x) cos L dx mean? 0 Example: square wave 2p nx 2p nx f ( x) a0 n1 an cos L n1 bn sin L L L L 2p nx 2p nx 2 1 2 a0 f ( x )dx an f ( x) cos dx bn f ( x)sin dx L 0 L 0 L L 0 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. 4 2p nx f ( x) n 1 np sin L (odd only) 4 2p x 4 6p x 4 10p x f ( x) sin sin sin ... p L 3p L 5p L Plots with Mathematica: Square wave, cont. Square wave, cont. nmax = 500 Square wave, 1.0 cont. 0.5 4 2 2 4 0.5 1.0 Why is it still not quite getting the corners right? Deriving the coefficient equations 2p nx 2p nx f ( x) a0 n1 an cos L n1 bn sin L L L L 2p nx 2p nx 2 1 2 a0 f ( x )dx an f ( x) cos dx bn f ( x)sin dx L 0 L 0 L L 0 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 2p nx f ( x) a0 n1 an cos L n1 bn sin L L L L 2p nx 2p nx 2 1 2 a0 f ( x )dx an f ( x) cos dx bn f ( x)sin dx L 0 L 0 L L 0 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

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