Announcements 10/20/10

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Announcements 10/20/10  Prayer  Term project proposals due on Saturday night! Email to me: proposal in body of email, 650 word max. See website for…
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Announcements 10/20/10  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   2p nx  f ( x)  a0 n1 an cos   L     n1 bn sin   L    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   2p nx  f ( x)  a0  n1  an cos   L     n1 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  n1  an cos   L    n1 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  L    5p  sin  L   ...  p   L   3p         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   2p nx  f ( x)  a0  n1  an cos   L    n1 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   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  2p nx   1 1  sin   2 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 0.4 Amplitude [m] 0.2 0 -0.2 -0.4 0 10 20 30 40 50 60 k [rad/m] The Spectrum of a Saw-tooth Wave 0.6 0 0.5 0.4 Amplitude [m] Phase [rad] -pi/4 0.3 0.2 0.1 -pi/2 0 0 10 20 30 40 50 60 k [rad/m] 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 -pi/4 0.4 Amplitude [m] Phase [rad] 0.3 -pi/2 0.2 -3 pi/4 0.1 0 -pi 0 10 20 30 40 50 60 k [rad/m] Low-Pass Filter – after filter 0.6 0 0.5 -pi/4 0.4 Amplitude [m] Phase [rad] 0.3 -pi/2 0.2 -3 pi/4 0.1 0 -pi 0 10 20 30 40 50 60 k [rad/m] Low Pass Filter 1 0.8 [m] 0.6 filtered y and y 0.4 0.2 0 0 0.5 1 1.5 2 2.5 3 x [m] Actual Data from Oscilloscope
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