Introduction to Fourier Analysis. CS 510 Lecture #5 February 2 nd 2015

Transcription

1 Introduction to Fourier Analysis CS 510 Lecture 5 February 2 nd 2015

2 Why? Get used to changing representa6ons! (and this par6cular transforma6on is ubiquitous and important) 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 2

3 Image Data as a Function g(x) = Intensity u coordinate Graphic from Computer Graphics:Principles and Practice by Foley, van Dam, Feiner Hughes. 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 3

4 Any repeating pattern can be built up from a sequence of sin/cosine waves. Graphic adapted from 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 4

5 In the extreme, a square wave Graphic from 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 5

6 Fourier Analysis Magic Many textbooks make is obscure, but Rewrite a function f(x) over a finite range. In effect, we pretend it repeats and reconstruct it as a sum of sinusoids For each sinusoid, we specify: A frequency An amplitude A phase 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 6

7 The Sine Wave - Basics Phase Amplitude Period (Frequency = 1/Period) 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 7

8 The Sine Wave (II) Frequency ( ) g( x) = acos fx + φ Amplitude Phase 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 8

9 Simplifying Phase Phase where wave crosses the x axis: If it crosses at 0 and -π, it s a sine wave. If it crosses at π/2 and - π/2, it a cosine wave. In general, if it crosses at φ and φ + π radians, it has phase φ - π/2 (i.e., cosine, not sine) φ = 0 cosine wave φ = π/2 sine wave A wave with phase φ can be expressed as: ( ) = αcos( x) + βsin( x) cos x + φ 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 9

10 Therefore A sinusoid of arbitrary phase can be written as the sum of a sine cosine: cos( x + φ) = αcos( x) + βsin( x) 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 10

11 Phase (II) Where: φ = tan 1 β ) ( + ' α * cos( θ + φ) = α 2 cos 2 ( θ) + β 2 sin 2 ( θ) (θ +φ) still indicates that the curve has been shifted by φ degrees. 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 11

12 Therefore g( x) = a 1 cos( f 1 x) + b 1 sin( f 1 x) + Remaining problems +a 2 cos( f 2 x) + b 2 sin( f 2 x) +a 3 cos( f 3 x) + b 3 sin( f 3 x) Now has twice as many terms (2 per frequency) Must specify amplitude and frequency for each 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 12

13 Visualization - Nice Applet bs.de/schaukasten/fourier/en_idx.html 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 13

14 Also, an enhanced version hwp://homepages.gac.edu/~huber/fourier/index.html 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 14

15 Another great resource 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 15

16 I forgot to mention 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 16

17 Now, Per Frequency For any given frequency f 1, we must calculate the amplitude of the cosine and sine functions g(x) is infinite (x can be anything), so we have to fit the cosines and sines to all the data: a 1 = + f ( x)cos( f 1 x)dx b 1 = f ( x)sin( f 1 x)dx + 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 17

18 Representing Frequencies Our function g(x) is infinitely repeating Assume, WLOG, the unit of repetition is 1 So g(0+x) = g(1+x) = g(2+x) Only cosines sines with periods that are integers make sense Otherwise is wouldn t repeat The cosine whose period is 1 is written as ( ) cos 2πx 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 18

19 Frequencies (cont). In general the frequency that repeats u times over the interval is written as ( ) cos 2πux Where u is any integer 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 19

20 So, to rewrite g(x) For every integer frequency u = 1, 2, 3, g( x) = a 1 cos 2π x + a 2 cos 2π2x + a 3 cos 2π3x + a 4 cos 2π4x + ( ) + b 1 sin( 2π x) ( ) + b 2 sin( 2π2x) ( ) + b 3 sin( 2π3x) ( ) + b 4 sin( 2π4x) a u = g( x)cos( 2πux)dx b u = g( x)sin( 2πux)dx 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 20

21 Fourier Transform Formally, the Fourier transform in 1D is: F u Where: + ( ) = f ( x) cos2πux isin2πux u is an integer in the range from 0 to i is used to create a 2D vector space F(u) = a u + ib u [ ]dx 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 21

22 The DC component What happens when u = 0? cos(0) = 1, sin(0) = 0 So ( ) = f ( x) cos2πux isin2πux F u = ( ) = f x [ ]dx dx This is the average value (or DC component ) of the function. For images, it is largely a function of lighting. 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 22

23 Inverse Fourier Transform What if I have F(u) for all u, and I want to recreate the original function g(x)? Well, sum it up for every u: f x + ( ) = F( u) [ cos( 2πux) + isin( 2πux) ]du 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 23

24 Discrete Fourier Transform Problem: an image is not an analogue signal that we can integrate. Therefore for 0 x < N and 0 u <N/2: ( ) = f ( x) cos% 2πux F u N 1 x =0 And the discrete inverse transform is: f ( x) = 1 N N 1 x =0 F u *, + ) + * $ N ( ) cos% 2πux $ N ( isin 2πux - % (/ ' $ N '. ( + isin 2πux, % (. ' $ N '- 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 24

25 Discrete vs. Continuous Summation replaces integration Division by N (the number of discrete samples) makes the unit of repetition 1. For any signal (continuous or discrete) G(x) is called the spatial domain F(u) is called the frequency domain 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 25

26 Spatial vs Frequency Spatial domain representation size? Given N samples, it is size N Frequency domain representation size? A total of N/2 frequencies A complex number (2 values) per frequency The DFT is invertible, so the two representations are equivalent: Exact same information and same size The FFT is O(n log n) 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 26

27 hwp://madebyevan.com/d`/ 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 27

28 hwp://madebyevan.com/d`/ Interactive trimming/filtering 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 28

29 Fourier as Linear Transform (N/2+2) x 1 (N/2+2) x 2N 2N x 1!! R F N 2 I F N " 2 R( F ( 0) ) I ( F ( 0) ) R( F ( 1) ) I ( F ( 1) ) ( ( )) ( ( ))! $ = % " cos(0) 0 cos(0) 0! cos(0) 0 0 sin 0 ( ) 0 sin( 0)! 0 sin( 0) ( cos 2π 0 + ( 2π 1+ ( 2π N 1 + * - 0 cos* - 0! cos* - 0 ) N, ) N, ), ( 0 sin 2π 0 + ( 2π 1+ ( 2π N 1 + * - 0 sin* -! 0 sin* - ) N, ) N, ),!!!! "!! ( ( 2π N + + ( ( * * - 0-2π N + + ( ( * * - 1-2π N + + * * - ( N 1) - ) 2, cos* - ) 2, 0 cos* - 0! cos* ) 2, - 0 N N N * - * - * - ), ), ), ( ( 2π N + + * * ) 2, 0 sin* - N * - ), ( ( 2π N + + * * - 1- ) 2, 0 sin* - N * - ), $! ( ) N ( ) N " ( ( 2π * N + + * - ( N 1) - ) 2,! 0 sin* - N * - ), % ( ) ( ) ( ) ( ) I 0 I 0 I 1 I 1! I N 1 ( ) ( ) I N 1 $ % 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 29

30 Fourier Properties Computing DFT takes O(n log(n)) By not implementing it as a matrix The Fast Fourier Transform (FFT) exploits patterns of zeros and ones But look at the matrix: Every row is orthogonal to every other Every row has length (N/2) It is a rotation and a scale of image space The inverse Fourier counter-rotates and counter scales But it rotates image space into a very special coordinate system 2/9/15 CSU CS 510, Ross Beveridge Bruce Draper 30