Filtering in the Frequency Domain

Filtering in the Frequency Domain

Outline Fourier Transform Filtering in Fourier Transform Domain 2/20/2014 2

Fourier Series and Fourier Transform: History Jean Baptiste Joseph Fourier, French mathematician and physicist (03/21/1768-05/16/1830) http://en.wikipedia.org/wiki/joseph_fourier Orphaned: at nine Egyptian expedition with Napoleon I: 1798 Governor of Lower Egypt Permanent Secretary of the French Academy of Sciences: 1822 Théorie analytique de la chaleur : 1822 (The Analytic Theory of Heat) 2/20/2014 3

Fourier Series and Fourier Transform: History Fourier Series Any periodic function can be expressed as the sum of sines and /or cosines of different frequencies, each multiplied by a different coefficients Fourier Transform Any function that is not periodic can be expressed as the integral of sines and /or cosines multiplied by a weighing function 2/20/2014 4

Fourier Series: Example 2/20/2014 5

Preliminary Concepts j = 1, a complex number C = R + ji the conjugate C* = R - ji 2 2 C = R + I and θ = arctan( I / R) C = C (cosθ + jsin θ) Using Euler's formula, j C = C e θ 2/20/2014 6

Fourier Series A function f( t) of a continuous variable t that is periodic with period, T, can be expressed as the sum of sines and cosines multiplied by appropriate coefficients where f() t = n= ce n 2π n j t T 2π n 1 T /2 j t T cn = f ( t) e dt for n = 0, ± 1, ± 2,... T T /2 2/20/2014 7

Impulses and the Sifting Property (1) A unit impulse of a continuous variable t located at t=0, denoted δ ( t), defined as if t = 0 δ () t = 0 if t 0 and is constrained also to satisfy the identity δ ( t) dt = 1 The sifting property f () t δ () t dt = f (0) f () t δ ( t t0) dt = f t0 2/20/2014 8 ( )

Impulses and the Sifting Property (2) A unit impulse of a discrete variable x located at x=0, denoted δ ( x), defined as 1 if x = 0 δ ( x) = 0 if x 0 and is constrained also to satisfy the identity x= δ ( x) = 1 The sifting property x= f( x) δ ( x) = f(0) x= f( x) δ ( x x ) = f( x ) 0 0 2/20/2014 9

Impulses and the Sifting Property (3) impulse train s T ( t), s T () t = δ ( t n T) n= 2/20/2014 10

Fourier Transform: One Continuous Variable The Fourier Transform of a continous function f ( t) j2πµ t F( µ ) { f()} t f() t e dt =I = The Inverse Fourier Transform of F( µ ) 1 2 f( t) { F( )} F( ) e j πµ t d =I µ = µ µ 2/20/2014 11

Fourier Transform: One Continuous Variable W /2 j2πµ t j2πµ t F( µ ) = f () t e dt = Ae dt W /2 A A = j = j W j2πµ t W /2 jπµ W jπµ W e e e 2πµ W /2 2π sin( πµ W ) = AW ( πµ W ) 2/20/2014 12

Fourier Transform: Impulses The Fourier transform of a unit impulse located at the origin: j2πµ t F( µ ) = δ() t e dt = e =1 j2πµ 0 The Fourier transform of a unit impulse located at t = t : j2πµ t F( µ ) = δ( t t0) e dt = e j2πµ t 0 =cos(2 πµ t ) jsin (2 πµ t ) 0 0 0 2/20/2014 13

Fourier Transform: Impulse Trains Impulse train s T(), t s T() t = δ ( t n T ) The Fourier series: where s () t = c e T n= n= 2π n j t T 2π n 1 T /2 j t T cn = s () T /2 T t e dt T n 2/20/2014 14

Fourier Transform: Impulse Trains 2πn 2πn 2πn j πµ t j πµ t I e = e e dt = e e dt 2πn 2πn c j 1 T/2 s 1 /2 () t e j t T j t T dt T n = () t e = T t j dt T/2 T δ T T t j t 2 2 T T T /2 1 0 1 = e = n Tj2 π ( µ T) t n T = e dt = δ ( µ ) T n n I = e du s T() t = c e = 1 2π n j t 2π n 2 ( ) T 1 ( j t ) j πµ δ µ δ µ t T n T= e T n= 2π nt T n= j T e 2/20/2014 15

Fourier Transform: Impulse Trains Let S( µ ) denote the Fourier transform of the periodic impulse train S ( t) T 1 S( µ ) S T () t e T n= 2π n j t { } T =I =I 2π n 1 j t T = I e T n= 1 n = δ ( µ ) T T n= 2/20/2014 16

Fourier Transform and Convolution The convolution of two functions is denoted by the operator f() t ht () = f( τ) ht ( τ) dτ I = j2 { () ()} ( τ) ( τ) τ πµ t f t ht f ht d e dt j2πµ t = f ( τ) h( t τ) e dt dτ j2πµτ = f( τ ) H( µ ) e dτ j2πµτ H f e d = ( µ ) ( τ) τ = H( µ ) F( µ ) 2/20/2014 17

Fourier Transform and Convolution Fourier Transform Pairs f() t ht () H( µ ) F( µ ) f() tht () H( µ ) F( µ ) 2/20/2014 18

Fourier Transform of Sampled Functions f () t = f() t s () t T = f() t δ ( t n T) n= 2/20/2014 19

Fourier Transform of Sampled Functions F ( µ ) =I { f () t } =I { f() t s ()} T t = F( µ )? S( µ ) F ( µ ) = F1 ( µ ) S( µ ) = n F( τ ) S( µ τ) dτ S( µ ) = δ ( µ ) 1 T n= T n = F( τ) δ ( µ τ ) dτ T T n= 1 n = F( τδµ ) ( τ ) dτ T T n= 1 n = F( µ ) T T n= 2/20/2014 20

Question The Fourier transform of the sampled function (shown in the following figure) is 1. Continuous 2. Discrete 2/20/2014 21

Fourier Transform of Sampled Functions A bandlimited signal is a signal whose Fourier transform is zero above a certain finite frequency. In other words, if the Fourier transform has finite support then the signal is said to be bandlimited. An example of a simple bandlimited signal is a sinusoid of the form, x( t) = sin(2 π ft + θ) 2/20/2014 22

Fourier Transform of Sampled Functions F ( µ ) = 1 T n= n F( µ ) T µ max µ max Over-sampling 2/20/2014 23 1 T > 2µ max Critically-sampling 1 T = 2µ max under-sampling 1 T < 2µ max

Nyquist Shannon sampling theorem We can recover f() tfrom its sampled version if we can isolate a copy of F ( µ ) from the periodic sequence of copies of this function contained in F ( µ ), the transform of the sampled function f () t Sufficient separation is guaranteed if 1 T 2µ Sampling theorem: A continuous, band-limited function can be recovered completely from a set of its samples if the samples are acquired at a rate exceeding twice the highest frequency content of the function > max 2/20/2014 24

Nyquist Shannon sampling theorem? 2 () ( ) j πµ t f t = F µ e dµ 2/20/2014 25

Aliasing If a band-limited function is sampled at a rate that is less than twice its highest frequency? The inverse transform will yield a corrupted function. This effect is known as frequency aliasing or simply as aliasing. 2/20/2014 26

Aliasing 2/20/2014 27

Aliasing 2/20/2014 28

Function Reconstruction from Sampled Data F( µ ) = H( µ ) F ( µ ) 1 { F µ } f() t = I ( ) = I 1 { H( µ ) F ( µ )} = ht () f () t [ ] f( t) = f( n T)sinc ( t n T) / n T n= 2/20/2014 29

The Discrete Fourier Transform (DFT) of One Variable M 1 j2 πµ xm / ( µ ) ( ), µ 0,1,..., 1 F = f xe = M x= 0 1 f x = F µ e x= M M M 1 j2 πµ xm / ( ) ( ), 0,1,2,..., 1 µ = 0 2/20/2014 30

2-D Impulse and Sifting Property: Continuous The impulse δ( tz, ), δ( tz, ) if t = z = 0 = 0 otherwise and δ ( t, z) dtdz = 1 The sifting property and f (, t z) δ (, t z) dtdz = f (0,0) f (, t z) δ ( t t, z z ) dtdz = f ( t, z ) 0 0 0 0 2/20/2014 31

2-D Impulse and Sifting Property: Discrete The impulse δ( xy, ), δ( xy, ) 1 if x= y = 0 = 0 otherwise The sifting property and x= y= x= y= f( xy, ) δ ( xy, ) = f(0,0) f( x, y) δ ( x x, y y ) = f( x, y ) 0 0 0 0 2/20/2014 32

2-D Fourier Transform: Continuous and j2 π ( µ t+ νz) F( µν, ) = f (, t z) e dtdz j2 π ( µ t+ νz) ftz (, ) = f(, ) e d d µν µ ν 2/20/2014 33

2-D Fourier Transform: Continuous j2 π ( µ t+ νz) F( µν, ) = f (, t z) e dtdz = T/2 Z/2 T/2 Z/2 Ae j2 π ( µ t+ νz) dtdz sin( πµ T) sin( πνt) = ATZ πµ T πνt 2/20/2014 34

2-D Sampling and 2-D Sampling Theorem 2 D impulse train: s (, tz) = δ ( t m Tz, n Z) T Z m= n= 2/20/2014 35

2-D Sampling and 2-D Sampling Theorem Function ftz (, ) is said to be band-limited if its Fourier transform is 0 outside a rectangle established by the intervals [- µ, µ ] and [- ν, ν ], that is max max F( µν, ) = 0 for µ µ and ν ν x max max Two-dimensional sampling theorem: A continuous, band-limited function ftz (, ) can be recovered with no error from a set of its samples if the sampling intervals are 1 1 T< and Z< 2µ 2ν max ma max max 2/20/2014 36

2-D Sampling and 2-D Sampling Theorem 2/20/2014 37

Aliasing in Images: Example In an image system, the number of samples is fixed at 96x96 pixels. If we use this system to digitize checkerboard patterns Under-sampling 2/20/2014 38

Aliasing in Images: Example Re-sampling 2/20/2014 39

Aliasing in Images: Example Re-sampling 2/20/2014 40

Moiré patterns Moiré patterns are often an undesired artifact of images produced by various digital imaging and computer graphics techniques e. g., when scanning a halftone picture or ray tracing a checkered plane. This cause of moiré is a special case of aliasing, due to under-sampling a fine regular pattern http://en.wikipedia.org/wiki/moiré_pattern 2/20/2014 41

Moiré patterns 2/20/2014 42

Moiré patterns A moiré pattern formed by incorrectly downsampling the former image 2/20/2014 43

Moire Pattern 2/20/2014 44

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2-D Discrete Fourier Transform and Its Inverse DFT: M 1N 1 F( µν, ) = f( xye, ) x= 0 y= 0 j2 π ( µ xm / + νyn / ) µ = 0,1,2,..., M 1; ν = 0,1,2,..., N 1; f( xy, ) is a digital image of size M N. IDFT: M 1N 1 1 f( xy, ) = F( µν, ) e MN x= 0 y= 0 j2 π ( µ xm / + νyn / ) 2/20/2014 47

Properties of the 2-D DFT relationships between spatial and frequency intervals Let T and Z denote the separations between samples, then the seperations between the corresponding discrete, frequency domain variables are given by µ = and ν = 1 M T 1 N Z 2/20/2014 48

Properties of the 2-D DFT translation and rotation and j2 π ( µ 0xM / + ν0yn / ) (, ) (, ) f xye f( x- x, y- y ) F( µν, ) e 0 0 Fµ µ ν ν 0 0 j2 π ( µ x / M+ νy / N) 0 0 Using the polar coordinates x= rcos θ y=rsin θ µ = ωcos ϕ ν= ωsinϕ results in the following transform pair: f(, r θ+ θ) F( ωϕ, + θ) 0 0 2/20/2014 49

Properties of the 2-D DFT periodicity 2 D Fourier transform and its inverse are infinitely periodic F( µν, ) = F( µ + km, ν ) = F( µν, + k N) = F( µ + km, ν+ k N) 1 2 1 2 f( x, y) = f( x+ k M, y) = f( x, y+ k N) = f( x+ k M, y+ k N) f xe 1 2 1 2 µ µ j 2 π ( µ 0xM ( ) / ) F( ) x µ 0 = M /2, f( x)( 1) F( µ M /2) x+ y f( xy, )( 1) F( µ M/2, ν N/2) 0 2/20/2014 50

Properties of the 2-D DFT periodicity 2/20/2014 51

Properties of the 2-D DFT Symmetry 2/20/2014 52

Properties of the 2-D DFT Fourier Spectrum and Phase Angle 2-D DFT in polar form Fourier spectrum Fuv (, ) = Fuv (, ) e jφ ( uv, ) Fuv = R uv + I uv Power spectrum 2 2 (, ) (, ) (, ) 1/2 Puv = Fuv = R uv + I uv 2 2 2 (, ) (, ) (, ) (, ) Phase angle Iuv (, ) φ(u,v)=arctan Ruv (, ) 2/20/2014 53

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Example: Phase Angles 2/20/2014 56

Example: Phase Angles and The Reconstructed 2/20/2014 57

2-D Convolution Theorem 1-D convolution 2-D convolution M 1 f( x) hx ( ) = f( mhx ) ( m) m= 0 M 1N 1 f( xy, ) hxy (, ) = f( mnhx, ) ( my, n) m= 0 n= 0 x = 0,1,2,..., M 1; y = 0,1,2,..., N 1. f( xy, ) hxy (, ) FuvHuv (, ) (, ) f( xyhxy, ) (, ) Fuv (, ) Huv (, ) 2/20/2014 58

An Example of Convolution Mirroring h about the origin Translating the mirrored function by x Computing the sum for each x 2/20/2014 59

An Example of Convolution It causes the wraparoun d error It can be solved by appending zeros 2/20/2014 60

Zero Padding Consider two functions f(x) and h(x) composed of A and B samples, respectively Append zeros to both functions so that they have the same length, denoted by P, then wraparound is avoided by choosing P A+B-1 2/20/2014 61

Zero Padding Let f(x,y) and h(x,y) be two image arrays of sizes A B and C D pixels, respectively. Wraparound error in their convolution can be avoided by padding these functions with zeros f h p p ( xy, ) ( xy, ) f ( x, y) 0 x A-1 and 0 y B -1 = 0 A x P or B y Q h( x, y) 0 x C -1 and 0 y D -1 = 0 C x P or D y Q Here P A+ C 1; Q B+ D 1 2/20/2014 62

Summary 2/20/2014 63

Summary 2/20/2014 64

Summary 2/20/2014 65

Summary 2/20/2014 66

The Basic Filtering in the Frequency Domain Why is the spectrum at almost ±45 degree stronger than the spectrum at other directions? 2/20/2014 67

The Basic Filtering in the Frequency Domain Modifying the Fourier transform of an image Computing the inverse transform to obtain the processed result gxy 1 (, ) = I { HuvFuv (, ) (, )} Fuv (, ) is the DFT of the input image Huv (, ) is a filter function. 2/20/2014 68

The Basic Filtering in the Frequency Domain In a filter H(u,v) that is 0 at the center of the transform and 1 elsewhere, what s the output image? 2/20/2014 69

The Basic Filtering in the Frequency Domain 2/20/2014 70

Zero-Phase-Shift Filters gxy = I 1 (, ) { HuvFuv (, ) (, )} Fuv (, ) = Ruv (, ) + jiuv (, ) gxy (, ) =I HuvRuv (, ) (, ) + jhuviuv (, ) (, ) 1 [ ] Filters affect the real and imaginary parts equally, and thus no effect on the phase. These filters are called zero-phase-shift filters 2/20/2014 72

Examples: Nonzero-Phase-Shift Filters Even small Phase changes angle is in the phase angle Phase angle can is have dramatic multiplied (usually by undesirable) effects multiplied on the by filtered output 0.5 0.5 2/20/2014 73

Summary: Steps for Filtering in the Frequency Domain 1. Given an input image f(x,y) of size MxN, obtain the padding parameters P and Q. Typically, P = 2M and Q = 2N. 2. Form a padded image, f p (x,y) of size PxQ by appending the necessary number of zeros to f(x,y) 3. Multiply f p (x,y) by (-1) x+y to center its transform 4. Compute the DFT, F(u,v) of the image from step 3 5. Generate a real, symmetric filter function*, H(u,v), of size PxQ with center at coordinates (P/2, Q/2) *generate from a given spatial filter, we pad the spatial filter, multiply the expadded array by (-1) x+y, and compute the DFT of the result to obtain a centered H(u,v). 2/20/2014 74

Summary: Steps for Filtering in the Frequency Domain 6. Form the product G(u,v) = H(u,v)F(u,v) using array multiplication 7. Obtain the processed image { 1 [ ] } (, ) (, ) ( 1) x + g y p x y = real I G u v 8. Obtain the final processed result, g(x,y), by extracting the MxN region from the top, left quadrant of g p (x,y) 2/20/2014 75

An Example: Steps for Filtering in the Frequency Domain 2/20/2014 76

Correspondence Between Filtering in the Spatial and Frequency Domains (1) Let H(u) denote the 1-D frequency domain Gaussian filter H ( u) = Ae 2 2 - u /2σ The corresponding filter in the spatial domain h( x) = 2πσ 2 2 2 2 x Ae π σ 1. Both components are Gaussian and real 2. The functions behave reciprocally 2/20/2014 77

Correspondence Between Filtering in the Spatial and Frequency Domains (2) Let Hu ( ) denote the difference of Gaussian filter 2 2 2 2 /2 σ1 u /2σ2 H ( u) = Ae Be -u - with A B and σ σ 1 2 The corresponding filter in the spatial domain 2 2 2 2 2 2 1 x π σ2 x h( x) = 2πσ Ae 2πσ Ae 2π σ 2 1 2 High-pass filter or low-pass filter? 2/20/2014 78

Correspondence Between Filtering in the Spatial and Frequency Domains (3) 2/20/2014 79

Correspondence Between Filtering in the Spatial and Frequency Domains: Example 600x600 2/20/2014 80

Correspondence Between Filtering in the Spatial and Frequency Domains: Example 2/20/2014 81

Generate H(u,v) f h p p ( xy, ) f ( x, y) 0 x 599 and 0 y 599 = 0 600 x 602 or 600 y 602 h( x, y) 0 x 2 and 0 y 2 ( xy, ) = 0 3 x 602 or 3 y 602 Here P A(600) + C(3) 1 = 602; Q B(600) + D(3) 1 = 602. 2/20/2014 82

Generate H(u,v) x+ y 1. Multiply h( xy, ) by (-1) to center the frequency domain filter p 2. Compute the forward DFT of the result in (1) 3. Set the real part of the resulting DFT to 0 to account for parasitic real parts u+ v 4. Multiply the result by (-1), which is implicit when hxy (, ) was moved to the center of h( xy, ). p 2/20/2014 83

Image Smoothing Using Filter Domain Filters: ILPF Ideal Lowpass Filters (ILPF) 1 if Duv (, ) Huv (, ) = 0 if Duv (, ) > D D 0 0 D is a positive constant and Duv (, ) is the distance between a point ( uv, ) 0 in the frequency domain and the center of the frequency rectangle Duv (, ) = ( u P/ 2) + ( v Q/ 2) 2 2 1/2 2/20/2014 84

Image Smoothing Using Filter Domain Filters: ILPF 2/20/2014 85

ILPF Filtering Example 2/20/2014 86

ILPF Filtering Example 2/20/2014 87

The Spatial Representation of ILPF 2/20/2014 88

Image Smoothing Using Filter Domain Filters: BLPF Butterworth Lowpass Filters (BLPF) of order with cutoff frequency Huv (, ) D 1 = 1 + (, )/ 0 [ Duv D] 0 2n n and 2/20/2014 89

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The Spatial Representation of BLPF 2/20/2014 91

Image Smoothing Using Filter Domain Filters: GLPF Gaussian Lowpass Filters (GLPF) in two dimensions is given Huv (, ) = e D 2 2 ( uv, )/2σ By letting σ = D 0 Huv (, ) = e 2 2 (, )/2 0 D uv D 2/20/2014 92

Image Smoothing Using Filter Domain Filters: GLPF 2/20/2014 93

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Examples of smoothing by GLPF (1) 2/20/2014 95

Examples of smoothing by GLPF (2) 2/20/2014 96

Examples of smoothing by GLPF (3) 2/20/2014 97

Image Sharpening Using Frequency Domain Filters A highpass filter is obtained from a given lowpass filter using H ( uv, ) = 1 H ( uv, ) HP LP A 2-D ideal highpass filter (IHPL) is defined as 0 if Duv (, ) D Huv (, ) = 1 if Duv (, ) > D 0 0 2/20/2014 98

Image Sharpening Using Frequency Domain Filters A 2-D Butterworth highpass filter (BHPL) is defined as Huv (, ) 1 = 1 + / (, ) [ D Duv] 2 0 n A 2-D Gaussian highpass filter (GHPL) is defined as Huv (, ) = 1 e 2 2 (, )/2 0 D uv D 2/20/2014 99

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The Spatial Representation of Highpass Filters 2/20/2014 101

Filtering Results by IHPF 2/20/2014 102

Filtering Results by BHPF 2/20/2014 103

Filtering Results by GHPF 2/20/2014 104

Using Highpass Filtering and Threshold for Image Enhancement BHPF (order 4 with a cutoff frequency 50) 2/20/2014 105

The Laplacian in the Frequency Domain Huv = π u + v 2 2 2 (, ) 4 ( ) Huv = u P + v Q 2 2 2 (, ) 4 π ( / 2) ( / 2) ) = 2 2 4 π D ( uv, ) The Laplacian image f( xy, ) =I HuvFuv (, ) (, ) 2 1 { } Enhancement is obtained gxy f xy c f xy c 2 (, ) = (, ) + (, ) = -1 2/20/2014 106

The Laplacian in the Frequency Domain The enhanced image gxy =I Fuv HuvFuv 1 (, ) (, ) (, ) (, ) 1 =I =I + { } {[ 1 Huv (, )] Fuv (, )} { 1 4 π D( uv, ) Fuv (, )} 1 2 2 2/20/2014 107

The Laplacian in the Frequency Domain 2/20/2014 108

Unsharp Masking, Highboost Filtering and High-Frequency-Emphasis Fitering g ( xy, ) = f( xy, ) f ( xy, ) mask Unsharp masking and highboost filtering gxy (, ) = f( xy, ) + k* g ( xy, ) LP f ( xy, ) = I H ( uvfuv, ) (, ) LP 1 [ ] LP mask [ LP ] [ ] { } gxy =I + k H uv Fuv 1 (, ) 1 * 1 (, ) (, ) 1 =I + { 1 k* H ( uv, ) Fuv (, )} HP 2/20/2014 109

Unsharp Masking, Highboost Filtering and High-Frequency-Emphasis Fitering 1 2 1 {[ ] } gxy (, ) =I k+ k * H ( uv, ) Fuv (, ) k 0 and k 0 1 2 HP 2/20/2014 110

Gaussian Filter D 0 =40 High-Frequency-Emphasis Filtering Gaussian Filter K1=0.5, k2=0.75 2/20/2014 111

Homomorphic Filtering f( xy, ) = ixyrxy (, ) (, ) [ f( xy, )] [ ixy (, )] [ rxy (, )] I I I =? zxy (, ) = ln f( xy, ) = ln ixy (, ) + ln rxy (, ) { zxy (, )} { ln f( xy, )} { ln ixy (, )} { ln rxy (, )} I =I =I +I Zuv (, ) = F( uv, ) + F( uv, ) i r 2/20/2014 112

Homomorphic Filtering Suv (, ) = HuvZuv (, ) (, ) sxy = HuvF (, ) ( uv, ) + HuvF (, ) ( uv, ) = I 1 (, ) (, ) { Suv} { HuvF (, ) i( uv, ) HuvF (, ) r( uv, )} { HuvF (, ) ( uv, )} { HuvF (, ) ( uv, )} 1 =I + =I +I 1 1 i = i'( xy, ) + r'( xy, ) i r r gxy e e e i xyr xy (, ) '(, ) '(, ) (, ) = sxy = i xy r xy = 0(, ) 0(, ) 2/20/2014 113

Homomorphic Filtering The illumination component of an image generally is characterized by slow spatial variations, while the reflectance component tends to vary abruptly These characteristics lead to associating the low frequencies of the Fourier transform of the logarithm of an image with illumination the high frequencies with reflectance. 2/20/2014 114

Homomorphic Filtering 2 2 c D ( uv, )/ D Huv (,) ( γ γ )1 e 0 = + γ H L L Attenuate the contribution made by illumination and amplify the contribution made by reflectance 2/20/2014 115

γ L Homomorphic Filtering γ H c = 1 D 0 = = 0.25 2 = 80 2/20/2014 116

Homomorphic Filtering 2/20/2014 117

Selective Filtering Non-Selective Filters: operate over the entire frequency rectangle Selective Filters operate over some part, not entire frequency rectangle bandreject or bandpass: process specific bands notch filters: process small regions of the frequency rectangle 2/20/2014 118

Selective Filtering: Bandreject and Bandpass Filters H ( uv, ) = 1 H ( uv, ) BP BR 2/20/2014 119

Selective Filtering: Bandreject and Bandpass Filters 2/20/2014 120

Selective Filtering: Notch Filters Zero-phase-shift filters must be symmetric about the origin. A notch with center at (u 0, v 0 ) must have a corresponding notch at location (-u 0,-v 0 ). Notch reject filters are constructed as products of highpass filters whose centers have been translated to the centers of the notches. H ( uv, ) = H( uvh, ) ( uv, ) NR k k k = 1 where H( uv, ) and H ( uv, ) are highpass filters whose centers are k at ( u, v ) and (- u,- v ), respectively. k k k k -k Q 2/20/2014 121

Selective Filtering: Notch Filters H ( uv, ) = H( uvh, ) ( uv, ) NR k k k = 1 where H( uv, ) and H ( uv, ) are highpass filters whose centers are k at ( u, v ) and (- u,- v ), respectively. k k k k -k A Butterworth notch reject filter of order n H D( uv, ) = ( u M/2 u) + ( v N/2 v) 2 2 k k k D ( uv, ) = ( u M/2 + u) + ( v N/2 + v) Q 3 1 1 ( uv, ) = [ D D uv] [ D D uv] NR 2n 2n k = 1 1 + 0k / k(, ) 1 + 0k / k(, ) 2 2 k k k 2/20/2014 122 1/2 1/2

Examples: Notch Filters (1) A Butterworth notch reject filter D =3 and n=4 for all notch pairs 0 2/20/2014 123

Examples: Notch Filters (2) 2/20/2014 124

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