1

• Low Pass Filter

• High Pass Filter

• Band pass Filter

• Blurring

• Sharpening

Image Processing

Image Operations in the Frequency Domain

Frequency Bands

Percentage of image power enclosed in circles(small to large) :

90, 95, 98, 99, 99.5, 99.9

Image Fourier Spectrum

Blurring - Ideal Low pass Filter

90% 95%

98% 99%

99.5% 99.9%

2

The Convolution Theorem

g = f * h g = f h

implies implies

G = F H G = F * H

Convolution in one domain is a multiplication in the other and vice

versa

Image Operation in the Frequency Domain

FilteredImage

Image Transform

Filtered

FFT

FFT-1NewImage

3

Low pass Filter

f(x,y) F(u,v)

g(x,y) G(u,v)

G(u,v) = F(u,v) • H(u,v)

spatial domain frequency domain

filter

•

f(x,y) F(u,v)

H(u,v)g(x,y)

H(u,v) - Ideal Low Pass Filter

u

v

H(u,v)

0 D0

1

D(u,v)

H(u,v)

H(u,v) = 1 D(u,v) ≤ D0

0 D(u,v) > D0

D(u,v) = √ u2 + v2

D0 = cut off frequency

Blurring - Ideal Low pass Filter

98.65%

99.37%

99.7%

Blurring - Ideal Low pass Filter

98.0%

99.4%

99.6% 99.7%

99.0%

96.6%

4

The Ringing Problem

G(u,v) = F(u,v) • H(u,v)

g(x,y) = f(x,y) * h(x,y)

Convolution Theorem

sinc(x)

h(x,y)H(u,v)

↑ D0 ↓ Ringing radius + ↓ blur

IFFT

The Ringing Problem

0 50 100 150 200 2500

50

100

150

200

250

Freq. domain

Spatial domain

H(u,v) - Gaussian Filter

D(u,v)0 D0

1

H(u,v)

uv

H(u,v)

D(u,v) = √ u2 + v2

H(u,v) = e-D2(u,v)/(2D20)

Softer Blurring + no Ringing

e/1

Blurring - Gaussain Lowpass Filter

96.44%

98.74%

99.11%

5

The Gaussian Lowpass Filter

Freq. domain

Spatial domain

0 50 100 150 200 250 3000

50

100

150

200

250

300

H(u,v) - Butterworth Filter

D(u,v)0 D0

1

H(u,v)

0.5

uv

H(u,v)

D(u,v) = √ u2 + v2

H(u,v) = 1 + (D(u,v)/D0)2n

1

Blurring - Butterworth Lowpass Filter

93.70%

95.95%

97.48%

The Butterworth Lowpass Filter

Freq. domain

Spatial domain

0 5 0 1 0 0 1 5 0 2 0 0 2 5 00

5 0

1 0 0

1 5 0

2 0 0

2 5 0

6

D0=1D0=2D0=3

Ideal Butterworth Gaussian

Low Pass Filters - Comparison Blurring in the Spatial Domain:

Averaging = convolution with 1 11 1

= point multiplication of the transform with sinc:

0 50 1000

0.05

0.1

0.15

-50 0 500

0.2

0.4

0.6

0.8

1

Gaussian Averaging = convolution with 1 2 12 4 21 2 1

= point multiplication of the transform with a gaussian.

Image Domain Frequency Domain

Original - 4 levelQuantized Image

OriginalNoisy Image

Smoothed Image

Smoothed Image

Low Pass Filtering - Image SmoothingImage Sharpening - High Pass Filter

H(u,v) - Ideal Filter

H(u,v) = 0 D(u,v) ≤ D0

1 D(u,v) > D0

D(u,v) = √ u2 + v2

D0 = cut off frequency

0 D0

1

D(u,v)

H(u,v)

u

v

H(u,v)

7

H(u,v)

D(u,v)0 D0

1

D(u,v) = √ u2 + v2

High Pass Gaussian Filter

u

v

H(u,v)

H(u,v) = 1 - e-D2(u,v)/(2D20)

e/11−

H(u,v)

D(u,v)0 D0

1

0.5

D(u,v) = √ u2 + v2

H(u,v) = 1 + (D0/D(u,v))2n

1

High Pass Butterworth Filter

u

v

H(u,v)

1 -

Ideal

Butterworth

Gaussian

High Pass Filters - ComparisonHigh Pass Filtering

Original High Pass Filtered

8

High Frequency Emphasis

Original High Pass Filtered

+

High Frequency Emphasis

Emphasize High Frequency.Maintain Low frequencies and Mean.

(Typically K0 =1)

H'(u,v) = K0 + H(u,v)

0 D0

1

D(u,v)

H'(u,v)

High Frequency Emphasis - Example

Original High Frequency Emphasis

Original High Frequency Emphasis

High Pass Filtering - Examples

Original High pass Butterworth Filter

High Frequency Emphasis

High Frequency Emphasis +

Histogram Equalization

9

Band Pass Filtering

H(u,v) = 1 D0- ≤ D(u,v) ≤ D0 +0 D(u,v) > D0 +

D(u,v) = √ u2 + v2

D0 = cut off frequency

u

v

H(u,v)

0

1D(u,v)

H(u,v)

D0- w2

D0+w2

D0

0 D(u,v) ≤ D0 -w2

w2

w2

w2

w = band width

H(u,v) = 1 D1(u,v) ≤ D0 or D2(u,v) ≤ D0

0 otherwise

D1(u,v) = √ (u-u0)2 + (v-v0)2

D0 = local frequency radius

Local Frequency Filtering

u

v

H(u,v)

0 D0

1

D(u,v)

H(u,v)

-u0,-v0 u0,v0

D2(u,v) = √ (u+u0)2 + (v+v0)2

u0,v0 = local frequency coordinates

H(u,v) = 0 D1(u,v) ≤ D0 or D2(u,v) ≤ D0

1 otherwise

D1(u,v) = √ (u-u0)2 + (v-v0)2

D0 = local frequency radius

Band Rejection Filtering

0 D0

1

D(u,v)

H(u,v)

-u0,-v0 u0,v0

D2(u,v) = √ (u+u0)2 + (v+v0)2

u0,v0 = local frequency coordinates

u

v

H(u,v)

+

+ =

=

Additive Noise Filtering

10

Local Reject Filter - Example

Original Noisy image Fourier Spectrum

Band Reject Filter

Local Reject Filter - Example

Original Noisy image Fourier Spectrum

Local Reject Filter

Homomorphic Filtering (multiplicative Noise Filtering)

Noise Model:

Original Image

Noise

Actual Image

i(x,y)

n(x,y)

f(x,y) = i(x,y) • n(x,y)

Goal: Clean multiplicative noise

( ) ( )( ) ( )( ) ( )( )yxnFyxiFyxnyxiF ,~,~,,~ ⋅≠⋅

Perform:

z(x,y) = log(f(x,y))

I(u,v) + N(u,v)

Apply filter H(u,v)

S(u,v) = H(u,v)•Z(u,v)-1

i'(x,y) + n'(x,y)

g(x,y) = exp(s(x,y))

Homomorphic Filtering:

log FFT-1FFT H(u,v) expimage image

log(i(x,y) • n(x,y)) = log(i(x,y)) + log(n(x,y))

=

Z(u,v) =

H(u,v)•I(u,v) + H(u,v)•N(u,v)=

s(x,y) =

exp(i'(x,y)) • exp(n'(x,y))=

11

Surface Reflectance r(x,y)

Illumination i(x,y)

Homomorphic Filtering - Example

Reflectance Model:

Brightness f(x,y) = r(x,y) • i(x,y)

Assumptions:

Illumination changes "slowly" across sceneIllumination ≈ low frequencies.

Surface reflections change "sharply" across scenereflectance ≈ high frequencies.

Illumination Reflectance Brightness

Homomorphic Filtering

Original Filtered

from: Jashmin K. Shah

More ExamplesPattern Matching in the

Freq. Domain

• Pattern Matching:– Finding occurrences of a particular

pattern in an image.

• Pattern:– Typically a 2D image fragment.– Much smaller than the image.

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Image Similarity Measures

• Image Similarity Measure:– A function that assigns a nonnegative

real value to two given images.– Small measure high similarity– Preferable to be a metric distance

(non-negative, identity, symmetric, trianglebinequality)

– Can be combined with thresholding:

d( - ) ≥ 0

1 ( , )( , )

0d P Q threshold

f P Qotherwise

<⎧ ⎫= ⎨ ⎬⎩ ⎭

The Matching Approach

• Scan the entire image, pixel by pixel.• For each pixel, evaluate the similarity

between its local neighborhood and the pattern.

The Euclidean Distance as a Similarity Measure

• Given:– k×k pattern P– n×n image I– kxk window of image I located at x,y - Ix,y

• For each pixel (x,y), we compute the distance:

• Complexity: 2 2( )O n k

( )

( ) ( )( )[ ]∑−

=

−++=

=−=

1

0,

22

2

,2,2

,,1

1,

k

ji

yxyxE

jiPjyixIk

PIk

PId

Improvement: FFT

Fixed 2( * )I P

• Convolution can be applied rapidly using FFT.

• Complexity: 2( log )O n n

( )2 * mask of 1'sI k k×

( ) 2

,2,2 1, PI

kPId yxyxE −=

( ) ( ) ( ) ( )jiPjyixIjiPjyixIk

ji,,2,, 2

1

0,

2 ++−+++= ∑−

=

13

Naïve and FFT Approaches: Performance

3.5 Sec.31.30 Sec.Run time for 64×643.5 Sec.4.86 Sec.Run time for 32×323.5 Sec.1.33 Sec.Run time for 16×16

NoYesInteger ArithmeticSpace Time Complexity

FFTNaïve2( log )O n n2 2( )O n k

2n2n

Performance table for a 1024×1024 image, on a 1.8 GHz PC:

0

0.5

1

1.5

2

x 107

Normalized Gray-scale Correlation

• NGC:– A similarity measure, based on a

normalized cross-correlation function.– Maps two given images to [0,1] (absolute

value).– Invariant to intensity scale and offset.

( ) ( ) ( )1111

,,

,, PPII

PPIIPId

yx

yxyxNC −−

−⋅−=

( ) ( ) ( )=

−−

−⋅−=

1111

,,,

,,, PPII

PPIIPId

yxyx

yxyxyxNC

( )( )22,

22,,

,2

,

PkPPIkII

PIkPI

yxyxyx

yxyx

−⋅−⋅

⋅−⋅=

( ) ( ) YXnYXYYXX −⋅=−⋅− 11Note that:

The above expression can be implemented efficiently using 3 convolutions.

and thus:

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0

0.5

1

1.5

0

0.5

1

1.5

2

x 107

Euclidean distance similarity measure

1-NGC similarity measure

Euclidean distance similarity measure

1-NGC similarity measure

10 20 30 40 50 60 70

10

20

30

40

50

60

0.2

0.4

0.6

0.8

1

1.2

10 20 30 40 50 60 70

10

20

30

40

50

60

1

2

3

4

5

6

7

8

9

10

11

x 106

Computer Tomographyusing FFT

CT Scanners• In 1901 W.C. Roentgen won the Nobel

Prize (1st in physics) for his discovery of X-rays.

• In 1979 Hounsfield & Cormack, won the Nobel Prize for developing the computer tomography.

• The invention revolutionized medical imaging.

1st prototype of CT scanner

15

Allan M. Cormack Godfrey N. Hounsfield

Wilhelm Conrad Röntgen

Computerized Tomography

Reconstruction from projections

f(x,y)

θ1

θ2

Projection & Sinogram

Sinogramt

θ

Sinogram: All projections

P(θ,t)

f(x,y)

t

θ

y

x

X-rays

Projection: All ray-sums in a direction

π

CT Image & Its Sinogram

K. Thomenius & B. Roysam

16

The Slice Theorem

f(x,y)

θ1

x

y

θ1

u

v

spatial domain frequency domainf(x,y) = objectg(x) = projection of f(x,y) parallel to the y-axis.

g(x) = ∫f(x,y)dy

F(u,v) = ∫ ∫ f(x,y) e -2πi(ux+vy) dxdyFourier Transform of f(x,y):

Fourier Transform at v=0 :F(u,0) = ∫ ∫ f(x,y) e -2πiuxdxdy

= ∫ [ ∫ f(x,y)dy] e -2πiuxdx

= ∫ g(x) e -2πiuxdx = G(u)

The 1D Fourier Transform of g(x)

Fourier Transform

u

vF(u,v)

Interpolations Method:Interpolate (linear, quadratic etc) in the frequency space to obtain all values in F(u,v).Perform Inverse Fourier Transform to obtain the image f(x,y).

Reconstruction from Projections - Example

Original simulated density image

8 projections-Frequency Domain

120 projections-Frequency Domain

8 projections-Reconstruction

120 projections-Reconstruction

Back Projection Reconstruction

g(x) is Back Projected along the line of projection.The value of g(x) is added to the existing values ateach point which were obtained from other back projections.

Note: a blurred version of the original is obtained.(for example consider a single point object is backprojected into a blurred delta).

17

Back Projection Reconstruction - Example

1 view 8 views

32 views 180 views

Filtered Back Projection - Example

1 view 2 views 4 views

8 view 16 views 32 views

180 views

FrequencySpatial

Filter

THE END