cse 185 introduction to computer vision

CSE 185 Introduction to Computer

VisionImage Filtering:

Frequency Domain

Image filtering

• Fourier transform and frequency domain– Frequency view of filtering– Hybrid images– Sampling

• Reading: Chapters 3• Some slides from James Hays, David

Hoeim, Steve Seitz, …

Gaussian Box filter

Gaussian filter

Hybrid images

• Filter one image with low-pass filter and the other with high-pass filter, and then superimpose them

• A. Oliva, A. Torralba, P.G. Schyns, “Hybrid Images,” SIGGRAPH 2006

Why do we get different, distance-dependent interpretations of hybrid images?

?

Why does a lower resolution image still make sense to us? What do we lose?

Thinking in terms of frequency

Jean Baptiste Joseph Fourier had crazy idea (1807):

Any univariate function can be rewritten as a weighted sum of sines and cosines of different frequencies.

• Don’t believe it? – Neither did Lagrange,

Laplace, Poisson and other big wigs

– Not translated into English until 1878!

• But it’s (mostly) true!– called Fourier Series– there are some subtle

restrictions

...the manner in which the author arrives at these equations is not exempt of difficulties

and...his analysis to integrate them still leaves something to be desired on the score

of generality and even rigour.

Laplace

LagrangeLegendre

A sum of sinesOur building block:

Add enough of them to get any signal g(x) you want!

xAsin(

Frequency spectra

• example : g(t) = sin(2πf t) + (1/3)sin(2π(3f) t)

= +

Frequency spectra

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Frequency spectra

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Frequency spectra

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Frequency spectra

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=

Frequency spectra

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=

Frequency spectra

= 1

1sin(2 )

k

A ktk

Frequency spectra

Example: Music

• We think of music in terms of frequencies at different magnitudes

Fourier analysis in images

Intensity Image

Fourier Image

Signals can be composed

+ =

More: http://www.cs.unm.edu/~brayer/vision/fourier.html

Fourier transform

• Fourier transform stores the magnitude and phase at each frequency– Magnitude encodes how much signal there is at a

particular frequency– Phase encodes spatial information (indirectly)– For mathematical convenience, this is often notated in

terms of real and complex numbers

22 )()( IRA )(

)(tan 1

R

IAmplitude: Phase:

Computing Fourier transform

Continuous

Discrete

k = -N/2..N/2

Fast Fourier Transform (FFT): NlogN

Phase and magnitude

• Fourier transform of a real function is complex– difficult to plot,

visualize– instead, we can think of

the phase and magnitude of the transform

• Phase is the phase of the complex transform

• Magnitude is the magnitude of the complex transform

• Curious fact– all natural images have

about the same magnitude transform

– hence, phase seems to matter, but magnitude largely doesn’t

• Demonstration– Take two pictures, swap

the phase transforms, compute the inverse - what does the result look like?

This is the magnitude transform of the cheetah pic

This is the phase transform of the cheetah pic

This is the magnitude transform of the zebra pic

This is the phase transform of the zebra pic

Reconstruction with zebra phase, cheetah magnitude

Reconstruction with cheetah phase, zebra magnitude

The convolution theorem• The Fourier transform of the convolution

of two functions is the product of their Fourier transforms

• Convolution in spatial domain is equivalent to multiplication in frequency domain!

]F[]F[]F[ hghg

]]F[][F[F* 1 hghg

Properties of Fourier transform

• Linearity

• Fourier transform of a real signal is symmetric about the origin

• The energy of the signal is the same as the energy of its Fourier transform

See Szeliski Book (3.4)

2D FFT

1,...,2,1,0,1,...,2,1,0for

),(),(1

0

1

0

)//(2

NyMx

evuFyxfM

u

N

v

NvyMuxj

• Fourier transform (discrete case)

• Inverse Fourier transform:

• u, v : the transform or frequency variables• x, y : the spatial or image variables

1,...,2,1,0,1,...,2,1,0for

),(1

),(1

0

1

0

)//(2

NvMu

eyxfMN

vuFM

x

N

y

NvyMuxj

Euler’s formula

2D FFT

Sinusoid with frequency = 1 and its FFT

2D FFT

Sinusoid with frequency = 3 and its FFT

2D FFT

Sinusoid with frequency = 5 and its FFT

2D FFT

Sinusoid with frequency = 10 and its FFT

2D FFT

Sinusoid with frequency = 15 and its FFT

2D FFT

Sinusoid with varying frequency and their FFT

Rotation

Sinusoid rotated at 30 degrees and its FFT

2D FFT

Sinusoid rotated at 60 degrees and its FFT

2D FFT

Image analysis with FFT

Image analysis with FFT

http://www.cs.unm.edu/~brayer/vision/fourier.html

Filtering in spatial domain

-101

-202

-101

* =

Filtering in frequency domain

FFT

FFT

Inverse FFT

=

FFT in Matlab

• Filtering with fft

• Displaying with fft

im = double(imread('cameraman.tif'))/255;[imh, imw] = size(im); hs = 50; % filter half-sizefil = fspecial('gaussian', hs*2+1, 10); fftsize = 1024; % should be order of 2 (for speed) and include paddingim_fft = fft2(im, fftsize, fftsize); % 1) fft im with paddingfil_fft = fft2(fil, fftsize, fftsize); % 2) fft fil, pad to same size as imageim_fil_fft = im_fft .* fil_fft; % 3) multiply fft imagesim_fil = ifft2(im_fil_fft); % 4) inverse fft2im_fil = im_fil(1+hs:size(im,1)+hs, 1+hs:size(im, 2)+hs); % 5) remove paddingfigure, imshow(im);figure, imshow(im_fil);

figure(1), imagesc(log(abs(fftshift(im_fft)))), axis image, colormap jet

Questions

Which has more information, the phase or the magnitude?

What happens if you take the phase from one image and combine it with the magnitude from another image?

FilteringWhy does the Gaussian give a nice smooth image, but the

square filter give edgy artifacts?

Gaussian Box filter

Gaussian

Gaussian filter

Box Filter

Box filter

Sampling

Why does a lower resolution image still make sense to us? What do we lose?

Throw away every other row and

column to create a 1/2 size image

Subsampling by a factor of 2

• 1D example (sinewave):

Aliasing problem

• 1D example (sinewave):

Aliasing problem

• Sub-sampling may be dangerous….• Characteristic errors may appear:

– “Wagon wheels rolling the wrong way in movies”

– “Checkerboards disintegrate in ray tracing”

– “Striped shirts look funny on color television”

Aliasing problem

Aliasing in video

Aliasing in graphics

Resample the checkerboard by taking one sample at each circle. In the case of the top left board, new representation is reasonable.

Top right also yields a reasonable representation.

Bottom left is all black (dubious) and bottom right has checks that are too big.

Sampling scheme is crucially related to frequency

Constructing a pyramid by taking every second pixel leads to layers that badly misrepresent the top layer

• When sampling a signal at discrete intervals, the sampling frequency must be 2 fmax

• fmax = max frequency of the input signal

• This will allows to reconstruct the original from the sampled version without aliasing

good

bad

v v v

Nyquist-Shannon Sampling Theorem

Anti-aliasing

Solutions:• Sample more often

• Get rid of all frequencies that are greater than half the new sampling frequency– Will lose information– But it’s better than aliasing– Apply a smoothing filter

Algorithm for downsampling by factor of 2

1. Start with image(h, w)2. Apply low-pass filter

im_blur = imfilter(image, fspecial(‘gaussian’, 7, 1))

3. Sample every other pixelim_small = im_blur(1:2:end, 1:2:end);

Sampling without smoothing. Top row shows the images, sampled at every second pixel to get the next; bottom row shows the magnitude spectrum of these images.

substantial aliasing

Magnitude of the Fourier transform of each image displayed as a log scale (constant component is at the center). Fourier transform of a resampled image is obtained by scaling the Fourier transform of the original image and then tiling the plane

Sampling with smoothing (small σ). Top row shows the images. We get the next image by smoothing the image with a Gaussian with σ=1 pixel, then sampling at every second pixel to get the next; bottom row shows the magnitude spectrum of these images.

Low pass filter suppresses high frequency components with less aliasing

reducing aliasing

Sampling with smoothing (large σ). Top row shows the images. We get the next image by smoothing the image with a Gaussian with σ=2 pixels, then sampling at every second pixel to get the next; bottom row shows the magnitude spectrum of these images.

Large σ less aliasing, but with little detailGaussian is not an ideal low-pass filter

lose details

Subsampling without pre-filtering

1/4 (2x zoom) 1/8 (4x zoom)1/2

Subsampling with Gaussian pre-filtering

G 1/4 G 1/8Gaussian 1/2

Why does a lower resolution image still make sense to us? What do we lose?

Why do we get different, distance-dependent interpretations of hybrid images?

?

Salvador Dali invented Hybrid Images?

Salvador Dali“Gala Contemplating the Mediterranean Sea, which at 30 meters becomes the portrait of Abraham Lincoln”, 1976

Application: Hybrid images

• Combine low-frequency of one image with high-frequency of another one

• Sad faces when looked closely• Happy faces when looked a few meters away• A. Oliva, A. Torralba, P.G. Schyns, “Hy brid Im ages,”

SIGGRAPH 2006

• Early processing in humans filters for various orientations and scales of frequency

• Perceptual cues in the mid-high frequencies dominate perception

• When we see an image from far away, we are effectively subsampling it

Early Visual Processing: Multi-scale edge and blob filters

Clues from human perception

Campbell-Robson contrast sensitivity curve

Detect slight change in shades of gray before they become indistinguishableContrast sensitivity is better for mid-range spatial frequencies

Hybrid image in FFT

Hybrid Image Low-passed Image High-passed Image

PerceptionWhy do we get different, distance-dependent

interpretations of hybrid images?

?

Things to remember• Sometimes it makes sense to

think of images and filtering in the frequency domain– Fourier analysis

• Can be faster to filter using FFT for large images (N logN vs. N2 for auto-correlation)

• Images are mostly smooth– Basis for compression

• Remember to low-pass before sampling

Practice question

1. Match the spatial domain image to the Fourier magnitude image

1 54

A

32

C

B

DE

Practice question

1. Match the spatial domain image to the Fourier magnitude image

1 54

A

32

C

B

DE

1 – D, 2 – B, 3 – A, 4 – E, 5 - C