Digital Image Processing_advanced Course

7/31/2019 Digital Image Processing_advanced Course

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Digital Image Processing

Transforms Advanced level


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Transforms Advanced level

Until now we considered discrete sinusoidal transforms. This concept should be familiar to someone that has BSc in

electrical engineering.

However, there are numerous other transforms developed

in the last 10 or 20 years that have found applications innumerous research areas.

Digital image processing is attractive application for thesetransforms.

Here, we will give and brief overview of advanced leveltransforms used in digital image processing.


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Digital image profile

Typical profile of the digital image (along singleimage line) is given below.

We can assume that im age has almost constant values w it hsmall variat ions in several neighbor pixels and aft er t hat w ehave abrupt changes of lum inance.

For image fi l tering it is import ant t hat image energy is

concentrated in t he smallest t ransformation coeff icients.Similar proper t y is desirable for image compression.


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Need for rectangular wavetransform

When we are performing 2D sinusoidal wave transformon profile of digital image (see slide 3) we need largenumber of coefficients to accurately describe abruptpoints since it is very difficult to represent abruptchanges with smooth functions.

Then several discrete rectangular wave transforms aredeveloped based on rectangular shaped periodicfunctions.

We will introduce some of them on several forthcoming

slides.


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Hadamard transform

Here, we will introduce the Hadamard transform forN=2n samples. There are some alternative forms ofHadamard transform for other number of samples.

Hadamard transform derivation begins with matrix:

2

1 1

1 1

=

Hadamard transform for N=2 is defined withtransformation matrix:

2 2

1

2= H


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Hadamard transform

Hadamard transform of larger dimensions can be definedusing recursively as:

1N N

N= H where / 2 / 2

/ 2 / 2

N N

N

N N

=

Example forN=8:

H8 12 2

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 11 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1

0

7

3

41

6

2

5

=

L

N

MMMM

MMMMMM

M

O

Q

PPPP

PPPPPP

P

Number on t heend of each rowis number ofsign changes in

the row ( f rom 1to 1 and viceversa).


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Walsh ordering

Very often instead of the Hadamard transform we areusing reordered form of the Hadamard transform (thisreordering is called Walsh ordering and sometimes thecorresponding transform is called the Walsh transform).

Reordering is performed in such manner that on the top ofthe transform matrix is put the row of the Hadamardtransform matrix with the smallest number of sign changes(from 1 to -1 and from -1 to 1) and below rows are

ordered in increasing order of sign changes. Hadamard and Walsh transforms are equivalent to each

other but the Walsh transform order has analogy with thesinusoidal transform with respect to increasing of

frequencies represented with corresponding coefficients.


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Walsh matrix - Example

1 1 1 1 1 1 1 1 0

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 2

1 1 1 1 1 1 1 1 31

1 1 1 1 1 1 1 1 42 21 1 1 1 1 1 1 1 5

1 1 1 1 1 1 1 1 6

1 1 1 1 1 1 1 1 7

Walsh transform matrix forN=8.

Your task (maybe it is not quite

simple) is to determine the inverseWalsh transform.

The simpler task is to graphicallyrepresent basis function of the

Hadamard and Walsh transforms.

Several alternative rectangular transforms are reviewed intextbook. Here, we will due to its importance just mention theHaar expansion.


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Columns of the Walsh matrix

Columns of t he Walsh matr ixare quit e similar t o t hesinusoidal function used insinusoidal t ransform s(expansions).

funct ion on frequency 0

funct ion on frequency 1

.

.

.

fr equency gradually increases


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Haar transform

Definition of the Haar transform is not quite simple.Consider integer k defined in domain 0 k N-1. It can bewritten as: k= 2p+q-1.

Integers p and q are selected in such a manner that p isthe largest integer that satisfy 2p k.

Values ofk represents rows of the Haar transform matrix(here the first row has index 0, next one has index 1while the last one has the index N-1).

The second index (corresponding to columns) is denoted

with i and this index is also defined within the range0 i N-1.

Now we can defined x as x = i / N.


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Haar transform

There is direct relationship between x and column indexi.

Elements in rows of the Haar transform are denoted ashk(x) where k is row index described on the previous

slide. These vectors are defined as:

h xN

0

1( ) =

/ 2

/ 2

1 1/ 22

2 2

1 1/ 2( ) 2

2 2

0 elsewhere

p

p p

p

k p p

q qx

q qh x x

N


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Haar transform - Example

In order to understand the Haar transform and role of(p,q) we give as an example matrix for N=8 and we givevalues of (p,q) next to each row.

1 1 1 1 1 1 1 1(0,0)

1 1 1 1 1 1 1 1 (0,1)

2 2 2 2 0 0 0 0 (1,1)

(1, 2)1 0 0 0 0 2 2 2 2

(2,1)8 2 2 0 0 0 0 0 0(2,2)0 0 2 2 0 0 0 0

(2,3)0 0 0 0 2 2 0 0

(2,4)0 0 0 0 0 0 2 2

=

Hr

There is naturalquest ion how itlooks like theinverse mat rix int his case?


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Haar transform (p,q)

Lets try to understand roles of (p,q) in transform. Obviously, p corresponds to power of elements

represented in matrix rows. This value is called thescale.

Coefficient q define position on non-zero elements in therows. Then it is called posit ion or shif t .

Haar transform will be used later for introducing quite

important concept ofwavelets and then we will explaindetails about this transform.


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Haar trans. 2 and 4-samples

Assume that we have 4 samples and assume that wetry to transform these 4 samples using HT for 2samples. Then this transform will perform operation

on the first pair of samples and after that on thesecond pair of samples. Results of operations appliedto pairs of samples are submitted in upper and lowerlines.

x(0)

x(1)HTN=2

(0) (0) 2 / 2 (1) 2 / 2y x x= +(1) (0) 2 / 2 (1) 2 / 2y x x=

Haar t ransformer for N= 2 iscreated in such manner t han onebranch is used to peek up tw oadjacent samples and t o addthem (w ith some mult iple) w hi lethe second branch is used t o

subt ract adj acent samples.


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Haar trans. 2 and 4-samples

In order to get the HT for N=4 samples we can notethat lower line already produces 2 samples of theoutput while we have to apply the HT for N=2 forupper line.

x(n), n=0,1,2,3 HTN=2

(0) 2 / 2 (1) 2 / 2, (2) 2 / 2 (3) 2 / 2x x x x+ +

(0) 2 / 2 (1) 2 / 2, (2) 2 / 2 (3) 2 / 2x x x x

x(n), n=0,1,2,3 HTN=2

HTN=2

y(0)

y(1)

y(2), y(3)


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Haar transf. N=8

Haar transform for larger number of samples can beobtained in similar manner. For example for N=8:

x(n), n=0,...,7 HTN=2

HTN=2

HTN=2

y(0)

y(1)

y(2), y(3)

y(4), y(5), y(6), y(7)


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Haar transf. Conclusion

It can be observed that lower lines of the Haartransform are not used for further processing.

For self learning session students should try to realizethe inverse Haar transformer.

The Haar transform follows general logic of theimage in transform domain. Namely, in the imagemore interesting things are on the lower frequencieswhere the high amount of energy is concentrated

than on higher frequencies. Then it is more importantto process samples on low than on high frequencies.


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Haar transf. Basis images

low frequenciesw it h image energy

high f requenciesw it h image details

Basis imagesfor N=8.


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Optimal transform

The Haar transform will be used for introduction of veryimportant class of transforms called wavelets. Thesetransforms have slightly different logic behind with respectto the introduced transforms. They are appropriate for

digital image processing since energy of images is hiddenon low frequencies while details are on high frequencies.

Before we proceed with wavelet transform description wewant to address the question what the ideal (optimal)

transform would be for us? Based on introduced facts we can conclude that the best

transform would be the one concentrating the signalenergy in the smallest number of transformation

coefficients.


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Optimal transform

Assume that we have transform with transformation matrixA. Then transformation can be written as:

X=A x Introduce important concept ofeigenvalues and

eigenvectors of matrix A. Eigenvalues i are solution of equation:

Let solutions of this equation be: i, i=1,...,N. These solutions are called eigenvalues.

| | 0

=A IUnit y matr ix of t he same

dimensions as matr ix A


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Eigenvalues

For a given i we should determine vector v isatisfying:Aq i= iq i

It is easy to show that under given conditions (wheni is solution of previously defined equation) thereare infinite number of vectors satisfying thisequation.

We can limit number of solutions to 1 bynormalization of vector q i in such a way that its

amplitude (square root of sum of squares of vectorelements) is equal to 1, | | q i| | = 1.


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Eigenvalues matrix

It holds: QTAQ= Q QT= A

Determine cross-correlation of transform X:CXX= E{ XX

T} = E{ AxxTAT} = AE{ xxT} AT

CXX= AcxxAT

For self -exercise review basicmat rix operat ions.

cross-correlat ion m atr ixof signal

Energy concentrated in transform can be connected with energy ofmatrix CXX and this energy has the smallest loss when CXX is a diagonal

matrix.


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Ideal transform

It is easy to conclude that ideal matrix CXX isproduced when A is matrix of eigenvalues ofcxx.

When cxx is known we can determine idealtransform ofA as a matrix of eigenvalues.

It can be shown that when we use just M the largesttransformation coefficients of vector obtained meansquared error is equal to:

1

N

i

i M

MSE= +

= eigenvalues corresponding t ot runcated coeff icient s, i.e.,sum of t he smallest N-Meigenvalues


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Ideal transform

It can be shown that this transform produces thesmallest MSE when we take M transformationcoefficients.

This transform is called the Karhunen-Loeve (KL) andit represents ideal transform that is goal for

compression and filtering applications. Why we are not using the KL? The answer is quite simple. The KL requires

calculation of the auto-correlation matrix and

eigenvalues. Both of these operations are quitedemanding.


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Eigenvalues in digital images

It would be great that the KL be applicable in digitalimages since most of images have structure that wouldproduces extremely small number of significanteigenvalues while huge number corresponds tocomponents of insignificant energy.

Fortunately, the DCT approaches to the KL transformfor numerous practical images!!!!

This is the reason why one of the most common image

compression standards is based on the DCT.


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Eigenvalues - Examples

We will review steps in calculation of eigenvalues byhand and how it can be implemented in MATLAB.

The following matrix is given:

1 2 1

3 1 0

1 0 1

=

R

I n MATLABR= [ 1 2 -1;3 1 0;1 0 -1]

3 2

1 2 1

det( ) 3 1 0 6 6

1 0 1

= = + + +

R

I n MATLABkar_j ed= poly(A)


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There are no direct approach for solving polynomialequation of higher order but to use numerical means.

Then it is quite usable MATLAB function eig(R) thatproduces eigenvalues.

In our case eigenvalues are:13.337 2,3-1.168 j0.658. Determine now one of the eigenvectors by hand.

It is eigenvector corresponding to the first eigenvalue

1.


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For this eigenvalue holds:11

21

31

2.337 2 1

3 2.337 0

1 0 4.337

q

q

q

=

0

q31=q11/4.337=0.231q11 q21=3/2.337q11=1.284q11

It is easy to prove that this expression satisfy the first

equation (accuracy is limited by small error in rounding)that is not independent on the other two.

-2.337q11+2q21-q31=0


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Eigenvalues - Example

Then the corresponding eigenvector is [q11, 1.284q11,0.231q11]. This vector can be normalized to modulo 1:

Then the eigenvector is [0.608,0.781,0.140].

MATLAB function can perform this in quite simple manner:

2 2 2 2

11 11 11 11(1.284 ) (0.231 ) 2.702 1q q q q+ + = =

11 0.608q =

[ Q,L] = eig([ 1 2 -1;3 1 0;1 0 -1] )matrix of

eigenvectors

diagonal matr ix of

eigenvalues


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Eigenvalues - Example The KL transform can be calculated for line of image as:

clearIm=imread('cameraman.tif');A=double(Im(180,:));A=A(113:142);

A=A-mean(A);[KL,L]=eig(A'*A) In numerous cases just one eigenvalue corresponds to

the most of the image energy and from entire matrix inthe KL transform it is enough to memorize just severaltransformation coefficients (eigenvalues). However, thisshould be paid with calculation complexity and withmemorizing eigenvectors for this transform (KL is signaldependent).


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Wavelet transform

Now, we can consider the Haar transform realized forN=8:

x(n), n=0,...,7 HTN=2

HTN=2

HT

N=2

y(0)

y(1)y(2), y(3)

y(4), y(5), y(6), y(7)

We noted t hat in t he low er branches t here are no import ant component s and weare not processing t hese branches fur t her. Then the Haar t ransform mat rix haslarge number of zero coeff icient s in t ransform ation matr ix.


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Wavelet with Haar wavelet

The first decomposition stage can be illustrated in thefollowing manner.

g(n)

h(n)

g(n) has impulse response

and 0 for other n

(0) 2 / 2g = (1) 2 / 2g =

h(n) has impulse response

and 0 for other n

(0) 2 / 2h =

(1) 2 / 2h =

x(n)

x(n)* g(n)

x(n)* h(n)

2

2

xg(2n)

xh(2n)

Circuit f ordecimation(downsampling).I t t akes evensamples of inpu tsignal.


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Haar wavelet

In upper branch we have samples:

while in the lower branch we have

0

(2 ) (2 1)( )

2

x n x nx n

+ +=

1

(2 ) (2 1)( )

2

x n x nx n

+=

n=[0,N/2-1)

n=[0,N/2-1)

How input (original) signal can be reconstructed?


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Signal reconstruction from theHaar wavelet

Let the first step in the procedure be upsampling defined as:

In the next step we are putting both signals through the same

filters as in the first stage (g(n) and h(n)). Finally, thesesignals should be summed.

Output of g(n) filter in upper branch is signal with samples:[(x(0)+x(1))/2,(x(0)+x(1))/2],

[(x(2)+x(3))/2,(x(2)+x(3))/2]... In lower branch we have: [(x(0)-x(1))/2,(-x(0)+x(1))/2],

[(x(2)-x(3))/2,(-x(2)+x(3))/2]...

(2 ) ( ) (2 1) 0 0,1i i ix n x n x n i= + = =


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Haar Wavelet - Scheme Decomposition (analysis) and reconstruction (synthesis) in

the case of the wavelet transform with Haar wavelet canbe given as.

g(n)

h(n)

x(n)2

2

Point t hat separat e analysisand synt hesis stage.

2

2

g(n)

h(n)

+

x(n)

There is a question why this transform is performed?


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Need for wavelet transform

In upper branch we are performing lowpass filtering withlowpass filter that average samples.

Highpass filtering is performed in lower branch.

This corresponds to the spectral characteristic of digital

images: different amounts of energy on low and highfrequencies.

Since low frequency region has huge amount of energywe are performing further decomposition of this part.


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Wavelet transform indecomposition

For exercise create system that performs synthesis of inputsignal from this decomposition.

In practice after wavelet decomposition we perform filteringof coefficients corrupted by noise or compression of signalwith removing from visual quality point of view un-important coefficients.

Then output signal is different from the input signal.

g(n)

h(n)

2

2

g(n)

h(n)

2

2

g(n)

h(n)

2

2

Element s on higher f requencies arecalled details.

elements on low er frequencies are

called scale (analysis part or r oughpart)


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General wavelet transform

The Haar wavelet was the first wavelet transform.

This wavelet (Haar) is good for modeling functionswith abrupt changes in the considered interval.

For slower variations this transform is not the best

one. Can we construct the wavelet transform that is

suitable for other slower variations in signal, i.e.,constructed using alternative filtering functions.

Model of general wavelet transform will beconsidered within this and the next lecture.


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Model of wavelet transformer

We are interested in conditions that should be satisfied withfilters h0(n) , h1(n) , g0(n) and g1(n) , in order that inputsignal be the same as the output one.

It is obvious that filters h0(n) and h1(n) should contain allfrequencies in order that we can reconstruct signal at theoutput.

h0(n)

h1(n)

2

2

2

2

g0(n)

g1(n)

+x(n)


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Filters in spectral domain

The simplest method forfilter bank analysis isusing the Z-transform.

|H0( )| |H1( )|

low

frequencieshigh

frequencies

( ) ( ) n

n

X z x n z

=

=

( ) (2 )dx n x n=

1/ 2 1/ 21

( ) [ ( ) ( )]2dX z X z X z= + Z-t ransform ofdow nsampling circuit :

For upsampling circuit:( / 2) 0,2,4,...

( )0 elsewhere

u

x n nx n

==

2( ) ( )uX z X z=

Check!


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Wavelet filters design

At the output of the wavelet system(aft er relat ively

simple mathematical derivat ions) we obtain:

0 0 1 1

0 0 1 1

1

( ) [ ( ) ( ) ( ) ( )] ( )2

1[ ( ) ( ) ( ) ( )] ( )

2

X z G z H z H z G z X z

G z H z H z G z X z

= +

+ +

it should beequal to 1

it should be 0

Assuming that we know h0(n) and h1(n)equation for determination of the g0(n)and g1(n) in the Z-domain is:

10

01

( )( ) 2

( )( ) det( ( ))m

H zG z

H zG z z

=

H


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Denote:0 0

1 1

( ) ( )( )

( ) ( )m

H z H zz

H z H z

=

H

Assume that det(Hm(z))=z-(2k+1) (delay only) and for =2we obtain:

0 1

1

1 0

( ) ( 1) ( )

( ) ( 1) ( )

n

n

g n h n

g n h n+

=

= For =-2 it follows:10 1

1 0

( ) ( 1) ( )

( ) ( 1) ( )

n

n

g n h n

g n h n

+=

=


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Under introduced assumptions it holds:0 0 1 1( ) ( ) ( ) ( ) ( )P z G z H z G z H z= =

Further we obtain:0 0 0 0( ) ( ) ( ) ( ) 2G z H z G z H z+ =

Calculating the inverse Z-transform it follows:

0 0 0 0( ) ( ) ( 1) ( ) ( ) 2 ( )n

k k

g k h n k g k h n k k

= =

+ =

This relationship represents the condition that should besatisfied by wavelet filters in time domain.

In the next lecture we will conclude design of waveletfilters, some comments related to the 2D wavelets will begiven and standard usage of wavelets explained.


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For self-exercise Write program for realization of the Hadamard and Walsh

transforms. Determine inverse of the Walsh and Hadamard transforms. To

which group they belong? Repeat the same procedure for theHaar transform.

Assume that we have signal of dimensions 8x8 with non-zero

coefficients x(1,2)=2, x(3,1)=1 and x(5,6)=1. Calculate theHadamard, Haar and Walsh transforms of this signal. For NxM=8x8 determine basis images for (p,q)=(1,2),

(p,q)=(3,3) and (p,q)=(6,7). Proj ect for self-exercise.Application of the Hadamard

transform in realization of the special purpose Hamming codes -Reed-Muller codes. Write program for the Hadamard and other rectangular

transforms of digital image. When image dimensions are notpower of 2 realize transforms with zero-padded signal.


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For self-exercise Give hardware realization of the inverse Haar transformers. For filters used in the Haar transform (or Haar wavelet)

determine spectral response. Since these filters have only 2non-zero samples of the impulse response perform the zero-padding for obtaining clear results.

Details of the eigenvalues decomposition and the KL transform

in signal processing could be learnt in details from the book: M. R. Stoji, M. S. Stankovi, R. S. Stankovi: Diskretne

transformacije u primjeni, Nauka, 1993 (strana 119).

For mini-project. Study singular value decomposition and caseswhen it can be used. Try to find proper references while

wikipedia can be used as a starting point in search related tothis topic.


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For self-exercise Write program that perform Wavelet Haar decomposition of the

signal of the given length (power of 2) given number of times(for example 3 times) in such a manner that in each stage ofdecomposition it is performed on lowpass signal.

Give graphical presentation of the signal decomposition usingthe wavelet transform. Wavelet decomposition is performed 3

times on the lowpass part. Give corresponding synthesis stage. Assume that in the lowpass branch of the wavelet we have ideal

cut-off filter with cut-off frequency equal to half of the maximalfrequency that corresponds to the sampling theorem. Determinefilter for highpass branch and corresponding filters on thesynthesis side. What is the main difference between this andthe system with Haar wavelet?


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For self-exercise Prove formulas for Z-transform of the downsampling and upsampling

circuit. Signal x(n) is transformed with downsampling and after that with

upsampling circuit. Determine output of this system in time and Z-domain.

Haar transform is determined. Try to determine methods for calculation ofthe Wavelet-Haar coefficients from the Haar transform.

Haar transform and wavelet with Haar filter are defined with the samebasic goal. Determine difference between these two transforms?

Determine relationship between transform coefficients in the Hadamardand the Walsh transform?

Mini-project. Consider several real images with their transforms (DFT,

DCT, DHT and introduced rectangular ones). For each transformdetermine how many coefficients are kept in 50%, 70%, 80%, 90%,95%, 98%, 99%, 99.5%, 99.9% of the signal power. Which of introducedtransform is the best in each considered case?


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For self-exercise Repeat previous experiments when image is divided in blocks 2nx2n.

Consider possibility to remove transformation coefficients that havesmall energy in function of n and used transforms. Can you make someconsistent conclusion?

Apply the wavelet-Haar decomposition on various images and try torepeat analysis from the previous two problems in this case formapplication of this transform in data compression.

Mini-project. Study wavelets starting from the continuous time form.For this research start from the textbook.


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Digital Image Processing

Wavelet transformFiltering of digital images - Intro


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Filters used in wavelet transform

We have proven that filters used in the wavelettransform should satisfy:

0 0 1 1

0 0 1 1

( ) ( ) ( ) ( ) 2

( ) ( ) ( ) ( ) 0

G z H z H z G z

G z H z H z G z

+ =

+ =

Under previously described conditions (filtersproduce pure time-delay) it holds:

0 0 1 1( ) ( ) ( ) ( ) ( )P z G z H z G z H z= =

0 0 0 0( ) ( ) ( ) ( ) 2G z H z G z H z+ =

0 0 1 0 0 1( ) ( ) ( ) ( ) ( ) ( ) 0G z H z G z G z H z G z + =


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Filters in wavelet transform

Inverse Z-transform of the first expression is:

0 0 0 0( ) ( ) ( ) ( ) 2G z H z G z H z+ =

0 0 0 0( ), (2( ) (2 ) ( ))k

g k h n k g k h n k n

= = =

0 1 0 1( ) ( ) ( ) ( ) 0H z G z H z G z + =

0 0 0 0( ) ( ) ( 1) ( ) ( ) 2 ( )n

k k

g k h n k g k h n k n

= = + =

It can be written as:

( ) ( ) ,k

x k y k x y

=

=I nner product .


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Similarly it can be proven that:

1 1( ), (2 ) ( )g k h n k n = 0 1( ), (2 ) 0g k h n k =

We can summarized these relationships as: t he w avelet f i l t ersshould sat isfy t he follow ing relat ionship in t he timedomain:

1 0( ), (2 ) 0g k h n k =

(2 ), ( ) ( ) ( ), , {0,1}i jh n k g k i j n i j = =


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When the following:

(2 ), ( ) ( ) ( ), , {0,1}i jh n k h k i j n i j = =

holds we have orthogonal wavelets (the same typeof wavelets used in analysis and synthesis stages).

Next we will discuss the realization of the wavelettransforms for 2D signals and after that we will givesome program realizations.


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2D Wavelets

How to perform wavelet decomposition for 2Dsignals?

In the same way as a generalization of all 1Dtransforms to 2D transforms we can perform

operation separately along rows and columns ofimages.

Therefore, we will perform wavelet transform alongcolumns (or rows) followed by wavelet transformsalong rows (or columns).


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2D Wavelet transforms

h0(n)

h1(n)

2

2

x(n,m)

applying f i l t ers tocolumns

t aking elements oneven positionsfrom each column

h0(m)

h1(m)

2

2

h0(m)

h1(m)

2

2

applying f i l ters t o row s

t aking elements oneven posit ions fr omeach row

a(n,m)

dH(n,m)

dV(n,m)

dD(n,m)


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2D Wavelet - Comments a(n,m) is low-pass image (sometimes called coarse

image). Notation a comes from the fact that it is productof analysis side of the wavelet transformers. In the nextstages we can proceed with its analysis using wavelettransform.

Images dV(n,m), dH(n,m) and dD(n,m) are called detailsin horizontal, vertical and diagonal directions respectively,These images are commonly not subject to furtherdecomposition.

Note that these images have dimensions N/2XN/2 andtotal dimension of data is not changed by this procedure.

What is the advantage of wavelet transforms and how torealize it?


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Wavelet realization

Very often instead of realization of the wavelet transform byusing definition, numerous scientists use well-developedsoftware tools.

We will use the MATLAB Wavelet Toolbox but on the Internetit is possible to find alternative tools and MATLAB toolboxes.

This is quite comprehensive and complicated toolbox and wewill not consider all its possibilities but we will just concentrateon several features that can illustrate our theory in step-by-step manner.

Note that there are single step functions in MATLAB toolboxable to perform very sophisticated operations. User shouldonly learn how to use these functions and knowledge aboutwavelets is not required. This is popular for usage but it issource of numerous mistakes in practice.


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Wavelet Toolbox The most important recent contributions to the Wavelet

Transform is given by Donoho and it is possible to find allMATLAB functions related to his research on his web-site.

For a brevity we will present the wavelet transform in step-by-step manner without giving all details. Only on several placeswe will describe useful shortcuts that can be used in

procedures. The first useful function has a format:

[LD,HD,LS,HS] = wfilters('haar') Output of this function are wavelet function used in

decomposition (analysis) stage LD and HD (lowpass andhighpass filters) as well as those used in synthesis stage LS andHS.

Argument of this function is used wavelet type. Here we usethe Haar wavelet but there are numerous alternativepredefined wavelet functions.


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Wavelet filters Instead of abrupt Haar wavelets we can use more smooth

functions such as the Daubechies class of wavelet functionscalleddbnwhere n is integer that can be selected withinrange from n=1 to n=45 (db1is in fact the Haar wavelet).Here, it will be useddb23.

Analysis stage can be performed using convolution of signalwith corresponding wavelet functions:clearx=wnoise(1,10);[LD,HD,LS,HS]=wfilters('db23');

xl=conv2(x,LD,'same');xh=conv2(x,HD,'same');

Test signal

Filtering with corresponding filters

(parameter same is introduced inorder that obtained signals havethe same length as an originalsignal the first argument. Un-required part of the signal is

truncated).

conv for 1D convolutionhas no argument same


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Decimation and reconstruction

xdl=xl(1:2:length(xl));

xdh=xh(1:2:length(xl));

xul=zeros(1,2*length(xdl));

xuh=zeros(1,2*length(xdl));

xul(2:2:length(xul))=xdl;

xuh(2:2:length(xul))=xdh;

xr=conv2(xul,LS,'same')+conv2(xuh,HS,'same');

Decimation

Upsampling

Filtering in the highpass and lowpass lines.

Reconstructed signal


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Wavelet transform - Comments

In the previous example we performed meaningless

operation: signal is decomposed and after thatreconstructed. Note that before the reconstruction stage lowpass part

can be decomposed several times.

Instead of showing the advantages of the wavelettransform on 1D we will do it on example of 2D signals. Here, we will use function dwt2 (there is similar function

dwt for 1D signals) for decomposition that can perform

it in more efficient manner than in the previous examplewhere realization is performed in step-by-step manner.


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Wavelet transform of images After reading image cameraman.tifwe perform three steps of

decomposition (wavelet transform is performed three times on lowpasspart of image).clear

x=imread('cameraman.tif');

[ca,ch,cv,cd] = dwt2(x,'db1','mode','sym');

[caa,cah,cav,cad]=dwt2(ca,'db1','mode','sym');

[caaa,caah,caav,caad]=dwt2(caa,'db1','mode','sym');

Slika=[caaa,caah;caav,caad];

Slika=[Slika,cah;cav,cad];

WavTransf=[Slika,ch;cv,cd];

Wavelet transform of image


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Wavelet transform of imageExample

Wavelet transform has

the same dimensionslike image but hugenumber of wavelettransform pixels isclose to 0. It meansthat we can neglectmany of them and that

we can reconstructimage with highaccuracy.


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Image Reconstruction - ExampleIn image reconstruction

we use 17% of thelargest waveletcoefficients (according toabsolute value). This is

basis image and imagesrepresenting edges,details and similarelements.

Reconstruction isperformed using idwt2command.


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Image reconstructionReconstructed

Cameraman based on17% of wavelet coefficients.

Due to the described

properties the wavelettransform has numerousapplication. Some of themwill be given on next slides.


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Wavelet transform application Here, we will present two the most important applications of

the wavelet transform. The first one is compression. It is clear that for relatively

small number of wavelet coefficients has significant energy.It means that only small number of coefficients has

importance for visual image quality. Wavelets are note usedalone for compression but they are combined with othercompressing strategies (for example RLE, Huffman etc) butthe wavelets are basis for some compression standards (forexample JPEG 2000). Furthermore, the wavelet transform

based compression is quite suitable since it can be applied indifferent manners for different channel state. For noisychannel with small capacity we are sending base imagewithout details but for better conditions in the channels weare sending images with details.


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Application in denoising Assume that image is corrupted in channel or during acquisition process

(for example medical images obtained using ultrasound devices could bevery noisy). It can be proved that noise in various wavelet coefficients is(almost) mutually independent and it is uniformly distributed (for relativelylarge number of decomposition steps) on all wavelet coefficients. Then wecan perform filtering independently for each coefficients. Two groups of

techniques are developed using wavelets: hard thresholding (allcoefficients below the threshold are set to zero and in reconstruction areused for reconstruction (threshold is determined based on estimated noisevariance)); soft thresholding (small wavelet coefficients are attenuatedproportionally more than large ones).

Obtained wavelet-based results are the last word of science in denoisingapplications. It is quite difficult to be better than wavelets in this area interms of quality of imaging and speed.


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Wavelet toolbox There are additional wavelet applications.

Check wavelet toolbox and demo programwavedemo.

There are single step functions in the toolbox forsignal filtering.

For example, check function wden with itsparameters.

Note that this function is used for the 1D denoising

and also see denoising 2D images.


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Drawbacks of wavelets When signal subject of transform is dominantly on low

frequencies with only details and edges of highfrequencies the wavelet transform is close to idealanalysis, synthesis, filtering and compression tool.

Digital images are commonly this signals of type and

today wavelet transform is quite commonly used in thedigital image processing field. However, when relatively large energy is concentrated

on high frequencies or when we have abrupt changes offrequencies in the signal, wavelet transforms are not

ideal tools.


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Optical interferograms These signals are not quite common in digital images but they

exists.

One of these signals are optical intereferograms commonlycaused by mistakes in design of optical communicationchannels or due to diffraction in optical microscopy (optical orlaser microscopes are cheaper than electronic counterparts but

they suffer from several drawbacks including interferograms).

Inteferogram: test example.

Wavelet transform is not suitable foranalysis and filtering of such signals.


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TF (S/SF) Transforms

Time-frequency t ransforms are designed for analysis of

1D signals with high frequency content.

The most important TFRs are: Short -t ime Fourier t ransform (STFT) ;

Wigner dist r ibut ion (WD) . For multidimensional signals, multidimensional generalization

of these transforms are used (sometimes calledSpace/Spatial-Frequency representations).

We described several representations from this group in thetextbook and here from the brevity reasons we will skip them.


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I mage f i l ter ing The most important part of our course is related to image

filtering and reconstruction. Under filtering we assume obtaining some image features (for

example in communications signal carrier of informationmodulated on some frequency).

Filtering in digital images is commonly assumed to bedenoising operation.

The goal of denoising process is obtaining signal as close aspossible to original image.

In addition we will consider in a brief manner image

reconstruction for images corrupted by noise and distortedwith some deterministic form of distortions.


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Spectral image characteristics

Recall several information related to the image spectral

characteristics. The 2D DFT of images is highly concentrated around origin

(0,0) and it decreases rapidly with frequency increasing.

Filters in the frequency domain can be divided in fourgroups: Lowpass filters that remove frequencies away from the origin.

Commonly, frequency response of these filters is symmetric aroundthe origin;

Highpass filters remove signal components around origin infrequency domain;

Bandpass filters allows frequency in the band between lowest andthe highest frequencies;

Stopband filters remove frequency band.


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Spectral characteristics of images

Before we define these filters we will perform a

simple experiment. Consider image Cameraman .

Perform lowpass filtering with 17x17 frequencysamples around the origin and perform highpassfiltering for a region outside this zone.

Results of this rough filtering scheme are given onthe next slide.


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Spectral characteristics of image

Lowpass filtered image is almost non-usable whilehighpass filtered counterpart is dark but recognizable(to be honest for better visualization we increased

energy of highpass image).

Here, energy

of lowpassimage islargerapproximately18 times!!!


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Spectral characteristics of images

Conclusion: The largest amount of energy is concentrated on low

frequencies, but it represents just image luminance andvisually not so important part of image.

Small energy on high frequency corresponds to visually very

important image features such as edges and details.

This is the reason why filtering of digital images arequite complicated task and one of reasons forintroducing the wavelet transform.


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Filters in frequency domain Filters in the frequency domain can be defined as

(here w e give them in analog form but it can be discretized inst raightforw ard manner):

1 22 11 22

1

(2

( ( ,) ))

)

, , ( H d dn Xf m

=

f i l t ered im age

2D DFT oforiginal image

frequency responseof f i l t er

Filtering can be performed for denosingpurposed, for removing distortions and/or forobtaining some important features from

digital images.


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Filters in frequency domain

Here, we will give the simplest filter types in frequency

domain with value 1 in pass region and 0 in stop region.These filters are called cut-off filters. In addition, we willassume rectangular filter forms. This is just examples andmore advanced filter forms (smooth) are commonly used inpractice.

1 1 2 2

1 2

1 | | | |( , )

0 elsewhere

H

=

Lowpass filter.


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Filters in frequency domain

For exercise, determine stopband filter and filters withcircle shape impulse responses.

In addition determine filters with the shape of theHanning window function in the pass part of thefrequency response.

1 1 2 2

1 2

0 | | | |( , ) 1 elsewhereH

= High-frequency filter.

11 1 12 21 2 22

1 2

1 | | | |

( , ) 0 elsewhereH

= Bandpass filter.


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Convolution and filtering

Up to now introduced filters are linear.

Linear space invariant filters can be realized in bothfrequency and space domain. Realization in spacedomain by using convolution:

0 0

0 0 0 0 ( , ( , ) ( , ) (( * *, )) , )

n

m

m

nx n n m mh n m h n m x n mf n m = =2D fil t er im pulse response.

Why we call this 2D filter impulse

response and how it is connectedwith H(1,2)?

2D convolut ion.

Determine 2D impulse response of introduced filtering functions.


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Convolution or FFT There are two alternatives for determination of the

linear space invariant systems outputs: convolutionor FFT in frequency domain.

What to use? The simpler technique.

At the first glance it is convolution in the spacedomain but in depth analysis can show that it is notthe case in general.

Let image x(n,m) has dimensions NxM and letimpulse response h(n,m) has dimensions N1xM1.

Convolution result has dimensions(N+N1-1)x(M+M1-1). Why?


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Convolution or FFT Under assumption that impulse response has smaller

duration than image for filtering using convolution isrequired: (N+N1-1)x(M+M1-1)xM1xN1 additions and the samenumber of multiplications.

For the FFT-based realization it is required 3 FFTs (for filter

response, input signal and inverse for output signal). Theseoperations require:(N+N1-1)x(M+M1-1)log2(N+N1-1)x(M+M1-1)

Number of pixels at the

filter output.

Number of operation forsingle pixels.

All signals are properly zero-padded in order to avoid

aliasing.


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Convolution or FFT In the FFT-based realization we need additional

(N+N1-1)x(M+M1-1) multiplications X(1,2)H(1,2). In order that realization using convolution is simpler

it is required to hold:(N+N1-1)x(M+M1-1)xM1xN1 |ey(n,m)| and when we want todetect edge along y-axis it is required that:|ey(n,m)|>|ex(n,m)|.

In addition, it is required that local maximum be largerth dj t l f d t t i


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For the mask based detectors (Sobel, Prewitt etc) wehave detector responses for x and y coordinates.There is possibility to estimate edge direction basedon ex(n,m) and ey(n,m) as

( , )

( , ) arctan ( , )

y

x

e x y

x y e x y

=

Kirch detectors Kirch designed additional masks for detection of edges

in other directions. For example detector that is

sensitive to edges in directions of x, y-axes and axesthat are giving angles of/4 and 3/4 with respect tomain axes can be designed using 4 matrices:


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main axes can be designed using 4 matrices:

1 1 10 0 0

1 1 1

1 0 11 0 1

1 0 1

0 1 1

1 0 1

1 1 0

1 1 0

1 0 1

0 1 1

Direction of edge is determined using the largestdetector response produced with these matrices.


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Laplace detector The logic behind the Laplace detector is quite

different than the mask-based detectors.

The Laplace differential operator is defined as:

2 1( ) ( ) [ ( 1 ) ( 1 ) ( 1) ( 1)]f n m f n m f n m f n m f n m f n m + + + + +


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( , ) ( , ) [ ( 1, ) ( 1, ) ( , 1) ( , 1)]4

f n m f n m f n m f n m f n m f n m + + + + +

Edge (abrupt change in luminance) can be detected as zeroin the Laplace operator (position with zero-crossing).

o X

If red is equal to 100 and green 50 Laplace detectorresponse for pixel X is: 100-350/4>0 while for o: 50-250/4


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There are numerous (more or less acceptable techniques)for removing noise in edge detection process. One of them

is to calculate local variance of pixels in local neighborhood:2

2

2

1( , ) ( , ) ( , )

(2 1)

n M n M

k n M l n M

n m f k l f n mM

+ +

= =

= +

local neighborhood

Laplace detector - drawbacks In the previous relation holds:

The Laplace detector introduces threshold ofvariance For small variance we assume that edge is

2

1( , ) ( , )

(2 1)

n M m M

k n M l m M

f n m f k l

M

+ +

= =

=

+


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variance. For small variance we assume that edge isnot detected (image is almost constant) while for

large variance and zero-crossing we assume thatedge is detected.

There is alternative that region where edges can bedetected be calculated based on difference:

( , ) max { ( , )} min { ( , )}A Aw n m f n m f n m=

Maxim al and m inimal luminance inlocal neighborhood

When w(n,m) is larger than athreshold we assume thatedge is detected.

Laplace-Gauss detector A technique for reducing noise influence in the Laplace

detector is filtering of image or detector response with

Gaussian filter (Gaussian blur). Both operations are linearand they can change order without affecting results.

Obtained detector is in literature called LoG (Laplacian of


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Obtained detector is in literature called LoG (Laplacian ofGaussian).

Matrix in the Gaussian is given in the shape:

2 2 2( , ) exp( ( ) / 2 )h n m n m= +

LoG detector The most common matrix for LoG detector is:

0 0 1 0 0

0 1 2 1 0

1 2 16 2 1

However, for various valuesof follows different matrixfor LoG but given matrix is


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1 2 16 2 1

0 1 2 1 00 0 1 0 0

for LoG but given matrix is

the most common in practice.

The Canny detector is a special form of the LoG detectorassumed to be one of the best (but not the simplest one).

Canny detector Three different filters are employed in the Canny detector

realization. The Gaussian with impulse response:

Filters with impulse response equal to partial derivativesof the Gaussian filter impulse response:

22 2 2 (( , 2) exp( ( 2 ) / )) /h n m n m= + adopted constant


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of the Gaussian filter impulse response:2 2 2 2

( , ) ( , ) / exp( ( ) / 2 ) /( )xh n m h n m n n n m= = + 2 2 2 2

( , ) ( , ) / exp( ( ) / 2 ) /( )y

h n m h n m m m n m= = +

Calculate the response of the Gaussian filter:

( , ) ( , )* * ( , )n m

g n m f n m h n m=

This operation is used t o reduce the noise inf luence.

Canny detector After this operation output of the Gaussian filter is filtered with filters

having responses equal to derivatives of the Gaussian function:

( , ) ( , ) * * ( , )x n m xe n m g n m h n m=( , ) ( , ) * * ( , )

y n m ye n m g n m h n m=


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Detector response is equal to:

2 2( , ) ( , ) ( , )x ye n m e n m e n m= + This value is normalizedcommonly to 1 by dividing it w it hmaximum of e(n,m) .

Commonly instead of e(n,m) we presentbinary image obtained by comparing it witha threshold.

Canny detector - Thresholds There are several steps in obtaining edges in the Canny detector

using e(n,m): Determine percentage of image that could belong to image. This

percentage corresponds to the largest values of e(n,m). Eliminate all points that are not local maximum along the corresponding

direction. In the Canny detector check it is performed also for diagonaldirections.Candidates for edges are pixels that are larger than the selected


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Candidates for edges are pixels that are larger than the selectedthreshold (threshold 1), that among the largest detector responses, and

that are local maximum. Sometimes we perform determination of the edges by additional

comparison with second threshold (larger than threshold 1) and finalresponse is obtained as some combination of edges obtained with thesetwo thresholds (larger number of pixels is recognized as edges withrespect to smaller threshold and larger than in the case of higher

threshold). HUHUHUH!!!

Detection of isolate points Isolated points can be detected by using mask function of

the shape:

When edge is not abrupt change in luminance but single

1 1 1

1 8 1

1 1 1


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When edge is not abrupt change in luminance but single

line passing over a background in other color we can usematrices of the form:

1 1 1

2 2 2

1 1 1

1 1 2

1 2 1

2 1 1

2 1 1

1 2 1

1 1 2

1 2 1

1 2 1

1 2 1

Detection of edges in color

images The simplest technique is to calculate edges for separate

channels and to connect them using logical operations:

A=imread('flowers.tif');er=edge(A(:,:,1));eg=edge(A(:,:,2));

Detectors for variouschannels calculated w it hdefault paramet ers of t heMATLAB edge function.


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eb=edge(A(:,:,3));

e=er|eg|eb;imshow(e)

Logical operat ion OR applied onedges detected in channels.

This is very simple and produces effective results but recently some moresophisticated techniques for color edge detection are proposed. Interestedstudents can take this techniques as a topic for self-excercise.

Other edge detectors We reviewed almost all important edge detectors

within our lecture.

Recently with a lot success several novel techniqueswhere size of mask is determined in adaptive manneris used. Theoretically almost exact values of theimage derivatives are determined in such a manner


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image derivatives are determined in such a manner.

This is significant advantage with respect to usage ofdifferences.

We have seen that threshold determination iscomplicated and commonly heuristic (ad-hoc)

technique are applied for this purpose.

Hough transform Images are in some application transformed to binary

form using edge detectors and after they are used forrecognition of geometrical objects primitives.

The simplest primitive object is straight line. It can be represented using linear function y=ax+b (or

m=an+b). The main goal is to determine if pixels belong


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) g p gto line and to determine parameters of line (a,b) .

The procedure is not quite simple. Consider (x1,y1) thatbelongs to edge but we do not know if this edge is straighline and if it is we do not know its parameters.

Hough transform

Algorithm We adopt domain of possible values a and b as Cartisian

product AxB where aiA and bjB are possible values

from these sets. Denote ai=amin+ia, bj=bmin+jb where amin and bmin are

minimal values from corresponding sets.

Let a are b maximal values from sets and let


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Let amax are bmax maximal values from sets and let

number of elements in sets is NA and NB respectively. Then: a=(amax-amin)/(NA-1) and b=(bmax-bmin)/(NB-1). Number of elements in sets should not be too large and

too small. Large number increases burden and sometimes

makes detection more difficult while small numberproduces not precise results.

Algorithm for Hough Transform Create matrix P(i,j)=P(ai,bj) where each element from

the set has value P(i,j).

Initialize all P(i,j) i=1,..., Na and j=1,...,Nb to zero. For each image pixel (xk,yk) with edge detector giving 1

(detected edge) e(xk,yk)=1 determine all straight lines to


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( g ) ( k,yk) gwhich this pixel can belong b

r= y

k-a

qx

kand increment

values of P corresponding to these lines:P(q,r)=P(q,r)+1.

Algorithm for Hough transform Procedure is repeated for each pixel from edges and after that

we detect maximum ofP and this value corresponds to linewith given parameters.

After that we remove this maximum and region around themaximum from matrix P and we repeat procedure for detectingnext maximum.Thi d i t d til d t t ll li h


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This procedure is repeated until we detect all lines or when

there are no dominant maxima in the matrix P. Even very simple the Hough transform is quite effective line

detector. The main problem in the Hough transform is related to lines

parallel to x-axis such as x+b=0 where coefficient a approachestoward infinity.

Problem in the Hough transform This problem can be overcome by the following

parameterization:

r=ycos+xsiny

Rr0 /2>-/2


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x

r

Limit of domain can be determinedby using image dimensions.

In this manner both parametersused for line parameterization have

limited domain.

Hough transform

Test exampleclear

%

N= 256;

B= zeros(N);% Lini je

for k= 60:150,B(k,k)= 1;end

for k= 20:110 B(k k+ 80)= 1;end

Realization is illustrative and withoutelements that can produce more efficientalgorithm or more robust result to noise

influence.Rectangle (four straight lines).


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for k= 20:110,B(k,k+ 80)= 1;end

for k= 60:-1:20,B(k ,120-k)= 1;endfor k= 150:-1:110,B(k ,300-k)= 1;end

M= 256;

P= zeros(M);

t acnost = 0.2;

R= linspace(-M,M,M);th= l inspace(-pi / 2,pi / 2,M);

Matrix P, initialized and space of parameters.

Additional parameter used in thealgorithm.

MATLAB Realization - Examplefor y= 1:Nfor x= 1:N

i f (B(x,y)= = 1)

for r= 1:Mfor p= 1:M

if(abs(R(r )-x * cos(t h(p)) -

y* sin(th(p)) )< tacnost) ,

If pixel belongs to edge...


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y sin(th(p)) )< tacnost) ,P(r ,p)= P(r,p)+ 1;end

end

end

end

end

endfigure(1),imshow(B)

figur e(2) ,pcolor( t h,R,P) ,shading int erp

... Increase for 1 elements from Pthat represents lines that passthrough the pixel.

Graphical representation.

Results


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Rectangle

Hough t ransform w ithdetected lines.

Hough transform - Conclusion This is very interesting, semi-intuitive, technique that can

be subject to numerous improvements (accuracyimprovement, robustness to noise influence, determinationof short lines) and even today it is one of the most popularalgorithms for detection of straight lines.

Comparative analysis has shown that it is better than other


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p y

algorithms almost on all criterion. It can be connected with the Radon transform.

It can be used in modified form for other shapes that canbe parameterized using 2 parameters or when they can be

reduced to case with 2 parameters in some acceptablemanner.

Hough transform Circle The circle can be represented using coordinates of

center and radius as:(x-a)2+(y-b)2=r2

Then for parameterization in the Hough algorithm isrequired three parameters (a,b,r).

All pixels belonging to the circuit can be given as:


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p b o g g o u a b g a

x=a+rcos y=b+rsin Strategies for simplification are developed in order to

reduce 3D search (for 3 parameters) to lessdemanding realization.

Circle parameters In the case of the circle parameterization we can use

edge direction (angle). Namely, we can employ someedge detection algorithm that produces edge e(x,y) andedge direction (angle) (x,y) .

In the case of the circle the edge direction is normal toradius and search for the circle center can be reduced to


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1D search along the line normal with respect to the edgedirection.

Unfortunately, edge direction can be determined up tolimited accuracy and we cannot employ this idea in

straightforward manner. Instead we will use segment of the circle in the search

procedure.

Circle parameters - illustration

detected edge direction


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Angular segment where we performsearch for circle center

For self-exercise Realize all introduced edge detectors: Roberts, Sobel, Prewitt,

Laplace, Gauss-Laplace, Canny detector. Realize described methods for mapping edge detector to binary

image. Study methodology for determination of the threshold for

various edge detectors (especially for Canny detector) in theMATLAB function edge.

For mini-project: Consider edge detectors for color images using


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p j g g g

the Canny and Kumani operators. In this case, detectors are notevaluated separately along channels but based on channelscombinations.

For mini-project:Application of the edge detector in realizationof adaptive filters.


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For self-exercise Establish relationship between the Hough and Radon transform. For mini-project: SLIDE algorithm for edge detection. For mini-project: Improvements of the Hough transform. For mini-project: Compare the improved Hough transform andthe SLIDE algorithm. Hint: References related to the SLIDE

algorithm can be found from the lecturer. Realizujte algoritam za prepoznavanje krugova i krunih lukova. Za miniprojekat: Algoritam za prepoznavanje elipsi


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Za miniprojekat:Algoritam za prepoznavanje elipsi. Kako se pomou Houghove transformacije moe prepoznati

poetak i kraj linije? Za miniprojekat: Realizacija algoritama za praenje ivica.

Pogledati skriptu.

Digital Image Processing_advanced Course

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