Formation et Analyse d’Images Session 2

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Formation et Analyse d’Images Session 2. Daniela Hall 26 September 2005. Course Overview. Session 1: Overview Human vision Homogenous coordinates Camera models Session 2: Tensor notation Image transformations Homography computation Session 3: Reflection models Color spaces - PowerPoint PPT Presentation

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Formation et Analyse d’ImagesSession 2

Daniela Hall

26 September 2005

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Course Overview

• Session 1: – Overview– Human vision – Homogenous coordinates– Camera models

• Session 2:– Tensor notation– Image transformations– Homography computation

• Session 3:– Reflection models– Color spaces

• Session 4:– Pixel based image analysis

• Session 5:– Gaussian filter operators– Scale Space

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Course overview

• Session 6:– Contrast description– Hough transform

• Session 7:– Kalman filter

• Session 8:– Tracking of regions, pixels, and lines

• Session 9:– Stereo vision

• Session 10:– Epipolar geometry

• Session 11: exercises and questions

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Trifocal Tensor

• A tensor is used in 3d position estimation with multiple cameras. The (trifocal) tensor encapsulates all the (projective) geometric relations between 3 camera views independent of the scene.

• Reference– book: R. Hartley, A.Zisserman: Multiple view geometry in

computer vision, Appendix 1, Cambridge University Press, 2000

– Exists on-line http://www.robots.ox.ac.uk/~vgg/hzbook/hzbook1.html

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Tensor notation

• In tensor notation a superscript stands for a column vector

• a subscript for a row vector (useful to specify lines)

• A matrix is written as

3

2

1

3,2,1)(

p

p

p

pP iii

),,( 321 lllLi

),,( 321

3

2

1

33

32

31

23

22

21

13

12

11

jjj

i

i

iji

MMM

M

M

M

mmm

mmm

mmm

M

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Tensor notation

• Tensor summation convention:– an index repeated as sub and superscript in a

product represents summation over the range of the index.

• Example: 3

32

21

1 plplplPL ii

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Tensor notation

• Scalar product can be written as

• where the subscript has the same index as the superscript. This implicitly computes the sum.

• This is commutative

• Multiplication of a matrix and a vector

• This means a change of P from the coordinate system i to the coordinate system j (transformation).

33

22

11 plplplPL i

i

iii

i LPPL

iji

j PMP

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Line equation

• In R2 a line is defined by the equation

• In homogenous coordinates we can write this as

• In tensor notation we can write this as

0 cbyax

0

1

),,(

y

x

cbaPLT

0iiPL

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The tensor operator Eijk and Eijk

• The tensor Eijk is defined for i,j,k=1,...,3 as

123 ofn permutatio oddan isrst if 1,-

123 ofn permutatioeven an isrst if 1,

distinct are t ands,r, unless 0

rstE

3 2 1 1 3 22 1 3

1 2 3 2 3 1 3 1 2

evenodd

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Determinant in tensor notation

jjjji mmmM 321 ,,

kjiijk

ji mmmEM 321)det(

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Cross product in tensor notation

bac

kjijkii baEbac )(

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Example

• Line equation in tensor notation

kjijki QPEL

1

,

1

2

1

2

1

q

q

Qp

p

P kj

0

),,(

321

12213

1131132

1

321

222332

lylxl

qpqpl

qpqpqpl

qpqpqpl

lllLi

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Example

• Intersection of two lines

• L: l1x+l2y+l3=0, M: m1x+m2y+m3=0

• Intersection:

• Tensor:

• Result:

1221

3113,

1221

2332

mlml

mlmly

mlml

mlmlx

kjijki MLEP

12213

31132

23321

mlmlp

mlmlp

mlmlp

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Translation

• Classic

• Tensor notation

• T is a transformation from the system A to B

• Homogenous coordinates

y

x

tyy

txx

12

12

1

1

1

100

10

01

1

2

2

y

x

ty

tx

y

x

1,2,3 BA, with ABA

B PTP

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Rotation

• Homogenous coordinates

1)cos(1)sin(

1)sin(1)cos(

2

2

yaxay

yaxax

1

1

1

100

0)cos()sin(

0)sin()cos(

1

2

2

y

x

aa

aa

y

x

• Classic

• Tensor notation1,2,3 BA, with AB

AB PRP

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Image transformation

For each position Pd in the destination image we searchthe pixel color I(Pd).

Source image Destination image

Tsd

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Image transformation

First we compute a position Ps in the source image.

dsd

s PTP

Source image Destination image

Tsd

1100

)cos()sin(

)sin()cos(

1d

d

yyx

xyx

s

s

y

x

tasas

tasas

y

x

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Image transformation

• P is not integer.• How do we compute I(Pd)=I(Ps)?• Answer: by a linear combination of the

neighboring pixels I(Psi) (interpolation).

Ps0 Ps1

Ps3Ps2

Ps

Pd

Tsd

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Interpolation methods

• 0th order: take value of closest neighbor– fast, applied for binary images

• 1st order: linear interpolation and bi-linear interpolation

• 3rd order: cubic spline interpolation

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1D linear interpolation

Ps0 Ps1

Ps3Ps2

Ps

position P

intensity I(P)

Ps0 Ps Ps1

)()()())(1(

)()()()(

)()(

1000

00

01

ssssss

ssssd

ss

PIPPPIPP

PIPPmPIPI

PIPIm

Gradient

Pixel color

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2D linear interpolation

Ps0 Ps1

Ps3Ps2

Ps

x

intensity I(P)

Ps0

Ps

Ps1

)()()()()(

)()(

)()(

000

02

01

sssyy

ssxx

sd

ssy

ssx

PIPPmPPmPIPI

PIPIm

PIPIm

Gradient

Pixel color

I(Ps)

y

Psx

Psy

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Bi-linear interpolationPs0 Ps1

Ps3Ps2

Ps

)()()()()(

,)1(

),1(),1)(1(

3210

0

0

sssss

yssy

xssx

PDIPCPBPAIPI

dxdyDdydxC

dydxBdydxA

PPdy

PPdx

The bilinear approach computes the weighted average of the four neighboring pixels. The pixels are weighted according tothe area.

AB

CD

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Higher order interpolation

• Cubic spline interpolation takes into account more than only the closest pixels.

• Result: more expensive to compute, but image has less artefacts, image is smoother.

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Homographie: projection from one plane to another

• Homographie HBA is bijective QB = HB

A PA

jj

jjB

jj

jjB

A

j

j

jAB

AB

B

PHPHy

PHPHx

P

H

H

H

PH

w

wy

wx

32

31

3

2

1

/

/

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Homography computation

• H can be computed from 4 point correspondences.

Ps1 Ps2

Ps3Ps4

Rd1 Rd2

Rd3Rd4Source image (observed)

Destination image (rectified)

4,3,2,1 and 2,1 with ,0 iSRHPRHP DiSD

SiDiSD

Si

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Homography computation

• H is 3x3 matrix and has 8 degrees of freedom (homogenous coordinates)

• gives 8 equations and one solution for H.

4,3,2,1 and 2,1 with ,0 iSRHPRHP DiSD

SiDiSD

Si

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Application: Rectifying images

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Applications