hw2o
play

HW2o Image Differentiation COMPSCI 527 Computer Vision COMPSCI - PowerPoint PPT Presentation

Dec 03, 2022 •17 likes •177 views

HW2o Image Differentiation COMPSCI 527 Computer Vision COMPSCI 527 Computer Vision Image Differentiation 1 / 16 Outline 1 The Meaning of Image Differentiation 2 A Conceptual Pipeline 3 Implementation 4 The Derivatives of a 2D Gaussian

  1. HW2o Image Differentiation COMPSCI 527 — Computer Vision COMPSCI 527 — Computer Vision Image Differentiation 1 / 16

  2. Outline 1 The Meaning of Image Differentiation 2 A Conceptual Pipeline 3 Implementation 4 The Derivatives of a 2D Gaussian 5 The Image Gradient COMPSCI 527 — Computer Vision Image Differentiation 2 / 16

  3. The Meaning of Image Differentiation What Does Differentiating an Image Mean? Values Derivatives in x 0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700 0 100 200 300 400 500 600 700 COMPSCI 527 — Computer Vision Image Differentiation 3 / 16

  4. The Meaning of Image Differentiation What Does Differentiating an Image Mean? is c Can we reconstruct 0 100 200 300 400 500 600 700 the black curve? COMPSCI 527 — Computer Vision Image Differentiation 4 / 16

  5. The Meaning of Image Differentiation Cameras Kass f eye IR camera aperture 1 1 2 3 C K focal distance Can I recover principal ray Ccny from I rt optical axis y lens in-focus plane I rig image plane if COMPSCI 527 — Computer Vision Image Differentiation 5 / 16

  6. A Conceptual Pipeline A Conceptual Pipeline I r c O 000 ∂ I(r, c) C (x, y) D (x, y) I c (r, c) i ∂x • Somehow reconstruct the continuous sensor irradiance C from the discrete image array I • Differentiate C to obtain D • Sample the derivatives D back to the pixel grid • Each would be hard to implement • Surprisingly, the cascade turns out to be easy! COMPSCI 527 — Computer Vision Image Differentiation 6 / 16

  7. A Conceptual Pipeline From Discrete Array to Sensor Irradiance EIen faint.EE ∂ I(r, c) C (x, y) D (x, y) I c (r, c) i ∂x d Cj PG What would the transformation from I to C look like formally, I if we could find one? Example: Linear interpolation x j FCK Icj P C x INTERPOLATION c 11.44 TER QIE7xm ei f t.IE tn sina.ci 9 l 7 T COMPSCI 527 — Computer Vision Image Differentiation 7 / 16

  8. A Conceptual Pipeline Linear Interpolation as a Hybrid Convolution C ( x , y ) = P ∞ P ∞ j = −∞ I ( i , j ) P ( x � j , y � i ) i = −∞ 2B in COMPSCI 527 — Computer Vision Image Differentiation 8 / 16

  9. A Conceptual Pipeline Gaussian Instead of Triangle L A • Noise ) : fit rather than interpolating • Noise ) : filter with a truncated Gaussian x 2 + y 2 • P ( x , y ) = G ( x , y ) / e − 1 σ 2 2 x j y I C i j G i C agg COMPSCI 527 — Computer Vision Image Differentiation 9 / 16

  10. Implementation Differentiating ∂ I(r, c) C (x, y) D (x, y) I c (r, c) i ∂x C ( x , y ) = P ∞ P ∞ j = −∞ I ( i , j ) G ( x � j , y � i ) i = −∞ (still don’t know how to do this, just plow ahead) D ( x , y ) = ∂ C ∂ P ∞ P ∞ ∂ x ( x , y ) = j = −∞ I ( i , j ) G ( x � j , y � i ) i = −∞ ∂ x D ( x , y ) = P ∞ P ∞ j = −∞ I ( i , j ) G x ( x � j , y � i ) i = −∞ • We transferred the differentiation to G , and we know how to do that ! (still don’t know how to implement a hybrid convolution) COMPSCI 527 — Computer Vision Image Differentiation 10 / 16

  11. Implementation Sampling ∂ I(r, c) C (x, y) D (x, y) I c (r, c) i ∂x D ( x , y ) = P ∞ P ∞ j = −∞ I ( i , j ) G x ( x � j , y � i ) i = −∞ • We are interested in the values of D ( x , y ) on the integer grid : x ! c and y ! r NOT I c ( r , c ) = P ∞ P ∞ j = −∞ I ( i , j ) G x ( c � j , r � i ) i = −∞ Wait! This is a standard, discrete convolution We know how to do that ! To differentiate an image array, convolve it (discretely) with the (sampled, truncated) derivative of a Gaussian COMPSCI 527 — Computer Vision Image Differentiation 11 / 16

  12. The Derivatives of a 2D Gaussian The Derivatives of a 2D Gaussian • The Gaussian function is separable: x 2 + y 2 G ( x , y ) / e − 1 = g ( x ) g ( y ) where 2 σ 2 x 2 g ( x ) = e − 1 2 σ 2 ∂ x = ∂ g G x ( x , y ) = ∂ G ∂ x g ( y ) = d ( x ) g ( y ) genes 00 d ( x ) = dg dx = � x σ 2 g ( x ) • Similarly, G y ( x , y ) = g ( x ) d ( y ) • Differentiate (smoothly) in one direction, smooth in the other • G x ( x , y ) and G y ( x , y ) are separable as well COMPSCI 527 — Computer Vision Image Differentiation 12 / 16

  13. The Derivatives of a 2D Gaussian The Derivatives of a 2D Gaussian G x ( x , y ) = d ( x ) g ( y ) and G y ( x , y ) = g ( x ) d ( y ) by COMPSCI 527 — Computer Vision Image Differentiation 13 / 16

  14. The Derivatives of a 2D Gaussian Normalization • Can normalize d ( c ) and g ( r ) separately • For smoothing, constants should not change: • We want k ⇤ g = k (we saw this before) • For differentiation, a unit ramp should not change: u ( r , c ) = c is a ramp • We want u ⇤ d = u (see notes for math) COMPSCI 527 — Computer Vision Image Differentiation 14 / 16

  15. The Image Gradient The Image Gradient  I x ( r , c ) � • Image gradient : r I ( r , c ) = ∂ I ∂ x = g ( r , c ) = I y ( r , c ) • View 1: Two scalar images I x ( r , c ) , I y ( r , c ) 118111 a It Is I l I te COMPSCI 527 — Computer Vision Image Differentiation 15 / 16

  16. The Image Gradient The Image Gradient • View 2: One vector image g ( r , c ) 0 • We can now measure changes of image brightness • Edges are of particular interest COMPSCI 527 — Computer Vision Image Differentiation 16 / 16

Recommend

More recommend