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2009.10.22

Digital Image Processing (Digitaalinen kuvankäsittely) Exercise 3. 2009.10.22. 1. Filter the following image with adaptive, local noise reduction filter in 3×3 window with estimated noise variance 0.5. Process only such pixels that have all the needed neighbors.

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2009.10.22

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  1. Digital Image Processing (Digitaalinen kuvankäsittely)Exercise 3 2009.10.22

  2. 1. Filter the following image with adaptive, local noise reduction filter in 3×3 window with estimated noise variance 0.5. Process only such pixels that have all the needed neighbors. The adaptive, local noise reduction filter is defined as : ( )

  3. ( ) Take point (1, 1) for example:

  4. The resulting image is:

  5. The Wiener filter is (Page 263): Thus, the equation becomes:

  6. ∵H(u,v) is a complex number ∴modulus2 = real part×imaginary part ∵ Real and positive with all u, v. ∴ imaginary part = modulus = real part Plug into the Wiener filter equation:

  7. a) nearest neighbor interpolation The coordinates are rounded to the nearest integer coordinates. (0.4,0.4) (0,0) 2 (1.4,1.7) (1,2) 5 0 1 2 0 1 2 (0.4, 0.4): interpolated value is 2. (1.4, 1.7): interpolated value is 5.

  8. b) bilinear interpolation The interpolated value (in red) is computed using values of four neighbors. Weights of the neighbors (in yellow) are proportional to overlapping area. (One way) 0 1 2 • (0.4, 0.4): • 0.6×0.6×2+ • 0.6×0.4×3+ • 0.4×0.6×3+ • 0.4×0.4×4 = • 2.8≈3 0 1 2 (0.4,0.4) ∑Areai×Weighti, i=1,2,3,4 2 3 3 4 • (1.4, 1.7): • 0.3×0.6×4+ • 0.7×0.6×5+ • 0.3×0.4×7+ • 0.7×0.4×3 = • 4.5≈5 (1.4,1.7) 4 5 7 3

  9. If the color balance is correct, rR ≈ rG ≈ rB for white and black test targets. In our case, histogram peak corresponding to black target is rR = 62; rG = 31; rB = 12, so the color balance correction is needed.

  10. Let us use a linear transformation: Take the red channel for example, in the input image, the black peak for red: rR = 62 And we want to get: sR = 16 For the white peak, rR = 242 and sR = 242 We get the equation pair: sR rR Solve this, gives:

  11. Next, check limits so that the output sR is in [0,255] , when input rR is between 0 and 255. sR = 1.2556 rR - 61.8444 [0,255] [0,255] sR (252.354,255) 255 So the color transformation for red channel: (0,0) rR (49.2566,0) Similarly, get green and blue channels:

  12. The Bayer array consists of alternating rows of red-green and green-blue filters.

  13. First,separate color channels. 2m+1 2m+1 2n+1 2n 2m 2m

  14. After the 1st phase, we have three separated channels: Second, interpolate the missing values by linear filtering. (Demo)

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