CSC 422 Lecture Intensity Transform, Spring 2012

Notre Dame University–Louaize · Introduction to Image Processing · 32 pages · Dr. Khaldoun El-Khaldi

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Intensity Transformations for Image Enhancement General Idea • Intensity transformation – Also called point operation – Zero-memory operation • Map a given gray level to another level typically monotonically increasing (but not always) input gray level r output gray level s s T(r) Intensity Histogram Intensity Histogram 1 3 8 6 6 8 6 11 8 10 8 9 9 7 10 10 0 1 2 3 4 k 6 7 8 9 10 11 12 13 14 15 0 1 0 1 0 0 3 1 4 2 3 1 0 0 0 0 5 H(k) • Example a 4x4, 4bits/pixel image  Intensity Histogram From [Gonzalez & Woods] Intensity Histogram From [Gonzalez & Woods] Fixed Intensity Transformations Basic Transformations From [Gonzalez & Woods] r L s    1 ) log(1 r c s   cr s  Negative: Log: Power-law:  cr  1 e s Inverse Log: …… Negative Transformation • Image negatives – For enhancing white or gray details embedded in dark regions of an image • Our visual systems prefer white background to dark background – No gain or lose of any information r L s    1 Log Transformation – Low gray-level values are mapped into a wider range of output levels – Useful when an image is dominated by dark pixels (ex: Fourier spectrum) Input gray level r Output gray level s Spreading gray levels Compressing gray levels From [Gonzalez & Woods] ) log(1 r c s   Power-law (Gamma) Transformation From [Gonzalez & Woods] s cr Power-law (Gamma) Transformation From [Gonzalez & Woods] s cr Power-law (Gamma) Transformation From [Gonzalez & Woods] s cr Application of Power Law Transformations • Gamma correction – Many devices respond according to a power law – Ex: The intensity-to-voltage response of a cathode ray tube (CRT)       m r if m if r c s 0 Thresholding From [Gonzalez & Woods] m : threshold Example: Fixed Intensity Transformation 1 3 8 6 6 8 6 11 8 10 8 9 9 7 10 10         2 15 1 ( ) r round T r s • A 4x4, 4bits/pixel image passes through an intensity transformation 0 1 4 2 2 4 2 8 4 7 4 5 5 3 7 7 1  round(0.0667) = 0; 3  round(0.6) = 1; 6  round(2.4) = 2; 7  round(3.2667) = 3; 8  round(4.2667) = 4; 9  round(5.4) =5; 10  round(6.6667) = 7; 11  round(8.0667) = 8; The resulting image is: Example: Histogram Change 1 3 8 6 6 8 6 11 8 10 8 9 9 7 10 10 0 1 4 2 2 4 2 8 4 7 4 5 5 3 7 7  Contrast Stretch General Idea: Make Best Use of the Dynamic Range From [Gonzalez & Woods] Contrast Stretch min max min )1 (2 r r r r s B                                         1 2 1 2 )1 (2 1 2 1 2 0 2 2 2 2 2 2 2 1 1 2 2 1 2 1 1 2 1 2 1 1 1 B B B B B r r r r s r r s r r r r r s r s r r r r s s r r r r s s max 2 r r  General form: Special case  Full-scale contrast stretch: min 1 r r  1 2 2   B s 1 0 s            min max min )1 (2 r r r r round s B Typically used: Example: Full-Scale Contrast Stretch 4 4 8 6 6 8 6 11 8 11 8 8 9 7 10 10 min 4 r 0 0 9 4 4 9 4 15 9 15 9 9 11 6 13 13 • Full-scale contrast stretch of a 4x4, 4bits/pixel image • Find rmax 11 15 1 2   B                              4) 7 ( 15 4 11 4 15 )1 2 ( min max min r round r round r r r r round s B 4  round(0) = 0; 6  round(4.29) = 4; 7  round(6.43) = 6; 8  round(8.57) = 9; 9  round(10.71) = 11; 10  round(12.86) = 13; 11  round(15) = 15; The resulting image is: Example: Histogram Change 4 4 8 6 6 8 6 11 8 11 8 8 9 7 10 10 0 0 9 4 4 9 4 15 9 15 9 9 11 6 13 13  Histogram Equalization Histogram Equalization From [Gonzalez & Woods] Histogram Equalization From [Gonzalez & Woods] Example 2 3 8 2 9 9 9 10 8 3 3 8 3 11 11 10 min 2 r 0 2 10 0 12 12 12 13 10 2 2 10 2 15 15 13 • A 4x4, 4bits/pixel image • First try: full-scale contrast stretch max 11 r                              2) 3 ( 5 2 11 2 15 )1 2 ( min max min r round r round r r r r round s B 2  round(0) = 0; 3  round(1.67) = 2; 8  round(10.00) = 10; 9  round(11.67) = 12; 10  round(13.33) = 13; 11  round(15) = 15; The resulting image is: Example: Histogram Change 2 3 8 2 9 9 9 10 8 3 3 8 3 11 11 10 0 2 10 0 12 12 12 13 10 2 2 10 2 15 15 13  big gap full-scale contrast stretch original Cumulative Histogram 0 1 2 3 4 k 6 7 8 9 10 11 12 13 14 15 0 0 2 4 0 0 0 0 3 3 2 2 0 0 0 0 5 H(k) 0 0 2 6 6 6 6 6 9 12 14 16 16 16 16 16 Q(k) 2 3 8 2 9 9 9 10 8 3 3 8 3 11 11 10 H(k) Q(k) Intermediate Image 2 3 8 2 9 9 9 10 8 3 3 8 3 11 11 10 2 6 9 2 12 12 12 14 9 6 6 9 6 16 16 14  0 1 2 3 4 k 6 7 8 9 10 11 12 13 14 15 0 0 2 4 0 0 0 0 3 3 2 2 0 0 0 0 5 H(k) 0 0 2 6 6 6 6 6 9 12 14 16 16 16 16 16 Q(k) intermediate image original 2 6 9 2 12 12 12 14 9 6 6 9 6 16 16 14 min 2 r 0 4 8 0 11 11 11 13 8 4 4 8 4 15 15 13 max 16 r                              2) 14 ( 15 2 16 2 15 )1 2 ( min max min r round r round r r r r round s B 2  round(0) = 0; 6  round(4.29) = 4; 9  round(7.50) = 8; 12  round(10.71) = 11; 14  round(12.86) = 13; 16  round(15) = 15; final result: histogram equalized image Full-Scale Contrast Stretch of Intermediate Image intermediat e image Histogram Comparison 4 4 8 6 6 8 6 11 8 11 8 8 9 7 10 10 0 4 8 0 11 11 11 13 8 4 4 8 4 15 15 13 0 2 10 0 12 12 12 13 10 2 2 10 2 15 15 13 original direct full-scale contrast stretch histogram-equalized more equalized Summary of the Histogram Equalization Algorithm original image histogram H(k) cumulative histogram Q(k) intermediate image full-scale contrast stretch histogram-equalized image