And Solutions: Image Processing Exam Questions

| r_k | freq | CDF | CDF_norm = CDF/8 | Equalized = round(15 × CDF_norm) | |-----|------|-----|------------------|----------------------------------| | 0 | 2 | 2 | 0.250 | 4 | | 1 | 0 | 2 | 0.250 | 4 | | 2 | 1 | 3 | 0.375 | 6 | | 3 | 0 | 3 | 0.375 | 6 | | 4 | 1 | 4 | 0.500 | 8 | | 5 | 0 | 4 | 0.500 | 8 | | 6 | 2 | 6 | 0.750 | 11 | | 7 | 0 | 6 | 0.750 | 11 | | 8-14| 0 | 6 | 0.750 | 11 | | 10 | 1 | 7 | 0.875 | 13 | | 14 | 1 | 8 | 1.000 | 15 |

10 10 20 10 10 20 10 10 20 Gx convolution at center: (-1×10)+(0×10)+(+1×20) + (-2×10)+(0×10)+(+2×20) + (-1×10)+(0×10)+(+1×20) = (-10+0+20) + (-20+0+40) + (-10+0+20) = 10 + 20 + 10 = 40. Gy = 0 (uniform vertically). Magnitude = 40 → strong vertical edge. Q8. Convolution and correlation are identical operations in image processing. Solution: False. In convolution, the kernel is flipped (rotated 180°) before applying; correlation does not flip. Image Processing Exam Questions And Solutions

10 12 12 14 16 12 10 12 14 16 12 12 10 14 16 14 14 14 10 18 16 16 16 18 20 Compute the output of a at center position (row 3, col 3) – 1-indexed (value=10). Use zero-padding. | r_k | freq | CDF | CDF_norm

Extract 3×3 neighborhood around row3,col3 (value=10) – rows 2-4, cols 2-4 (1-indexed): In convolution, the kernel is flipped (rotated 180°)