Image Processing Exam Questions And Solutions -

Final mapping: 0→4, 2→6, 4→8, 6→11, 10→13, 14→15 Q7. Explain the steps to perform edge detection using the Sobel operator. Include masks and a brief example.

| 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 |

b) Middle value after sorting the 9 neighbors – definition of median filter. Section B: Short Answer Q3. What is histogram equalization? Write its main advantage and one limitation.

Sobel operator approximates gradient using two 3×3 masks: Image Processing Exam Questions And Solutions

| Spatial Domain | Frequency Domain | |----------------|------------------| | Operates directly on pixels | Operates on Fourier transform of image | | Uses masks/kernels (e.g., Sobel, averaging) | Uses filters (low-pass, high-pass) | | Faster for small kernels | Faster for large kernels (using FFT) | | Intuitive for local operations | Better for periodic noise removal | Q5. Given a 5×5 image region (pixel values):

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.

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. Final mapping: 0→4, 2→6, 4→8, 6→11, 10→13, 14→15

Here’s a useful, structured piece covering for an undergraduate-level Image Processing course. It includes multiple-choice, short answer, and problem-solving formats with explained solutions. Image Processing: Exam Questions & Solutions Section A: Multiple Choice (concepts) Q1. Which operation is not a point operation? a) Log transformation b) Histogram equalization c) Median filtering d) Gamma correction

c) Median filtering – it is a spatial operation using a neighborhood, not a point operation. Q2. In a 3×3 median filter applied to a grayscale image, the output pixel value is: a) Mean of the 9 neighbors b) Middle value after sorting the 9 neighbors c) Most frequent value d) Weighted sum of neighbors

Output pixel = Q6. Perform histogram equalization on a 4-bit image (0-15) with histogram: Gray level: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Frequency: 2 0 1 0 1 0 2 0 0 0 1 0 0 0 1 0 Total pixels = 8 | r_k | freq | CDF | CDF_norm

Extract 3×3 neighborhood around row3,col3 (value=10) – rows 2-4, cols 2-4 (1-indexed):

10 12 14 12 10 14 14 14 10 Flatten: [10,12,14,12,10,14,14,14,10] Sorted: [10,10,10,12,12,14,14,14,14] Median (5th value) =