Question

A 3×3 image (Image A) has been linearly stretched to get the maximum contrast in an 8-bit display system. Digital Number (DN) values of the pixels in Image A are shown. The value of the pixel marked as '?' in the output Image B after linear stretching is ________
\includegraphics[width=0.5\linewidth]{63.png}

Hint:

To perform linear contrast stretching, identify the minimum and maximum pixel values in the original image and stretch the pixel values to the full range of the display system (0 to 255 for an 8-bit image).

Answer 1

Step 1: Understanding linear contrast stretching. Linear contrast stretching is used to increase the contrast of an image by linearly transforming the range of pixel values. For an 8-bit system, the pixel values are stretched to span the full range from 0 to 255. The formula for linear contrast stretching is: \[ DN_{B} = \frac{(DN_A - DN_{min})}{(DN_{max} - DN_{min})} \times 255 \] Where:
\( DN_A \) is the original pixel value in Image A,
\( DN_{min} \) is the minimum value in Image A,
\( DN_{max} \) is the maximum value in Image A,
\( DN_{B} \) is the pixel value after contrast stretching in Image B.
Step 2: Identifying the minimum and maximum values in Image A.
From Image A, the minimum value (\( DN_{min} \)) is 30 and the maximum value (\( DN_{max} \)) is 180. Step 3: Applying the formula to calculate the pixel value for '?'.
Let the pixel value marked as '?' be the same as the one at the position with a value of 100 in Image A. Using the linear contrast stretching formula: \[ DN_{B} = \frac{(100 - 30)}{(180 - 30)} \times 255 \] \[ DN_{B} = \frac{70}{150} \times 255 \] \[ DN_{B} = 0.4667 \times 255 \approx 51 \] Thus, the pixel value marked as '?' in Image B after linear stretching is \( 51 \).