Question:easy

Which one of the following is NOT a linear raster operation?

Show Hint

A linear raster operation must be expressible as convolution with a fixed kernel; check which option instead depends on the image's own histogram.
Updated On: Jul 20, 2026
  • Histogram equalization
  • Averaging filter
  • Convolution with a Gaussian kernel
  • Laplacian edge detection
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Use the superposition test directly. $T$ is linear only if, for any two images $f_1, f_2$ and constants $a, b$: $T(af_1+bf_2) = aT(f_1)+bT(f_2)$.
Step 2: Check the convolution-based options. An averaging filter, a Gaussian-kernel convolution, and Laplacian edge detection are all implemented as $g(x,y) = \sum_{s,t} w(s,t) f(x-s, y-t)$ for a fixed weight mask $w$. Convolution with a fixed mask is inherently linear, since scaling or adding input images scales or adds the output by the same amounts. So options (B), (C) and (D) all pass the linearity test.
Step 3: Check histogram equalization by trying the same test. Histogram equalization builds its mapping from the histogram of the specific image being processed. If every pixel value of the input is doubled ($f_2 = 2f_1$), the histogram, and hence the CDF-based mapping computed for $f_2$, is completely different from that computed for $f_1$; the outputs are not simply related by a factor of 2. This directly violates $T(2f_1) = 2T(f_1)$, so the test fails.
Step 4: Match against the answer choices. Since the only operation among the four that fails the superposition/fixed-weighting test is histogram equalization, that is the one that is NOT linear. \[ \boxed{\text{Histogram equalization}} \]
Was this answer helpful?
0