Question

High boost filtered image is expressed as: fhb = A f(x, y) – flp(x, y), where f(x, y) the input image, A is a constant and flp(x, y) is the lowpass filtered version of f(x, y). Which of the following fact validates if A=1?

(a) High-boost filtering reduces to regular Highpass filtering

(b) High-boost filtering reduces to regular Lowpass filtering

(c) All of the mentioned

(d) None of the mentioned

I got this question during an interview.

This intriguing question originated from Unsharp Masking, High-boost filtering and Emphasis Filtering topic in chapter Intensity Transformations and Spatial Filtering of Digital Image Processing

Answer 1

Right choice is (a) High-boost filtering reduces to regular Highpass filtering

The explanation: High boost filtered image is modified as: fhb = (A-1) f(x, y) +f(x, y) – flp(x, y)

i.e.  fhb = (A-1) f(x, y) + fhp(x, y). So, when A=1, High-boost filtering reduces to regular Highpass filtering.