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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?

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? Correct Answer High-boost filtering reduces to regular Highpass filtering

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.

Related Questions

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(s) validates if A increases past 1?
Which of the following gives an expression for high boost filtered image fhb, if f represents an image, f blurred version of f, fs unsharp mask filtered image and A ≥ 1?
If we use a Laplacian to obtain sharp image for unsharp mask filtered image fs(x, y) of f(x, y) as input image, and if the center coefficient of the Laplacian mask is negative then, which of the following expression gives the high boost filtered image fhb, if ∇2 f represent Laplacian?
If, Fhp(u, v)=F(u, v) – Flp(u, v) and Flp(u, v) = Hlp(u, v)F(u, v), where F(u, v) is the image in frequency domain with Fhp(u, v) its highpass filtered version, Flp(u, v) its lowpass filtered component and Hlp(u, v) the transfer function of a lowpass filter. Then, unsharp masking can be implemented directly in frequency domain by using a filter. Which of the following is the required filter?
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In frequency domain terminology, which of the following is defined as “obtaining a highpass filtered image by subtracting from the given image a lowpass filtered version of itself”?