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 andHlp(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?(a) Hhp(u, v) = Hlp(u, v)(b) Hhp(u, v) = 1 + Hlp(u, v)(c) Hhp(u, v) = – Hlp(u, v)(d) Hhp(u, v) = 1 – Hlp(u, v)I had been asked this question during an interview.My query is from Unsharp Masking, High-boost filtering and Emphasis Filtering in section Intensity Transformations and Spatial Filtering of Digital Image Processing

If, Fhp(u, v)=F(u, v) – Flp(u, v) and Flp(u, v) = Hlp(u, v)F(u, v),

1.	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 andHlp(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?(a) Hhp(u, v) = Hlp(u, v)(b) Hhp(u, v) = 1 + Hlp(u, v)(c) Hhp(u, v) = – Hlp(u, v)(d) Hhp(u, v) = 1 – Hlp(u, v)I had been asked this question during an interview.My query is from Unsharp Masking, High-boost filtering and Emphasis Filtering in section Intensity Transformations and Spatial Filtering of Digital Image Processing
Answer» Correct choice is (d) Hhp(u, v) = 1 – Hlp(u, v) For explanation: Unsharp masking can be implemented DIRECTLY in frequency DOMAIN by USING a composite FILTER: Hhp(u, v) = 1 – Hlp(u, v).

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