If unsharp masking can be implemented directly in frequency domain by using a composite filter: Hhp(u, v) = 1 – Hlp(u, v), where Hlp(u, v) the transfer function of a lowpass filter. Then, the composite filter for High-boost filtering is __________(a) Hhb(u, v) = 1 – Hhp(u, v)(b) Hhb(u, v) = 1 + Hhp(u, v)(c) Hhb(u, v) = (A-1) – Hhp(u, v), A is a constant(d) Hhb(u, v) = (A-1) + Hhp(u, v), A is a constantThis question was addressed to me in exam.The above asked question is from Unsharp Masking, High-boost filtering and Emphasis Filtering in portion Intensity Transformations and Spatial Filtering of Digital Image Processing

If unsharp masking can be implemented directly in frequency domain by

1.	If unsharp masking can be implemented directly in frequency domain by using a composite filter: Hhp(u, v) = 1 – Hlp(u, v), where Hlp(u, v) the transfer function of a lowpass filter. Then, the composite filter for High-boost filtering is __________(a) Hhb(u, v) = 1 – Hhp(u, v)(b) Hhb(u, v) = 1 + Hhp(u, v)(c) Hhb(u, v) = (A-1) – Hhp(u, v), A is a constant(d) Hhb(u, v) = (A-1) + Hhp(u, v), A is a constantThis question was addressed to me in exam.The above asked question is from Unsharp Masking, High-boost filtering and Emphasis Filtering in portion Intensity Transformations and Spatial Filtering of Digital Image Processing
Answer» CORRECT option is (d) Hhb(u, v) = (A-1) + Hhp(u, v), A is a constant Easiest EXPLANATION: For given COMPOSITE filter of unsharp MASKING Hhp(u, v) = 1 – Hlp(u, v), the composite filter for High-boost filtering is Hhb(u, v) = (A-1) + Hhp(u, v).

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