If, Fhp(u, v)=F(u, v) ‚Äö√Ñ√∂‚àö√ë‚àö¬® 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?$

Hhp(u, v) = ‚Äö√Ñ√∂‚àö√ë‚àö¬® Hlp(u, v)

Hhp(u, v) = 1 ‚Äö√Ñ√∂‚àö√ë‚àö¬® Hlp(u, v)

If, Fhp(u, v)=F(u, v) ‚Äö√Ñ√∂‚àö√ë‚àö¬® Flp(u,

1.	If, F_hp(u, v)=F(u, v) ‚Äö√Ñ√∂‚àö√ë‚àö¬® F_lp(u, v) and F_lp(u, v) = H_lp(u, v)F(u, v), where F(u, v) is the image in frequency domain with F_hp(u, v) its highpass filtered version, F_lp(u, v) its lowpass filtered component and H_lp(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.	H<sub>hp</sub>(u, v) = H<sub>lp</sub>(u, v)
B.	H<sub>hp</sub>(u, v) = 1 + H<sub>lp</sub>(u, v)
C.	H<sub>hp</sub>(u, v) = ‚Äö√Ñ√∂‚àö√ë‚àö¬® H<sub>lp</sub>(u, v)
D.	H<sub>hp</sub>(u, v) = 1 ‚Äö√Ñ√∂‚àö√ë‚àö¬® H<sub>lp</sub>(u, v)
Answer» E.

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