**1 Answer**

written 8.1 years ago by |

In Homomorphic filter multiplicative noise is changed to an additive noise. To attain this, the illumination reflectance model is used where the image is modeled as the product of illumination i(x, y) and reflectance r(x, y) components.

Let f(x, y) = i(x, y)r(x, y) where $0\lti(x, y)\lt∞ and 0\ltr(x, y)\lt1$

Take logarithm of gray level of image at each pixel. Now f(x, y)=i(x, y)r(x, y)

ln[f(x, y)]=ln[i(x, y)]+ln[r(x, y)]

Let, f1(x, y) = ln[i(x, y)]+ln[r(x, y)]

Take DFT of f1(x, y) and then suppress the noise by passing it through filter then take inverse of output of filter.

Now, f1(x, y) = ln[i(x, y)]+ln[r(x, y)]

By DFT,

DFT{f1(x, y)} = DFT{ln[i(x, y)]} + DFT{ln[r(x,y)]}

F1(U,V) = I(U,V) + R(U,V)

F2(U,V) = H(U,V).F1(U,V)

F2(U,V)=H(U,V){I(U,V)+R(U,V)}

F2(U,V)=H(U,V)I(U,V) + H(U,V)R(U,V)

By Inverse DFT,

f2(x, y) = i’(x, y) + r’(x, y)

Take antilog at every (x, y) and scale them to 0-255 to obtain the enhanced image f0(x, y)

Now f2(x, y) = i’(x, y) + r’(x, y)

Let f0(x, y) = Antilog[f2(x, y)]

=exp[f2(x, y)]

=exp[i’(x, y) + r’(x, y)]

=exp[i’(x, y)]exp[r’(x, y)]

**f_0 (x,y)=i_0 (x,y) r_0 (x,y)**