Getting the contrast from DFT

1 回表示 (過去 30 日間)
Steve
Steve 2012 年 7 月 12 日
Hello Experts,
Given the spectrum image:
F(u,v) = (1/N^2) * Sum(from x=0 to N-1)Sum(from y = 0 to N-1) [f(x,y) * exp(-i* (2*pi/N)*(u*x + v*y))]
I need to find using Parseval equation: Double Sum x=0..N-1,y=0..N-1 [f^2(x,y)] = (1/N^2) * Double Sum v=0..N-1, u=0..N-1 [F^2(u,v)].
That I can get c^2 = (1/N^2)* Double Sum v=0..N-1, u=0..N-1 [f(x,y) - B]^2
Where B = 1/N^2 Double Sum x=0..N-1, y = 0..N-1 [f(x,y)] = F(0,0)
This is all part of image processing, B is brightness, c is contrast.
Please help me to get the contrast from the spectrum image.
Thanks a lot in advance. Or maybe you know where can I see the proof.

回答 (0 件)

カテゴリ

Help Center および File ExchangeModify Image Colors についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by