RE: [WM]: On DFT/FFT-based Watermarking

["Martin, Christopher \(cmarti02\)" <Cmarti02@harris.com>]

You must maintain the original Phase angle between the Re and Im components.
Otherwise you will get harsh artifacts. 

To increase the magnitude (M)  by some amount 'm', add m*cos(theta) to the
Real conponent and m*sin(theta) to the Imaginary component.


Chris

-----Original Message-----
From: Editha A. DeLuna [mailto:edith@circuit.nagaokaut.ac.jp]
Sent: Wednesday, July 11, 2001 7:00 AM
To: watermarking@watermarkingworld.org
Subject: [WM]: On DFT/FFT-based Watermarking


Hello everyone.

I am confuse with one very simple question.

On the DFT based watermark embedding, for instance, the watermark is
embedded into the 1000 largest DFT Magnitude coefficient [Cox. et al]. 

So, for the color image, the process is:
(1) Transform RGB into YCrCb components
(2) FFT the Y component
(3) Compute the magnitude & phase which is
   mag = sqrt (Re*Re + Im*Im): MAGNITUDE
   theta = atan2(Im,Re)  :PHASE
(4) Embed the watermark into the magnitude
(5) Compute for the new Re, Im components  after modifying the mag into
mag'
(6) IFFT 
(7) YCrCb --> RGB

My question is in no. (5). I don't know how to compute the Re and Im
components of the modified magnitudes, I assumed that it can be computed
by:

 Re'= sqrt((mag'*mag')/ 1+ tan(theta)*tan(theta))
    since phase: tan(theta) = Im/Re didn't change.

 Im' = tan(theta)* Re'

But I am getting all (+) values for the Re, which is because I think my
program takes only the positive results calculated from the square root.
Thus, the resulted image is quite different from the original image.

Is my calculation wrong? Please advise me on this matter!

Thank you in advance.

Editha
--------------o0o-------------------
Editha A. DeLuna 
Graduate School of Electronics Engg.
Nagaoka University of Technology
Niigata, JAPAN
email: edith@circuit.nagaokaut.ac.jp


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