tag:blogger.com,1999:blog-20076457.post8990414964274113190..comments2022-10-05T14:17:04.229-07:00Comments on Rants of a spud: Investigating the GIMP Fourier transformBiospudhttp://www.blogger.com/profile/13304428707742568072noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-20076457.post-65502464015216328012009-08-16T00:31:46.955-07:002009-08-16T00:31:46.955-07:00Quote: "The bottom line here is that is seems...Quote: "The bottom line here is that is seems to me that the Fourier transform should have twice as much data as the original image"<br /><br />Well in one sense as two images are produced for a FFT, their is twice as much data.<br /><br />But in a practical sense that data is repeated twice! Only half (say the left half) of each magnitude and phase image is needed, as the other half is simply a repeat of the other (rotated 180 degrees) about the origin.<br /><br />Note that origin itself only needs the magnitude component (DC value) as the Phase value for this component is always zero (it is the constant average color of the image).<br /><br />So while twice as much data is generated, only half is actually needed, or used.<br /><br />Anthony ThyssenAnthony Thyssenhttp://www.imagemagick.org/Usage/fourier/noreply@blogger.comtag:blogger.com,1999:blog-20076457.post-6664540068126311122009-06-26T13:26:31.082-07:002009-06-26T13:26:31.082-07:00Hi,
What Daniel means is that half of the transfor...Hi,<br />What Daniel means is that half of the transformed data is superfluous, because it must cancel out the other half in order to end up with a purely real image when you transform back.<br /><br />Another way to think of this is that you are giving it the same amount of information, except the "imaginary channel" of the original image is all zeros.<br /><br />Good luck!<br />AdrianoAdriano Ferrarihttps://www.blogger.com/profile/18358356461276325279noreply@blogger.comtag:blogger.com,1999:blog-20076457.post-1407975652138615642009-02-27T13:29:00.000-08:002009-02-27T13:29:00.000-08:00"Thus you can find a representation of the FT whic..."Thus you can find a representation of the FT which as well is real only."<BR/><BR/>You flatter me Daniel. I'm sure you will agree that the Fourier transform of a real-valued function has complex values. But I do not know how to find a "representation" of that Fourier transform that is real only. Unless you mean that the complex Fourier coefficients can be represented as either (phase, amplitude) or (r cos theta, r sin theta) or (real, imaginary). But that would be using twice the number of data values. Which is exactly what you are saying is unnecessary. Please explain further. I would like to know how to "represent" the Fourier transform of a real valued function using only real values.Biospudhttps://www.blogger.com/profile/13304428707742568072noreply@blogger.comtag:blogger.com,1999:blog-20076457.post-21849057777059976142009-02-27T12:54:00.000-08:002009-02-27T12:54:00.000-08:00TLDR, but no, the fourier transform will not conta...TLDR, but no, the fourier transform will not contain 'twice as much information', it will contain exactly the same amount of information. The key is that the imaginary component of the input data is always zero (images are real only). Thus you can find a representation of the FT which as well is real only.Danielhttps://www.blogger.com/profile/14406761733484637804noreply@blogger.comtag:blogger.com,1999:blog-20076457.post-34580025614627015312008-11-11T07:40:00.000-08:002008-11-11T07:40:00.000-08:00Hey vegas. What you see here in these posts is pr...Hey vegas. What you see here in these posts is pretty much as far as I got. I hope to return to this project one of these days.Biospudhttps://www.blogger.com/profile/13304428707742568072noreply@blogger.comtag:blogger.com,1999:blog-20076457.post-85245603763245759592007-10-22T14:15:00.000-07:002007-10-22T14:15:00.000-07:00hey, I'd be interested to hear how this ended up f...hey, I'd be interested to hear how this ended up for you.Alexander Fairleyhttps://www.blogger.com/profile/16271647580060960779noreply@blogger.com