tag:blogger.com,1999:blog-11295132.post2698212297987135360..comments2017-08-12T12:44:59.674-07:00Comments on A Neighborhood of Infinity: An elementary way to approach Fourier transformsDan Piponihttps://plus.google.com/107913314994758123748noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-11295132.post-44816033880991101042012-12-13T13:28:21.098-08:002012-12-13T13:28:21.098-08:00Stricktly speaking: A "Constant" signal ...Stricktly speaking: A "Constant" signal (a "Cosine" and many other signals) do not have Fourier Transform, because it is not a finite energy signal. But we can work around this by "taking the Fourier Transform to the limit". In doing so, the Fourier Transfom (in the limit) for a "Constant" signal is a: "Dirac Delta"Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-85595953576033938522011-07-21T06:52:07.903-07:002011-07-21T06:52:07.903-07:00@m I sold all my books on Fourier transforms a whi...@m I sold all my books on Fourier transforms a while ago so I'm not sure what books to recommend.<br /><br />Many of the elementary books tend to harp on about how the Fourier transform decomposes a signal into frequencies. But this gives no intuition for why the important theorems hold. But a web search on "fourier transform diagonalizes translation" will give you lots of useful hits.<br /><br />From a pure mathematical perspective, the DFT is a special (in fact, easy) case of representation theory for finite groups: http://en.wikipedia.org/wiki/Representation_theory_of_finite_groups#Discrete_Fourier_transformsigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-53290913867072976382011-07-21T06:18:22.477-07:002011-07-21T06:18:22.477-07:00"Approaching Fourier transforms through the p..."Approaching Fourier transforms through the properties I listed is common in the more advanced mathematical literature." Could you provide some references? Any favorite textbook using this kind of approach?mhttps://www.blogger.com/profile/06076269543269046558noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-82983801874307548682011-07-05T08:32:53.659-07:002011-07-05T08:32:53.659-07:00I'm confused by the use of the word "cons...I'm confused by the use of the word "constant" in this discussion. I didn't use the word constant and at no point did I talk about images that have the same value for each pixel.<br /><br />BTW A shift is convolution with an image that has *one* non-zero pixel, not an image with a row of non-zero pixels. The two coordinates of the pixel give the amount of shift along the two different axes.<br /><br />A and B aren't constant images in the sense you describe. They're just images. You can see what they are because I give code to compute them (in two different ways). You use Octave to examine them and see they don't have the same value for each pixel. Given any choice of Fourier transform (eg. Octave's fft2 function) there is a pair of images, A and B, that have the property I give.<br /><br />Once I figure out what the confusion is here I'll try to go back and edit the original text to try to avoid it in future.sigfpehttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-78106994322221006422011-07-05T08:20:11.961-07:002011-07-05T08:20:11.961-07:00By a "constant," here, I mean an array t...By a "constant," here, I mean an array that has the same value in all its slots. The transform of an array that has all ones in a single row or a single column and zeros is not, in general, a constant. For example, in 1 dimension, Fourier of {1,0,0,0} is {.5,.5,.5,.5}, a constant, but Fourier of {0,1,0,0} is {.5,.5i,-.5,-.5i}, not a constant. So I'm confused about the transform of a shift, which is convolution with {0,1,0,0} or {0,0,1,0}, etc., being the product of a constant with the transform of the original image.A Breaking Changehttps://www.blogger.com/profile/00115031720080635093noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-2662222999943985962011-07-04T22:34:37.052-07:002011-07-04T22:34:37.052-07:00@A Breaking Change: The transform of a constant is...@A Breaking Change: The transform of a constant is obviously a constant. Why do you say "but it's not"? What definition of "constant" are you using?<br /><br />("Constant" here is the opposite of "variable". It doesn't mean a scalar, nor has sigfpe claimed that it's the <i>same</i> constant for every shift.)displaynamehttps://www.blogger.com/profile/09068351772472305473noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-74744480256606904562011-06-27T19:36:56.929-07:002011-06-27T19:36:56.929-07:00I got really excited for "an elementary appro...I got really excited for "an elementary approach to Fourier transforms". And then I got really confused, really quickly.<br /><br />Seriously... what?<br /><br />Can someone please explain this explanation?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-11295132.post-36634160179694395362011-06-27T07:05:00.715-07:002011-06-27T07:05:00.715-07:00Something seems fishy, here.
The convolution theo...Something seems fishy, here.<br /><br />The convolution theorem is that the transform of a convolution is the (pointwise) product of the transforms.<br /><br />I think a shift is convolution of image A with a special image B that has all ones in a single row (or column).<br /><br />Apply convolution theorem: transform of a shift is Transform(A) .* Transform(B that has all ones in a single row [or column]). If that's a constant .* Transform(A) (as in your axiom #2) then Transform(B that has all ones in a single row [or column]) should be a constant, but it's not.<br /><br />Where am I going wrong?A Breaking Changehttps://www.blogger.com/profile/00115031720080635093noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-86846297251366107702011-06-27T05:04:55.597-07:002011-06-27T05:04:55.597-07:00@Anonymous: The post is written in "literate ...@Anonymous: The post is written in "literate Octave", which means you're supposed to have Octave/Matlab and actually run the code as you read. You'll see the image if you do.displaynamehttps://www.blogger.com/profile/09068351772472305473noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-59370870108278251932011-06-26T01:59:46.193-07:002011-06-26T01:59:46.193-07:00So what does the test image look like with bokeh a...So what does the test image look like with bokeh applied?Anonymousnoreply@blogger.com