[TROLLING] From: FFT Analysis for Dummies, From Topic ID: 79806 
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[TROLLING] From: FFT Analysis for Dummies, From Topic ID: 79806 
Jan 15 2012, 01:26
Post
#1


Group: Members Posts: 1418 Joined: 9January 05 From: In the kitchen Member No.: 18957 
Some points, an FFT is not an approximation, nor is it a model. It is a precise transform with a precise inverse, one that obeys power and amplitude conservation both in the time and shortterm frequency domain. I take exception to this. An FFT, like anything else, is a model and/or an approximation if it is used as such. Sometimes 3 is an approximation of pi. There's no fantastic mathematical property that can stop something from being used as a very blunt instrument. No, an FFT is an orthonormal projection, no matter how it's used. It's not the FFT's fault (it has no violition, so how can it be) if somebody uses it in a particular way.  
J. D. (jj) Johnston 


Jan 15 2012, 11:25
Post
#2


Group: Members Posts: 51 Joined: 11December 11 From: United Kingdom Member No.: 95728 
Some points, an FFT is not an approximation, nor is it a model. It is a precise transform with a precise inverse, one that obeys power and amplitude conservation both in the time and shortterm frequency domain. I take exception to this. An FFT, like anything else, is a model and/or an approximation if it is used as such. Sometimes 3 is an approximation of pi. There's no fantastic mathematical property that can stop something from being used as a very blunt instrument. No, an FFT is an orthonormal projection, no matter how it's used. It's not the FFT's fault (it has no violition, so how can it be) if somebody uses it in a particular way. I don't disagree with that in any way, but the point still stands that the statement is somewhere between vacuous and disingenuous. Similarly, I could calculate the area of a circle by multiplying the square of the radius by three. I could then assert that three is not an approximation because it is the precise midpoint between 2 and 4, and that it is also the sum of 1 and 2, and that it's even itself to the first power. It is precisely what it is. 


Jan 15 2012, 13:23
Post
#3


Group: Developer Posts: 1191 Joined: 29April 11 From: Austria Member No.: 90198 
Sorry but the above makes no sense to me, at all.
 "we are having an educated and deep technical discussion"amirm



Jan 15 2012, 18:55
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#4


Group: Members Posts: 51 Joined: 11December 11 From: United Kingdom Member No.: 95728 



Jan 16 2012, 11:26
Post
#5


Group: Developer Posts: 1318 Joined: 20March 04 From: Göttingen (DE) Member No.: 12875 
There's nothing "approximation"esque about it. Do you think it would help to repeat it a few more times? As long as you appear to make little sense and don't seem to understand what people are telling you, repetition and rewording is among the few options that could help, yes. QUOTE (Gumboot) [...] in fact, I can't think of many uses of the FFT where it isn't acting as an approximation of some other idealised operation which is either not feasible or not adequately characterised. It's acting always the same. It turns N complex samples into other N complex samples in a linear and invertable way. But I guess, when you hear "FFT" you're thinking of more than just the FFT. You're thinking of spectral analysis. Many collegues of mine also don't make this distinction. When they say "do an FFT on that signal" they really mean "divide your signal into overlapping blocks, window them, perform an FFT per each block, take the absolute value squared and average them to approximate the power spectral density" (Welch's method). That's pretty much the difference between an engineer and a mathematician. Example1: performing a fast convolution using the FFT has nothing to do with approximations. Example2: using the FFT to compute the typeIV DCT efficiently has nothing to do with approximations. Cheers! SG This post has been edited by SebastianG: Jan 16 2012, 11:28 


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