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Topic: Is the Fourier Transform or Series of a Square Wave more Accurate? (Read 39664 times) previous topic - next topic
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Is the Fourier Transform or Series of a Square Wave more Accurate?

Reply #25 –
Quote from: jensend on
Quote from: saratoga on

Quote from: jensend on
It's not some special collection of finite functions which have fourier series; given any finite interval and any function whatsoever on that interval, we can easily compute its Fourier series.

Yes, but as I said above, and you disagreed with, if the function isn't finite, you have a problem.


No. What you said above is that the fourier transform has a problem if the function isn't finite, which is precisely the opposite of correct.


Why are you changing the subject?  You said above that the Fourier Series is defined for finite functions and I asked you for an example of one.  This should be a very simple thing to do.

Quote from: jensend on
The Fourier series has a problem, which is the entire point of why the fourier transform was invented in the first place- to extend the idea of trigonometric representation to functions defined on infinite domains (so as to solve the heat equation for infinite initial conditions).


Which is what I said above and you told me was "the opposite of correct"  (I assume that is "wrong").

Quote from: jensend on
You can only take the fourier transform of a finite signal by considering it to be an infinite signal which just happens to have compact support.


What?  Why can't I just evaluate the Fourier integral?  If the function is finite, it stands a very good chance of converging. 

Quote from: jensend on
And the fourier transform of a nonzero function with compact support will not have compact support, e.g. rectangle<->sinc.


Obviously. 

Quote from: jensend on
What???? That's nonsense. As I said above, the more precise way to put this is that periodic functions aren't infinite in the relevant sense, they're just functions on the set R mod (tau) which is (isomorphic to) the finite interval [0,tau] with endpoints identified (equivalently, a circle of circumference tau).


This is the very definition of splitting hairs.

Quote from: jensend on
Quote from: saratoga on

Quote from: jensend on

Again, sin(x) doesn't have an energy since it's not square integrable.
Hence my suggestion that the Fourier series is more appropriate for analyzing it.
That's silly. Again, if a function is periodic and we know its period beforehand, the fourier series and the transform accomplish exactly the same thing; the fourier series is just the coefficients of the dirac comb which is the transform. Neither is more appropriate than the other in such a case and square integrability has nothing to do with it.


I don't agree for the reason I explained above.  Namely, that you are turning a signal with infinite area into a distribution that does not.  There may be a way to make the math work on this, as I said above I have no idea.  But it is certainly more appropriate when analyzing signals to avoid non-physical solutions like this.  Using the appropriate transform is an entirely appropriate way to do this.

Is the Fourier Transform or Series of a Square Wave more Accurate?

Reply #26 –
Much of the problem with this conversation was terminological; when you say "finite signal" you apparently mean a function on an infinite domain which is defined as zero everywhere outside an interval (i.e. it has compact support).

This is counter to the normal conventional use of the terms; normally a finite signal is a function with a domain of finite length. If a signal was nonzero for only one second before falling to zero for a millennium, the fact that it is still zero for the first second of the next millennium is new information; even if its value is zero out to infinity it is still an infinite signal. When we compute the fourier transform of a function, at no point does the question of whether it is compactly supported or not come up at all; our integrals are over the full real line.

The way you're talking about it doesn't allow for any domains of finite size. I bring up signals of finite-sized domain (which of course have a fourier series expansion) and you say "that's not a finite signal" because you're jumping to the periodic extension to a full real line.

Your claim that fourier series do not apply to 'finite signals' in your terminology doesn't contradict my claim that they apply just fine to finite signals, meaning functions whose domain has finite length. Not only that, it's a special case of what I mean when I say that in a sense they only really apply to finite signals-- since to apply it to an infinite periodic signal we first have to know a period and we thus reduce the signal to a function on a finite interval. Our integrals have to be over an interval of period length and in essence we have done away with its existence outside that interval by modding out (tau).

Anyhow, under your nonstandard terminology, then, fourier series apply neither to finite signals nor to any but a vanishingly slim minority of infinite ones, while the fourier transform applies to all signals except for those that grow faster than polynomially. Even if you want to exclude fourier transforms which are distributions and not functions, every function that's square integrable, along with some others that aren't (e.g. the Cauchy distribution), have fourier transforms that are plain old functions, regardless of whether or not they are compactly supported.

Quote from: saratoga on
I don't agree for the reason I explained above.  Namely, that you are turning a signal with infinite area into a distribution that does not.  There may be a way to make the math work on this, as I said above I have no idea.  But it is certainly more appropriate when analyzing signals to avoid non-physical solutions like this.  Using the appropriate transform is an entirely appropriate way to do this.
There's nothing about the fourier transform of a periodic signal that's any more "non-physical" than the sequence of coefficients is (when was the last time you saw a sequence as an analog signal?). And the "reason you explained above" was *energy* not area and as I've said a number of times now both sin(x) and a delta function have infinite energy.

The fact that the fourier transform turns this function that's bounded but has infinite support into a distribution that's unbounded but has finite support and vice versa shouldn't be surprising; the gabor limit/uncertainty principle stuff is very basic to what goes on with the fourier transform, and area is most definitely not conserved.
 

Is the Fourier Transform or Series of a Square Wave more Accurate?

Reply #27 –
Quote from: saratoga on
I don't agree for the reason I explained above.  Namely, that you are turning a signal with infinite area into a distribution that does not.


If you want to get "physical", why don't just think of Dirac's delta as a limit case of the modified sinc function, sin(c*pi*x)/(pi*x) as c tends to infinity. This should straighten out everything, doesn't it? You have energy preservation, and intuitively straightforward mapping from shifted sinc in frequency domain to a windowed sine wave in time domain. As the time window length extends to infinity, sinc function in frequency domain narrows to a singular pulse.

Or am I missing something here?