Whatever the original data for a DFT, the result is a DC term, a fundamental frequency at some phase, and a series of harmonics of that fundamental, each at some phase and amplitude. Can we all agree that that is the output that a DFT provides? If it is not, that, what is it? A DC term, a fundamental frequency at some phase, and a series of harmonics of that fundamental, each at some phase and amplitude, is the description of a periodic function. We may know that the function which provided the original data is not in fact periodic, but that doesn't make the description one of a non-periodic function. It merely means that the description is to some extent inaccurate. Jerry -- Engineering is the art of making what you want from things you can get.
I think I know what R.B-J. is saying
Started by ●February 15, 2011
Reply by ●February 15, 20112011-02-15
On Mon, 14 Feb 2011 20:44:27 -0800 (PST), Jerry Avins <jya@ieee.org> wrote:>Whatever the original data for a DFT, the result is a DC term, a fundamenta= >l frequency at some phase, and a series of harmonics of that fundamental, e= >ach at some phase and amplitude. Can we all agree that that is the output t= >hat a DFT provides? If it is not, that, what is it?That's the general interpretation, and it's a very useful one, but it does assume that the spectral coefficients describe the magnitude and phase of component sinusoids with extent outside of the N-dimensional space operated on by the DFT. A signal made up of spectral components with infinite extent (or however long) will certainly be as long as the components. This is not the only way to look at it, though. It is not necessary to assume that basis functions exist outside of the boundaries of the transform matrix.>A DC term, a fundamental frequency at some phase, and a series of harmonics= > of that fundamental, each at some phase and amplitude, is the description = >of a periodic function. We may know that the function which provided the or= >iginal data is not in fact periodic, but that doesn't make the description = >one of a non-periodic function. It merely means that the description is to = >some extent inaccurate. > >Jerry >--=20 >Engineering is the art of making what you want from things you can get.Even the continuous FT produces an estimate, so it's hard to speak in absolutes, so the description may be inaccurate regardless of which FT or viewpoint is used. And you're exactly right that if the DFT spectral terms are coefficients of infinite sinusoids with period of N or integer factors thereof, then the coefficients describe a signal of infinite extent with period of N or integer factors thereof. I think a really cool thing is that if you don't assume that the basis functions extend beyond the transform length, and apply the N-space map viewpoint of the DFT to the same input, you'll get exactly the same output as you do if you look at it as describing periodic extension. The DFT neither knows nor cares. Only people do. Eric Jacobsen http://www.ericjacobsen.org http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php
Reply by ●February 15, 20112011-02-15
Eric, When you see cos(x), don't you expect it to be meaningful for all values of x? Whether or not there is meaning for (say) negative values of x, the mathematics allows it. A projectile leaves a tower at time = zero and velocity = V. Choose any angle you like and compute the time it reaches the level ground below. Two times, one of them negative come out. One can (and usually does) discard the negative-time intercept, but it is nevertheless a solution to the second-order equation that describes the trajectory. That is not a point-of-view issue. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●February 15, 20112011-02-15
On Tue, 15 Feb 2011 05:21:38 -0800 (PST), Jerry Avins <jya@ieee.org> wrote:>Eric, > >When you see cos(x), don't you expect it to be meaningful for all values of= > x? Whether or not there is meaning for (say) negative values of x, the mat= >hematics allows it.cos(x) is actually a good example because it relates back to the circle. The argument for cos is an angle in the range of 0 to 2pi, or -pi to pi, or 0 to 360 degrees or however one wishes to describe the angle over the full sweep of a circle. It is modulo to a full sweep of the circle so more sweeps will get the same answers as the first sweep. We can exploit that to extend it's meaning to many multiples of the phase argument, in either direction, but it's still just really defined around one rotation of a circle. We can describe a continuous wave with the cos() function by allowing the phase to keep repeating modulo 2pi, and it is easy to understand that the continued rotation of the phase vector around the circle can be seen as a periodic extension of one rotation around the circle. We use this trick routinely to describe continuous waves, so, yes, when one sees a wave described using cos() or sin() we don't have a problem interpreting it as a continuous wave that can repeat to infinity if we wish to imagine it as such. We seldom calculate the phase of 2.849563x10^23 degrees, though. Note that if one so desires, it is not necessary to view multiple trips around the circle as periodic extension of the circle. It can be viewed as the same circle processed in a modulo fashion. There is no difference in the result. The DFT, however, defines only a length-N segment of waveform in the transform matrix. It is easy to imagine that the basis functions in the matrix extend beyond the matrix, and doing so allows a lot of nice analysis using periodic extension. Restricting one's view to only what is contained in the matrix, though, to the N-space it occupies, the DFT gets exactly the same result as if one imagines it with periodically extended basis functions. Analysis done this way also holds, and makes no assumptions about what happens outside the N-space occupied by the signal and basis functions. Both points of view work just fine, I see them as duals and complements of each other, and where one is clunky the other is usually more elegant. Using both is very effective. Claiming one or the other is not useful is a mistake IMHO.>A projectile leaves a tower at time =3D zero and velocity =3D V. Choose any= > angle you like and compute the time it reaches the level ground below. Two= > times, one of them negative come out. One can (and usually does) discard t= >he negative-time intercept, but it is nevertheless a solution to the second= >-order equation that describes the trajectory. That is not a point-of-view = >issue. > >Jerry >--=20 >Engineering is the art of making what you want from things you can get.You lost me on that example. We do throw away results that don't make sense or are out of the support region of the problem (like negative time), I don't see an issue with that. That's more or less how I'm seeing Robert's example B; I couldn't see how it fit the problem since it seemed like, "If you write it this way, with the window outside the N-space, it doesn't work." Okay, then discard that answer, because there's another way to write it that -does- work to make the exact same point. Eric Jacobsen http://www.ericjacobsen.org http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php
Reply by ●February 15, 20112011-02-15
On 02/14/2011 09:17 PM, Eric Jacobsen wrote:> On Mon, 14 Feb 2011 20:44:27 -0800 (PST), Jerry Avins<jya@ieee.org> > wrote: > >> Whatever the original data for a DFT, the result is a DC term, a fundamenta= >> l frequency at some phase, and a series of harmonics of that fundamental, e=<< snip >>> > Even the continuous FT produces an estimate, so it's hard to speak in > absolutes, so the description may be inaccurate regardless of which FT > or viewpoint is used.Uh uh. For functions that are absolutely integrable (I think that's the right term), the Fourier transform is exact. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●February 15, 20112011-02-15
On Tuesday, February 15, 2011 1:48:17 PM UTC-5, Eric Jacobsen wrote: ...> You lost me on that example. We do throw away results that don't > make sense or are out of the support region of the problem (like > negative time), I don't see an issue with that. That's more or less > how I'm seeing Robert's example B; I couldn't see how it fit the > problem since it seemed like, "If you write it this way, with the > window outside the N-space, it doesn't work." Okay, then discard > that answer, because there's another way to write it that -does- work > to make the exact same point.Again, we agree in almost everything but what we call things. Indeed, we do throw away solutions that don't make sense or are out of the support region. By discarding them, we implicitly acknowledge their existence. The argument in these threads seems to me to be between those who acknowledge their existence and those who deny it. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●February 15, 20112011-02-15
Tim Wescott <tim@seemywebsite.com> wrote: (snip, someone wrote)>> Even the continuous FT produces an estimate, so it's hard to speak in >> absolutes, so the description may be inaccurate regardless of which FT >> or viewpoint is used.> Uh uh. For functions that are absolutely integrable (I think that's the > right term), the Fourier transform is exact.I thought I remember square integrable, but I am not so sure now where I am remembering that from. -- glen
Reply by ●February 15, 20112011-02-15
On Tue, 15 Feb 2011 12:19:03 -0800 (PST), Jerry Avins <jya@ieee.org> wrote:>On Tuesday, February 15, 2011 1:48:17 PM UTC-5, Eric Jacobsen wrote: > > ... > >> You lost me on that example. We do throw away results that don't >> make sense or are out of the support region of the problem (like >> negative time), I don't see an issue with that. That's more or less >> how I'm seeing Robert's example B; I couldn't see how it fit the >> problem since it seemed like, "If you write it this way, with the >> window outside the N-space, it doesn't work." Okay, then discard >> that answer, because there's another way to write it that -does- work >> to make the exact same point. > >Again, we agree in almost everything but what we call things. Indeed, we do= > throw away solutions that don't make sense or are out of the support regio= >n. By discarding them, we implicitly acknowledge their existence. The argum= >ent in these threads seems to me to be between those who acknowledge their = >existence and those who deny it. > >Jerry >--=20 >Engineering is the art of making what you want from things you can get.It's been observed here before that often these long debates boil down to some sort of semantic issue. In this case I think it's a bit more than that, since it seems that it's not just what things are called but how people think of them. I guess it's a related problem since it's hard to convey how one thinks of things without using the limited vocabulary that we have to describe them. Eric Jacobsen http://www.ericjacobsen.org http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php
Reply by ●February 15, 20112011-02-15
On Tue, 15 Feb 2011 20:27:11 +0000 (UTC), glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:>Tim Wescott <tim@seemywebsite.com> wrote: > >(snip, someone wrote) >>> Even the continuous FT produces an estimate, so it's hard to speak in >>> absolutes, so the description may be inaccurate regardless of which FT >>> or viewpoint is used. > >> Uh uh. For functions that are absolutely integrable (I think that's the >> right term), the Fourier transform is exact. > >I thought I remember square integrable, but I am not so sure now >where I am remembering that from. > >-- glenEither is not the general case for signals encountered in DSP applications. And actually IIRC both are important conditions for certain cases (like power or energy spectrum estimates), but one usually doesn't need to worry about it because there are workarounds. This seems to be why the PSD is usually defined as the transform of the autocorrelation function rather than the signal itself, and even then there are assumptions about stationarity. Much of this stuff is not exact or absolute in the details, which is, IMHO, why it's hard to speak with absolute certainty at the subtle levels and why the details are often unimportant in practice. As engineers I think half the trick is knowing how much one really needs to know and not letting the rest interfere with getting work done. ;) Eric Jacobsen http://www.ericjacobsen.org http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php
Reply by ●February 15, 20112011-02-15
