Talk:Short-time Fourier transform

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n is discrete but omega is continuous, which may be why it is sometimes written

$S T F T [n,ω)$ ?

specgram is |stft|^2

http://www.ece.mcmaster.ca/~ibruce/courses/COE4TL4_lecture27.pdf

1 Heisenberg uncertainty principle?
2 Wavelet transform?
3 Omega vs W
4 notation
5 resolution
6 Missing "Archived" Peer Review

[edit] Heisenberg uncertainty principle?

Now I hear that the resolution problem IS the same as the Heisenberg uncertainty principle, but i have read enough over there to know that they are not actually the exact same thing. a sound/voltage/whatever wave can have instantaneous frequency, for instance. Can someone who understands in depth please explain this? Omegatron 17:05, Feb 24, 2004 (UTC)

(as in these paragraphs on the HUP talk page):

That's not true. It is not the same theorem. Standard QM uncertaintaty relations are derived from a theorem about operators on a hilbert space. These are not things that occur in classical wave mechanics. This idea about frequency and intervals is what underscores the time-energy relations, which is precisly why I cautioned against talking about them in the same language as standard QM relations. -- Matthew

There is a theorem relating the "uncertainty" in a function and the uncertainty in its Fourier transform. In standard wave mechanics, the Fourier transform is a Hilbert space automorphism which translates between the position observable x and the momentum observable -i d/dx. So the Fourier theorem can actually be used to prove the space-momentum uncertainty relation. --AxelBoldt

Omegatron 17:13, Feb 24, 2004 (UTC)

On Heisenberg - perhaps not just one mathematical formulation that says it. There is a version that says 'you can't localise a function and its Fourier transform in a way that is better than with Gaussians'. The reason that all these versions are really the same, in essence, is hidden in some group representation theory (as I'd explain it). But I think it's fine to allude to this question in passing - the reasons no one can get round the issue are surely the same in both cases.

Charles Matthews 17:46, 26 Feb 2004 (UTC)

Basically, the Heisenberg incertainty principle lies in the fact that the wider your window is, the more precise your frequency measure will be, but at the same time the less precise your time measure will be. Take as an example a 440 Hz wave sampled at 44KHz. What is the measurable frequency of, say, 1 sample? You know from the definition that it will be 440 Hz, but the measure won't tell it to you. Same for 2 or slightly more consecutive samples. So when your measure time is very precise, the measure of the frequency is at best very unprecise. Now if you measure over more of a hundred samples, you will get the frequency, but you time measure will be far less precise. That basically means that a very precise time, there is no real frequency. For a comparison with quantum mechanics, I am not a quantum physicist, but I think that quantum variables are modelized using waves, and are entangled within them. For example, position and speed of a particle are modelized using frequency and X-axis. That means that the more precise you measure one of them, the more you lose precision on the other one. Again, I have no deep knowledge in the field, but that's what I have understood. Fafner 14:05, 7 November 2006 (UTC)

I'm aware that they're analogous, but are they really the same principle, or just similar in concept? — Omegatron 14:49, 7 November 2006 (UTC)

In quantum mechanics the fourier transform is often used to convert between a position-based representation and a momentum-based one. This trick can be used to derive the momentum-position uncertainty principle from the property of the STFT. Also, the standard treatment of the uncertainty principle really is a derivation of this property of the STFT from the Cauchy-Schwartz inequality. Because of this, I think they are essentially the same principle 69.234.75.107 20:58, 29 December 2006 (UTC)

[edit] Wavelet transform?

Is it true that the STFT is a special case of a wavelet transform? -- GWO 13:59, 26 Feb 2004 (UTC)

I'm not certain, but I really don't think so. Rather, I think they are both special cases of the time-frequency-scale transform. This is like a bunch of STFTs stacked up on top of each other like sheets of paper to make a volume, and each STFT has a different windowlength, increasing in one direction. An STFT of a certain windowlength is a slice in one dimension, and a wavelet transform is a slice in a perpendicular dimension. But I really don't know much about that. Just skimmed through some articles a while ago. :-) Omegatron 15:00, Feb 26, 2004 (UTC)

I know that this is a really old comment, but I had to comment to correct. The advantages of the DWT (discrete wavelet transform)is that it doesn't "slice" the time frequency domain.

A single application of the DWT will give coefficients for very localized in time (but not in frequency) high frequency "events" and very localized in frequency (but not in time) low frequency "events". I'm only calling them events because I don't know what else to call them. You're basically dividing the time-frequency domain into layers of different shaped (but equal in area) blocks.

This is only what I've read so don't take my word as if I'm an expert in this area. Root⁴(one) 15:59, 27 March 2007 (UTC)

[edit] Omega vs W

Hello,

Wikipedia is rendering ω $(ω)$ and w $(w)$ the same way. The equation got really confusing. It would be nice if the writer changed, at least temporarily, say, the w to W. Also, admins should try to fix the bug. --Hdante 23:13, 9 February 2006 (UTC)

In the TeX equations or in the HTML? If ω and w look the same to you, that's a browser issue. It might be helpful to:

Put all the math in this article inside math tags
Use a letter other than w for the windowing function — Omegatron 02:53, 10 February 2006 (UTC)

No, I meant in TeX. I can see ω and w here. Anyway, I could see that they are different in the TeX rendering. However, they are too much similar, only differing by a small curved stroke in the left side and small thickness differences. I still think it should be changed. --Hdante 19:36, 10 February 2006 (UTC)

[edit] notation

When referring to the imaginary unit, can we please change the j to an i? Only electrical engineers use j, and the justification there (beyond "tradition!") is silly, since current (I) is talked about much more than current density (i). --Aaron Denney 22:49, 5 March 2007 (UTC)

[edit] resolution

the wikilink to resolution to point to a more exact page. Resolution (logic) makes sense to me but I could be wrong. STHayden ^{[ Talk ]} 02:30, 22 August 2006 (UTC)