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Jun 16, 2022 at 18:37 history edited OverLordGoldDragon CC BY-SA 4.0
deleted 67 characters in body
Jun 16, 2022 at 18:03 comment added Eddy Piedad Edited finally. I got confused with the overlap and hop. Hehe.
Jun 16, 2022 at 18:02 history edited Eddy Piedad CC BY-SA 4.0
added 2 characters in body
Jun 16, 2022 at 18:00 comment added OverLordGoldDragon What I mean is, $L - 1$ takes longest time, not least. So "when you have an overlap of $0$".
Jun 16, 2022 at 17:55 history edited Eddy Piedad CC BY-SA 4.0
clarification
Jun 16, 2022 at 17:53 comment added Eddy Piedad Oops, I think I made a mistake. Yes you're right. Edited.
Jun 16, 2022 at 17:49 comment added OverLordGoldDragon Didn't you mean "hop size" instead of "overlap" in "least possible computation time is when you have an overlap of $L−1$"?
Jun 16, 2022 at 17:39 comment added Eddy Piedad Edited. What I mean by my answer is that $L-1$ is the maximum overlapping size, that is, the minimum hop size is 1.
S Jun 16, 2022 at 17:37 review First answers
Jun 16, 2022 at 19:31
S Jun 16, 2022 at 17:37 history edited Eddy Piedad CC BY-SA 4.0
clarification of nonoverlapping STFT
Jun 16, 2022 at 17:35 comment added OverLordGoldDragon Invertibility breaks for $\geq L + 1$, yes.
Jun 16, 2022 at 16:20 comment added Eddy Piedad Do you mean the nonoverlapping STFT where $L$ is the hop size?
Jun 16, 2022 at 16:06 comment added OverLordGoldDragon Nice visual. PS maximum possible hop size is $L$ (also "overlap of" -> "hop size of"). It does depend on window but it's not $L - 1$ in general either.
S Jun 16, 2022 at 15:48 review First answers
Jun 16, 2022 at 16:29
S Jun 16, 2022 at 15:48 history answered Eddy Piedad CC BY-SA 4.0