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Mar 7, 2023 at 13:09 answer added mins timeline score: 1
Aug 22, 2018 at 20:08 history edited Laurent Duval CC BY-SA 4.0
minor typo editing
Aug 22, 2018 at 19:47 answer added aman deep timeline score: 2
Jan 4, 2012 at 19:10 vote accept gallamine
Oct 30, 2011 at 15:40 history edited Jason R CC BY-SA 3.0
rephrased title as a question
Oct 28, 2011 at 8:20 history tweeted twitter.com/#!/StackSignals/status/129789596379774976
Oct 26, 2011 at 17:22 comment added Dilip Sarwate If you pass the last $N$ of the $N+100$ samples to your $N$-point FFT subroutine, you will get the same FFT as you got before. No difference whatsoever. If you pass the first $N$ of the $N+100$ samples (with the first $100$ samples being $0$) to your $N$-point FFT subroutine, you will get things that are difficult to interpret. Read the Answer by @JasonR carefully which tells you that if the first $100$ samples are filled from your data via a circular or cyclic shift, then you will see the delay reflected in the phase of the samples.
Oct 26, 2011 at 17:12 comment added gallamine @Dilip I'm looking for a more general answer. Perhaps an explanation of what would change in those scenarios would be helpful?
Oct 26, 2011 at 16:10 answer added Spacey timeline score: 3
Oct 26, 2011 at 13:44 answer added Jason R timeline score: 27
Oct 26, 2011 at 13:37 comment added Dilip Sarwate It depends on what you mean by FFT. Say your original signal had $N$ time samples. Suppose the delay is $100$ samples. So now you have $N+100$ samples with the first $100$ being $0$. Are you computing the FFT of the first $N$ samples (same as before)? of the $N+100$ samples? of the last $N$ of the $N+100$ samples? The answer will depend on what you mean by FFT...
Oct 26, 2011 at 12:26 history asked gallamine CC BY-SA 3.0