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Signal Processing

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    $\begingroup$ 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... $\endgroup$ Commented Oct 26, 2011 at 13:37
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    $\begingroup$ @Dilip I'm looking for a more general answer. Perhaps an explanation of what would change in those scenarios would be helpful? $\endgroup$ Commented Oct 26, 2011 at 17:12
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    $\begingroup$ 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. $\endgroup$ Commented Oct 26, 2011 at 17:22