This question is derived by this question and the corresponding answer. The problem was that I had a $sech(x)$-function in a specific time interval, and I applied a fft on it. But when I increased the amount of points in the time interval, the resulting function got narrower. The answer was that I am shifting the original function to higher frequencies (by keeping it at the same width), and therefore reducing the size of the fft-result.
As far as I know, that is not correct. Is there a mathematical explanation why I get a narrower fft-function when I am increasing the amount of points in the time interval, but keeping the FWHM of the original function constant?
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arc_lupus
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The fourier transform of your function has not gotten narrower. It is exactly as it was before. You simply have appended more empty space to the size because the frequency bandwidth is limited.
Addition: If you expand the time interval, the frequency domain will get denser. If sample the same time interval more densely, the frequency domain will expand.
Gappy Hilmore
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In my example I kept the whole time interval the same, just increased the amount of points in this time interval. – arc_lupus Jun 09 '15 at 17:08
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That is what said. The effects are swapped in time and frequency domain: "If sample the same time interval more densely, the frequency domain will expand" – Gappy Hilmore Jun 09 '15 at 17:24
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But is there a final solution? Or is it always depending on the number of sampling points? – arc_lupus Jun 09 '15 at 20:10
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It will always depend on the number of sampling points. To understand this, you need to under stand https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem You need to sample the time signal with twice the frequency of the highest frequency component. That will suffice. – Gappy Hilmore Jun 09 '15 at 20:17