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I have a question: I am doing some research on the FFT and had an inquiry. If I have a fixed number of samples per period, what is the effect of increasing my data points, say from 128 to 256? I know that my resolution increases since it is the sampling frequency divided by my number of data points. In other words, the distance in HZ between two adjacent data points decreases. Is there anything else? Or am I looking at this wrong. Any clarification would be greatly appreciated. Thanks!

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    \$\begingroup\$ What do you mean by "fixed number of samples per period"? Do you mean a fixed sampling rate? Or a fixed total sampling window? \$\endgroup\$ – Kevin White Feb 20 '17 at 3:09
  • \$\begingroup\$ @KevinWhite a fixed sampling rate. I apologize for the poor choice in words. \$\endgroup\$ – Alex Ponce Feb 20 '17 at 3:12
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    \$\begingroup\$ I want to take out the dsp tag (which is obvious) and add a new tag: NYQUIST - You must sample more than twice the rate of the highest frequency that FFT will properly analyze. A fixed scanning rate can be divided by 2 to figure out what your top frequency will be properly represented in the data. \$\endgroup\$ – SDsolar Feb 20 '17 at 4:58
  • \$\begingroup\$ @SDsolar -- Nyquist is not a necessary tag -- or rather, it's not any more necessary than it is on ANY question where data is sampled. \$\endgroup\$ – Scott Seidman Feb 20 '17 at 16:29
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Yes you will get better resolution as the total sampling period will have doubled. You will also get a reduced noise level in each bin as the quantization noise is spread over more bins.

kevin

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If you increase the number of sampling points 2X keeping the same sampling rate, you will have the same bandwidth covered (top frequency remains the same in accord with Nuquist-Shannon-Kotelnikov sampling theorem), but you will have 2X better resolution of signal spectrum details.

If you increase the number of points 2X but sample 2X faster, you will have the same frequency resolution, but cover 2X of bandwidth.

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