- #1
Wox
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If I have a signal, sampled at [itex]N[/itex] data points with a time-interval of [itex]T[/itex], does this restrict the frequency resolution I can obtain in Fourier space?
I understand that from the Nyquist-Shannon sampling theorem it follows that all information on the Fourier transform of a [itex]T[/itex]-sampled signal is mapped in the frequency interval [itex][-\frac{1}{2T},\frac{1}{2T}][/itex]. Hence the sampling rate restricts the maximal frequency that can be measured (if there are higher frequencies, you get aliasing effects).
However it is not clear to me how the frequency resolution is limited by [itex]N[/itex] and/or [itex]T[/itex]. Typical FFT algorithms return as many data points in Fourier space as there are data points in real space. Taking into account the [itex][-\frac{1}{2T},\frac{1}{2T}][/itex] limits it follows that the frequency resolution is in this case [itex]\frac{1}{NT}[/itex]. But is there a fundamental reason for having as many data points in Fourier space as there are data points in real space, or is this just convenient to implement?
I understand that from the Nyquist-Shannon sampling theorem it follows that all information on the Fourier transform of a [itex]T[/itex]-sampled signal is mapped in the frequency interval [itex][-\frac{1}{2T},\frac{1}{2T}][/itex]. Hence the sampling rate restricts the maximal frequency that can be measured (if there are higher frequencies, you get aliasing effects).
However it is not clear to me how the frequency resolution is limited by [itex]N[/itex] and/or [itex]T[/itex]. Typical FFT algorithms return as many data points in Fourier space as there are data points in real space. Taking into account the [itex][-\frac{1}{2T},\frac{1}{2T}][/itex] limits it follows that the frequency resolution is in this case [itex]\frac{1}{NT}[/itex]. But is there a fundamental reason for having as many data points in Fourier space as there are data points in real space, or is this just convenient to implement?