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?(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Fourier transform frequency resolution

Loading...

Similar Threads for Fourier transform frequency |
---|

I Repetitive Fourier transform |

I Proving the Linear Transformation definition |

I Units of Fourier Transform |

A Help with Discrete Sine Transform |

**Physics Forums | Science Articles, Homework Help, Discussion**