Continuous and Discrete Fourier Transform at the Nyquist frequency

  • #1
Hi there,

A quick question concerning the FFT. Let's say I explicitly know a 2D function [tex]\tilde{f}\left(\xi_1,\xi_2 \right)[/tex] in the frequency domain.

If I want to know the values of [tex]f\left(x_1,x_2 \right)[/tex] in the time domain at some specific times, I can calculate [tex]\tilde{f}[/tex] at [tex]N_j[/tex]discrete frequencies (i.e. [tex]\xi_j=0, \xi_j=1/(N_j \Delta_j),...,\xi_j=\pm 1/(2 \Delta_j),...,\xi_j=-1/(N_j \Delta_j)[/tex]) and then use the inverse DFT.

My problem is the following, at the Nyquist frequencies (if [tex] \xi_1=\pm 1/(2 \Delta_j)[/tex] and/or [tex] \xi_2=\pm 1/(2 \Delta_j)[/tex]), what frequency values do I have to use to calculate [tex]\tilde{f}[/tex] ? [tex] +1/(2 \Delta_j)[/tex] or [tex] -1/(2 \Delta_j)[/tex] ?

This choice matters since they are not the same... For instance, if the frequency is not correctly chosen, then [tex]f[/tex] is not real though [tex]\tilde{f}\left(\xi_1,\xi_2 \right)=\tilde{f}\left(-\xi_1,-\xi_2 \right)[/tex]
 
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Answers and Replies

  • #2
14,168
11,468
Maybe this helps:
The same wrap-around occurs for negative frequencies. When the real-valued time series contains a component sine wave with a frequency of 100 Hz, it implicitly also contains a frequency of -100Hz. This -100Hz component also appears in the result of the FFT, but instead of mapping to a negative bin, it wraps around and appears in the second half the the spectrum. Hence, the result of FFT can be divided into a first half and a second half. For a band-limited signal with all frequencies below the Nyquist frequency, the first half of the spectrum corresponds to positive frequencies, the second half of the spectrum is the negative frequencies.
http://wiki.analytica.com/FFT
 

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