Continuous and Discrete Fourier Transform at the Nyquist frequency

[CharlesMareau]

Hi there,

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

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

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

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

 

[fresh_42]

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