How to get information on low frequencies of a signal using FFT?

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The discussion focuses on applying FFT to analyze low frequencies in a fringe pattern captured by a CCD camera. It highlights the confusion regarding obtaining temporal frequency measurements from an image's FFT, as images typically represent spatial frequencies. The contributor notes that the width of the fringes suggests a spatial frequency of 10, not 1/10, which is crucial for understanding the FFT output. Additionally, it clarifies that the lowest positive frequency in the FFT corresponds to the total duration of the signal, emphasizing the relationship between time and frequency resolution. Overall, the conversation underscores the importance of correctly interpreting spatial versus temporal frequencies in FFT analysis.
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Hello. I've made a lab experiment of interference and with a CCD camera photographed fringe pattern which one can understand as a periodic signal.
My question is: When I apply an FFT to the image is there any limit to the amplitude information I may get from lower frequencies?
I read in the internet that for a 1 sec signal, one may be able to obtain information about only freqs above 1 Hz.
Why do I ask this? Because the width of the fringes is of about 1/10 the size of the total image, and I am interested in low frequencies.

Now I am processing the image in ImageJ in order to get some information.

Thanks in advance.
 
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I'm not quite sure how you're getting temporal measurements (hz) from the FFT of an image (usually in vector space), unless you're asking about contributions from low-frequencies in general.

When you mention fringes being 1/10 the size of the image, you may have things reversed: the contribution from that should actually have a spatial frequency of 10, not 1/10 (think 10 fringes per image):
http://sharp.bu.edu/~slehar/fourier/fourier.html#harmonics
 
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The lowest positive frequency represented in the FFT output will be 1 cycle over the length of data. So, for a 1 second signal, there would be information for ..., 0 Hz, 1Hz, ...

1 * fs/N = 1/(Ts * N) = 1/(Total time)
 
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