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Spatial Frequencies of the Fourier Transform

  1. Jul 16, 2015 #1
    The Fourier Transform transforms a function of space into a function of frequency. Considering a function ##f\left(x, y\right)##, the Fourier Transform of such a function is ##\mathcal{F}\left\{f\left(x, y\right)\right\} = F\left(p, q\right)##, where ##p## and ##q## are the spatial frequencies.

    In numerical simulations, the function ##f## can easily be transformed by using an algorithm (built-in in the software). However, I am concerned on acquiring the spatial frequencies ##p## and ##q##. Is there a way to do it?

    Thank you in advance.
     
  2. jcsd
  3. Jul 16, 2015 #2

    blue_leaf77

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    If the function in space domain is sampled ##N## times with sampling interval ##\Delta x##, then the frequency resolution in the frequency domain is given by ##2\pi/N\Delta x##.
     
  4. Jul 16, 2015 #3
    I don't know if I got this right. Is the value of ##N## the number of points, ##\Delta x## is the interval between two adjacent points, and the interval between two points in the frequency domain is ##\frac{2 \pi}{N \Delta x}##?
     
  5. Jul 16, 2015 #4

    blue_leaf77

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    Yes, exactly.
     
  6. Jul 16, 2015 #5
    Thank you.

    I have another question, how will I know the boundaries (maximum and minimum values) of the frequency domain?
     
  7. Jul 16, 2015 #6

    Dale

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  8. Jul 16, 2015 #7

    boneh3ad

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    I'll also throw in that a "spatial frequency" is more commonly called a "wavenumber".
     
  9. Jul 19, 2015 #8
    The wavenumber of what? What is ##\lambda## in this case?
     
  10. Jul 19, 2015 #9

    olivermsun

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    The wavenumber of "whatever the spatial frequency belongs to."

    The usual way of defining the wavenumber is exactly analogous that for frequency:
    ##k = 2\pi/\lambda## in radians per length, or ##\nu = 1/\lambda## if the units are cycles per length.
     
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