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I have some questions:

1. What is the horizontal and vertical frequency range (and the steps) of
the FFT amplitudes of an image?

2. What is the relevant axes range for the phase image?

3. Which sector is physically relevant?

May be you can explain it for the following image:

enter image description here

of which the FFT amplitude image is:

enter image description here

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2 Answers 2

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  1. What is the horizontal and vertical frequency range (and the steps) of the FFT amplitudes of an image?

Unlike FFT of time varying signals whose frequency are represented in hertz, for images the frequency will be represented in cycles/pixel. 0.5 cycles per pixel is the maximum possible spatial frequency. So the FFT will be ranging from 0 to 0.5. Due to negative frequency component the value ranges from -0.5 to 0.5 for both fx and fy. From this you can calculate the step. Step_x = 1/Width(cycles/pixel), Step_y = 1/Height(cycles/pixel)

  1. What is the relevant axes range for the phase image?

For both magnitude and phase, you will plot for the same frequency range(-0.5 to +0.5)

  1. Which sector is physically relevant?

Zero frequency is DC component.So the image will vary from low frequency to high frequency as you move outwards from center of the image.

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  • $\begingroup$ I'd like to add that the DFT of a sampled time-varying signal is not Hertz, either, by itself, but "cycles per sample". The interpretation of that as some amount of "Hertz" is a multiplication with the sampling rate, just as you can convert cycles/pixel to "repetitions per millimeter" if you know the geometric properties of the imaging system. $\endgroup$
    – Marcus Müller
    Nov 23, 2016 at 12:42
  • $\begingroup$ @Navin Prashath: Thank you for your help. By sector I meant the sector when I divide the FFT image in 4 parts (part 1: fx=0 to 0.5 and fy=0 to 0.5, part 2: fx=0 to -0.5 and fy=0 to 0.5, ... ) which part has a physical meaning only part 1?. $\endgroup$
    – lio
    Nov 23, 2016 at 12:54
  • $\begingroup$ @lio . I guess you have doubt on whether the negative frequencies have any physical interpretation. Kindly refer to this link $\endgroup$
    – Navin Prashath
    Nov 23, 2016 at 13:41
  • $\begingroup$ @MarcusMüller .Based on my understanding "cycles per sample" to "cycles per second" in time varying signal is equivalent to "cycles per sample" to "cycles per pixel" in spatially varying signal. "Repetitions per millimeter" or "line pairs per image height" is preferred in film cameras whereas "cycles per pixel" in preferred in digital cameras. $\endgroup$
    – Navin Prashath
    Nov 23, 2016 at 13:46
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The FFT calculates the discrete Fourier transform (DFT) of an image. Since it is a separable transform (calculated on rows and columns independently) the explanation is the same for 1D signals.

The DFT of a signal, $X[k]$, is the projection of the signal $x[n]$ onto a complex frequency: $X[k]=\sum_n x[n]e^{-2\pi jnk/N}$.

$k/N$ can be considered to be the frequency. Since $k=0...N-1$ it starts with the lowest frequency (DC or the mean of the signal, $e^0$) and gets the highest frequency - a basis function $[-1,1,-1,1,-1,1,...]$, on the central frequency, $k=N/2$: $e^{2\pi j \frac{1}{2}n }=e^{j\pi n}$. In your example the result has been shifted (using fftshift) so that the lowest frequency is in the middle of the image, and the highest frequencies are in the border.

So, the range is the spatial frequency $0..N-1$ for each dimension. Since the result is complex you can present it using a magnitude image and a phase image, but the axes are the same. A certain point in the 2D transform image $(u,v)$ represents a frequency $e^{2\pi j(xu/N+vy/M)}=e^{2\pi j ux/N}e^{2\pi j vy/M}$ which turns out to be a spatial frequency with a certain direction and "speed".

E.g., $X[u,v]$ where $u=5, v=10, N=M=80$ represents the projection of the image onto the following basis function (here shown is the real part):

basis function with $u=v,v=10,N=M=80$

In your image, you can see most of the frequencies are concentrated near the origin (i.e. low frequencies), because most of the image is flat. Since there are edges (high frequencies) you can see some high magnitude lines on the transform image. You can compare it to an image with high frequency content, such as white noise, in which there will be much more high frequency content.

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