The Fast Fourier Transform of an array of lines is?

Fast Fourier Transform (FFT) is a powerful way of analyzing (and filtering) images. In one of the presentations today at the Royal Microscopical Society Frontiers in Bioimaging, it was proposed to evaluate and compare the resolution of various superresolution techniques. In the context of the stripy controversy, there has been some confusion over the (apparently very simple) question of what is the FFT of an array of lines.

That confusion is apparent in the response of Miao Yu and Francesco Stellacci (see below), but it is particularly stunning in this 2009 comment by one of the Nature Materials referee [Nature Materials considered and then rejected Stripy Nanoparticles Revisited in 2009]:

Arrays of parallel straight ripples are clearly visible in each nanoparticle, and it is well known that the Fourier Transform of an array of parallel straight lines is again an array of parallel straight lines. [empasis mine]

I do not know how ‘well known’ it is, but as with many ‘well known’ things, it is false. Let’s first consider the array of nice and smooth lines below.

Array of lines (sinusoidal modulation)

Array of lines (sinusoidal modulation)

Here is its FFT:

FFT of sinusoidal lines

FFT of sinusoidal lines

That is a pretty boring FFT! First thing to notice is that the FFT is constant except on one horizontal line. The reason is quite simple: the original image is invariant along the vertical axis: there is no need for vertical frequency components to describe that image. Let’s look more closely at this single line then:

Graph1We can see two nice peaks corresponding to the frequency of the oscillations above. So, in short, the FFT of an array of lines is two sharp peaks. Is it that simple? Yes, but the careful reader will have noticed the ‘nice and smooth’ above. What happens if the lines are instead steps, like this?

square lines

square lines

The FFT is still zero everywhere except on a single line, but now higher harmonics are required to account for these sharp transitions, so instead of two peaks, we get lots of them:

FFT of square lines (and line profile)

FFT of square lines (and line profile)

In the response to stripy revisited, Miao Yu and Francesco Stellacci write:

As shown in Figure S1 in the Supporting Information, a series of aligned bands (with a very well-defined wavelength λ ) shows two parallel bands in the FT, questioning this interpretation of the FT images.

How is this possible? Simple, they present clipped bands, i.e. the image is not invariant anymore; the bands have horizontal sharp edges and therefore a range of high frequency modes is required to represent those sharp edges.

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7 comments

  1. Great post, Raphael. The misconception that the Fourier transform of an array of lines is another array of lines is fairly widespread, in my experience. I taught a module on Fourier analysis for five years and it was intriguing that quite a few students “intuitively” felt that lines in real space would transform to lines in Fourier space.

    I love Fourier analysis, by the way – it’s such elegant mathematics/mathematical physics. One of the very few “Eureka!” moments I had as an undergraduate student was when I realised that the Heisenberg uncertainty principle is nothing more than a Fourier transform.

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    1. Indeed! The (Fraunhofer) diffraction pattern seen in the Young’s slits experiment is the Fourier transform of the aperture function which describes the width and separation of the slits.

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  2. Well the Fourier transform of an array of infinitely narrow lines (a Dirac comb) is indeed another Dirac comb. But if the lines have some width, then the output Dirac comb is multiplied by the Fourier transform of the line profile. In a diffraction grating this gives rise to the phenomenon of “missing orders” for example.

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    1. I agree that the FT of a Dirac comb is indeed another comb. But the Dirac delta ‘function’ isn’t really a function in the true sense of the term and it doesn’t represent a “line” as described in Raphael’s post. The Dirac delta can only really be considered within the context of the integral (Fourier, Laplace, etc..) in question.

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