As a newbie in the world of signal processing, I am having a hard time in appreciating image 2-D fourier transforms.

I am fully able to appreciate the concept of 1-D Fourier transform. Essentially, given a random causal signal, it can be decomposed into sinusoids. We cross-corelate known sinusoids (Basis functions) using FT and obtain the frequency spectrum. This is just perfect.

Please help me understand why in image processing, transformation along both axes is needed.

Given a greyscale image, like the Lena image, stored in a matrix $M \times N$, I infer the following:

  • Pixel intensity varies from left to right
  • Pixel intensity also varies top to bottom
  • A FT performed on 1 Row, Left to Right, gives us a spectrum of frequencies belonging to that row
  • Similarly, such a FT can be individually performed on each of the rows.
  • So in the end, we end up with M frequency lists of frequencies. e.g., Freq-List[Row-0] = {f1, f2, f4 ... fj}, Freq-List[Row-5] = {f2, f11 ..}

With this data, the Row frequency lists, will we not be able to tell how each pixel is affected by frequencies of that row? Shouldn't the row frequencies be sufficient?

Will the frequency lists along the columns also have a bearing on the pixels?


2 Answers 2


Please help me understand why in image processing, transformation along both axes is needed.

A one dimensional signal describes how does a quantity vary across, usually, time. Time, commonly represented by the symbol $t$, is the only parameter required to describe completely the signal at $t$.

A two dimensional signal describes how does a quantity vary across two parameters that are absolutely required to describe it completely. When referring to images, the quantity that is described is, usually, radiant flux. That is, how much "light" (more generally, radiation) is received by the sensor. In common handheld cameras, each pixel of an image describes how much visible light is received by some point in the scene that is viewed.

The complete set of pixels of an image, describes the variation of visible light across the surface of the camera's sensor.

When applying the Fourier Transform to a one dimensional signal, the dimension of time is transformed to a dimension of frequency and the transform breaks the signal down to a sum of sinusoids.

When applying the Fourier Transform to a two dimensional signal, its two spatial dimensions are decomposed into sums of orthogonal spatial sinusoids. To cut a long story short, if the basis functions of the Fourier Transform were not orthogonal, this trick of decomposing and recomposing would not be possible. How does this look like? It looks like a carton of eggs:

enter image description here

The higher the spatial frequency, the smaller the eggs (more of them fit in the same length) and vice versa.

More formally, it looks like this:

enter image description here

And it doesn't even have to be "symmetrical", that is, each one of its dimensions may be supporting a different spatial frequency:

enter image description here

In this last image, we have many more sinusoidal cycles across the $x$ dimension than across the $y$ dimension.

Therefore, whereas in the one dimensional case, a signal as complicated as the voice of a singer is decomposed into a set of simple "whistles", in two dimensions, an image as complicated as Lena is decomposed into a set of elementary little blobs. In the first case, the signal is correlated with a series of progressively increasing frequency sinusoids, in the second case, exactly the same thing happens only now, the signal is a patch of pixels and the "sinusoid" is a patch of spatial frequencies that could vary differently across the $x$ and $y$ dimension.

Now, in terms of expressing this process via the one dimensional Fourier Transform (that performs this correlation process with one set of sinusoids), the same, is applied twice.

Consider Lena. And apply the Fourier Transform across each one of its rows. What do you get? You get a set of rows in the frequency domain. These describe the set of sinusoids that visible light across the rows of the image varies. But!!! at this point, we know nothing about the set of sinusoids that describe visible light variation along the vertical.

Another way to "visualise" this, is to consider the DC bin of the row Fourier Transforms (frequency is zero). This tells you the average brightness of the pixels in each row but it still varies along the column direction! (i.e. we know nothing about the DC along the columns).

In other words, where we had an $f(x,y)$, we pass it through the Fourier Transform along the rows and we now have an $f(F_x, y)$. We are now in an intermediate state where one dimension is frequency and the other is still space.

For this reason, you apply the Fourier Transform once more along the columns of the Fourier Transform of the rows. (And to get back to the DC example, you now get one DC coefficient that describes the average brightness along the rows and the columns, that is, you get the average brightness of the image.)

Now, remember, the one dimensional Fourier Transform decomposes a signal ($x(t)$) into two series of "strength" coefficients, one for the strengths of the $\sin$ and one for the strength of the $\cos$ at different frequencies. The two dimensional Fourier Transform does exactly the same thing but now the "strength" coefficients are two dimensional. That is, some coefficient at pixel $i,j$ (in the Fourier Transformed image, i.e. the Frequency Domain) describes the contribution of a "carton of eggs" with different number of cycles in the $x$ and $y$ dimension to the whole image.

Generalising to three and higher dimensions is done similarly.

Hope this helps.

(Note, all images retrieved via Google Images and linked, rather than uploaded, to the post)

  • $\begingroup$ Thank you very much for such a marvelous answer. I will continue to read your answer until I reach my Aha moment. Most text books describe the 2-D FT as F(u,v) equating to an expression containing u and v. In plain english, what does the pair u-v signify? By plugging in u=10Hz and v=15Hz into the expression, how do we interpret the value returned by the expression? Please help. $\endgroup$
    – Raj
    Commented Sep 18, 2016 at 12:04
  • 1
    $\begingroup$ You are welcome. $u,v$ are the equivalent of $x,y$ in the frequency domain. In the spatial domain each pixel quantifies the amount of radiation at a $x,y$ location on the sensor. In the spatial frequency domain, each "pixel" describes the contribution of a spatial frequency of $u$ cycles along the $x$ direction and $v$ cycles along the $y$ direction. $u,v$ are the indices to the "strengths" I mention above. A 10,15 sinusoid "plate" looks like the third image (frequency domain) in my response $\endgroup$
    – A_A
    Commented Sep 18, 2016 at 12:51
  • $\begingroup$ Please forgive my lack of understanding. In our example, we have M x N pixels. By indexing into the image matrix by plugging in values for x and y, we obtain the corresponding pixel intensity. How do I know what pair of frequencies contributed most to the intensity? Alternatively, given a pair of frequencies, how can I find out the pair's effect on a specific x,y pair? $\endgroup$
    – Raj
    Commented Sep 18, 2016 at 13:34
  • $\begingroup$ I think I get it now. The bounds of u and v are in fact M and N. F(u,v) gives the contribution of "a" (u,v) pair for "ALL" pixels of the image. $\endgroup$
    – Raj
    Commented Sep 18, 2016 at 14:55
  • $\begingroup$ Indeed. You could decompose the image to $M \times N$ spatially delayed "spikes" and obtain the FT of each spike, to find out the contribution of each pixel to the spectrum. BUT, again, each one of these FFTs produces a "surface" of spatial frequencies. So, each pixel, contributes to the whole spectrum and each coefficient of the spectrum contributes to all pixels of the image. $\endgroup$
    – A_A
    Commented Sep 18, 2016 at 16:06

First of all, the image is a 2-D function, e.g., f(x,y). Then, if you could not represent it as a separable form, e.g., f(x,y) = f1(x) * f2(y), you could not say that a given row will not affect other rows. In contrast, the 2-D DFT kernel is separable, so you can choose the order freely as you like.

I know this explanation is not clear for better visualization.


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