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have a look at this video of Fourier Decomposition of an image (otherwise you can also refer the image, which shows few plots of different extracted waves from an image). We also know that a Fourier series is given as

$$\frac {a_0} 2 + \sum \limits _{m=1} ^\infty (a_m \cos \frac {2 \pi m t} T + b_m \sin \frac {2 \pi m t} T)$$

  1. In the given video at bottom middle there is a plot of extracted waves. My doubt is whether these waves means cosine (or sine) functions from the Fourier series formula.

  2. Whether these images (plot of extracted waves) are called as basis functions in mathematics?If not,can anybody tell what are basis functions in the Fourier series formula shown above?

Note: Although the above Fourier formula is for 1D signals and video is of a 2D image, please keep in mind Fourier series formula for 2D signal.

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  • $\begingroup$ 1. They're phase plots of the 2-D waves that comprise the 2-D FT. If you look at the general form the waves that are the basis functions, $e^{j(2\pi \omega_m m + 2\pi \omega_n n)}$, what those plots show is the wave's angle, which will be constant along lines of a given orientation. 2. Yes, the 2-D waves are the basis functions of the 2-D FT. $\endgroup$ – AnonSubmitter85 Apr 29 '15 at 23:56
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I am far from an expert in signal-processing or image-processing. As far as I can tell, for each $n$, the wave and conjugate wave with the largest magnitude response is extracted.

These waves amplitude responses are then plotted as a visual representation of the wave. As $\sin$ or $\cos$ wraps around the unit circle, the phase response increases until they switch sign and suddenly drop. This is represented by the alternating colors while the intensity of the color measures the magnitude response. So in some sense, the extracted wave represents both trigonometric functions and the coefficients.

The algorithm simultaneously grabs the conjugates and plots them as a single "extracted wave" so this set of extracted waves wouldn't constitute a basis for a reasonable space of greyscale images.

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I've look at the linked video and I'm quite sure the bottom middle section is supposed to show the (am,bm) values from your formula somehow, it is however not clear how this is done and most of the time it just looks plain gray to me.

So yes, the Fourier transform is the (am, bm) values from your formula. The m value typically indicates the number of wave periods across the picture and the combination and amplitude of the values indicate the intensity and phase of the wave.

And yes, the waves described in your formula forms an orthogonal basis for the transform. It means that each picture has a unique transform that can be reverse transformed back to the picture without loss. The orthogonality of the basis means that a picture containing exactly one of the basis waves is transformed into a single value, while all other output values of the transform are zeroes.

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