Undergrad What Causes Repetition in Fourier Transforms of Audio and Visual Data?

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SUMMARY

The discussion centers on the properties of Fourier transforms, specifically the behavior of repeated transformations. It is established that applying the Fourier transform twice to a function results in a mirrored version of the original function, denoted as F(F(f)) = f' where f'(x) = f(-x). Additionally, it is confirmed that applying the Fourier transform four times returns the original function, F(F(F(F(f)))) = f, provided the function is sufficiently smooth. Functions such as Gaussians exhibit this self-Fourier property, and the discussion highlights the importance of understanding the conventions of forward and inverse transforms.

PREREQUISITES
  • Understanding of Fourier transforms and their mathematical definitions
  • Familiarity with complex numbers and exponential functions
  • Knowledge of the properties of Gaussian functions in the context of Fourier analysis
  • Basic grasp of signal processing concepts, including the transform and inverse transform domains
NEXT STEPS
  • Research the properties of self-Fourier transforms and their applications
  • Learn about the implications of the Fourier transform in signal processing
  • Explore the mathematical proofs related to the behavior of repeated Fourier transforms
  • Investigate the role of Gaussian functions in Fourier analysis and their significance
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Mathematicians, signal processing engineers, audio and visual data analysts, and anyone interested in the theoretical aspects of Fourier transforms and their applications in various fields.

mertcan
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I would like to express that when I am viewing the repetitive Fourier transform on Internet I encounter that for instance twice Fourier transform may lead the same value at the end of first Fourier transform. When does repetitive( twice or third... consecutively)fourier transform be same with the first Fourier transform? Or what kind of functions have this property when they are transformed according to fourier??
 
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Fourier transform and inverse transform have almost identical kernels e^{itx} \ and\ e^{-itx}. That might explain your concern.
 
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mertcan said:
I would like to express that when I am viewing the repetitive Fourier transform on Internet I encounter that for instance twice Fourier transform may lead the same value at the end of first Fourier transform. When does repetitive( twice or third... consecutively)fourier transform be same with the first Fourier transform? Or what kind of functions have this property when they are transformed according to fourier??

@mathman is right. In detail:

If for any function ##f## from reals to complex numbers, we define ##F(f)## to be that function ##\tilde{f}## such that:

##\tilde{f}(y) = \frac{1}{\sqrt{2\pi}}\int f(x) e^{-iyx} dx##

Then ##F(F(f))## is that function ##f'## such that ##f'(x) = f(-x)##.

So a double transform returns you almost to where you started, except the mirror-image. ##F(F(F(F(f))))## will always be equal to ##f##. (Well, if ##f## is sufficiently smooth, anyway.)
 
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The Fourier transform of a Gaussian gives a Gaussian. There may be other functions that have this property but I have never seen one. Are you looking for a proof of a functional form that is preserved under repeated Fourier transformation operations? Maybe a mathematician on this forum can point you to literature on the subject.

Peace
Fred
 
A quick google search for self Fourier transforms gives a host of references...

There are multiple functions that have this property.
 
Just it's inverse... sort of at least, the only difference between a forward transform and an inverse transform is the sign of the exponential and the initial data... So apply Euler's formula and have X[k] = x[n]*(cos(-2.0*k*n*pi/frame_size) + i*sin(-2.0*k*n*pi/frame_size)) or for the inverse, x[n] = X[k]*(cos(2.0*k*n*pi/frame_size) + i*sin(2.0*k*n*pi/frame_size)). Video sample here:



The example is slight more complicated because a z transform is used...

I'm taking things slow, because I would like to understand the butterfly idea with going fast; anyone got any hints for me?

EDIT:
I'm sort of casual with my explaining. I presume that you already understand this material like me.

A forward Fourier transform will bring you to the transform domain. The transform domain's data is the magnitude and phase of the signal. The magnitude is displayed in red above.

An inverse Fourier transform takes the transform data and brings it back to the time domain; shown above as the yellow signal.
 
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just for fun:

 
yet another:

 
mathman said:
Fourier transform and inverse transform have almost identical kernels
eitx and e−itxeitx and e−itx​
e^{itx} \ and\ e^{-itx}. That might explain your concern.

A forward transform has a negative sign, the inverse is positive...
 
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ADDA said:
A forward transform has a negative sign, the inverse is positive...
Yes, but that is a matter of convention, just like there are two conventions for the factor of (2\pi)^{-n} in front of the integral.
 
  • #11
Dr Transport said:
Yes, but that is a matter of convention, just like there are two conventions for the factor of (2\pi)^{-n} in front of the integral.

It might be a convention as to whether you use a + sign or - sign in the definition of the Fourier transform, but it's absolutely necessary that the inverse uses the opposite sign.
 
  • #13
just for fun:

60HZ frame rate audio or a 44.1k sample rate

3HZ visual frames

The green signal would be a representation of the left channel or the data received by your stereo or phone. The background data transforms that green signal to the top yellow signal or original audio in blue.

 

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