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Is there any practical application for performing a double Fourier transform? …or an inverse Fourier transform on a time-domain input?

Fourier transform 4 times = original function (from Bracewell book)understanding complex fft resultsDiscrete Fourier transform of complex time seriesDouble size inverse fourier transformFourier transform of a Fourier transformHow do I interpret the result of a Fourier Transform?Analyzing a particular discrete-time LTI system for input signal $x[n]=(1/3)^n$ for *all* $n$Result of inverse FFT is sometimes shifted in real spaceInstantaneous velocity and displacement from acceleration signal using a proper filtering methodWhy is the size of results from FFT half the size of the input, while that is not the case in image processing?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

1

$begingroup$

In mathematics you can take the double derivative, or double integral of a function. There are many cases where performing a double derivative models a practical real-world situation, like finding the acceleration of an object.

Since the Fourier transform takes a real or complex signal as an input, and produces a complex signal as an output, there is nothing stopping you from taking that output and applying the Fourier transform a second time... Are there any practical uses for doing this? Does it help to model some complex real-world situations?

With the same logic, nothing would stop you from taking the inverse Fourier transform of your original time-domain input signal... would this ever be useful? Why or why not?

fft signal-analysis fourier-transform theory

share|improve this question

asked 8 hours ago

tjwrona1992tjwrona1992

12066 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

In mathematics you can take the double derivative, or double integral of a function. There are many cases where performing a double derivative models a practical real-world situation, like finding the acceleration of an object.

Since the Fourier transform takes a real or complex signal as an input, and produces a complex signal as an output, there is nothing stopping you from taking that output and applying the Fourier transform a second time... Are there any practical uses for doing this? Does it help to model some complex real-world situations?

With the same logic, nothing would stop you from taking the inverse Fourier transform of your original time-domain input signal... would this ever be useful? Why or why not?

fft signal-analysis fourier-transform theory

share|improve this question

asked 8 hours ago

tjwrona1992tjwrona1992

12066 bronze badges

$endgroup$

add a comment | 

$begingroup$

In mathematics you can take the double derivative, or double integral of a function. There are many cases where performing a double derivative models a practical real-world situation, like finding the acceleration of an object.

Since the Fourier transform takes a real or complex signal as an input, and produces a complex signal as an output, there is nothing stopping you from taking that output and applying the Fourier transform a second time... Are there any practical uses for doing this? Does it help to model some complex real-world situations?

With the same logic, nothing would stop you from taking the inverse Fourier transform of your original time-domain input signal... would this ever be useful? Why or why not?

fft signal-analysis fourier-transform theory

share|improve this question

asked 8 hours ago

tjwrona1992tjwrona1992

12066 bronze badges

$endgroup$

share|improve this question

asked 8 hours ago

tjwrona1992tjwrona1992

12066 bronze badges

share|improve this question

asked 8 hours ago

tjwrona1992tjwrona1992

12066 bronze badges

asked 8 hours ago

tjwrona1992tjwrona1992

12066 bronze badges

asked 8 hours ago

tjwrona1992tjwrona1992

12066 bronze badges

2 Answers
2

active

oldest

votes

3

$begingroup$

No, taking the Fourier transform twice is equivalent to time inversion (or inversion of whatever dimension you're in). You just get $x(-t)$ times a constant which depends on the type of scaling you use for the Fourier transform.

The inverse Fourier transform applied to a time domain signal just gives the spectrum with frequency inversion. Have a look at this answer for more details.

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Matt L.Matt L.

53.5k22 gold badges3939 silver badges9999 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You just recursively blew my mind.
  $endgroup$
  – tjwrona1992
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do I illustrate what Matt. L said but in 2D with my code? i.e. we get f(-x,-y).
  $endgroup$
  – Machupicchu
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Machupicchu, yes that looks right.
  $endgroup$
  – tjwrona1992
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  haha then you can select my answer ad the top one ^^ (he has 53K rep so it doe not mak any diff for him haha)
  $endgroup$
  – Machupicchu
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right after I said that I realized that there's probably easier ways than a double Fourier transform to time invert a signal haha
  $endgroup$
  – tjwrona1992
  7 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

2D Fourier transform (2D DFT) is used in image processing since an image can be seen as a 2D signal. E.g. for a grayscale image $I$, $I(x,y)=z$, that means that at the coordinates $x$ and $y$ the image has intensity value z. Look at this for example:

https://ch.mathworks.com/help/matlab/ref/fft2.html

Try this:

x=imread('cameraman.tif');
X=fft2(fft2(x));
imagesc(abs(X));

and compare to :

x=imread('cameraman.tif');
X= ifft2(fft2(x));
imagesc(abs(X));

rather like that. I applied fft2 to times, not ifft2 the second time. I think this illustrates what @Matt L. said:

"taking the Fourier transform twice is equivalent to time inversion",

you can see the image is inverted because of the of the -i imaginary negative instead of positive in ifft().

I also did it for a 1D signal (e.g. temporal):

share|improve this answer

edited 7 hours ago

answered 8 hours ago

MachupicchuMachupicchu

13111 silver badge1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm aware there is such a thing as a 2D Fourier transform, but that isn't the same as taking an input signal and running it through the algorithm then taking the output of that run and running it through again.
  $endgroup$
  – tjwrona1992
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Fourier transform is separable.
  $endgroup$
  – Machupicchu
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My question would also apply for a 2D Fourier transform. You could in theory take a 2D input signal, apply the 2D Fourier transform, then take the 2D output signal and use it as an input and apply the 2D Fourier transform again.
  $endgroup$
  – tjwrona1992
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  look in Matlab what happends if you do the following: cf. I updated my answer
  $endgroup$
  – Machupicchu
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the code I just uploaded illustrates what Mall L said, right?
  $endgroup$
  – Machupicchu
  7 hours ago
  

  
  
  
  

 | 
show 6 more comments

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3

$begingroup$

No, taking the Fourier transform twice is equivalent to time inversion (or inversion of whatever dimension you're in). You just get $x(-t)$ times a constant which depends on the type of scaling you use for the Fourier transform.

The inverse Fourier transform applied to a time domain signal just gives the spectrum with frequency inversion. Have a look at this answer for more details.

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Matt L.Matt L.

53.5k22 gold badges3939 silver badges9999 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You just recursively blew my mind.
  $endgroup$
  – tjwrona1992
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do I illustrate what Matt. L said but in 2D with my code? i.e. we get f(-x,-y).
  $endgroup$
  – Machupicchu
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Machupicchu, yes that looks right.
  $endgroup$
  – tjwrona1992
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  haha then you can select my answer ad the top one ^^ (he has 53K rep so it doe not mak any diff for him haha)
  $endgroup$
  – Machupicchu
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right after I said that I realized that there's probably easier ways than a double Fourier transform to time invert a signal haha
  $endgroup$
  – tjwrona1992
  7 hours ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

No, taking the Fourier transform twice is equivalent to time inversion (or inversion of whatever dimension you're in). You just get $x(-t)$ times a constant which depends on the type of scaling you use for the Fourier transform.

The inverse Fourier transform applied to a time domain signal just gives the spectrum with frequency inversion. Have a look at this answer for more details.

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Matt L.Matt L.

53.5k22 gold badges3939 silver badges9999 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You just recursively blew my mind.
  $endgroup$
  – tjwrona1992
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do I illustrate what Matt. L said but in 2D with my code? i.e. we get f(-x,-y).
  $endgroup$
  – Machupicchu
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Machupicchu, yes that looks right.
  $endgroup$
  – tjwrona1992
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  haha then you can select my answer ad the top one ^^ (he has 53K rep so it doe not mak any diff for him haha)
  $endgroup$
  – Machupicchu
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right after I said that I realized that there's probably easier ways than a double Fourier transform to time invert a signal haha
  $endgroup$
  – tjwrona1992
  7 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

No, taking the Fourier transform twice is equivalent to time inversion (or inversion of whatever dimension you're in). You just get $x(-t)$ times a constant which depends on the type of scaling you use for the Fourier transform.

The inverse Fourier transform applied to a time domain signal just gives the spectrum with frequency inversion. Have a look at this answer for more details.

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Matt L.Matt L.

53.5k22 gold badges3939 silver badges9999 bronze badges

$endgroup$

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Matt L.Matt L.

53.5k22 gold badges3939 silver badges9999 bronze badges

answered 8 hours ago

Matt L.Matt L.

53.5k22 gold badges3939 silver badges9999 bronze badges

answered 8 hours ago

Matt L.Matt L.

53.5k22 gold badges3939 silver badges9999 bronze badges

1

$begingroup$

2D Fourier transform (2D DFT) is used in image processing since an image can be seen as a 2D signal. E.g. for a grayscale image $I$, $I(x,y)=z$, that means that at the coordinates $x$ and $y$ the image has intensity value z. Look at this for example:

https://ch.mathworks.com/help/matlab/ref/fft2.html

Try this:

x=imread('cameraman.tif');
X=fft2(fft2(x));
imagesc(abs(X));

and compare to :

x=imread('cameraman.tif');
X= ifft2(fft2(x));
imagesc(abs(X));

rather like that. I applied fft2 to times, not ifft2 the second time. I think this illustrates what @Matt L. said:

"taking the Fourier transform twice is equivalent to time inversion",

you can see the image is inverted because of the of the -i imaginary negative instead of positive in ifft().

I also did it for a 1D signal (e.g. temporal):

share|improve this answer

edited 7 hours ago

answered 8 hours ago

MachupicchuMachupicchu

13111 silver badge1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm aware there is such a thing as a 2D Fourier transform, but that isn't the same as taking an input signal and running it through the algorithm then taking the output of that run and running it through again.
  $endgroup$
  – tjwrona1992
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Fourier transform is separable.
  $endgroup$
  – Machupicchu
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My question would also apply for a 2D Fourier transform. You could in theory take a 2D input signal, apply the 2D Fourier transform, then take the 2D output signal and use it as an input and apply the 2D Fourier transform again.
  $endgroup$
  – tjwrona1992
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  look in Matlab what happends if you do the following: cf. I updated my answer
  $endgroup$
  – Machupicchu
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the code I just uploaded illustrates what Mall L said, right?
  $endgroup$
  – Machupicchu
  7 hours ago
  

  
  
  
  

 | 
show 6 more comments

1

$begingroup$

2D Fourier transform (2D DFT) is used in image processing since an image can be seen as a 2D signal. E.g. for a grayscale image $I$, $I(x,y)=z$, that means that at the coordinates $x$ and $y$ the image has intensity value z. Look at this for example:

https://ch.mathworks.com/help/matlab/ref/fft2.html

Try this:

x=imread('cameraman.tif');
X=fft2(fft2(x));
imagesc(abs(X));

and compare to :

x=imread('cameraman.tif');
X= ifft2(fft2(x));
imagesc(abs(X));

rather like that. I applied fft2 to times, not ifft2 the second time. I think this illustrates what @Matt L. said:

"taking the Fourier transform twice is equivalent to time inversion",

you can see the image is inverted because of the of the -i imaginary negative instead of positive in ifft().

I also did it for a 1D signal (e.g. temporal):

share|improve this answer

edited 7 hours ago

answered 8 hours ago

MachupicchuMachupicchu

13111 silver badge1111 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm aware there is such a thing as a 2D Fourier transform, but that isn't the same as taking an input signal and running it through the algorithm then taking the output of that run and running it through again.
  $endgroup$
  – tjwrona1992
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The Fourier transform is separable.
  $endgroup$
  – Machupicchu
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My question would also apply for a 2D Fourier transform. You could in theory take a 2D input signal, apply the 2D Fourier transform, then take the 2D output signal and use it as an input and apply the 2D Fourier transform again.
  $endgroup$
  – tjwrona1992
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  look in Matlab what happends if you do the following: cf. I updated my answer
  $endgroup$
  – Machupicchu
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the code I just uploaded illustrates what Mall L said, right?
  $endgroup$
  – Machupicchu
  7 hours ago
  

  
  
  
  

 | 
show 6 more comments

$begingroup$

2D Fourier transform (2D DFT) is used in image processing since an image can be seen as a 2D signal. E.g. for a grayscale image $I$, $I(x,y)=z$, that means that at the coordinates $x$ and $y$ the image has intensity value z. Look at this for example:

https://ch.mathworks.com/help/matlab/ref/fft2.html

Try this:

x=imread('cameraman.tif');
X=fft2(fft2(x));
imagesc(abs(X));

and compare to :

x=imread('cameraman.tif');
X= ifft2(fft2(x));
imagesc(abs(X));

rather like that. I applied fft2 to times, not ifft2 the second time. I think this illustrates what @Matt L. said:

"taking the Fourier transform twice is equivalent to time inversion",

you can see the image is inverted because of the of the -i imaginary negative instead of positive in ifft().

I also did it for a 1D signal (e.g. temporal):

share|improve this answer

edited 7 hours ago

answered 8 hours ago

MachupicchuMachupicchu

13111 silver badge1111 bronze badges

$endgroup$

x=imread('cameraman.tif');
X=fft2(fft2(x));
imagesc(abs(X));

x=imread('cameraman.tif');
X= ifft2(fft2(x));
imagesc(abs(X));

share|improve this answer

edited 7 hours ago

answered 8 hours ago

MachupicchuMachupicchu

13111 silver badge1111 bronze badges

answered 8 hours ago

MachupicchuMachupicchu

13111 silver badge1111 bronze badges

answered 8 hours ago

MachupicchuMachupicchu

13111 silver badge1111 bronze badges