Fourier transform of a fourier transform

In summary: Frequency method:-Peak detection is observing the square wave pulse as one signal overlaps another. -Peak detection is relying on the fact that the peak is a result of additions of a pulses, or, in other words, they are points of change with infinite bandwidth, where a pulse in the frequency domain is a sinc function, and since a pulse is finite in time, it has infinite bandwidth. -Using the method I described above with extracting the comb filter, you are finding the frequency component of the comb filter in the 2nd Fourier transform, and the 2nd
  • #1
DragonPetter
830
1
Does this ever have meaningful data to it? What are its applications?

I am measuring a signal that will have phase-shifted echoes, which means it will have a comb filter waveform multiplied by the original signals Fourier transform because of the phase-shifting.
explanation here: http://www.webcitation.org/5pBNPaijA

The comb's lobes repeat periodically proportional to the time delay. If I can divide out the original signal's Fourier transform, I should be left with the comb filter applied by the phase-shifts. If I take a Fourier transform of this frequency spectrum, I think that the comb filter's periodicity will look like a spike in the second Fourier transform, and will make it very easy to identify phase shifts. This will be even more relevant when there are multiple time delays and so multiple combs that need to be identified (from echoes).

Does anyone know if this is a good method for finding time delays in signals where images of itself are added with time-delays?
 
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  • #2
What do you mean by "images of its self" ?
 
  • #3
MisterX said:
What do you mean by "images of its self" ?

Say you have a unit square pulse, and it is high from 0 to 1 seconds. The image with a time delay of say .5 seconds will be a second unit square pulse, and it is high from 0.5 seconds to 1.5 seconds.

The signal will consist of the original pulse and its delayed image added together. so it will be 1 unit high from 0 to .5 seconds, 2 units high from .5 second to 1 second, and 1 unit high from 1 second to 1.5 seconds.

I guess image is bad terminology, but I wanted to make sure its understood that the original signal and the time delayed signal are both present in the signal-its not just a signal with a time delay. It really is just a reflection added onto the original signal.
 
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  • #6
Okay supposing the signal is the sum of 1 second square pulses and you want to find the time delay between them?

Why not something like the following?

1. Apply a matched filter for a 1s pulse
2. Detect peaks in the output
3. Get the time difference between consecutive peaks
 
  • #7
MisterX said:
Okay supposing the signal is the sum of 1 second square pulses and you want to find the time delay between them?

Why not something like the following?

1. Apply a matched filter for a 1s pulse
2. Detect peaks in the output
3. Get the time difference between consecutive peaks

I think this is the way it is done most often actually, but I'm not so sure, when you have many sound echoes that also are undergoing path attenuation, how much peak difference you will actually see and how many echoes you can attribute to 1 peak difference. This application will be underwater with more than one reflection.

This is digitally so I will need very high sampling frequency to see the very small time differences. I guess frequency or time domain, it doesn't make a difference since the information is in both, but I have a hard time trusting a easy way to analyze lots of peak time differences easily with time domain.
 
  • #8
DragonPetter said:
I think this is the way it is done most often actually, but I'm not so sure, when you have many sound echoes that also are undergoing path attenuation, how much peak difference you will actually see and how many echoes you can attribute to 1 peak difference. This application will be underwater with more than one reflection.

This is digitally so I will need very high sampling frequency to see the very small time differences. I guess frequency or time domain, it doesn't make a difference since the information is in both, but I have a hard time trusting a easy way to analyze lots of peak time differences easily with time domain.

I've been playing around in matlab, and I actually think the frequency method has a distinct advantage over the peak detection method:

Peak detection is observing the square wave pulse as one signal overlaps another. Peak detection is relying on the fact that the peak is a result of additions of a pulses, or, in other words, they are points of change with infinite bandwidth, where a pulse in the frequency domain is a sinc function, and since a pulse is finite in time, it has infinite bandwidth. This means that to get perfect peak detection, I'd want infinite bandwidth.

Using the method I described above with extracting the comb filter, you are finding the frequency component of the comb filter in the 2nd Fourier transform, and the 2nd transform of the comb filter only has a bandwidth proportional to the time delay, rather than the infinite bandwidth of a sinc function. Even using a bandwidth of only with a few periods of the comb filter's frequency response, at the location of highest frequency (center of the sinc), the spikes should show up in the 2nd Fourier transform.

So I think I don't need as much bandwidth to look at the frequency information as I would need to look at the time domain peaks.
 

What is a Fourier transform of a Fourier transform?

A Fourier transform of a Fourier transform is a mathematical operation that involves taking the Fourier transform of another Fourier transform. This process is also known as a double Fourier transform.

What is the purpose of taking a Fourier transform of a Fourier transform?

The purpose of taking a Fourier transform of a Fourier transform is to analyze the frequency components of a signal in both the time and frequency domains. This can provide a more detailed understanding of the signal and help with various applications such as signal processing and image reconstruction.

How is a Fourier transform of a Fourier transform calculated?

A Fourier transform of a Fourier transform is calculated by taking the Fourier transform of the initial signal, and then taking the Fourier transform of the resulting Fourier transform. This can be represented mathematically as F[F[f(x)]] or F²[f(x)], where f(x) is the initial signal and F represents the Fourier transform operation.

What is the relationship between a Fourier transform and a Fourier transform of a Fourier transform?

A Fourier transform of a Fourier transform is essentially the same as taking two Fourier transforms in succession. In other words, it is the result of applying the Fourier transform operation twice. This relationship can also be seen in the mathematical representation F²[f(x)].

What are some common applications of a Fourier transform of a Fourier transform?

A Fourier transform of a Fourier transform has various applications in fields such as signal processing, image reconstruction, and data compression. It can also be used in the analysis of complex systems and in solving differential equations. Additionally, it plays a crucial role in the development of technologies such as MRI and radar.

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