Can the convolution operator be diagonalized using the Fourier transform?

In summary: Thanks for asking!In summary, the Fourier transform is a useful tool for analyzing LTI systems and diagonalizing the convolution operator. It can be shown that the operator convolution also has an eigenfunction basis, similar to the operator \frac{d^2}{dt^2}. A paper by A. B. Aleksandrov titled "Eigenfunctions of Convolution Operators" may provide further insight on this topic.
  • #1
math771
204
0


Hi there,

I am also familiar with Hilbert spaces and Functional Analysis and I find your question very interesting. I agree that the Fourier transform is a powerful tool for analyzing LTI systems and diagonalizing the convolution operator. As for your question about whether the same can be shown for the operator convolution, I believe it is possible.

I did a quick search and found a reference that might be helpful for you. It's a paper titled "Eigenfunctions of Convolution Operators" by A. B. Aleksandrov. I haven't read it in depth, but from the abstract it seems to address your question.

I hope this helps! Let me know if you have any other questions or if you find any other useful references.
 
Physics news on Phys.org
  • #2
Hi everyone,

I have some knowledge of Hilbert spaces and Functional Analysis and I have the following question.

I ofter have read that "Fourier transform diagonalize the convolution operator". So, we can say that for LTI systems (that can always be described with a convolution and "live" in [itex]L^{2}(-\infty +\infty)[/itex]) the Fourier transform is a formidable tool because it permits to write a problem (that involves the convolution operator and the signals) in the eigenfunction basis (that are [itex]e^{i \omega t}[/itex] ), expanding both the convolution operator and the signals of [itex]L^{2}(-\infty +\infty)[/itex].

Now, I know that, for example, the operator [itex]\frac{d^2}{dt^2}[/itex] on [itex]L^{2}(-\infty +\infty)[/itex] defines an eigenvalue problem and from this we can get an eigenfunction basis [itex]e^{i \omega t}[/itex] that describes the whole space.

It is possible to show the same for the operator convolution?
Do you know where I can find any reference?
Thank you in advance!
 
  • #3
This simply means that if ##\mathcal F## is the Fourier transform, and ##T## is a convolution operator, then ##\mathcal F T \mathcal F^{-1} ## is a multiplication operator. It follows from the convolution theorem, ##\mathcal F(f*g) = \mathcal F g \mathcal F f ##, see for example Wikipedia
 

1. What is the Convolution Operator Spectrum?

The Convolution Operator Spectrum is a mathematical concept used in signal processing to analyze the frequency components of a signal before and after applying a convolution operation. It is a representation of the frequency response of the convolutional operation and can be used to understand the effects of the operation on the original signal.

2. How is the Convolution Operator Spectrum calculated?

The Convolution Operator Spectrum is calculated by taking the Fourier Transform of both the input signal and the convolution kernel, and then multiplying the two transforms together. The resulting spectrum represents the frequency response of the convolution operation.

3. What is the significance of the Convolution Operator Spectrum?

The Convolution Operator Spectrum is important because it allows us to analyze the frequency components of a signal before and after applying a convolution operation. This can help us understand the effects of the operation on the original signal and can be used to optimize the operation for specific applications.

4. Can the Convolution Operator Spectrum be used to design filters?

Yes, the Convolution Operator Spectrum can be used to design filters by analyzing the frequency response of the convolution operation and determining the desired frequency components to be filtered or enhanced. This information can then be used to design a filter kernel that will achieve the desired results.

5. Are there any limitations to using the Convolution Operator Spectrum?

While the Convolution Operator Spectrum is a powerful tool for analyzing frequency components and designing filters, it does have some limitations. It assumes linearity and time-invariance in the signal, which may not always hold true in real-world scenarios. Additionally, it may not accurately represent the effects of non-linear or non-stationary signals.

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
9
Views
3K
  • Linear and Abstract Algebra
2
Replies
43
Views
5K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Replies
9
Views
2K
Replies
4
Views
2K
  • Linear and Abstract Algebra
Replies
16
Views
8K
  • Other Physics Topics
Replies
6
Views
3K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
4
Views
2K
  • Calculus
Replies
8
Views
4K
  • Other Physics Topics
Replies
1
Views
1K
Back
Top