Calculate the Fourier transform of a product of three functions

In summary, the conversation discusses the calculation of the Fourier transform of a product of three functions and how it relates to convolution. The process is similar to extending the product or sum of two numbers to three, where the convolution of three functions can be written as the convolution of two functions. The confusion arises when calculating the second convolution, where one expects to see two integrals instead of just one. More clarification is needed to understand the process.
  • #1
parton
83
1
I have a problem understanding the following:

I should calculate the Fourier transform of a product of three functions:

[tex] \mathcal{F} \left[ f(x_{1}) g(x_{2}) h(x_{1} + x_{2}) \right] = \int dx_{1} dx_{2} f(x_{1}) g(x_{2}) h(x_{1} + x_{2}) e^{i p x_{1} + i q x_{2}} [/tex]

okay, and this goes over into a convolution:

[tex] = \dfrac{1}{2 \pi} \hat{f}(p) \ast \hat{g}(q) \ast \hat{h}(p+q) = \dfrac{1}{2 \pi} \int dk \hat{f}(p-k) \hat{g}(q-k) \hat{h}(k) [/tex].

I know how to calculate the convolution between two functions, but here we have three and I don't undestand how to get to the last line here. Could somebody explain that to me, please?
 
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  • #2


Pretty much the same way you extend the product of two numbers to three or the sum of two numbers to three. Writing f #g for the convolution of two functions, the convolution of three can be written (f#g)#h= f#(g#h).
 
  • #3


ok, I start with:

[tex] \hat{g}(q) \ast \hat{h}(p+q) = \int dk ~ \hat{g}(p+q-k) \hat{h}(k) [/tex]

But if I go further and calculate the second convolution, i.e. [tex] \hat{f}(p) \ast \hat{g}(q) \ast \hat{h}(p+q) [/tex], I would get a second integral. But in the equation above there is just one. I don't understand that. Maybe it is trivial, but I just don't see what I have to do?
 

1. What is the Fourier transform of a product of three functions?

The Fourier transform of a product of three functions is the mathematical operation that decomposes the product into its constituent frequencies and their corresponding amplitudes. It is a powerful tool used in signal processing and image analysis.

2. Why is the Fourier transform of a product of three functions important?

The Fourier transform of a product of three functions allows us to analyze and understand the frequency content of a signal or image. This can help in identifying patterns and features that may not be visible in the time or spatial domain.

3. How do you calculate the Fourier transform of a product of three functions?

The Fourier transform of a product of three functions can be calculated by taking the individual Fourier transforms of each function, multiplying them together, and then taking the inverse Fourier transform of the product. This process can be simplified using mathematical formulas and properties of the Fourier transform.

4. Can the Fourier transform of a product of three functions be applied to any type of signal or image?

Yes, the Fourier transform of a product of three functions can be applied to any continuous and finite signal or image. It is a versatile tool that is widely used in various fields such as engineering, physics, and mathematics.

5. Are there any limitations to using the Fourier transform of a product of three functions?

One limitation of the Fourier transform of a product of three functions is that it assumes the signal or image is continuous and finite. This means that it may not be as effective when applied to signals or images with discontinuities or infinite boundaries. In addition, the Fourier transform may not accurately capture high-frequency components in signals with sharp spikes or edges.

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