# Fourier Transforms - The Convolution Theorem.

• binbagsss
In summary, the convolution theorem can be written as F(h(x) * g(x)) = H(k)G(k), where F represents a Fourier transform. However, in a problem from a book, the formula is written as F(h(x)g(x)) = H(k) * G(k), where H(k) = F(h(x)) and G(k) = F(g(x)). It may seem like these formulas are not equivalent, but the last page of the provided source gives an explanation for why they are in fact equivalent.
binbagsss
Ok so I've seen the convolution theorem written as:

F(h(x)$\otimes$g(x))=H(k)G(k)

(And this is how it appears when I have a quick google).

My book then does a problem in which is uses:

F(h(x)g(x))=H(k)$\otimes$G(k)

Where H(k)=F(h(x)) and similarly G(k)=F(g(x)),
and F represents a Fourier transform

My question
- I can't see how these are equivalent at all?

Many Thanks to anyone who can help shed some light !

binbagsss:
Please post your math questions in one of the Math HW forums. This helps to organize the questions so that posters can receive appropriate responses to questions.

binbagsss said:
Ok so I've seen the convolution theorem written as:

F(h(x)$\otimes$g(x))=H(k)G(k)

(And this is how it appears when I have a quick google).

My book then does a problem in which is uses:

F(h(x)g(x))=H(k)$\otimes$G(k)

Where H(k)=F(h(x)) and similarly G(k)=F(g(x)),
and F represents a Fourier transform

My question
- I can't see how these are equivalent at all?

Many Thanks to anyone who can help shed some light !

Take a look at:
http://ugastro.berkeley.edu/infrared09/PDF-2009/convolution2.pdf

It gives an argument for the formula on the last page.

## 1. What is a Fourier Transform?

A Fourier Transform is a mathematical tool used to break down a complex signal into its individual frequency components. It converts a function of time or space into a function of frequency, which can then be analyzed and manipulated in the frequency domain.

## 2. How does the Convolution Theorem relate to Fourier Transforms?

The Convolution Theorem states that the Fourier Transform of a convolution of two functions is equal to the product of their individual Fourier Transforms. This means that instead of performing a time-domain convolution, which can be computationally expensive, we can simply multiply the Fourier Transforms of the two functions in the frequency domain to achieve the same result.

## 3. What are the advantages of using the Convolution Theorem for Fourier Transforms?

Using the Convolution Theorem can greatly reduce the computational complexity of certain operations, making them more efficient and faster to perform. It also allows for easier manipulation of signals in the frequency domain, as multiplication is a simpler operation compared to convolution.

## 4. Are there any limitations to using the Convolution Theorem for Fourier Transforms?

The Convolution Theorem is only applicable to linear, time-invariant systems. If the system is nonlinear or time-varying, the theorem cannot be used. Additionally, the theorem may not apply to some special cases, such as when the functions being convolved have infinite support.

## 5. How is the Convolution Theorem used in practical applications?

The Convolution Theorem is widely used in a variety of fields, including signal processing, image processing, and communication systems. It allows for efficient and accurate analysis of signals and data, making it a valuable tool in many scientific and engineering applications.

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