- #1
Dustinsfl
- 2,281
- 5
I am trying to prove the convolution of the Fourier Transform
$$
(\widehat{f\star g})(\xi) = 2\pi\hat{f}(\xi)\hat{g}(\xi)
$$
$$
(\widehat{f\star g})(\xi) = 2\pi\hat{f}(\xi)\hat{g}(\xi)
$$
$$Ackbach said:So, what have you got so far?
Ackbach said:And what do you have for the LHS?
It isn't the definition of convolution that is throwing me for a loop but the transform of it.Ackbach said:Well, what is the definition of convolution?
dwsmith said:It isn't the definition of convolution that is throwing me for a loop but the transform of it.
$$
(f\star g)(x) = \int_{-\infty}^{\infty}f(s)g(x-s)ds
$$
Ackbach said:Great. So the result of this convolution is a function of $x$, correct? Let's say that
$$h(x):=(f\star g)(x) = \int_{-\infty}^{\infty}f(s)g(x-s)\,ds.$$
Now, can you write down the Fourier Transform of $h$?
$$Ackbach said:I would try putzing around with a change of variable, say, $z=x-s$, and then maybe changing the order of integration. See what that gets you.
\begin{alignat}{3}dwsmith said:$$
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)ds\right)e^{-i\xi x}dx=
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(z)ds\right)e^{-i\xi (z+s)}dx
$$
Even with a change of integration order, I don't see how it will get to the right result.
But now we haveAckbach said:Try this instead:
\begin{align*}\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)\,ds\right)e^{-i\xi x}\,dx&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s)g(x-s)e^{-i\xi x}\,ds\,dx\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s)g(x-s)e^{-i\xi x}\,dx\,ds\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(s)g(z)e^{-i\xi (z+s)}\,dz\,ds.
\end{align*}
Where can you go from here?
dwsmith said:But now we have
$$
\frac{1}{2\pi}\hat{f}(\xi)\hat{g}(\xi) = (\widehat{f\star g})(\xi)
$$
The purpose of convolving a transform is to combine two functions and produce a new function that represents the overlap between the two original functions. This allows for the analysis and manipulation of signals or data in the frequency domain.
The convolution of a transform is calculated by multiplying the two functions in the frequency domain and then taking the inverse Fourier transform of the product. This results in a new function that represents the convolution of the two original functions.
The convolution theorem states that convolution in the time domain is equivalent to multiplication in the frequency domain. This means that the convolution of two functions can be simplified by performing a multiplication of their transforms and then taking the inverse transform.
Convolution of a transform has various applications in signal processing, image processing, and data analysis. It is commonly used in filtering, deconvolution, and correlation of signals or images. It is also used in operations such as edge detection and feature extraction.
Yes, the convolution of a transform can be performed on both linear and non-linear functions. However, the convolution theorem only applies to linear systems. Non-linear systems require a more complex approach for convolution, such as using numerical methods.