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[SOLVED] Convolution of a transform

dwsmith

Well-known member
Feb 1, 2012
1,673
I am trying to prove the convolution of the Fourier Transform
$$
(\widehat{f\star g})(\xi) = 2\pi\hat{f}(\xi)\hat{g}(\xi)
$$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
With which definition of the Fourier Transform are you working?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\hat{f}(\xi) = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-i\xi x}dx
$$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
So, what have you got so far?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
So, what have you got so far?
$$
2\pi\hat{f}(\xi)\hat{g}(\xi) = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-i\xi x}g(x)e^{-i\xi x}dx
$$
 
Last edited by a moderator:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
And what do you have for the LHS?
 

dwsmith

Well-known member
Feb 1, 2012
1,673

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Well, what is the definition of convolution?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Well, what is the definition of convolution?
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

Indicium Physicus
Staff member
Jan 26, 2012
4,193
It isn't the definition of convolution that is throwing me for a loop but the transform of it.
$$
(f\star g)(x) = \int_{-\infty}^{\infty}f(s)g(x-s)ds
$$
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 $h$?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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 $h$?
$$
\hat{h}(\xi)=\frac{1}{2\pi}\int_{-\infty}^{\infty}h(x)e^{-i\xi x}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)ds\right)e^{-i\xi x}dx
$$
but how do I get to the RHS now?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
$$
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)ds\right)e^{-i\xi x}dx=
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(z)ds\right)e^{-i\xi (z+s)}dx
$$
Even with a change of integration order, I don't see how it will get to the right result.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
If $z=x-s$, then $x=z+s$. Don't forget to change the differential as well.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(x-s)ds\right)e^{-i\xi x}dx=
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(z)ds\right)e^{-i\xi (z+s)}dx
$$
Even with a change of integration order, I don't see how it will get to the right result.
\begin{alignat}{3}
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(s)g(z)ds\right)e^{-i\xi (z+s)} d(z + s) & = &
\left(\int_{-\infty}^{\infty}f(s)e^{-i\xi s} ds\right)\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}g(z)e^{-i\xi z}d(z + s)\right)\\
& = & \hat{f}(\xi)\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}g(z)e^{-i\xi z}d(z + s)\right)
\end{alignat}
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
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

Well-known member
Feb 1, 2012
1,673
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?
But now we have
$$
\frac{1}{2\pi}\hat{f}(\xi)\hat{g}(\xi) = (\widehat{f\star g})(\xi)
$$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
But now we have
$$
\frac{1}{2\pi}\hat{f}(\xi)\hat{g}(\xi) = (\widehat{f\star g})(\xi)
$$
Check your constants again. To get two Fourier Transforms on the RHS, you need two factors of $2\pi$ in the denominator. How many do you actually have?