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- Thread starter dwsmith
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- Jan 26, 2012

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With which definition of the Fourier Transform are you working?

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- Jan 26, 2012

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So, what have you got so far?

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$$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

$$

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- Jan 26, 2012

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And what do you have for the LHS?

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Not really sure on how to handle the LHSAnd what do you have for the LHS?

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- #8

- Jan 26, 2012

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Well, what is the definition of convolution?

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It isn't the definition of convolution that is throwing me for a loop but the transform of it.Well, what is the definition of convolution?

$$

(f\star g)(x) = \int_{-\infty}^{\infty}f(s)g(x-s)ds

$$

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- #10

- Jan 26, 2012

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Great. So the result of this convolution is a function of $x$, correct? Let's say thatIt 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

$$

$$h(x):=(f\star g)(x) = \int_{-\infty}^{\infty}f(s)g(x-s)\,ds.$$

Now, can you write down the Fourier Transform of $h$?

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- #11

$$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?

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- #13

$$

\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.

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- #14

- Jan 26, 2012

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If $z=x-s$, then $x=z+s$. Don't forget to change the differential as well.

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\begin{alignat}{3}$$

\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.

\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}

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- #16

- Jan 26, 2012

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\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?

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- #17

But now we have

\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?

$$

\frac{1}{2\pi}\hat{f}(\xi)\hat{g}(\xi) = (\widehat{f\star g})(\xi)

$$

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- #18

- Jan 26, 2012

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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?But now we have

$$

\frac{1}{2\pi}\hat{f}(\xi)\hat{g}(\xi) = (\widehat{f\star g})(\xi)

$$