fourier transform convolutio theorem

**silversand** · April 24th 2009, 05:35 AM

show that $\int_{-\infty}^{\infty}{\frac{\sin(bk)e^{ikx}}{k(a^{2}+k^ {2})}dk}=\frac{\pi \sinh(ab)}{a^{2}}e^{-ax}$
by considering fourier transforms of
$f(x)=e^{-a|x|} \ \ \ and \ \ \ g(x)=1$ when x>b

**InvisibleMan** · April 24th 2009, 06:35 AM

My advice, is obviously to calculate the fourier transform of these functions, and then use the convolution formula, but in this case use the fact that exp (i b k)=cos(bk)+isin(bk) to replace this with sin(bk) in the integrand.
if f(k),g(k) are the fourier transforms of f(x),g(x), and h(z) is the convolution of f and g then, then h(x)=F^-1(f(k)g(k)) where F^-1 is the inverse fourier transform.

**silversand** · April 24th 2009, 07:55 AM

i know we should use convolution. and i found the transform of f and g.

but i cant see they can make that expression in the integrand

anyone can make it more clear?

**Opalg** · April 24th 2009, 12:41 PM

Originally Posted by silversand

show that $\int_{-\infty}^{\infty}{\frac{\sin(bk)e^{ikx}}{k(a^{2}+k^ {2})}dk}=\frac{\pi \sinh(ab)}{a^{2}}e^{-ax}$
by considering fourier transforms of
$f(x)=e^{-a|x|} \ \ \ and \ \ \ g(x)=1$

I think that the function g(x) ought to be a rectangular function. In other words, it is 1 on some interval ([–1,1], [–1/2,1/2] or [-π,π], depending on which definition of the Fourier transform you are using), and 0 elsewhere. (g obviously can't be 1 on the whole real line because then it would not have a Fourier transform.)

The integral $\int_{-\infty}^{\infty}{\frac{\sin(bk)e^{ikx}}{k(a^{2}+k^ {2})}dk}$ represents the value at x of (a multiple of) the inverse Fourier transform of the product of two functions, namely $\hat{g}(k) = \text{sinc}(bk) := \frac{\sin(bk)}{bk}$ and $\hat{f}(k) = \frac{2a}{a^2+k^2}$ . But the sinc function is the Fourier transform of a rectangular function, and the function $\hat{f}$ is the Fourier transform of $f(x) = e^{-a|x|}$ . So the inverse Fourier transform of their product will be the convolution of the rectangular function and that exponential function.

**CaptainBlack** · April 24th 2009, 01:29 PM

Originally Posted by Opalg

(g obviously can't be 1 on the whole real line because then it would not have a Fourier transform.)

Not in the conventional sense, but in another existence its FT is $\delta(\omega)$ (give or take a constant multiplier).

CB

**silversand** · April 25th 2009, 12:32 AM

So how can you get that $sinh(ab)$ on the right?

**Opalg** · April 25th 2009, 08:36 AM

If $g(x) = \begin{cases}1&\text{if }|x|\leqslant b\\ 0&\text{otherwise}\end{cases}$ and $f(x) = e^{-ax}/{2a}$ then $(g*f)(x) = \int_{-b}^b\frac1{2a}e^{-a(x-t)}dt = \Bigl[\tfrac1{2a^2}e^{-ax}e^{at}\Bigr]_{-b}^b = \frac{e^{-ax}(e^{ab}-e^{-ab})}{2a^2} = \frac{e^{-ax}\sinh(ab)}{a^2}$ .

That looks promising, because it gets the right answer apart from a constant multiple. However, I have cheated by using the function $e^{-ax}$ instead of $e^{-a|x|}$ . That doesn't matter provided that x is positive. This means that, in the integral, we want x–t to be positive whenever $|t|\leqslant b$ ; and that in turn means that I am assuming that $x\geqslant|b|$ . I claim that this is a reasonable assumption. Indeed, if it does not hold then the function $\frac{e^{-ax}\sinh(ab)}{a^2}$ will not have a Fourier transform (in the conventional sense: as the Captain points out, it will have a distributional Fourier transform of some sort), and so this whole method of evaluating the integral will become suspect.

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