rect([itex]\frac{x}{5})[/itex][itex]\otimes[/itex]([itex]\delta[/itex](x+3)-[itex]\delta[/itex](x-3))

=F[itex]^{\infty}_{\infty}[/itex] [itex]\int[/itex]rect([itex]\frac{x’}{5}[/itex])([itex]\delta[/itex](x+3-x’)-[itex]\delta[/itex](x-3-x’)) dx’ [1]

**– where ** F **represents a fourier transform.**

**My Issue**

*Okay I am fine doing this using the convolution theorem, that the fourier transform of a convultion is given by the product of the two individual fourier transforms, but I am having trouble doing it explicitly*

So from [1] integrating over each delta function, I deduce that the first term collapses everywhere except x’=x+3, and the second everywhere except x’=x-3, . So I get:

F(rect[itex]\frac{x+3}{5}[/itex]-rect[itex]\frac{x-3}{5})[/itex]

= (5sinc[itex]\frac{5k}{2}[/itex]exp[itex]^{\frac{3ik}{5}}[/itex]exp[itex]^{\frac{-3ik}{5}}[/itex])

*using the properties that F(rect([itex]\frac{x}{1}[/itex]))=asinc([itex]\frac{ka}{2}[/itex]) and that F(f(x+a))=F(f(x))exp[itex]^{ika}[/itex]*

**Which does not agree with the convultion theorem were I get : **

5sinc[itex]\frac{5k}{2}[/itex]exp[itex]^{3ik}[/itex]exp[itex]^{-3ik}[/itex]

**Thanks alot in advance for any assistance !**

http://ift.tt/1f8Mgqq