Fourier Transformation - Convolution quick question

[binbagsss]

Okay the question is to find the Fourier transform of:

rect(\frac{x}{5})\otimes(\delta(x+3)-\delta(x-3))

=F^{\infty}_{\infty} \intrect(\frac{x'}{5})(\delta(x+3-x')-\delta(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\frac{x+3}{5}-rect\frac{x-3}{5})
= (5sinc\frac{5k}{2}exp^{\frac{3ik}{5}}exp^{\frac{-3ik}{5}})
using the properties that F(rect(\frac{x}{1}))=asinc(\frac{ka}{2}) and that F(f(x+a))=F(f(x))exp^{ika}

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

5sinc\frac{5k}{2}exp^{3ik}exp^{-3ik}

Thanks a lot in advance for any assistance !

 

[vela]

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:
$$F\left[\mathrm{rect }\left(\frac{x+3}{5}\right) -\mathrm{rect }\left(\frac{x-3}{5}\right)\right] =
5\mathrm{sinc }\left(\frac{5k}{2}\right)e^{3ik/5}e^{-3ik/5}$$ using the properties that $F\left[\mathrm{rect }\left(\frac{x}{a}\right)\right]=a\mathrm{sinc }\left(\frac{ka}{2}\right)$ and that $F[f(x+a)] = F[f(x)]e^{ika}$. This does not agree with the convolution theorem where I get:
$$5\mathrm{sinc }\left(\frac{5k}{2}\right)e^{3ik}e^{-3ik}$$

Why are the exponentials being multiplied in both cases? In fact, if they were supposed to be multiplied, they'd cancel in both cases, leaving you with identical but incorrect answers.

Your mistake is in determining the shift. You're shifting rect(x/5) by 3, not 3/5, because to go from rect(x/5) to rect((x±3)/5), you replace x by x±3, not x±3/5.