Finding the Inverse Fourier Transform of e^-5w*sinc(2w)?

Luongo
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1. find the inverse Fourier transform of f(w)=e-i5wsinc(2w)



2. I set up the integral to be from defn of sinc: 1/2pi*integral from -infinity to infinity (sin(2w)/2w)*e^-5w



3. i have no idea how to solve this integral, is there a better way to do this?
i know that rect(t) has a F.T. of sinc(w/2) but how do i go the other way if it's 2w, not w/2?
 
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The way that one normally does this sort of thing it to use countour integration.
 
Luongo said:
1. find the inverse Fourier transform of f(w)=e-i5wsinc(2w)



2. I set up the integral to be from defn of sinc: 1/2pi*integral from -infinity to infinity (sin(2w)/2w)*e^-5w



3. i have no idea how to solve this integral, is there a better way to do this?
i know that rect(t) has a F.T. of sinc(w/2) but how do i go the other way if it's 2w, not w/2?

Yes, there is a better way to do it. Use the shifting property of FT. If we denote the transform of f(t) by F(ω), one of the shifting properties gives:

f(t-t_0) \leftrightarrow e^{-i\omega t_0}F(\omega)
 
There's also property that relates the Fourier transforms of f(t) and f(at). (I'll leave it to you to look it up in your textbook.) You should be able to solve the problem using that property along with the shifting property LCKurtz mentioned.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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