Fourier Transform of v(t)=A*e(-t) for t≥0

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Homework Statement


I need to find the Fourier transform of v(t)=A*e(-t) such that t≥0.


Homework Equations


∫v(t)*e(-j*2*∏*f*t)dt,t,0,∞)


The Attempt at a Solution



∫v(t)*e(-j*2*∏*f*t)dt,t,0,∞)=A/(4*f2*∏2+1)-i*(2*A*f*∏)/(4*f2*∏2+1)

the answer should be A/(1+i*2*∏*f). It seems like this problem should be straight forward so I'm wondering if I'm just missing an identity of some sort or something? This is actually only one part of the problem but I know how to get the rest of the answer once I get this part done.
 
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Ok so I realize these are the same answer, but how do I simplify complex numbers like the one I got in my answer? I've been trying to multiply it by the complex conjugate but haven't been getting much luck :/
 
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Combine the two terms and then factor the denominator.
 
I've combined the two equations to get: (A-j*2*A*f*∏)/(1+4*f2*∏2) but it doesn't seem that the denominator can be factored at all. I can factor the numerator to get A*(1-j*2*f*∏)/(1+4*f2*∏2) but that doesn't seem to help me much yet.
 
Last edited:
Oh I got it now
 
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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