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What is the Fourier Transform of : $$\sum_{n=1}^N A_ne^{\large-a_nt} u(t)~?$$

This is a time domain function, how can I find its Fourier Transform (continuous not discrete) ?

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I tried to attach a picture but the blog didn't let me. –  marina Jun 28 '12 at 4:12
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Hello marina, welcome to math.SE. I've used LaTeX to typeset your equation; did I capture your intended meaning? –  Zev Chonoles Jun 28 '12 at 4:13
    
The n's suppose to be the subscripts of A and a. If you can fix them I will be thankful. –  marina Jun 28 '12 at 4:18
    
Perfect. That is the correct expression. Thanks –  marina Jun 28 '12 at 4:31

1 Answer 1

up vote 1 down vote accepted

Tips:

  1. The Fourier transform is linear; $$\mathcal{F}\left\{\sum_l a_lf_l(t)\right\}=\sum_l a_l\mathcal{F}\{f_l(t)\}.$$
  2. Plug $e^{-ct}u(t)$ into $\mathcal{F}$ and then discard part of the region of integration ($u(t)=0$ when $t<0$):

$$\int_{-\infty}^\infty e^{-ct}u(t)e^{-2\pi i st}dt=\int_0^\infty e^{(c-2\pi is)t}dt=? $$

Now put these two together..

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Two questions:1)What happen with the exponential that its evaluated in infinite? I mean the result is infinite - 1, over the denominator. 2) It doesn't matter that my summation start in n=1 and I am integrating from zero because of the limits of u(t)? –  marina Jun 28 '12 at 4:54
    
@marina (1) Remember, $\int_0^\infty:=\lim_{N\to\infty}\int_0^N$. Can you see that the limit of the antiderivative tends to $0$ as $N\to\infty$? (2) I'm not sure what your question here is, but remember that $\int_a^b=\int_a^c+\int_c^b$; do this with $a=-\infty,c=0,b=+\infty$; then note that $\int_{-\infty}^0 0dt =0$. –  anon Jun 28 '12 at 4:59
    
Sorry but its been a while since I took my calculus class. About the 2nd question. Here you put the summation with the A outside of the integral (because of the linearity). That summation go from n=1 to N. The integral go from t=0 to t=inf. First I evaluate the integral with the limits of t? and then I evaluate the result in the summation from n=1 to N? –  marina Jun 28 '12 at 5:10
    
Yes, but I don't think one will be able to "evaluate" the resulting summation; you will need to be satisfied with the summation as it is. –  anon Jun 28 '12 at 5:11
    
Its true, I can not "evaluate" the summation because is until "N" so I have to leave the expression with the integral result. This is just part of one exercise. The whole problem consist in multiply this Fourier Transform with transfer function (in Fourier Transform) that I found. Thanks anon –  marina Jun 28 '12 at 5:20

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