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Hi: I'm reading a text "Fourier Transforms for Pedestrians" and it's a nice text but it skips steps that I sometimes don't understand.

The current example that I don't follow is one where the lorentzian is convoluted with the cosine. The convolution is denoted as

$$H(\omega) = \int_{-\infty}^{\infty} \frac{\sigma}{\pi} \frac{1}{\omega^2 + \sigma^2} \cos(\omega - \omega_{0})t \,d\omega .$$

It is assumed that $h(t) = f(t) g(t)$. So, therefore, using the fact that the convolution $ H(\omega) = \frac{1}{2 \pi}F(\omega) \ast G(\omega)$, the terms can be picked out of the integral so that

$F(\omega) = \frac{\sigma}{\pi} \frac{1}{\omega^2 + \sigma^2} $

and

$$G(\omega_{0}) = 2 \pi \cos(\omega_{0}).$$

The question is to find $h(t)$.

I can show that

$$g(t) = 2 \pi \left(\frac{(\delta(t_{0} - t)}{2} + \frac{\delta(t_{0} + t)}{2}\right).$$

but I can't get $f(t)$ which means that I can't get $h(t)$. The answer is that

$f(t) = \exp(-\sigma t_{0})$. Thanks for your help in advance.

Cameron Williams
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mark leeds
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1 Answers1

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I think this one is easier evaluated directly. Write

$$h(t) = \frac{\sigma}{\pi} \operatorname{Re}{\left [e^{-i \omega_0 t} \int_{-\infty}^{\infty} d\omega \frac{e^{i \omega t}}{\omega^2+\sigma^2} \right ]} $$

Note that we are now just dealing with the direct, FT of the Lorentzian, which is simply $(\pi/\sigma) e^{-\sigma |t|}$. Thus,

$$h(t) = e^{-\sigma |t|} \cos{\omega_0 t}$$

Ron Gordon
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  • giusppe: yes, the F() or G( ) denote the fourier transforms. Ron: I'm sorry but I don't understand how you got the initial term or the $\frac{\pi}{\sigma} \exp^{-\sigma |t|}$. I'm just starting to learn this material on my own so my bad for cluelessness and thanks. – mark leeds May 30 '14 at 19:30
  • ron: also, your answer is the same as the answer in the book except that the book doesn't have an absolute value. but there are typos in the book so my guess is that it's a book typo. thanks again. – mark leeds May 30 '14 at 19:36
  • ron: I found the following link which gave some more details on your result so I'm going to check your answer. thanks a lot. http://math.stackexchange.com/questions/377073/calculate-the-fourier-transform-of-rm-b-leftx-right-1-leftx2-a2 – mark leeds May 31 '14 at 16:29