# Fourier transform of x(t)= d/dt e^-|t|

• Dec 7th 2008, 05:19 PM
lorenzo
Fourier transform of x(t)= d/dt e^-|t|
How can I apply fourier tranform to this EQ

x(t)= d/dt e^-|t|

Thank you for help
• Dec 7th 2008, 08:14 PM
mr fantastic
Quote:

Originally Posted by lorenzo
How can I apply fourier tranform to this EQ

x(t)= d/dt e^-|t|

Thank you for help

Here is an approach that requires minimal thinking and effort:

Theorem: $FT\left[ \frac{df}{dt}\right] = i \omega FT\left[ f(t) \right]$.

Theorem: $FT\left[ e^{-\alpha |t|} \right] = \frac{2 \alpha}{\alpha^2 + \omega^2}$.

Other approaches are possible.
• Dec 7th 2008, 09:15 PM
lorenzo
Quote:

Originally Posted by mr fantastic
Here is an approach that requires minimal thinking and effort:

Theorem: $FT\left[ \frac{df}{dt}\right] = i \omega FT\left[ f(t) \right]$.

Theorem: $FT\left[ e^{-\alpha |t|} \right] = \frac{2 \alpha}{\alpha^2 + \omega^2}$.

Other approaches are possible.

Hey,.... Did you get to use any software to obtain this particular solution?
• Dec 7th 2008, 09:20 PM
lorenzo
So, Ok you mean that the answer would be something like this thanks to the Theorems...

jwFT((2a)/(a^2+w^2))
• Dec 7th 2008, 10:28 PM
mr fantastic
Quote:

Originally Posted by lorenzo
Hey,.... Did you get to use any software to obtain this particular solution?

No. I looked at a standard table of Fourier Transforms in a textbook.

Quote:

Originally Posted by lorenzo
So, Ok you mean that the answer would be something like this thanks to the Theorems...

jwFT((2a)/(a^2+w^2))

OK, I lied about the minimal thinking ..... The answer in your case is $j \omega \left( \frac{2}{1 + \omega^2}\right) ~$ (since $\alpha = 1$).