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

1. ## 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

2. 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.

3. 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?

4. So, Ok you mean that the answer would be something like this thanks to the Theorems...

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

5. 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.

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 $\alpha = 1$).