# Thread: Fourier Transform of a derivative of a decaying exponent function?

1. ## Fourier Transform of a derivative of a decaying exponent function?

Hello folks, new to the forum and first post here Hope this is the correct subforum. Tried the search option but did not find anything to help me out. Here's the problem:

I have a decaying exponential function f(t) = exp(-at); where a>0 and t>=0 [and f(t)=0 when t<0)] and I need to find the Fourier transformation of it's derivative.

I can do it using the Derivative Theorem:
F {f'(t)} = iw F(w);
and I get
F {f'(t)} = (iw)/(a+iw).

However, if I just take d/dt of the original f(t), I get f'(t)= -a exp(-at) = -a f(t).
And since a is a constant, calculating the Fourier transform for this one should go like:
F {f'(t)} = F {-a f(t)} = -aF(w) = (-a)/(a+iw).

These 2 results are not the same (unless, for some reason a=-iw), so obviously I'm not doing it right. Just cannot get my head around it. Any thoughts & pointing out where I go wrong appreciated.

2. I don't think you're taking the derivative correctly. Your function is actually

$\displaystyle f(t)=U(t)\,e^{-at},$

where $\displaystyle U(t)$ is the Heaviside step function. The Heaviside step function actually has a derivative (though the result is technically a distribution or a measure, not a function): the Dirac delta function. Hence, the derivative looks more like this:

$\displaystyle f'(t)=U'(t)\,e^{-at}-aU(t)\,e^{-at}=\delta(t)\,e^{-at}-aU(t)\,e^{-at}.$

Try taking the Fourier Transform of this result, and see if it doesn't match up with what you got before.