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

• Nov 22nd 2010, 02:15 AM
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. :)
• Nov 22nd 2010, 05:56 AM
Ackbeet
I don't think you're taking the derivative correctly. Your function is actually

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

where $U(t)$ 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:

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

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