0
$\begingroup$

The Fourier transform of right handed exponential function $e^{-\alpha t}$ is given by

$$\mathcal{F}\left[ e^{-\alpha t}\right] = \frac{1}{\alpha + j \omega}.$$

Its derivative is

$$\frac{d}{dt} e^{-\alpha t} = -\alpha e^{-\alpha t}.$$

So the Fourier transform of derivative is

$$\mathcal{F}\left[ \frac{d}{dt}e^{-\alpha t}\right] = -\frac{\alpha}{\alpha + j \omega}.$$

But if I use the Fourier transform derivative formula

$$j \omega F(\omega) = \frac{j \omega}{\alpha + j \omega}.$$

These results are different. Where is my mistake?

$\endgroup$
1
$\begingroup$

Your mistake is that the derviative is

$$\dfrac{d}{dt}e^{-\alpha t}u(t) = e^{-\alpha t}\delta(t) -\alpha e^{-\alpha t}u(t)$$

From there, the rest follows and agrees with the derivative theorem of the Fourier transform.

$\endgroup$
  • $\begingroup$ thanks a lot for your answer. Now, everything is clear. I am sorry but I could not vote up because my reputation is too low. So, I could only accept the answer. $\endgroup$ – Stef1611 Jan 20 at 9:46
  • $\begingroup$ @Stef1611: you're welcome. $\endgroup$ – Andy Walls Jan 20 at 10:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.