2
$\begingroup$

I'm trying to calculate the antitransform of:

$\frac{1}{2\cdot(1+5w)^2}$

Now I know the antitransform of $\frac{1}{(1+5w)^2} = t \cdot e^{-5t} u(t) $

But in this case I got that divided by 2. I assumed I had to use the scaling property which says:

$F[f(ax)] = \frac{1}{|a|} \hat{f}(\frac{w}{a})$

Now I'm not really sure how to apply this. Could anyone help?

| improve this question | |
$\endgroup$
  • $\begingroup$ Scaling property only applies is you multiply the time/frequency variable. In this case, you only need to multiply the inverse with 1/2. $\endgroup$ – Hilmar Dec 12 '19 at 5:04
1
$\begingroup$

If $h(t)$ is the inverse Fourier transform of $H(\omega)$, then by linearity the inverse Fourier transform of $aH(\omega)$ is simply $ah(t)$. This has nothing to do with the scaling property you mentioned, because the latter refers to the scaling of the argument of the function.

| improve this answer | |
$\endgroup$

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.