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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?

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  • $\begingroup$ 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
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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.

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