Currently, I am dealing with the sampling problems and I don't understand how to calculate inverse Fourier transform of a scaling impulse function

$\textrm{IFT}\{\delta(\Omega T)\} = ?$, $T$ is sampling period but we can see it as a constant.

It comes from this equation since I have to find the $y(t)$ of reconstructing signal

$$Y_r(j\Omega) = j\omega_0\pi\delta(\Omega T - \omega_0) - j\omega_0\pi\delta (\Omega T + \omega_0) $$

The solution for $\textrm{IFT}\{\delta(\Omega T)\}$ is $\frac{1}{2\pi T}$, but I don't know how to get them.

I got stuck when trying to expand and calculate it by integral of inverse FT.


1 Answer 1


This is just a matter of variable substitution in the integral. The inverse Fourier transform of $\delta(\Omega T)$ is

$$\mathcal{F}^{-1}\big\{\delta(\Omega T)\big\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\Omega T)\,e^{j\Omega t}\,d\Omega\tag{1}$$

Substituting $\alpha=\Omega T$ gives $d\alpha=d\Omega\cdot T$, and $(1)$ becomes

$$\mathcal{F}^{-1}\big\{\delta(\Omega T)\big\}=\frac{1}{2\pi T}\int_{-\infty}^{\infty}\delta(\alpha)\,e^{j\alpha t/T}\,d\alpha=\frac{1}{2\pi T}\tag{2}$$

The last equation is a consequence of the sifting property of the Dirac delta impulse.

  • $\begingroup$ Thank you so much, I already thought that but i still stick to standard inverse fourier transform, and I got stuck with dw and eˆjwt. Btw, thank you so much. $\endgroup$
    – PhilN
    Oct 8, 2019 at 15:32

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