Using the duality property, I guess it happens whenever the original signal is composed of a sum of dirac delta functions spaced at equal intervals of time.

I conclude this as the Fourier transform of a periodic function is a number of dirac delta functions (of size $2\pi a_k$) at intervals of $\frac{2\pi k}{T}$ What is the actual condition?

  • $\begingroup$ It depends on what you consider a "function". If you want an authentic function, say in $L^2(\mathbb{R})$, then I guess the answer is "when the function is (almost) $0$". $\endgroup$ Commented Apr 2, 2016 at 9:52

1 Answer 1


You are right. A periodic Fourier transform can be written as a Fourier series:

$$F(\omega)=\sum_kc_ke^{i2\pi k\omega/\omega_0}\tag{1}$$

where $\omega_0$ is the period. Taking the inverse transform of $(1)$ we get

$$f(t)=\mathcal{F}^{-1}\{F(\omega)\}=\sum_kc_k\mathcal{F}^{-1}\{e^{i2\pi k\omega/\omega_0}\}=\sum_kc_k\delta\left(t+\frac{2\pi k}{\omega_0}\right)\tag{2}$$


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