Prove that $\sum_n e^{-iωnT} = \cfrac{2π}{T}\sum_n δ(ω-\cfrac{2πn}{T}) $

Question

From the Fourier Transform Table I found that the transform of $\sum_{n=-\infty}^{\infty}δ(t-nT)$ is equal to $S_2$. Using the definition of Fourier Transform (integration) I found that it is equal to $S_1$, where

$$ S_1=\sum_{n=-\infty}^{\infty}\exp(-iωnT) $$ $$ S_2=\cfrac{2π}{T}\sum_{n=-\infty}^{\infty}δ(ω-\cfrac{2πn}{T}) $$ How can we prove that $S_1=S_2$?

Thank you in advance

Answer 1

This is the distributional way to define the Fourier series theorem (that $C^1$ periodic functions are equal to their Fourier series).

For $f\in C^1_c(a,a+1)$ and $F$ its $1$-periodization

$$\langle \sum_n \delta(t-n),f\rangle \overset{def}=\sum_n f(n)=F(0)$$

$$\langle \sum_n e^{-2i\pi nt},f\rangle \overset{def}=\sum_n c_n(F),\qquad c_n(F)=\int_0^1 F(t)e^{-2i\pi nt}dt=\int_{-\infty}^\infty f(t)e^{-2i\pi nt}dt$$ The Fourier series theorem is that $F(0)=\sum_n c_n(F)$, thus by definition $ \sum_n \delta(t-n)=\sum_n e^{-2i\pi nt}$ in the sense of distributions.

Be careful that $\sum_n e^{-2i\pi nt}$ converges only in the sense of distributions, it diverges in the sense of functions.

Answer 2

Referencing a Fourier table such as the one published here: https://uspas.fnal.gov/materials/11ODU/FourierTransformPairs.pdf, We observe the time shift property of the Fourier transform:

$$F\left(t-t_{o}\right) \longleftrightarrow F\left(\omega\right)e^{-i \omega t_{o}} $$

which when applied to the train of impulses gives the following transform pair:

$$\sum^{\infty}_{n=-\infty} \delta\left(t-n T\right)\longleftrightarrow \sum^{\infty}_{n=-\infty} e^{-i \omega n T} $$

Notice the pair item in the frequency domain has a form similar to a Fourier Series expansion of a periodic function in $\omega$, shown below:

$$ G\left(\omega\right)= \sum^{\infty}_{n=-\infty} c_{n} e^{-i \frac{2\pi n \omega}{P} } $$

Where $c_{n}$ is the Fourier series coefficient and $P$ is the period in the $\omega$ domain. Comparing sums we conclude that for all $n$: $$c_{n} = 1$$ and, $$\frac{2\pi}{P}=T$$ Now, recall that: $$c_{n}=\frac{1}{P}\int_{P} G\left(\omega\right) e^{i \frac{2\pi n \omega}{P}}d\omega $$ which can only be identically 1 for all n if $G\left(\omega\right)$ is the dirac distribution $P\delta\left(\omega\right)$ over period $P$. Therefore: $$\sum^{\infty}_{n=-\infty} e^{-i \omega n T}=P\sum^{\infty}_{n=-\infty}\delta\left(\omega-n P\right)=\frac{2\pi}{T}\sum^{\infty}_{n=-\infty}\delta\left(\omega-n \frac{2\pi}{T}\right)$$