Dirac Comb

Geometric Series of Exponentials

To calculate the sum $\sum _{k=-\infty }^{\infty }\exp (-j\omega kT)$, let $r=\exp (-j\omega T)$, and then our series can be written as $\sum _{k=-\infty }^{\infty }{r}^k$. We can split this into two parts:

$$\sum _{k=-\infty }^{\infty }{r}^k=\sum _{k=-\infty }^{-1}{r}^k+r^0+\sum _{k=1}^{\infty }{r}^k$$

The right-hand sum $\sum _{k=1}^{\infty }{r}^k$ is a geometric series with first term $a=r$ and common ratio $r$. The sum of this series (if $|r|<1$) is $\sum _{k=1}^{\infty }{r}^k=\frac{r}{1-r}$. Similarly, the left-hand sum is $\sum _{k=1}^{\infty }(r^{-1}{)}^k=\frac{r^{-1}}{1-r^{-1}}=\frac{1}{r-1}$. Putting it all together:

$$\sum _{k=-\infty }^{\infty }{r}^k=\frac{1}{r-1}+1+\frac{r}{1-r}=0$$

This result is valid when $r\ne 1$, which means $\omega T\ne 2\pi n$ for any integer $n$. When $r=1$ (i.e., $\omega T=2\pi n$), the series diverges. This sum can be expressed by the Dirac Delta Function

$$\sum _{k=-\infty }^{\infty }\exp (-j\omega kT)=\frac{2\pi }{T}\sum _{n=-\infty }^{\infty }\delta \left(\omega -n\frac{2\pi }{T}\right)$$

The factor $2\pi$ in front of the sum ensures that the integral of each impulse over frequency is unity

This representation is known as the Dirac comb or Shah function. It's a very important concept in signal processing, sampling theory, and Fourier analysis, as it represents the frequency-domain equivalent of a periodic impulse train in the time domain.

Dirac Comb

$${Ш}_T(t):=\sum _{k=-\infty }^{\infty }\delta (t-kT)$$

Unit Dirac Comb

$$Ш(t):=\sum _{k=-\infty }^{\infty }\delta (t-k)$$

and

$${Ш}_T(t)=\frac{1}{T}Ш\left(\frac{t}{T}\right)$$