Why is the term for $k_0=0$ the only remaining one for this Discrete Fourier Transform?

Question

We want to take the Discrete Fourier Transform of the following: $$(1) \ \hat{x}[k]=p \sum_{0}^{N/2-1} \omega_{N/2}^{kr} + \omega_{N}^{k}q \sum_{0}^{N/2-1} \omega_{N/2}^{kr}$$

where $\omega_N = e^{-2\pi i/N}$. Using the fact that the DFT of $e^{2\pi i n k_0/N}$ is $N\delta[k-k_0]$, we get:

$$(2) \ \hat{x}(k) = \frac{pN}{2}\delta_{N/2}[k]+ \omega_N^k \frac{qN}{2}\delta_{N/2}[k]$$

where $\delta_{N/2}[k]$ has Period $N/2$

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I don't see why the sum suddenly disappeared. Why is the only remaining term the one for $k_0=0$ ? Does it have something to do with the geometric sum ? I'm a bit lost and would be grateful for hints/ advices.

Thanks for your help !

Answer 1

By calculating the DFT of $e^{\frac{i 2 \pi n k_0}{N}}=\left(\frac{1}{\omega_N}\right)^{n k_0}=\omega_N^{-n k_0}$, we obtain

$\sum_{r=0}^{N-1} e^{\frac{i 2 \pi r k_0}{N}}\omega_{N}^{rk}=\displaystyle\sum_{r=0}^{N-1} \omega_N^{-r k_0}\omega_{N}^{rk}$

$=\sum_{r=0}^{N-1} \omega_{N}^{r(k-k_0)}$

We have

$\delta_N[k-k_0]=\displaystyle\frac{1}{N}\sum_{r=0}^{N-1} e^{-i 2\pi \frac{(k-k_0)}{N}r}$

$=\displaystyle\frac{1}{N}\sum_{r=0}^{N-1} \omega_{N}^{(k-k_0)r}$

So The first sum:

$\displaystyle\sum_{r=0}^{N/2-1} \omega_{N/2}^{kr}$

$=\frac{N}{2}\frac{1}{N/2}\displaystyle\sum_{r=0}^{N/2-1} \omega_{N/2}^{kr}$

$=\frac{N}{2}\delta_{N/2}[k]$

(replacing $k_0=0$)