I am having some trouble finding the period of this function:
$$W(\omega) = \frac{\sin[(2N +1)\omega \Delta t / 2]}{(2N + 1)\sin[\omega \Delta t /2]}$$
Here $N$ is an integer, $\omega$ is angular frequency, and $\Delta t$ is a sample rate.
The correct answer should be $2 \pi/ \Delta t$, but I don't see how this is derived. In this function, $\Delta t$ is kept constant. Thus since we have that $\omega = 2 \pi f$, we must have:
$$\omega \Delta t /2 = 2 \pi$$
$$2 \pi f \Delta t /2 = 2 \pi$$
$$2 \pi \Delta t / 2T = 2 \pi$$
$$\Delta t / 2T = 1$$
$$2T = \Delta t$$
$$T = \Delta t /2$$
which of course is different from the answer I'm supposed to get. If anyone can explain what is wrong with my reasoning, and inform me how the correct period is derived, I would appreciate it greatly!
It may help to know that $$ W(\omega)=\frac{1}{2N+1}\left[1+2\sum_{n=1}^N\cos(n\omega\Delta t)\right] $$
So the period should just be the period of the $n=1$ term.