I've encountered an infinite geometric sum while working a question: $$ (1 - e^{-6wi}) \sum_{0}^{\infty} {e^{-iwn}} $$
According to the answer sheet, this should resolve to: $$ \frac{(1 - e^{-6wi})}{(1 - e^{-wi})} $$
The context of this question is applying a Discrete Fourier Transform on the signal x[n] = u[n] - u[n-6] where u[n] is the step function.
I'm having a hard time understanding this step, could someone elaborate?
On
You are right to have some doubt indeed
$$\sum_{0}^{\infty} {e^{-iwn}}=\sum_{0}^{\infty} {(e^{-iw})^n}=\frac1{1-e^{-iw}}$$
only holds if $|e^{-iw}|<1$.
Therefore for $\omega\in \mathbb R$ the identity doesn't hold and for $\omega =x+iy$ we have
$$e^{-iw}=e^{-i(x+iy)}=e^y \cdot e^{ix}$$
and in this case the identity holds for $|e^y \cdot e^{ix}|=|e^y|<1$ that is $y<0$.
I've given it some thought and managed to wiggle my way to the solution using the Z-transform rather than geometric progression.
We know the Z-Transform of a unit complex number is equivalent to the Discrete Time Fourier Transform.
Hence:
x[n] = u[n] - u[n-6]
Applying the Z-transform gives us: $$\frac{1 - z^{-6}}{1 - z^{-1}}$$
Setting |z| = 1 gives us the answer we're looking for.