I am studying for my final and doing some practice questions, but I am confused by something:
Here the solution says k at 0 we get N/2, but there is no way that answer is correct. If k is at 0 the $e^{-j2pik}$ should evaluate to 1, which should give us 0.
And here is the thing, there are several similar problems given by the book that all solves it this way. So I don't think its a typo error.
Can someone help me out with this? It is Q8.5a of Digital Signal Processing by oppenhiem and schaffer if you have that book and want to look it up.
thanks
For any integer $k$, the numerator is $1 - 1 = 0$.
If $k$ is also a multiple of $2N$ then the bottom is also $1-1 = 0$. This is indeterminate.
(It seems their $N/2$'s should be $2N$'s.)
Whenever $k$ is not a multiple of $2N$, the exponential on the bottom is not equal to one, so the expression on the bottom is nonzero, this whole expression is zero since the numerator is.
Here's what they must have meant in the indeterminate case:
$\frac{1-e^{-j2\pi k}}{1-e^{-j\pi k/N}}$
is of the form
$\frac{1 - x^{2N}}{1-x} = 1 + x + \cdots + x^{2N-1}$
for $x = e^{-j\pi k/N}$, and this sum must be what they really mean.
If $k$ is a multiple of $2N$ then $e^{-j\pi k/N}$ is one, so each term in this sum is one. The sum has $2N$ terms, so its total sum is $2N$.
I imagine this sum is what they're actually interested in, as this problem is about discrete fourier transforms.
If the problem were $\frac{1-e^{-j2\pi k}}{1-e^{-4j\pi k/N}}$, then their $N/2$'s would be correct: it would be of the form $\frac{1-x^{N/2}}{1-x}$ for $x =e^{-4j\pi k/N}$, which is one precisely when $k$ is a multiple of $N/2$, and there are also $N/2$ terms, all equal to one, if you write it as a summation.