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I want to obtain the $N$-point IDFT of $X(k)=\frac{3}{5-4 \cos \left(2 \pi \frac{k}{N}\right)}$. My idea was to go with the discrete Poisson summation formula. $$ x_{s}(n)=\sum_{m=-\infty}^{\infty} x(n-m N)=\frac{1}{N} \sum_{k=0}^{N-1} X(k) e^{i 2 \pi k n / N} $$ but I didn't manage to solve it.

Any suggestions??

metamorphy
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why_me
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1 Answers1

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Following this answer of mine, introduce $\omega=\exp(2\pi i/N)$ and consider $$S=\sum_{k=0}^{N-1}\frac{\exp(2nk\pi i/N)}{1-2z\cos(2k\pi/N)+z^2}=\sum_{k=0}^{N-1}\frac{\omega^{kn}}{(1-z\omega^k)(1-z\omega^{-k})}$$ where $|z|\neq 1$ for simplicity (you're computing $3S/N$ at $z=2$). Now use $$\sum_{p=0}^{N-1}(z\omega^{\pm k})^p=\frac{1-(z\omega^{\pm k})^N}{1-z\omega^{\pm k}}=\frac{1-z^N}{1-z\omega^{\pm k}}$$ to get $$(1-z^N)^2 S=\sum_{k=0}^{N-1}\omega^{kn}\sum_{p=0}^{N-1}(z\omega^k)^p\sum_{q=0}^{N-1}(z\omega^{-k})^q=\sum_{p,q=0}^{N-1}z^{p+q}\sum_{k=0}^{N-1}\omega^{k(n+p-q)}.$$

The inner sum is $N$ if $n+p-q$ is a multiple of $N$ (assuming $\color{blue}{0\leqslant n<N}$, this happens iff $0\leqslant p<N-n$ and $q=p+n$, or $N-n\leqslant p<N-n$ and $q=p+n-N$).

Otherwise, this sum is $0$. And finally $$S=\frac{N}{(1-z^N)^2}\left(\sum_{p=0}^{N-n-1}z^{2p+n}+\sum_{p=N-n}^{N-1}z^{2p+n-N}\right)=\frac{N(z^n+z^{N-n})}{(1-z^2)(1-z^N)}$$ (note that $n=N$ is also acceptable here, so that the result holds for $0\leqslant n\leqslant N$).

metamorphy
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  • Hi, @metamorphy! Is there any way to generalize the formula for $S$ for half-integer n? – Sl0wp0k3 Feb 01 '22 at 10:00
  • @Sl0wp0k3: I think this deserves a dedicated post. – metamorphy Feb 01 '22 at 11:05
  • Just trying to reduce the problem that I have to something that I already know, and one of the ways is to reduce it to this sum $S$, but then the $n$ parameter is out of the bounds (it can be $<0$, and $>N$, that is easy), and is half-integer (it is harder to generalize). Basically, it is directly related to the question of mine https://math.stackexchange.com/questions/4364976/estimating-the-following-finite-discrete-sum-involving-periodic-functions If you have ideas/hints for this half-integer n case It will be enough as I have done the rest already. – Sl0wp0k3 Feb 01 '22 at 11:52
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    I did it, even though its not about half-integer $n$. Actually, I needed to find the sum like $\sum\limits_{k=1}^{N-1} \frac{\exp(\pi k n/N)}{\sin(\pi k/N)}$, and decided to do it by computing the $\sum\limits_{k=0}^{N-1} \dfrac{\exp (2 \pi k n/N)}{1 - i 2 z \sin(2 \pi k / N) -z^2}$ then subtracting $k=0$, $N \to 2N$, and getting the limit $z \to 1$. – Sl0wp0k3 Feb 03 '22 at 12:36
  • Actually, I was only able to find it for odd $N$, for even it is tricky :). I would be happy if you can briefly see through it, and mb advice something https://math.stackexchange.com/questions/4376118/a-peculiar-limit-appearing-for-sum-limits-j-1n-1-frac-exp-left-i-pi – Sl0wp0k3 Feb 07 '22 at 13:25