How is the DTFT of a periodic, sampled signal linked to the DFT? I am trying to understand the connection between FT, DTFT and ultimately the DFT. But I am failing to link the DTFT to the DFT.
This is how far I am getting: Say I have a function $f(t)$, and its Fourier Transform $F(t)=\mathcal{F}\{f(t)\}(\nu)$ defined as $F(t)= \int_{-\infty}^{\infty} dt \, f(t)\exp(-i \,2\pi\nu t)$. Lets then say we have ideally sampled the signal at time intervals $\Delta T$ like so:
$$\bar{f}(t)=f(t)\sum_{n=-\infty}^{\infty} \delta(t-n\Delta T).$$
I understand that the FT of this signal leads me to the DTFT:
$$\bar{F}(\nu) = \mathcal{F}\{\bar{f}(t)\}(\nu) = \sum_{n=-\infty}^{\infty} f_n \exp(-i\,2\pi \nu n\,\Delta T), \, f_n = f(n \, \Delta T)$$
I also understand that $\bar{F}(\nu)$ is linked to the (continous time) Fourier Transform via:
$$\bar{F}(\nu) = \frac{1}{\Delta T}\sum_{n=-\infty}^{\infty}F\left(\nu - \frac{n}{\Delta T}\right),$$
which is a periodic function with period $1/\Delta T$. If $f(t)$ is bandlimited then the values of the DTFT at discrete frequencies $\nu_m = m/M \cdot \Delta T, \, m=0,1,...,M-1$ will give me a value proportional to the FT of $f$ at these frequencies. I also understand that I only need $M$ samples in Fourier space due to the periodicity of $\bar{F}$
Where I am struggling is to understand how to get from here to the DFT. I'll plug the discrete frequencies into the DTFT above and I further assume that the sequence $f_n$ is $M$-perdiodic such that $f_{n-kM}=f_{n}, \, k\in \mathbb{Z}$. Following this wiki article I write:
$$\begin{eqnarray}
\bar{F}(\nu_m)=\bar{F}_m &amp;=&amp; \sum_{n=-\infty}^{\infty} f_n \exp\left(-i \,2\pi \frac{m}{M}n\right) \\
&amp;=&amp; \sum_{k=-\infty}^{\infty} \left(\sum_{n=0}^{M-1}\exp\left(-i \,2\pi \frac{m}{M}n\right) f_{n-kM}\right) \\
&amp;=&amp; \sum_{n=0}^{M-1} \exp\left(-i \,2\pi \frac{m}{M}n\right)  \sum_{k=-\infty}^{\infty}f_{n-kM}
\end{eqnarray}$$
The wikipedia article then calls the inner sum $\tilde{f}[n]$, which leads us to the DFT. But I fail to see how this infinite sum of a periodic series relates to the cyclic extension of a finite series. I just don't see it.
I know that this topic is a perpetual source of confusion for people and we have excellent answers e.g. here, here, and here. But I don't quite get it yet.
I think that your misunderstanding comes from the fact that you assume that $f[n]$ is periodic. If $f[n]$ were periodic then its DTFT would be a Dirac comb. What is happening here is that you want to find the $N$-periodic sequence whose DFT coefficients equal equidistant samples of the DTFT of $f[n]$. And it turns out that that sequence is given by
$$\tilde{f}[n]=\sum_{k=-\infty}^{\infty}f[n-kN]\tag{1}$$
which is - as expected - an aliased version of the original sequence. Remember that sampling in one domain corresponds to aliasing in the other domain.
A derivation of $(1)$ is given in this answer.