Difference between discrete time fourier transform and discrete fourier transform
alright, i'm gonna answer this with an argument that &quot;opponents&quot; to my rigid nazi-like position regarding the DFT have.
first of all, my rigid, nazi-like position:  the DFT and Discrete Fourier Series is one-and-the-same.  the DFT maps one infinite and periodic sequence, $x[n]$ with period $N$ in the &quot;time&quot; domain to another infinite and periodic sequence, $X[k]$, again with period $N$, in the &quot;frequency&quot; domain.  and the iDFT maps it back.  and they're &quot;bijective&quot; or &quot;invertible&quot; or &quot;one-to-one&quot;.
DFT:
$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi nk/N} $$
iDFT:
$$ x[n] = \frac{1}{N} \sum\limits_{k=0}^{N-1} X[k] e^{j 2 \pi nk/N} $$
that is most fundamentally what the DFT is.  it is inherently a periodic or circular thing.
but the periodicity deniers like to say this about the DFT.  it is true, it just doesn't change any of the above.
so, suppose you had a finite-length sequence $x[n]$ of length $N$ and, instead of periodically extending it (which is what the DFT inherently does), you append this finite-length sequence with zeros infinitely on both left and right.  so
$$ \hat{x}[n] \triangleq \begin{cases}
x[n] \qquad &amp; \text{for } 0 \le n \le N-1 \\
\\
0 &amp; \text{otherwise}
\end{cases} $$
now, this non-repeating infinite sequence does have a DTFT:
DTFT:
$$ \hat{X}\left(e^{j\omega}\right) = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} $$
$\hat{X}\left(e^{j\omega}\right)$ is the Z-transform of $\hat{x}[n]$ evaluated on the unit circle $z=e^{j\omega}$ for infinitely many real values of $\omega$.
now, if you were to sample that DTFT $\hat{X}\left(e^{j\omega}\right)$ at $N$ equally spaced points on the unit circle, with one point at $z=e^{j\omega}=1$, you would get
$$ \begin{align} 
\hat{X}\left(e^{j\omega}\right)\Bigg|_{\omega = 2 \pi\frac{k}{N}} &amp; = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} \Bigg|_{\omega = 2 \pi\frac{k}{N}} \\
&amp; = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j 2 \pi k n/N}  \\
&amp; = \sum\limits_{n=0}^{N-1} \hat{x}[n] e^{-j 2 \pi k n/N}  \\
&amp; = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi k n/N}  \\
&amp; = X[k]  \\
 \end{align} $$
that is precisely how the DFT and DTFT are related.  sampling the DTFT at uniform intervals in the &quot;frequency&quot; domain causes, in the &quot;time&quot; domain, the original sequence $\hat{x}[n]$ to be repeated and shifted by all multiples of $N$ and overlap-added.  that's what uniform sampling in one domain causes in the other domain.  but, since $\hat{x}[n]$ is hypothesized to be $0$ outside of the interval $0 \le n \le N-1$, that overlap-adding does nothing.  it just periodically extends the non-zero part of $\hat{x}[n]$, our original finite-length sequence, $x[n]$.