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Alright, I'm gonna answer this with an argument that "opponents" to my rigid nazi-like position regarding the DFT have.

First of all, my rigid, nazi-like position:

The Discrete Fourier Transform and Discrete Fourier Series are one-and-the-same.

The DFT maps one infinite and periodic sequence, $x[n]$ with period $N$ in the "time" domain to another infinite and periodic sequence, $X[k]$, again with period $N$, in the "frequency" domain. And the iDFT maps it back. and they're "bijective" or "invertible" or "one-to-one".

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.

$$ x[n+N]=x[n] \qquad \forall n \in \mathbb{Z} $$ $$ X[k+N]=X[k] \qquad \forall k \in \mathbb{Z} $$

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

$$ \hat{x}[n] \triangleq \begin{cases} x[n] \qquad & \text{for } 0 \le n \le N-1 \\ \\ 0 & \text{otherwise} \end{cases} $$

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

$$ \begin{align} \hat{X}\left(e^{j\omega}\right)\Bigg|_{\omega = 2 \pi\frac{k}{N}} & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} \Bigg|_{\omega = 2 \pi\frac{k}{N}} \\ & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi k n/N} \\ & = X[k] \\ \end{align} $$

That is precisely how the DFT and DTFT are related. Sampling the DTFT at uniform intervals in the "frequency" domain causes, in the "time" 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, $x[n]$.

Alright, I'm gonna answer this with an argument that "opponents" to my rigid nazi-like position regarding the DFT have.

First of all, my rigid, nazi-like position:

The Discrete Fourier Transform and Discrete Fourier Series are one-and-the-same.

The DFT maps one infinite and periodic sequence, $x[n]$ with period $N$ in the "time" domain to another infinite and periodic sequence, $X[k]$, again with period $N$, in the "frequency" domain. And the iDFT maps it back. and they're "bijective" or "invertible" or "one-to-one".

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.

$$ x[n+N]=x[n] \qquad \forall n \in \mathbb{Z} $$ $$ X[k+N]=X[k] \qquad \forall k \in \mathbb{Z} $$

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

$$ \hat{x}[n] \triangleq \begin{cases} x[n] \qquad & \text{for } 0 \le n \le N-1 \\ \\ 0 & \text{otherwise} \end{cases} $$

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

$$ \begin{align} \hat{X}\left(e^{j\omega}\right)\Bigg|_{\omega = 2 \pi\frac{k}{N}} & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} \Bigg|_{\omega = 2 \pi\frac{k}{N}} \\ & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi k n/N} \\ & = X[k] \\ \end{align} $$

That is precisely how the DFT and DTFT are related. Sampling the DTFT at uniform intervals in the "frequency" domain causes, in the "time" 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, $x[n]$.

Alright, I'm gonna answer this with an argument that "opponents" to my rigid nazi-like position regarding the DFT have.

First of all, my rigid, nazi-like position:

The Discrete Fourier Transform and Discrete Fourier Series are one-and-the-same.

The DFT maps one infinite and periodic sequence, $x[n]$ with period $N$ in the "time" domain to another infinite and periodic sequence, $X[k]$, again with period $N$, in the "frequency" domain. And the iDFT maps it back. and they're "bijective" or "invertible" or "one-to-one".

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

$$ \hat{x}[n] \triangleq \begin{cases} x[n] \qquad & \text{for } 0 \le n \le N-1 \\ \\ 0 & \text{otherwise} \end{cases} $$

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

$$ \begin{align} \hat{X}\left(e^{j\omega}\right)\Bigg|_{\omega = 2 \pi\frac{k}{N}} & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} \Bigg|_{\omega = 2 \pi\frac{k}{N}} \\ & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi k n/N} \\ & = X[k] \\ \end{align} $$

That is precisely how the DFT and DTFT are related. Sampling the DTFT at uniform intervals in the "frequency" domain causes, in the "time" 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, $x[n]$.

Alright, I'm gonna answer this with an argument that "opponents" to my rigid nazi-like position regarding the DFT have.

First of all, my rigid, nazi-like position:

The Discrete Fourier Transform and Discrete Fourier Series are one-and-the-same.

The DFT maps one infinite and periodic sequence, $x[n]$ with period $N$ in the "time" domain to another infinite and periodic sequence, $X[k]$, again with period $N$, in the "frequency" domain. And the iDFT maps it back. and they're "bijective" or "invertible" or "one-to-one".

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.

$$ x[n+N]=x[n] \qquad \forall n \in \mathbb{Z} $$ $$ X[k+N]=X[k] \qquad \forall k \in \mathbb{Z} $$

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

$$ \hat{x}[n] \triangleq \begin{cases} x[n] \qquad & \text{for } 0 \le n \le N-1 \\ \\ 0 & \text{otherwise} \end{cases} $$

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

$$ \begin{align} \hat{X}\left(e^{j\omega}\right)\Bigg|_{\omega = 2 \pi\frac{k}{N}} & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} \Bigg|_{\omega = 2 \pi\frac{k}{N}} \\ & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi k n/N} \\ & = X[k] \\ \end{align} $$

That is precisely how the DFT and DTFT are related. Sampling the DTFT at uniform intervals in the "frequency" domain causes, in the "time" 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, $x[n]$.

alright, i'm gonna answer this with an argument that "opponents" 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 "time" domain to another infinite and periodic sequence, $X[k]$, again with period $N$, in the "frequency" domain. and the iDFT maps it back. and they're "bijective" or "invertible" or "one-to-one".

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

$$ \hat{x}[n] \triangleq \begin{cases} x[n] \qquad & \text{for } 0 \le n \le N-1 \\ \\ 0 & \text{otherwise} \end{cases} $$

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

$$ \begin{align} \hat{X}\left(e^{j\omega}\right)\Bigg|_{\omega = 2 \pi\frac{k}{N}} & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} \Bigg|_{\omega = 2 \pi\frac{k}{N}} \\ & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi k n/N} \\ & = X[k] \\ \end{align} $$

that is precisely how the DFT and DTFT are related. sampling the DTFT at uniform intervals in the "frequency" domain causes, in the "time" 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, $x[n]$.

Alright, I'm gonna answer this with an argument that "opponents" to my rigid nazi-like position regarding the DFT have.

First of all, my rigid, nazi-like position:

The Discrete Fourier Transform and Discrete Fourier Series are one-and-the-same.

The DFT maps one infinite and periodic sequence, $x[n]$ with period $N$ in the "time" domain to another infinite and periodic sequence, $X[k]$, again with period $N$, in the "frequency" domain. And the iDFT maps it back. and they're "bijective" or "invertible" or "one-to-one".

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

$$ \hat{x}[n] \triangleq \begin{cases} x[n] \qquad & \text{for } 0 \le n \le N-1 \\ \\ 0 & \text{otherwise} \end{cases} $$

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

$$ \begin{align} \hat{X}\left(e^{j\omega}\right)\Bigg|_{\omega = 2 \pi\frac{k}{N}} & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j \omega n} \Bigg|_{\omega = 2 \pi\frac{k}{N}} \\ & = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} \hat{x}[n] e^{-j 2 \pi k n/N} \\ & = \sum\limits_{n=0}^{N-1} x[n] e^{-j 2 \pi k n/N} \\ & = X[k] \\ \end{align} $$

That is precisely how the DFT and DTFT are related. Sampling the DTFT at uniform intervals in the "frequency" domain causes, in the "time" 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, $x[n]$.