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alright, the way to think about the rectangular window (as well as all of the other windows) is that you begin with an infinite sequence of samples and you multiply the samples outside the window with 0 and the samples within the window with something non-zero (and usually, but not always, positive).

so the DFT (of which the FFT is fast method of doing the DFT) does periodically extend the $N$ samples passed to it. but there are a couple of different ways of looking at it. one way to look at it is with just the definitions of the DFT and iDFT:

$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-j 2 \pi nk/N} $$

$$ x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k] \ e^{j 2 \pi nk/N} $$

the $x[n]$ that the DFT $X[k]$ is aware of is simply a periodic sequence with period of $N$. both $$ x[n+N]=x[n] $$ and $$ X[k+N]=X[k] $$ are true.

now the alternative way to think about the DFT is that it is the sampling of the spectrum of the Discrete-Time Fourier Transform (DTFT):

$$ \hat{X}\left( e^{j \omega} \right) = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] \ e^{-j \omega n} $$

where $ \hat{x}[n] = \begin{cases} 0, & n < 0 \\ x[n], & \text{for }0 \le n < N \\ 0, & N \le n \end{cases} $

and this is the same as multiplying by a rectangular window.

$$ \hat{x}[n] \ = \ x[n] \ w[n] $$

where $ w[n] = \begin{cases} 0, & n < 0 \\ 1, & \text{for }0 \le n < N \quad \quad \\ 0, & N \le n \end{cases} $ (length-$N$ rectangular window)

but here you cannot tell that you extracted $\hat{x}[n]$ from a periodic $x[n]$ or if those $N$ samples came from an infinitely long and non-repeating stream of samples. whatever it is, $\hat{x}[n]$ is non-repeating, infinite in length (with all those zeros padded), and you can apply the DTFT to it.

it doesn't have to be the rectangular window, it could be any length-$N$, finite-valued window for $w[n]$ and the math that follows is the same.

now with that definition of $\hat{x}[n]$ and $\hat{X}\left( e^{j \omega} \right)$, it's not hard to show that

$$ \hat{X}\left( e^{j \omega} \right) \Bigg|_{\omega = 2 \pi \frac{k}{N}} = \hat{X}\left( e^{j 2 \pi k/N} \right) = X[k] $$

now, what happens in either domain, is that when you uniformly sample the function (say, in this case, the spectrum $\hat{X}\left( e^{j \omega} \right)$) in one domain, it causes in the other domain, the signal ($\hat{x}[n]$) to be repeatedly shifted (by $N$ samples) and overlap-added:

$$ x[n] = \sum\limits_{m=-\infty}^{+\infty} \hat{x}[n+mN] $$

it is here where the discontinuity between $x[N-1]$ and $x[N]$ (which is the same as $x[0]$) becomes a concern, and tapered windowing to reduce or eliminate this discontinuity is perhaps indicated.

Alright, the way to think about the rectangular window (as well as all of the other windows) is that you begin with an infinite sequence of samples and you multiply the samples outside the window with 0 and the samples within the window with something non-zero (and usually, but not always, positive).

So the DFT (of which the FFT is fast method of doing the DFT) does periodically extend the $N$ samples passed to it. But there are a couple of different ways of looking at it. One way to look at it is with just the definitions of the DFT and iDFT:

$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-j 2 \pi nk/N} $$

$$ x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k] \ e^{j 2 \pi nk/N} $$

The $x[n]$ that the DFT $X[k]$ is aware of is simply a periodic sequence with period of $N$. both $$ x[n+N]=x[n] $$ and $$ X[k+N]=X[k] $$ are true.

Now the alternative way to think about the DFT is that it is the sampling of the spectrum of the Discrete-Time Fourier Transform (DTFT):

$$ \hat{X}\left( e^{j \omega} \right) = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] \ e^{-j \omega n} $$

where $ \hat{x}[n] = \begin{cases} 0, & n < 0 \\ x[n], & \text{for }0 \le n < N \\ 0, & N \le n \end{cases} $

and this is the same as multiplying by a rectangular window.

$$ \hat{x}[n] \ = \ x[n] \ w[n] $$

where $ w[n] = \begin{cases} 0, & n < 0 \\ 1, & \text{for }0 \le n < N \quad \quad \\ 0, & N \le n \end{cases} $ (length-$N$ rectangular window)

But here you cannot tell that you extracted $\hat{x}[n]$ from a periodic $x[n]$ or if those $N$ samples came from an infinitely long and non-repeating stream of samples. Whatever it is, $\hat{x}[n]$ is non-repeating, infinite in length (with all those zeros padded), and you can apply the DTFT to it.

It doesn't have to be the rectangular window, it could be any length-$N$, finite-valued window for $w[n]$ and the math that follows is the same.

Now with that definition of $\hat{x}[n]$ and $\hat{X}\left( e^{j \omega} \right)$, it's not hard to show that

$$ \hat{X}\left( e^{j \omega} \right) \Bigg|_{\omega = 2 \pi \frac{k}{N}} = \hat{X}\left( e^{j 2 \pi k/N} \right) = X[k] $$

Now, what happens in either domain, is that when you uniformly sample the function (say, in this case, the spectrum $\hat{X}\left( e^{j \omega} \right)$) in one domain, it causes in the other domain, the signal ($\hat{x}[n]$) to be repeatedly shifted (by $N$ samples) and overlap-added:

$$ x[n] = \sum\limits_{m=-\infty}^{+\infty} \hat{x}[n+mN] $$

It is here where the discontinuity between $x[N-1]$ and $x[N]$ (which is the same as $x[0]$) becomes a concern, and tapered windowing to reduce or eliminate this discontinuity is perhaps indicated.

alright, the way to think about the rectangular window (as well as all of the other windows) is that you begin with an infinite sequence of samples and you multiply the samples outside the window with 0 and the samples within the window with something non-zero (and usually, but not always, positive).

so the DFT (of which the FFT is fast method of doing the DFT) does periodically extend the $N$ samples passed to it. but there are a couple of different ways of looking at it. one way to look at it is with just the definitions of the DFT and iDFT:

$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-j 2 \pi nk/N} $$

$$ x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k] \ e^{j 2 \pi nk/N} $$

the $x[n]$ that the DFT $X[k]$ is aware of is simply a periodic sequence with period of $N$. both $$ x[n+N]=x[n] $$ and $$ X[k+N]=X[k] $$ are true.

now the alternative way to think about the DFT is that it is the sampling of the spectrum of the Discrete-Time Fourier Transform (DTFT):

$$ \hat{X}\left( e^{j \omega} \right) = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] \ e^{-j \omega n} $$

where $ \hat{x}[n] = \begin{cases} 0, & n < 0 \\ x[n], & \text{for }0 \le n < N \\ 0, & N \le n \end{cases} $

and this is the same as multiplying by a rectangular window.

$$ \hat{x}[n] \ = \ x[n] \ w[n] $$

where $ w[n] = \begin{cases} 0, & n < 0 \\ 1, & \text{for }0 \le n < N \quad \quad \\ 0, & N \le n \end{cases} $ (length-$N$ rectangular window)

but here you cannot tell that you extracted $\hat{x}[n]$ from a periodic $x[n]$ or of those $N$ samples came from an infinitely long and non-repeating stream of samples. whatever it is, $\hat{x}[n]$ is non-repeating, infinite in length (with all those zeros padded), and you can apply the DTFT to it.

it doesn't have to be the rectangular window, it could be any length-$N$, finite-valued window for $w[n]$ and the math that follows is the same.

now with that definition of $\hat{x}[n]$ and $\hat{X}\left( e^{j \omega} \right)$, it's not hard to show that

$$ \hat{X}\left( e^{j \omega} \right) \Big|_{\omega = 2 \pi \frac{k}{N}} = \hat{X}\left( e^{j 2 \pi k/N} \right) = X[k] $$

now, what happens in either domain, is that when you uniformly sample the function (say, in this case, the spectrum $\hat{X}\left( e^{j \omega} \right)$) in one domain, it causes in the other domain, the signal ($\hat{x}[n]$) to be repeatedly shifted (by $N$ samples) and overlap-added:

$$ x[n] = \sum\limits_{m=-\infty}^{+\infty} \hat{x}[n+mN] $$

it is here where the discontinuity between $x[N-1]$ and $x[n]$ (which is the same as $x[0]$) becomes a concern, and tapered windowing to reduce or eliminate this discontinuity is perhaps indicated.

alright, the way to think about the rectangular window (as well as all of the other windows) is that you begin with an infinite sequence of samples and you multiply the samples outside the window with 0 and the samples within the window with something non-zero (and usually, but not always, positive).

so the DFT (of which the FFT is fast method of doing the DFT) does periodically extend the $N$ samples passed to it. but there are a couple of different ways of looking at it. one way to look at it is with just the definitions of the DFT and iDFT:

$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-j 2 \pi nk/N} $$

$$ x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k] \ e^{j 2 \pi nk/N} $$

the $x[n]$ that the DFT $X[k]$ is aware of is simply a periodic sequence with period of $N$. both $$ x[n+N]=x[n] $$ and $$ X[k+N]=X[k] $$ are true.

now the alternative way to think about the DFT is that it is the sampling of the spectrum of the Discrete-Time Fourier Transform (DTFT):

$$ \hat{X}\left( e^{j \omega} \right) = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] \ e^{-j \omega n} $$

where $ \hat{x}[n] = \begin{cases} 0, & n < 0 \\ x[n], & \text{for }0 \le n < N \\ 0, & N \le n \end{cases} $

and this is the same as multiplying by a rectangular window.

$$ \hat{x}[n] \ = \ x[n] \ w[n] $$

where $ w[n] = \begin{cases} 0, & n < 0 \\ 1, & \text{for }0 \le n < N \quad \quad \\ 0, & N \le n \end{cases} $ (length-$N$ rectangular window)

but here you cannot tell that you extracted $\hat{x}[n]$ from a periodic $x[n]$ or if those $N$ samples came from an infinitely long and non-repeating stream of samples. whatever it is, $\hat{x}[n]$ is non-repeating, infinite in length (with all those zeros padded), and you can apply the DTFT to it.

it doesn't have to be the rectangular window, it could be any length-$N$, finite-valued window for $w[n]$ and the math that follows is the same.

now with that definition of $\hat{x}[n]$ and $\hat{X}\left( e^{j \omega} \right)$, it's not hard to show that

$$ \hat{X}\left( e^{j \omega} \right) \Bigg|_{\omega = 2 \pi \frac{k}{N}} = \hat{X}\left( e^{j 2 \pi k/N} \right) = X[k] $$

now, what happens in either domain, is that when you uniformly sample the function (say, in this case, the spectrum $\hat{X}\left( e^{j \omega} \right)$) in one domain, it causes in the other domain, the signal ($\hat{x}[n]$) to be repeatedly shifted (by $N$ samples) and overlap-added:

$$ x[n] = \sum\limits_{m=-\infty}^{+\infty} \hat{x}[n+mN] $$

it is here where the discontinuity between $x[N-1]$ and $x[N]$ (which is the same as $x[0]$) becomes a concern, and tapered windowing to reduce or eliminate this discontinuity is perhaps indicated.

alright, the way to think about the rectangular window (as well as all of the other windows) is that you begin with an infinite sequence of samples and you multiply the samples outside the window with 0 and the samples within the window with something non-zero (and usually, but not always, positive).

so the DFT (of which the FFT is fast method of doing the DFT) does periodically extend the $N$ samples passed to it. but there are a couple of different ways of looking at it. one way to look at it is with just the definitions of the DFT and iDFT:

$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-j 2 \pi nk/N} $$

$$ x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k] \ e^{j 2 \pi nk/N} $$

the $x[n]$ that the DFT $X[k]$ is aware of is simply a periodic sequence with period of $N$. both $$ x[n+N]=x[n] $$ and $$ X[k+N]=X[k] $$ are true.

now the alternative way to think about the DFT is that is is a sampling of the DTFT:

$$ \hat{X}\left( e^{j \omega} \right) = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] \ e^{-j \omega n} $$

where $ \hat{x}[n] = \begin{cases} 0, & n < 0 \\ x[n], & \text{for }0 \le n < N \\ 0, & N \le n \end{cases} $

(the rectangular window).

now with that definition of $\hat{x}[n]$ and $\hat{X}\left( e^{j \omega} \right)$, it's not hard to show that

$$ \hat{X}\left( e^{j \omega} \right) \Big|_{\omega = 2 \pi \frac{k}{N}} = \hat{X}\left( e^{j 2 \pi k/N} \right) = X[k] $$

now, what happens in either domain, is that when you uniformly sample the function (say, in this case, the spectrum $\hat{X}\left( e^{j \omega} \right)$) in one domain, it causes in the other domain, the signal ($\hat{x}[n]$) to be repeatedly shifted (by $N$ samples) and overlap-added:

$$ x[n] = \sum\limits_{m=-\infty}^{+\infty} \hat{x}[n+mN] $$

it is here where the discontinuity between $x[N-1]$ and $x[n]$ (which is the same as $x[0]$) becomes a concern, and tapered windowing to reduce or eliminate this discontinuity is perhaps indicated.

alright, the way to think about the rectangular window (as well as all of the other windows) is that you begin with an infinite sequence of samples and you multiply the samples outside the window with 0 and the samples within the window with something non-zero (and usually, but not always, positive).

so the DFT (of which the FFT is fast method of doing the DFT) does periodically extend the $N$ samples passed to it. but there are a couple of different ways of looking at it. one way to look at it is with just the definitions of the DFT and iDFT:

$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-j 2 \pi nk/N} $$

$$ x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k] \ e^{j 2 \pi nk/N} $$

the $x[n]$ that the DFT $X[k]$ is aware of is simply a periodic sequence with period of $N$. both $$ x[n+N]=x[n] $$ and $$ X[k+N]=X[k] $$ are true.

now the alternative way to think about the DFT is that it is the sampling of the spectrum of the Discrete-Time Fourier Transform (DTFT):

$$ \hat{X}\left( e^{j \omega} \right) = \sum\limits_{n=-\infty}^{+\infty} \hat{x}[n] \ e^{-j \omega n} $$

where $ \hat{x}[n] = \begin{cases} 0, & n < 0 \\ x[n], & \text{for }0 \le n < N \\ 0, & N \le n \end{cases} $

and this is the same as multiplying by a rectangular window.

$$ \hat{x}[n] \ = \ x[n] \ w[n] $$

where $ w[n] = \begin{cases} 0, & n < 0 \\ 1, & \text{for }0 \le n < N \quad \quad \\ 0, & N \le n \end{cases} $ (length-$N$ rectangular window)

but here you cannot tell that you extracted $\hat{x}[n]$ from a periodic $x[n]$ or of those $N$ samples came from an infinitely long and non-repeating stream of samples. whatever it is, $\hat{x}[n]$ is non-repeating, infinite in length (with all those zeros padded), and you can apply the DTFT to it.

it doesn't have to be the rectangular window, it could be any length-$N$, finite-valued window for $w[n]$ and the math that follows is the same.

now with that definition of $\hat{x}[n]$ and $\hat{X}\left( e^{j \omega} \right)$, it's not hard to show that

$$ \hat{X}\left( e^{j \omega} \right) \Big|_{\omega = 2 \pi \frac{k}{N}} = \hat{X}\left( e^{j 2 \pi k/N} \right) = X[k] $$

now, what happens in either domain, is that when you uniformly sample the function (say, in this case, the spectrum $\hat{X}\left( e^{j \omega} \right)$) in one domain, it causes in the other domain, the signal ($\hat{x}[n]$) to be repeatedly shifted (by $N$ samples) and overlap-added:

$$ x[n] = \sum\limits_{m=-\infty}^{+\infty} \hat{x}[n+mN] $$

it is here where the discontinuity between $x[N-1]$ and $x[n]$ (which is the same as $x[0]$) becomes a concern, and tapered windowing to reduce or eliminate this discontinuity is perhaps indicated.