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I'm given a continuous-time analog signal $x_a(t) = \cos(2\pi f_1t)+\sin(2\pi f_2t)$, for some frequency $f_1, f_2$. I'm asked to sample $x_a(t)$ at $F_s=1024\textrm{ Hz}$, apply a 128-point Hamming Window, and take the DFT(Discrete Fourier Transform) of the resulting windowed, sampled signal.

The expression for a sampled signal is:

$$x[n] = x_a\left(\dfrac{n}{F_s}\right)$$

To apply the hamming window, we multiply $x[n]$ by the 128-point window function (courtesy of ML):

$$x_w[n]=x[n].*\mathrm{hamming}[128]$$

We then take the DFT of the windowed function $x_w[n]$:

$$X_w[k] = \mathrm{DFT}\left\{x_w[n]\right\}$$

I've read that the DTFT(Discrete-time Fourier transform) is a continuous spectrum of $x[n]$ and that the DFT of $x[n]$ (in a nutshell) is a sampling of that spectrum. When we take the $N$-point DFT of $x[n]$, we are taking $N$ samples of the DTFT of $x[n]$ where frequencies are unique (i.e., we sample within $2\pi$). That is:

\begin{align} \mathrm{DTFT}\left\{x[n]\right\} &= X(\omega)\\ \mathrm{DFT}\left\{x[n]\right\} &= X\left(\omega = \frac{2\pi k}{N}\right) = X[k],\quad \textrm{ for } k=0,1,...,N-1. \end{align}

My questions:

  1. If we plot $X[k]$, we are still plotting the sample of the spectrum against frequency, right? Is the frequency, given $k$, denoted by $2\pi k/N $ radians?

  2. My professor asked me to plot the FFT versus $k/F_s$ where $k=0,\ldots,127$. What does $k/F_s$ represent?

  3. A homework problem asks (given application of the 128-point Hamming window, above) to find the "continuous-time frequency spacing $\Delta F$ between DFT samples"? What does this mean? Isn't this just $2\pi/N$?

    My professor told me that $2\pi/N$ is the discrete-time frequency spacing in radians and that I have to multiply that by $Fs/(2 \pi)$. I'm not sure how to make sense of $Fs/N$.

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  • $\begingroup$ I thought about Q3 some more. I recalled that $2\pi F_s$ represents the highest frequency that can be represented when sampling a signal at Fs Hertz. Therefore, Fs/N should represent the spacing between two samples. We divide by (2*pi) to convert units: rads/sample to Hz/Sample. Is this right? $\endgroup$ – Minh Tran Mar 7 '16 at 21:36
  • $\begingroup$ any tick for the answer ? $\endgroup$ – Gilles Aug 10 '16 at 8:58
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As you've noted, for a length-$N$ time-domain sequence $x[n]$, the $\mathrm{DTFT}$ (aka Fourier transform) is given by the equation:

$$ \mathrm{DTFT}\left\{x[n]\right\}=X(e^{j\omega})=X(\omega), \quad 0\leq n\leq N - 1.\tag{1} $$

And the $N$-point $\mathrm{DFT}$ is given by uniformly sampling equation $(1)$ to obtain $N$ equally spaced frequencies $\omega_k = 2\pi k/N$:

$$ \mathrm{DFT}\left\{x[n]\right\}=X(e^{j\omega_k})=X\left(\frac{2\pi k}{N}\right)=X[k], \quad 0\leq k\leq N - 1.$$

  • If you plot $X[k]$, you're plotting the $N$ equally spaced samples from the $\mathrm{DFT}$ equation not against the continuous frequency $\omega$ but against the index number $k$ which in frequency corresponds to $2\pi k/N$.
  • The $\mathrm{FFTs}$ are algorithms to compute the $\mathrm{DFT}$. I don't know what's the goal in plotting the $\mathrm{DFT}$ against $k/F_s$. If you check the units you see that $k$ is just a bin number, unit-less, but $F_s$ is in Samples per second. So, that division has units in Seconds per samples...
  • The spacing between the discrete frequencies is indeed $2\pi/N$ radians. By the continuous-time frequency spacing, I think he meant the corresponding frequency spacing in $\rm Hz$ and this is the frequency resolution: $$f_k=k\cdot \frac{F_s}{N}$$ From here you can guess what's the frequency spacing since two consecutive bins from the $\mathrm{DFT}$ samples are at $\Delta k=1$.
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