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