Does the DFT calculate spectral components up to half the sampling frequency, $f_s/2$?

Question

This question is prompted by a statement made in this response (reproduced below):

The DFT calculates spectral components up to $f_s/2$, no matter what the input signal is.

A book I'm reading explains that the DFT of a signal $x(n)$ is the result of the inner product of $x(n)$ with each of the sampled complex sinusoidal basis vectors $s_k(n)$. Here is the author's definition of the DFT:

The value $k$ is called the bin number and $\omega_k$ are harmonics of the sampling frequency $f_s$ (expressed in angular unit of radians, alternatively represented as $f_k = \frac{k}{N}f_s$).

Since $k$ is taken from $0$ to $N-1$, $X(\omega_k)$ is defined up to and including $f_{N-1} = \frac{N-1}{N}f_s$. Why would Deve (answer author) say $f_s/2?$

Answer 1

Depends a bit on the indexing convention, but in typically you would interpret the frequency interval as $[-f_s/2, f_s/2]$ and not as $[0,f_s]$. The DFT is periodic in N so you you have

$$X(N-1) = X(-1) $$

Keep in mind that that the sampling theorem requires the signal to be band limited to $f_s/2$ so assuming that you have actual independent information about frequencies higher than $f_s/2$ is misleading.

For real signals, it's conjugate symmetric anyway, i.e. $f_{-k} = f^*_k$ so there is only independent information on half of the spectrum anyway.

Answer 2

You are right, the DFT bins correspond to frequencies $f_k=\frac{k}{N}f_s$. In order to see this let's consider finite length signals with potentially non-zero elements in the index range $[0,N-1]$. In this case, the DFT is just a sampled version of the DTFT (discrete-time Fourier transform):

$$\textrm{DTFT:}\quad X(e^{j\omega})=\sum_{n=0}^{N-1}x[n] e^{-jn\omega}\tag{1}$$

$$\textrm{DFT:}\quad \hat{X}[k]=\sum_{n=0}^{N-1}x[n] e^{-jn\frac{2\pi k}{N}}\tag{2}$$

From $(1)$ and $(2)$ we get

$$\hat{X}[k]=X(e^{j\omega_k}),\qquad\omega_k=\frac{2\pi k}{N}\tag{3}$$

So the index $k=0$ corresponds to DC, $k=N/2$ (if $N$ is even) corresponds to Nyquist, and $k=N-1$ corresponds to just below $f_s$ (actually, $\frac{N-1}{N}f_s$).