Why does the bandwidth of a signal need to be half of the sampling rate?

Question

Suppose I perform a DFT on some function with sampling rate of $\frac{1}{\Delta t}$. According to this page, the bandwidth, which is the maximum frequency that can be analyzed when performed the DFT, is half of the sampling rate, i.e. $\frac{1}{2\Delta t}$. Why is this true and how can we prove this mathematically?

Answer 1

This is copied from another answer. It needs editing. Just be patient.

we can uniformly sample $x(t)$ if the sample rate, $f_\text{s} \triangleq \tfrac{1}{T} > 2B $, is sufficiently high

$$\begin{align} x_\text{s}(t) &= x(t) \cdot T \, \mathbf{III}_T(t) \\ &= x(t) \cdot T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ &= T \sum\limits_{n=-\infty}^{\infty} x(t) \, \delta(t - nT) \\ &= T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \delta(t - nT) \\ &= T \sum\limits_{n=-\infty}^{\infty} x[n] \, \delta(t - nT) \\ \end{align}$$

it is also true that the sampling function is periodic and has a Fourier series.

$$\begin{align} T \, \mathbf{III}_T(t) &\triangleq T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ &= \sum\limits_{k=-\infty}^{\infty} e^{j 2 \pi k f_\text{s} t} \\ \end{align}$$

turns out that all of the Fourier series coefficients are 1. this means that the uniform sampled function is

$$\begin{align} x_\text{s}(t) &= x(t) \cdot T \, \mathbf{III}_T(t) \\ &= x(t) \cdot T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ &= x(t) \sum\limits_{k=-\infty}^{\infty} e^{j 2 \pi k f_\text{s} t} \\ &= \sum\limits_{k=-\infty}^{\infty} x(t) \, e^{j 2 \pi k f_\text{s} t} \\ \end{align}$$

accordingly, taking the continuous Fourier Transform, the spectrum of the sampled signal is

$$\begin{align} X_\text{s}(f) & \triangleq \mathscr{F} \Big\{ x_\text{s}(t) \Big\} \\ &= \mathscr{F} \left\{ \sum\limits_{k=-\infty}^{\infty} x(t) \, e^{j 2 \pi k f_\text{s} t} \right\} \\ &= \sum\limits_{k=-\infty}^{\infty} \mathscr{F} \Big\{ x(t) \, e^{j 2 \pi k f_\text{s} t} \Big\} \\ &= \sum\limits_{k=-\infty}^{\infty} X(f - k f_\text{s}) \\ \end{align}$$

and we know, as long as $B < \tfrac12 f_\text{s}$, that there is no overlap in the adjacent shifted spectra of $X(f)$ and the original $X(f)$ can be recovered from the $k=0$ term of the summation.

$$\begin{align} X(f) &= \Pi\left( \tfrac{f}{f_\text{s}} \right) \, \sum\limits_{k=-\infty}^{\infty} X(f - k f_\text{s}) \\ &= \Pi\left( \tfrac{f}{f_\text{s}} \right) \, X_\text{s}(f) \\ \end{align}$$

where $\Pi(u)$ (sometimes "$\operatorname{rect}(u)$") is the rectangular function

$$\Pi(u) \triangleq \begin{cases} 1 \qquad & \text{ if } |u| < \tfrac12 \\ \tfrac12 \qquad & \text{ if } |u| = \tfrac12 \\ 0 \qquad & \text{ if } |u| > \tfrac12 \\ \end{cases}$$

and we know that the inverse Fourier transform is

$$ \mathscr{F}^{-1} \left\{ \Pi\left( \tfrac{f}{f_\text{s}} \right) \right\} = f_\text{s} \, \operatorname{sinc}(f_\text{s} t) $$

where the sinc function is

$$\operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} \qquad & \text{ if } |u| \ne 0 \\ 1 \qquad & \text{ if } |u| = 0 \\ \end{cases}$$

then, remembering that $f_\text{s}=\tfrac1T $, we know that the output of the brickwall reconstruction filter is

$$\begin{align} X(f) &= \Pi\left( \tfrac{f}{f_\text{s}} \right) \, X_\text{s}(f) \\ & \iff \\ x(t) &= f_\text{s} \, \operatorname{sinc}(f_\text{s} t) \ \circledast \ x_\text{s}(t) \\ &= f_\text{s} \, \operatorname{sinc}(f_\text{s} t) \ \circledast \ T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \delta(t - nT) \\ &= f_\text{s} \, T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \big( \operatorname{sinc}(f_\text{s} t) \ \circledast \ \delta(t - nT) \big) \\ &= \sum\limits_{n=-\infty}^{\infty} x(nT) \, \operatorname{sinc}\big( f_\text{s} (t - nT) \big) \\ &= \sum\limits_{n=-\infty}^{\infty} x(nT) \, \operatorname{sinc}\big( f_\text{s} t - n \big) \\ \end{align}$$

that's how we reconstruct out original $x(t)$ out of the samples $x(nT)$. so much for the sampling and reconstruction theorem.