Electrical – Sampling rate for a real signal, which is not band-limited

You can google that exact question to find several variations of this answer:

A noiseless channel can carry an arbitrary large amount of information, no matter how often it is sampled.

Just send a lot of data per sample.

For 4KHz channel, make 1000 samples/sec. If each sample is 16 bits, the channel can send 16 Kbps.

If each sample is 1024 bits, the channel can send 1000 samples/sec * 1024 bits = 1024 Mbps.

The key word here is “noiseless”. With a normal 4 KHz channel, Shannon limit would not allow this.

For the 4 KHz channel we can make 8000 samples/sec. In this case if each sample is 1024 bits this channel can send 8.2 Mbps.

"even a simple UART samples a digital signal at the same speed..." the UART doesn't need to reconstruct the analog square wave signal that carries the digital information, so it doesn't take the theorem into account.

The Shannon-Nyquist theorem actually talks about the perfect representation of an analog signal. Perfect representation here means that knowing only the samples of the signal you could reconstruct perfectly the time-domain analog signal that was sampled.

Of course this is only possible in theory. In fact the reconstruction formula involves a series of "sinc" functions (\$ \mathrm{sinc}(x) = \frac{\sin(\pi x)}{\pi x}\$), which aren't time limited (they extend from \$-\infty\$ to \$+\infty\$), so they are not really implementable perfectly in hardware. High end scopes use a truncated form of that sinc function to achieve higher bandwidth capability with lower sampler rates, i.e. more MHz with less samples, because they don't simply "join the dots", so they don't need much oversampling.

But still they need some oversampling, because the sampling rate must be greater than 2B, where B is the bandwidth, and the fact that they use a truncated sinc function in the reconstruction doesn't allow to get too close to that 2B figure.