In the quest for a not so expensive PC scope/logic analyzer, I have found a nice little device it looks very well done and I know it will do the job.
However looking at the specifications, I encountered this:
Bandwidth vs Sample Rate
In order to accurately record a signal, the
sample rate must be sufficiently higher in order to preserve the
information in the signal, as detailed in the Nyquist–Shannon sampling
theorem. Digital signals must be sampled at least four times faster
than the highest frequency component in the signal. Analog signals
need to be sampled ten times faster than the fastest frequency
component in the signal.
And consequently it has a sampling rate of 500MSPs but a bandwidth(filter) of 100MHz so a ratio 1:5 for digital signals and a sampling rate of 50MSPs and a bandwith(filter) of 5MHz so a ratio 1:10 for analog signals
As far as I understand Niquist-Shannon only talks about sampling at twice the maximum frequency (in theory), It is of course good not to push the limits and there are no perfect filters. but even a simple UART samples a digital signal at the same speed than the baudrate!
So is this a usual rule of thumb for sampling? or is this something someone from sales may have written? It lets me somehow clueless I have never heard about this.
Best Answer
"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.