NRZ-OOK Unipolar signal. 1bit/symbol. For bit rate: Rb=500Mb/s,

Is the bandwidth Rb/2=250MHz?

What is the minimum sampling rate required at the receiver?

Is it 2B (500Msamples/s) or at least 2Rb (1Gsamples/s)?

Cheers

bandwidth of 250MHz is enough to carry info. actual bandwidth is higher but you filter it to 250 (pulse shaping).

minimum Fs at receiver is 2B if signal is centred on zero

OK, thanks.

2B (1Rb) if the sampling is synchronized to the incoming bits, yes. But if you need to do bit synchronization then you need to sample faster, and the slower you sample the better your analog filters need to be.

The post is not about Fs at Tx but BW.

Normally pulse shaping filter is used which implies upsampling.

At Rx still the minimum is 2B but a higher Fs means less burden on processing the signal.

If the signal is 500Mbps NRZ, and you sample the receiver at 500Msps and you don't happen to get the receive samples lined up with the peaks of the symbols, then you won't be receiving optimally. If you happen to get the receive samples lined up with the transitions, you won't be able to receive useful information at all.

I'm not sure why you discard this so cavalierly. What sort of a system are you considering where sampling at the correct instants just happens by magic?

read about timing and symbol recovery in BPSK/QPSK receivers and in particular fractional interpolation

NRZ and OOK are mutually contradictory. Is your signal on (not zero) and off (return to zero), or is it something else?

Non-polar Non Return to Zero (NRZ) means that the signal toggles for one logic level input (like 1) and stays the same for the other logic level input. So you could have On-Off Keying with NRZ encoding. Granted, I don't know what the point would be but you could do it.

Seems like there are a few variations on the theme:

https://en.wikipedia.org/wiki/Non-return-to-zero