# What is the relationship between Symbol Duration and Sampling Frequency in a 256 MPSK radio link

I'm struggling to work out the relationship between Symbol Duration and Sampling Frequency in an MPSK link.

For instance, if the Sampling Frequency is 10MHz this means that a sample is taken every 10ms and this would suggest one bit every 10ms

If we have an Alphabet Size of 256, that requires log2(256) bits to encode, so 8 bits. Does this mean that the Symbol Duration is 80ms?

Do I need to consider Shannon-Nyquist? In which case my understanding is we need 2 samples per bit, and this doubles the Symbol Duration (and halves the data rate)

Where this gets even more confusing, is where the link is described in terms of its bandwidth (eg 1Gbps). In this case (as far as I can tell) the Symbol Duration is log2(256)/1e9. I've not even started to think what this means in terms of sample frequency.

Thanks - from a very confused student.

There is a difference between the information bits that you are sending and the bits that represent the samples that you obtain at the receiver. Each sample should be represented by high enough number of bits in order to capture the analog signal. Nyquist states that for a signal of bandwidth $$B$$ one should use at least $$2B$$ sample per second to represent the analog signal reliably in the digital domain. Low number of bits to represent each sample will result in higher quantification noise.
Now it happens to be that theoretically, a link with $$X$$ Gbaud will occupy a bandwidth of $$X$$ GHz. In reality it is a little bit more than that and the excess with depend on the modulation type, the shaping filters, etc