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.


1 Answer 1


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.

On the other hand, the symbol duration is more linked to the data rate. If you a have a source that outputs 10 Gbits per second, and you use a qpsk (2 bits per symbol) this will lead to a symbol rate of 5Gsymbol per second or 5 Gbaud.

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

  • $\begingroup$ Thanks - that makes sense, What I'm struggling with is the symbol duration and the impact on bandwidth of different systems $\endgroup$
    – BivvyH
    Nov 26, 2023 at 16:07
  • $\begingroup$ Thanks - that makes sense, What I'm struggling with is the symbol duration and the impact on bandwidth of different systems With a 20MHz sampling frequency, Shannon says the bit rate 10Mbps Now with QPSK, we have 2 bits per symbol, so is the symbol duration 2/10M? Increasing the alphabet to 4 bits per symbol, is the symbol duration 4/10M? I'm trying to model this in Matlab and the bandwidth of the signal seems to increase as I reduce the alphabet size which seems counter-intuitive $\endgroup$
    – BivvyH
    Nov 26, 2023 at 16:13
  • $\begingroup$ eg, with a constellation of 10 bits per symbol, 10 Gbit/s will result in 1 Gsymbol/s, as each symbol contains 10 bits. Ideally, a 10 Gbaud signal will occupy more or less 10 GHz. Now, when you decrease the alphabet of the modulation, say 2 bits per symbol, you end up with a symbol rate of 5GBaud, which will occupy more or less 5GHz of bandwidth. At fixed bit rate, the higher the modulation, the smaller the occupied bandwidth becomes, but this comes at the expense of the performance, as packing 10 bits per symbol will result in a constellation which is more sensitive to noise $\endgroup$
    – Trektrek
    Dec 11, 2023 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.