0
$\begingroup$

So I have a very basic on which I couldn't find much details anywhere.

I'll start with an example:

I have to transmit 100Mbps of BPSK data. Roll-off factor is 0.25. This data will be processed at baseband. For the receiver recovery algorithms, lets say I need a minimum 2.5 samples per symbol. At TX: I will first pulse shape the bits using SRRC, and then interpolate to get a final rate of 250MSamples per second.

After this is where i get lost:

  1. Is the bandwidth of the signal 100MHz or 125MHz or 250Mhz, and is it talked about in terms of baseband or passband after the analog filtering?
  2. Where does the nyquist criteria come into picture and how, baseband or passband?
  3. Do I need a DAC with sampling rate of 250 MSPS or 500 MSPS, how is it calculated?
  4. Is it possible for me to use a DAC corresponding to only needing 2 samples per symbol, while at the ADC side I can increase the sampling rate and sample 2.5 or even 4 samples per symbol?
$\endgroup$

1 Answer 1

1
$\begingroup$

The two-sided bandwidth would be 125MHz. (+/-62.5 MHz). This is given as the symbol rate (which is the data rate for BPSK) times one plus the roll-off factor (so $100(1.25)=125$). At RF the occupied bandwidth would be 125 MHz and at baseband as a low pass signal extending from DC to 62.5 MHz which is referred to as the one-sided bandwidth.

If sampling at baseband, with two ADC’s for receiver (for I and Q) and one DAC for transmitter (only need I if BPSK) then the minimum sampling rate to just meet Nyquist would be 125 MHz. In practice we need more than that allow for realizable filtering (20% more or so as a trade with filter complexity). Given the waveform is BPSK, you can use I only for the transmitter (single DAC) but you will need two in the receiver given inevitable offsets in phase and frequency between the transmitter and receiver. The complex receiver baseband would allow for phase measurement from symbol to symbol and with that carrier offset correction can be implemented. Another option for the receiver is to use a real digital IF frequency and the do the downconversion to baseband digitally. In this case the IF frequency must be greater than 62.5 MHz with additional margin to allow for filtering. A 125 MHz digital IF and 500 MHz sampling rate would be a reasonable choice.

The minimum sampling rate in the Tx is 125 MHz plus room for filtering as described. Two samples per symbol (200 MHz) would be a reasonable choice, and may simplify the transmitter timing given the integer multiple of the symbol rate. This decision is a trade between digital complexity and power consumption and the analog reconstruction filter after the DAC. Using a lower number of samples per symbol has the advantage of a longer time span in the pulse shaping filter, resulting in a higher stopband rejection given the same number of coefficients, but together with that a tighter analog reconstruction filter is also needed given the closer proximity of images at the output of the DAC. For a specific demonstration of this, see DSP.SE #51088. One common solution is to interpolate to a higher rate after the pulse shaping filter as the interpolation filter there can often be simpler in complexity than the pulse shaping filter used.

The sampling rate used in the Rx can be completely different than Tx if desired. (In between the waveform is analog so the receiver can sample that waveform to meet its own requirements regardless of how it was sampled in the transmitter).

$\endgroup$
3
  • $\begingroup$ Thanks a lot for the response. It clears up a lot of things. How does the nyquist criteria, samples per symbol and the sampling rate of DAC relate? Correct me if I am wrong, nyquist criteria dictates the minimum sampling frequency, that is 125MSPS, generally for easing the analog filtering we modify the nyquist criteria from 2 to a 2.5(156.25MSPS). Now the samples per symbol of the DSP are not related to this "Nyquist frequency", for BPSK I have 100MSymbols per second, I interpolate this to 200MSamples per second. Hence, 156MSPS < DAC = 200MSPS(sufficient for this example)? $\endgroup$
    – Rituj
    Mar 13 at 6:25
  • $\begingroup$ Yes 156.25 MSamples per second would be a reasonable sampling rate. If the symbol rate is 100 MSymbols/sec and we sample at 156.25 M Samples/sec, then the number of Sample per Symbol is 156.25 Samples/sec divided by 100 M Symbols / sec. If you keep track of all the variables you have Samples/sec times sec/Symbol and you get 1.5625 Samples/Symbol $\endgroup$ Mar 13 at 14:33
  • $\begingroup$ You could sample the DAC at 100 samples per symbol and sample the ADC at 2 samples per symbol. They are completely independent. You just need to meet Nyquist in both cases (assuming we don't have perfect timing and synchronization in the receiver at the point where the received waveform is sampled). $\endgroup$ Mar 13 at 14:36

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.