1
$\begingroup$

I am trying to design a matched filter at the receiver side to a half sine pulse shaping function sent by the transmitter of an OQPSK transceiver. I am not sure I understand the effect of varying the over sampling factor of the ADC right before the matched filter (the minimum is twice the symbol rate). I think it may have something to do with the quality of the timing synchronization. I'm using an early-late gate synchronization algorithm, which needs at least 3 samples per symbol. I've also read in literature that the over sampling is typically a factor of 8 for QPSK systems. Does it have other effects, maybe the BER of the system?

$\endgroup$
1
$\begingroup$

Perhaps the answer to a similar question may be helpful.

I've spent most of my career in SONAR, so I want to see a lot of oversampling After a matched filter.

A matched filter should be sampled at the exact time corresponding to it's delay, which is unknown but can be within a range of time that is often called a range gate. Any filter output that corresponds to a maximum deflection inside of that range gate is a possible return. The nearer those samples are in time the more likely a match in delay will occur, and the nearer to the actual time, the greater the expected deflection. So, in this case matched filter miss match can be mitigated with a higher sample rate.

The hard part is calculating the performance because the range gate is moving in time so there are several chances to pick a maximum that is a signal and several chances to produce a false alarm and the tests are on overlapping data, so one's tests are not independent. The detector also has memory because the probability of a threshold exceeding is different if the previous test exceeded the threshold or did't exceed a threshold. It does not produce simple closed form expressions.

So oversampling should increase performance but showing by how much, is typically done by simulation, and results will depend on the details of the filter. Doppler introduces more complication.

Hopefully, the answer to a question you didn't ask may help.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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