0
$\begingroup$

I was reading MATLAB's documentation on dpskmod. They gave an example of an $\pi/8$-DQPSK. I felt that this is strange as I have seen $\pi/4$-DQSPK and $\pi/8$-D8PSK, but I have not seen $\pi/8$-DQPSK.

Attached is a screenshot of the documentation showing $\pi/8$-DQPSK with 16 clusters in the constellation:

screenshot of matlab dpskmod documentation

My question is: Is $\pi/8$-DQPSK actually used in industry or is this just a demonstration of the flexibility of MATLAB's functions?

$\endgroup$
1
$\begingroup$

I believe the correct name is pi/8 D8PSK. The ETSI TETRA Release 2 uses pi/8 D8PSK (digital trunked mobile standard for Professional Mobile Radio and Public Access Mobile Radio). The purpose of pi/4 DQPSK and pi/8 D8PSK is to have lower peak to average power ratios by not allowing the trajectories to go through the origin. Lower peak to average ratios means you can drive power amplifiers in the transmitter further into saturation, resulting in higher power efficiency (more transmit power from available DC power). pi/8 provides for higher data rates at the expense of the more SNR that would be needed to get the same error rate performance.

More info on TETRA:

(see page 40 for details on pi/8 modulation which is identical to what the OP has shown): https://www.etsi.org/deliver/etsi_tr/102500_102599/102580/01.01.01_60/tr_102580v010101p.pdf

Motorola TEDS: https://www.motorolasolutions.com/en_xu/products/tetra/teds.html

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the answer! I have checked page 40 of the ETSI pdf you have attached but that seems to show a pi/8-D8PSK instead of a pi/8-DQPSK. Are they the same thing? $\endgroup$ – tpl May 8 at 1:51
  • $\begingroup$ Yes, compare the constellation to what you posted—- it’s identical. I believe the proper name for it is D8PSK and not DQPSK $\endgroup$ – Dan Boschen May 8 at 1:58
  • $\begingroup$ I see. The constellation is the same indeed but in the matlab documentation they said that they are simulating a DQPSK. The input data are between 0 and 3. Additionally, M=4. But because they use a pi/8 phase rotation, they get 16 clusters in the constellation instead of 8. I dont feel that this example is of a 8PSK. $\endgroup$ – tpl May 8 at 2:18
  • $\begingroup$ With pi/4 DQPSK you get 8 clusters in the constellation, so this is no different- the doubling of the clusters is because of the added rotation between each symbol) which serves the purpose of avoiding the origin). With pi/8 there is a pi/8 rotation between each symbol so you will still get a physical 16PSK constellation with only 4 symbols used. This is clear if you review the modulation details on page 40, paying particular attention to how the mapping changes whether k is even or odd for any given symbol S(k) $\endgroup$ – Dan Boschen May 8 at 11:07
  • $\begingroup$ (Table 6.4 specifically- you would get the same thing with two bits at a time but the symbol you select depends on the previous symbol used; as expected for “Differential”) $\endgroup$ – Dan Boschen May 8 at 11:13

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.