Equalizer and Timing recovery

Started by Unknown March 10, 2009
On Wed, 11 Mar 2009 22:14:51 -0500, "steveu" <steveu@coppice.org>

>>On Wed, 11 Mar 2009 00:32:01 -0500, "steveu" <steveu@coppice.org> >>wrote: >> >>>>On Tue, 10 Mar 2009 13:13:15 -0500, Vladimir Vassilevsky >>>><antispam_bogus@hotmail.com> wrote: >>>> >>>>> >>>>> >>>>>Eric Jacobsen wrote: >>>>> >>>>> >>>>>>>Of course. QPSK has less variation from symbol to symbol then BPSK, >so >>> >>>>>>>the timing detector produces less of useful signal and more of the >>>noise. >>>>>> >>>>>> >>>>>> Uh...how can this be true? The channels are orthogonal and >>>processed >>>>>> identically (typically) between QPSK and BPSK. Wouldn't a QPSK >>>>>> signal just be two BPSK signals with orthogonality? >>>>> >>>>>Same power, same symbol rate: (x^2 + y^2) is less then (x+y)^2 >>>> >>>>That's a bit odd way of looking at it, I think. The question was >>>>synchronization rather than BER. I've never noticed a difference in >>>>the ability to achieve timing synch between BPSK and QPSK. >>>> >>>>In many systems the timing detector only looks at one of either I or >>>>Q, so it really can't tell whether the system is BPSK or QPSK (and >>>>I've never seen it make a difference in a BER vs Eb/N0 plot, i.e., the >>>>implementation loss was the same). >>> >>>Why is it odd? The BER is worse because the noise is bigger in >proportion >>>to the signal, and this affects both carrier recovery and symbol timing >>>recovery (at least in terms of getting a quick initial lock) just as >much >>>as it does the BER. >>> >>>Steve >> >>As I suspect you know, BER vs Eb/No is the same for BPSK and QPSK. >>Likewise I've not seen a case where a properly designed modem has a >>different implementation loss whether BPSK or QPSK is selected, and >>this is with the exact same TED and timing loop for both cases. >> >>That suggests to me that the timing synchronization doesn't really >>care whether the signal is BPSK or QPSK. That makes sense to me, >>since the TED sees the same waveform in either case. > >You are assuming the two components of QPSK are orthogonal. That is only >true after the equaliser has cleaned them the incoming signal. The TED >normally occurs before the equalizer, in a domain where things may be far >from orthogonal. > >Steve
That's certainly true. In the presence of distortion the Q channel may become an interferer to I and vice-versa. In AWGN or a channel that didn't cause cross-channel distortion, though, I don't think it would matter. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
On Wed, 11 Mar 2009 21:25:46 -0700 (PDT), fpgaasicdesigner@gmail.com

>On Mar 10, 2:05&#4294967295;pm, Eric Jacobsen <eric.jacob...@ieee.org> wrote: >> On Mon, 9 Mar 2009 20:49:42 -0700 (PDT), fpgaasicdesig...@gmail.com >> wrote: >> >> >Hi, >> >> >First question: >> >I wonder if I should place the equalizer before or after the Gardner >> >Timing Recovery loop. >> >It is about a BPSK/QPSK digital demod. >> >If the equalizer goes before the TED, it should works at 2 samples/ >> >symbol and after it can works at 1 sample by symbol. >> >All this will be done before carrier recovery. >> >I guess if I do the equalizer before it will help the TED to lock and >> >have good jitter. >> >> As has been pointed out, generally the EQ comes after timing >> synchronization. &#4294967295; Otherwise they tend to fight each other as both try >> to steer timing. &#4294967295; If you have the time and inclination try putting >> your EQ prior to timing synchronization and just see what happens. It >> can be fun to see these things for yourself. >> >> >My second question is about the Gardner loop (without any equalizer): >> >I have a good 15ns jitter on BPSK and with the same loop filter I have >> >100ns for QPSK. With 500ns symbol period. >> >Why this ? I have a 0.25 roll-off RRC filter (Tx and Rx). Perhaps I >> >cannot have the same tracking performance in BPSK and QPSK with the >> >same filter ? >> >> Depending on which Gardner detector you're using (there are more than >> one, IIRC), there may be a phase sensitivity that gets worse with QPSK >> since the phase changes are smaller. &#4294967295; Are you locking the carrier >> phase or not? &#4294967295; If you are, there should be no difference between QPSK >> and BPSK, if you're not, then the smaller phase deviations in the >> rotating constellation can cause events that perturb some timing error >> detectors. >> >> Many, if not most, timing error detectors that claim to be >> "rotationally invariant" really aren't. >> >> Eric Jacobsen >> Minister of Algorithms >> Abineau Communicationshttp://www.ericjacobsen.org >> >> Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php > >I talk about x[k-0.5](Signx[k-1]-x[k+1]); x=I+jQ >Is there other TED ?
In a practical modem the signum function creates enough noise in the detector when the constellation dwells near the coordinate axes that synchronization will be lost. Been there, done that. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php