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. 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 ? Any idea ? Suggestion, paper, book... THX.
Equalizer and Timing recovery
Started by ●March 10, 2009
Reply by ●March 10, 20092009-03-10
On Mar 9, 11:49�pm, 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.Synchronization is highly recommended to be before the equalizer.> > 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 ?Did you remember to conjugate the correct complex-valued symbol(s)?> Any idea ? Suggestion, paper, book...Myer / Moeneclaey / Fechtel is the best.
Reply by ●March 10, 20092009-03-10
fpgaasicdesigner@gmail.com wrote:> Hi, > > First question: > I wonder if I should place the equalizer before or after the Gardner > Timing Recovery loop.Equalizer should be synchronized, i.e. after the loop. If the equalizer is placed before the loop, it needs higher oversampling then x2. Also, the update procedure will be rather nontrivial.> It is about a BPSK/QPSK digital demod.BPSK/QPSK generally need no equalizer.> 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. > > 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 ?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.> 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 ?The more complex the modulation is, the more difficult is the synchronization. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●March 10, 20092009-03-10
On Tue, 10 Mar 2009 09:19:08 -0500, Vladimir Vassilevsky <antispam_bogus@hotmail.com> wrote:> > >fpgaasicdesigner@gmail.com wrote: > >> Hi, >> >> First question: >> I wonder if I should place the equalizer before or after the Gardner >> Timing Recovery loop. > >Equalizer should be synchronized, i.e. after the loop. If the equalizer >is placed before the loop, it needs higher oversampling then x2. Also, >the update procedure will be rather nontrivial. > >> It is about a BPSK/QPSK digital demod. > >BPSK/QPSK generally need no equalizer.Unless, of course, the channel is distorted and you want that performance loss back.>> 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. >> >> 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 ? > >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?>> 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 ? > >The more complex the modulation is, the more difficult is the >synchronization.Sometimes, but as modulation orders increase so does SNR. Sometimes that makes it easier. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
Reply by ●March 10, 20092009-03-10
On Mon, 9 Mar 2009 20:49:42 -0700 (PDT), fpgaasicdesigner@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. Otherwise they tend to fight each other as both try to steer timing. 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. Are you locking the carrier phase or not? 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 Communications http://www.ericjacobsen.org Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
Reply by ●March 10, 20092009-03-10
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 Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●March 10, 20092009-03-10
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)^2That'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). Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
Reply by ●March 11, 20092009-03-11
>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 thenoise.>>> >>> >>> Uh...how can this be true? The channels are orthogonal andprocessed>>> 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
Reply by ●March 11, 20092009-03-11
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. > >SteveAs 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. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
Reply by ●March 12, 20092009-03-12
>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 inproportion>>to the signal, and this affects both carrier recovery and symbol timing >>recovery (at least in terms of getting a quick initial lock) just asmuch>>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
