On Mar 10, 7:30�am, julius <juli...@gmail.com> wrote:> 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.Why do you mean by conjugate the complex-valued symbols ? Why do I need to complex conjugate the symbols ?
Equalizer and Timing recovery
Started by ●March 10, 2009
Reply by ●March 12, 20092009-03-12
Reply by ●March 12, 20092009-03-12
On Mar 10, 10:19�am, Vladimir Vassilevsky <antispam_bo...@hotmail.com> wrote:> fpgaasicdesig...@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 Consultanthttp://www.abvolt.comWhy BPSK/QPSK needs no equalizer ? If I have a delay spread of multi- paths, and receive symbols several symbols (that can be inverted) later, in a wireless case, that will mess up my symbols...
Reply by ●March 12, 20092009-03-12
On Mar 10, 2:01�pm, Eric Jacobsen <eric.jacob...@ieee.org> wrote:> On Tue, 10 Mar 2009 09:19:08 -0500, Vladimir Vassilevsky > > > > <antispam_bo...@hotmail.com> wrote: > > >fpgaasicdesig...@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?Yes but you don't have the same data on I and Q in QPSK case, in BPSK you have same I and Q, so I guess they can have more transition in QPSK (both channels)...> > >> 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 Communicationshttp://www.ericjacobsen.org > > Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
Reply by ●March 12, 20092009-03-12
On Mar 10, 2:05�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. � 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 Communicationshttp://www.ericjacobsen.org > > Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.phpI don't lock the IF. And effectively Gardner said and we can see it mathematically, the rotation of I and Q should not affect the TED. Really, I can see that it affects a lot (in terms of locking time, jitter, ...)
Reply by ●March 12, 20092009-03-12
On Mar 10, 10:02�pm, Eric Jacobsen <eric.jacob...@ieee.org> wrote:> On Tue, 10 Mar 2009 13:13:15 -0500, Vladimir Vassilevsky > > <antispam_bo...@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). > > Eric Jacobsen > Minister of Algorithms > Abineau Communicationshttp://www.ericjacobsen.org > > Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.phpperhaps my issue is in the fact I have a 0.25 roll-off factor. And becomes harder for the synchro in QPSK case. Because a 0/1 alternation in I and Q cause with this 0.25 factor less amplitude than a long 1 or 0... And add me jitter, because differents amplitude values at the TED output...
Reply by ●March 12, 20092009-03-12
On Mar 10, 10:19�am, Vladimir Vassilevsky <antispam_bo...@hotmail.com> wrote:> fpgaasicdesig...@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 Consultanthttp://www.abvolt.comBut how the EQ can work at one sample by symbol ? The symbol synchronizer takes a decision on a symbol without equalization ? And if we put the EQ before the synchro, we can remove multi-path and feed the synchro with "cleaned" symbols.
Reply by ●March 12, 20092009-03-12
On Mar 10, 2:05�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. � 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 Communicationshttp://www.ericjacobsen.org > > Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.phpI talk about x[k-0.5](Signx[k-1]-x[k+1]); x=I+jQ Is there other TED ?
Reply by ●March 12, 20092009-03-12
>On Mar 10, 10:02=A0pm, Eric Jacobsen <eric.jacob...@ieee.org> wrote: >> On Tue, 10 Mar 2009 13:13:15 -0500, Vladimir Vassilevsky >> >> <antispam_bo...@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 thenoi=>se. >> >> >> Uh...how can this be true? =A0 The channels are orthogonal andprocess=>ed >> >> identically (typically) between QPSK and BPSK. =A0 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. =A0 The question was >> synchronization rather than BER. =A0 I've never noticed a differencein>> 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 Communicationshttp://www.ericjacobsen.org >> >> Blog:http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php > >perhaps my issue is in the fact I have a 0.25 roll-off factor. And >becomes harder for the synchro in QPSK case. Because a 0/1 alternation >in I and Q cause with this 0.25 factor less amplitude than a long 1 or >0... And add me jitter, because differents amplitude values at the TED >output...In general Gardner needs substantial excess bandwidth to work well, especially if used for high order QAM. However, for simple QPSK modulation Gardner should work OK with 0.25 roll-off. Steve
Reply by ●March 12, 20092009-03-12
On Mar 11, 11:20�pm, fpgaasicdesig...@gmail.com wrote:> > Why do you mean by conjugate the complex-valued symbols ? > Why do I need to complex conjugate the symbols ?Why don't you post your Gardner TED formula for QPSK so that we can check?
Reply by ●March 12, 20092009-03-12
Eric Jacobsen wrote:> 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.IIRC in general "Gardner" looks at the variation of the power of the signal at the output of the matched filter vs time. This power is supposed to be at maximum at the proper sampling instants and at minimum in the middle between the sampling instants. So, the detector is agnostic to rotation. The devil is in the details; there is 1001 of simplified methods to compute the derivative of power. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
