Hi, I have setup a simulation of an interpolation-based timing recovery loop with Gardner's 2 s/s error detector. The loop seems to work fine, but I am stuck with this question: Out of the two samples produced from the interpolator, how do we know which one is the correct one ? Basically, the TED, which is given by: e = (y(n) - y(n-2))*y(n-1) would produce zero error on both the sample points within a symbol (under locked condition). Do we need to use any outside information (such as sync/FEC lock) to determine the correct symbol location ? Thanks, Vijay.
Question about Symbol Timing Recovery (Gardner Method).
Started by ●May 5, 2007
Reply by ●May 6, 20072007-05-06
On May 5, 5:31 pm, BERT <callm...@gmail.com> wrote:> Hi, > > I have setup a simulation of an interpolation-based timing recovery > loop with Gardner's 2 s/s error detector. The loop seems to work fine, > but I am stuck with this question: > > Out of the two samples produced from the interpolator, how do we know > which one is the correct one ? Basically, the TED, which is given by: > > e = (y(n) - y(n-2))*y(n-1) > > would produce zero error on both the sample points within a symbol > (under locked condition). Do we need to use any outside information > (such as sync/FEC lock) to determine the correct symbol location ? > > Thanks, > Vijay.OK. I just found a bug in my implementation. The TED operates every T seconds instead of T/2 seconds (T = symbol rate). I noticed that if the TED operates every T/2 seconds, the error output oscillates between positive and negative values. But, the original ambiguity still remains. If we sample exactly at the zero crossings, then the TED would still produce a zero output. How to resolve this ambiguity ? Thanks, Vijay.
Reply by ●May 6, 20072007-05-06
BERT wrote:> On May 5, 5:31 pm, BERT <callm...@gmail.com> wrote: >> Hi, >> >> I have setup a simulation of an interpolation-based timing recovery >> loop with Gardner's 2 s/s error detector. The loop seems to work fine, >> but I am stuck with this question: >> >> Out of the two samples produced from the interpolator, how do we know >> which one is the correct one ? Basically, the TED, which is given by: >> >> e = (y(n) - y(n-2))*y(n-1) >> >> would produce zero error on both the sample points within a symbol >> (under locked condition). Do we need to use any outside information >> (such as sync/FEC lock) to determine the correct symbol location ? >> >> Thanks, >> Vijay. > > OK. I just found a bug in my implementation. The TED operates every T > seconds instead of T/2 seconds (T = symbol rate). I noticed that if > the TED operates every T/2 seconds, the error output oscillates > between positive and negative values. > > But, the original ambiguity still remains. If we sample exactly at the > zero crossings, then the TED would still produce a zero output. How to > resolve this ambiguity ?In order to resolve frequency F, you need to sample *faster* that 2F. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●May 6, 20072007-05-06
BERT wrote:> On May 5, 5:31 pm, BERT <callm...@gmail.com> wrote: >> Hi, >> >> I have setup a simulation of an interpolation-based timing recovery >> loop with Gardner's 2 s/s error detector. The loop seems to work fine, >> but I am stuck with this question: >> >> Out of the two samples produced from the interpolator, how do we know >> which one is the correct one ? Basically, the TED, which is given by: >> >> e = (y(n) - y(n-2))*y(n-1) >> >> would produce zero error on both the sample points within a symbol >> (under locked condition). Do we need to use any outside information >> (such as sync/FEC lock) to determine the correct symbol location ? >> >> Thanks, >> Vijay. > > OK. I just found a bug in my implementation. The TED operates every T > seconds instead of T/2 seconds (T = symbol rate). I noticed that if > the TED operates every T/2 seconds, the error output oscillates > between positive and negative values. > > But, the original ambiguity still remains. If we sample exactly at the > zero crossings, then the TED would still produce a zero output. How to > resolve this ambiguity ?You have three samples, spaced by half a symbol. You run this error expression e = (y(n) - y(n-2))*y(n-1) Assuming the signal is PSK, you will get an "e" that is positive for advance, negative for retard, and zero when spot on. The actual size of "e" doesn't say a whole lot - only its sign conveys much meaning. You generally need to filter "e" somewhat, do damp out noise effects. When the filtered "e" settles around zero the y(n-1) sample should be mid way between symbols, and y(n) and y(n-2) bang in the centre of symbols. For QAM this will also work, as long as you filter "e" pretty hard, and you have sufficient excess bandwidth in the signal. This isn't obvious, as QAM transitions don't generally go through zero between symbols. However, it statistically works out, so with heavy damping of "e" it forms a usable metric for symbol timing error. Regards, Steve
Reply by ●May 6, 20072007-05-06
> > In order to resolve frequency F, you need to sample *faster* that 2F. > > JerryI agree. But, the Gardner TED is almost ubiquitously used, and it seems to work fine ... So, I am wondering if I am missing something here. Even my MATLAB simulations always converge on the correct sampling instants and never converge on the zero-crossing points (even with no additive noise). I will experiment with this a little bit more ...
Reply by ●May 6, 20072007-05-06
> Assuming the signal is PSK, you will get an "e" that is positive for > advance, negative for retard, and zero when spot on. The actual size of > "e" doesn't say a whole lot - only its sign conveys much meaning. You > generally need to filter "e" somewhat, do damp out noise effects. When > the filtered "e" settles around zero the y(n-1) sample should be mid way > between symbols, and y(n) and y(n-2) bang in the centre of symbols. >Good point about "e". So, is it prudent to implement a "decision- directed" approach to computing the error, where y(n) and y(n-2) are now quantized to +/-1 (let's assume BPSK/QPSK) ? So, in the presence of noise, it is impossible to converge on the ambiguous sampling point, i.e., exactly at the zero-crossings (because it will produce a random component that can filtered out by the loop filter).
Reply by ●May 8, 20072007-05-08
On May 6, 12:17 pm, BERT <callm...@gmail.com> wrote:> > Good point about "e". So, is it prudent to implement a "decision- > directed" approach to computing the error, where y(n) and y(n-2) are > now quantized to +/-1 (let's assume BPSK/QPSK) ? So, in the presence > of noise, it is impossible to converge on the ambiguous sampling > point, i.e., exactly at the zero-crossings (because it will produce a > random component that can filtered out by the loop filter).By now you realize that the problem is solved in two steps: 1. How to extract a "error signal" 2. What to do with that "error signal", i.e. filtering. For the first one, typically one considers the "S-curve", that is, the map between actual error and estimated error. Usually it's defined as the expected value of the timing error detector for a given actual timing error, assuming unit amplitude. For the second one, given the S-curve it's possible to do a linearization of the problem, and use your usual analysis of feedback filters. So given the above, do you think that it's possible to get to the ideal sampling instance? Julius
Reply by ●May 8, 20072007-05-08
On May 8, 12:42 am, julius <juli...@gmail.com> wrote:> On May 6, 12:17 pm, BERT <callm...@gmail.com> wrote: > > > > > Good point about "e". So, is it prudent to implement a "decision- > > directed" approach to computing the error, where y(n) and y(n-2) are > > now quantized to +/-1 (let's assume BPSK/QPSK) ? So, in the presence > > of noise, it is impossible to converge on the ambiguous sampling > > point, i.e., exactly at the zero-crossings (because it will produce a > > random component that can filtered out by the loop filter). > > By now you realize that the problem is solved in two steps: > 1. How to extract a "error signal" > 2. What to do with that "error signal", i.e. filtering. > > For the first one, typically one considers the "S-curve", that is, > the map between actual error and estimated error. > > Usually it's defined as the expected value of the timing error > detector for a given actual timing error, assuming unit amplitude. > > For the second one, given the S-curve it's possible to do a > linearization of the problem, and use your usual analysis of > feedback filters. > > So given the above, do you think that it's possible to get to > the ideal sampling instance? > > JuliusThanks for your response Julius. I have never heard of S-curves before, but I think I understand your point. The "S-curve" plot for the zero-crossing point would have a slope that is reverse from that of the point where the matched filter output is maximum. Hence, if we match the loop to track the ideal sampling point, this would make the other sampling point kind of "unstable" - i.e., it is still possible to stay locked at that point, but any small amount of noise/ interference would force it out of the locked state because of positive feedback. So, did I get anywhere near the right explanation ?
Reply by ●May 8, 20072007-05-08
On May 8, 4:09 pm, BERT <callm...@gmail.com> wrote:> Thanks for your response Julius. I have never heard of S-curves > before, but I think I understand your point. The "S-curve" plot for > the zero-crossing point would have a slope that is reverse from that > of the point where the matched filter output is maximum. Hence, if we > match the loop to track the ideal sampling point, this would make the > other sampling point kind of "unstable" - i.e., it is still possible > to stay locked at that point, but any small amount of noise/ > interference would force it out of the locked state because of > positive feedback. So, did I get anywhere near the right explanation ?I think you're jumping a little bit ahead of yourself here! The S-curve is simply a way to see how the actual timing error maps to the output of a timing error detector/estimator. Hopefully it at least tells you whether you need to delay or advance your sampling instance. So what you said about zero-crossing etc may be true for certain timing error detectors, but is not necessarily a desirable property. Think of what the ideal S-curve would look like: I think I would want it to be a straight line through the origin, so that my timing error detector will tell me exactly how far I am from my ideal sampling time, at least subject to a gain factor). In my opinion it is not so much the slope of the S-curve as the actual value itself. Of course, no such S-curve exists because the effect of delay on the observation is not linear in general.>From there one can assume small error and approximate theS-curve as a line crossing the origin, with a given slope that is usually called the "gain" of the timing error detector. Now we have a linearized model, which is basically an IIR filter. Then everything that you know about filters apply here: notions of bandwidth, gain, stability, etc. Then from here you can deduce whether it is possible to get what you want in your previous question. The above is pretty much the way that we can obtain a rather simplified but insightful analysis of the performance of timing synchronization systems. Of course, there's a whole array of research work in coming up with more accurate approximations by taking advantage of the structure of the modulation scheme, etc. Hope that helps. Julius
Reply by ●May 8, 20072007-05-08
On May 8, 4:42 pm, julius <juli...@gmail.com> wrote:> Now we have a linearized model, which is basically an IIR > filter. Then everything that you know about filters apply here: > notions of bandwidth, gain, stability, etc. Then from here you > can deduce whether it is possible to get what you want in > your previous question.Instead of IIR filter I should have used the term "linear feedback control system", i.e. PID controllers. Julius
