FPGA Central

Hi!

I have a query regarding the computation of Gardner Timing Error
detector (TED). Dr. Gardner on his article "A BPSK/QPSK Timing Error
Detector for Sampled Receivers" states that Timing Error (for BPSK)
can be computed as:

U(r) = x(r-1/2)[x(r)-x(r-1)]

He calls the samples x(r) as "STROBE" values and those like x(r-1/2)
as "MIDWAY"
between strobes [like x(r-1/2) is midway between x(r) and x(r-1)].

Its also mentioned that this algorithm uses only 2 samples/symbol to
compute the error. I need to ask that:

1)What he means by these STROBE and MIDWAY values ,x(r) and x(r-1/2) ,
and how to compute them from the available 2 samples/symbol?

2) For the complete timing recovery loop, there is also an
interpolator. If the interpolator is fed with 2 samples/symbol, then
what is it computing? is it computing the midway value x(r-1/2) or
what?

Regards
Rider

"rider" <shabana_rizvi@yahoo.com> wrote in message
news:ca3a68c8.0403312046.7db78add@posting.google.c om...
> Hi!
>
> I have a query regarding the computation of Gardner Timing Error
> detector (TED). Dr. Gardner on his article "A BPSK/QPSK Timing Error
> Detector for Sampled Receivers" states that Timing Error (for BPSK)
> can be computed as:
>
> U(r) = x(r-1/2)[x(r)-x(r-1)]
>
> He calls the samples x(r) as "STROBE" values and those like x(r-1/2)
> as "MIDWAY"
> between strobes [like x(r-1/2) is midway between x(r) and x(r-1)].
>
> Its also mentioned that this algorithm uses only 2 samples/symbol to
> compute the error. I need to ask that:
>
> 1)What he means by these STROBE and MIDWAY values ,x(r) and x(r-1/2) ,
> and how to compute them from the available 2 samples/symbol?
>
> 2) For the complete timing recovery loop, there is also an
> interpolator. If the interpolator is fed with 2 samples/symbol, then
> what is it computing? is it computing the midway value x(r-1/2) or
> what?
>
> Regards
> Rider

You are sampling two samples per symbol, so if the loop is locked and you
are sampling at the symbol instants and midway between the samples, then
x(r-1), x(r-1/2), and x(r) will be three successive samples. These three
would comprise two strobes and one midway or one strobe and two midway
samples.

The TED can be used to adjust a PLL that controls the sample clock. Then no
interpolation is required; the sample clock is just adjusted so that the ADC
samples at the correct points. That's mostly an older method. The other
type of loop uses an ADC that samples at a constant rate, and then a
resampling filter controlled by the loop interpolates the value at the
sampling instant, which will be between the ADC samples. Before the loop is
locked, the interpolator will not be sampling at the symbol instants, but
the error value will be large and will push the loop until the interpolated
values occur at the symbol instants and the midway points.
-Kevin

"Kevin Neilson" <kevin_neilson@removethiscomcast.net> wrote in message news:<8oOac.52137$K91.127528@attbi_s02>...
> > You are sampling two samples per symbol, so if the loop is locked and you
> are sampling at the symbol instants and midway between the samples, then
> x(r-1), x(r-1/2), and x(r) will be three successive samples. These three
> would comprise two strobes and one midway or one strobe and two midway
> samples.
>
> The TED can be used to adjust a PLL that controls the sample clock. Then no
> interpolation is required; the sample clock is just adjusted so that the ADC
> samples at the correct points. That's mostly an older method. The other
> type of loop uses an ADC that samples at a constant rate, and then a
> resampling filter controlled by the loop interpolates the value at the
> sampling instant, which will be between the ADC samples. Before the loop is
> locked, the interpolator will not be sampling at the symbol instants, but
> the error value will be large and will push the loop until the interpolated
> values occur at the symbol instants and the midway points.
> -Kevin

Why do you call tuned sampling an older method? It is really a matter
of circumstances, rather than age. If you are, say, building modems to
run in the PCM telephone network you have no control over the
sampling, and have to fiddle things around by filtering. If you have
control over the sampler, especially for very high speed channels,
tuning the sampling time to match the symbols simplies things quite a
bit.

Regards,
Steve

"Kevin Neilson" <kevin_neilson@removethiscomcast.net> wrote in message news:<8oOac.52137$K91.127528@attbi_s02>...
> "rider" <shabana_rizvi@yahoo.com> wrote in message
> news:ca3a68c8.0403312046.7db78add@posting.google.c om...
> > Hi!
> >
> > I have a query regarding the computation of Gardner Timing Error
> > detector (TED). Dr. Gardner on his article "A BPSK/QPSK Timing Error
> > Detector for Sampled Receivers" states that Timing Error (for BPSK)
> > can be computed as:
> >
> > U(r) = x(r-1/2)[x(r)-x(r-1)]
> >
> > He calls the samples x(r) as "STROBE" values and those like x(r-1/2)
> > as "MIDWAY"
> > between strobes [like x(r-1/2) is midway between x(r) and x(r-1)].
> >
> > Its also mentioned that this algorithm uses only 2 samples/symbol to
> > compute the error. I need to ask that:
> >
> > 1)What he means by these STROBE and MIDWAY values ,x(r) and x(r-1/2) ,
> > and how to compute them from the available 2 samples/symbol?
> >
> > 2) For the complete timing recovery loop, there is also an
> > interpolator. If the interpolator is fed with 2 samples/symbol, then
> > what is it computing? is it computing the midway value x(r-1/2) or
> > what?
> >
> > Regards
> > Rider
>
> You are sampling two samples per symbol, so if the loop is locked and you
> are sampling at the symbol instants and midway between the samples, then
> x(r-1), x(r-1/2), and x(r) will be three successive samples. These three
> would comprise two strobes and one midway or one strobe and two midway
> samples.
>
> The TED can be used to adjust a PLL that controls the sample clock. Then no
> interpolation is required; the sample clock is just adjusted so that the ADC
> samples at the correct points. That's mostly an older method. The other
> type of loop uses an ADC that samples at a constant rate, and then a
> resampling filter controlled by the loop interpolates the value at the
> sampling instant, which will be between the ADC samples. Before the loop is
> locked, the interpolator will not be sampling at the symbol instants, but
> the error value will be large and will push the loop until the interpolated
> values occur at the symbol instants and the midway points.
> -Kevin


Thanks Kevin!

Yes i m trying to follow the new method in which the sampling clock of
the receiver is fixed at Fs and an all-digital interpolation/TED loop
is use to recover timing. Like in the fig below:


(~) Fixed Clk --/--Fs= Nfb (fb=symbol clock , Fs=sampling clock)

Fs=1/Ts|-------------| Fi=1/Ti |---|
x(mTs)----|Interpolator |--------->|TED|
----^-------- ---
| |
|-------| |---------|
|Control|<-------------|Loop Fil |
--------- -----------


However, i still cant understand a point . When the Gardner algorithm
says it uses 2 samples/symbol, then does it mean that Fs in the above
figure SHOULD exactly be 2fb? In my work, i have chosen fb=1khz and
Fs=8khz. Do i have to decimate this 8khz signal to 2 Khz before
applying it to the interpolator or i can feed this 8Khz signal
directly?

Thanks

"rider" <shabana_rizvi@yahoo.com> wrote in message
news:ca3a68c8.0404012151.576d0986@posting.google.c om...
> "Kevin Neilson" <kevin_neilson@removethiscomcast.net> wrote in message
news:<8oOac.52137$K91.127528@attbi_s02>...
> > "rider" <shabana_rizvi@yahoo.com> wrote in message
> > news:ca3a68c8.0403312046.7db78add@posting.google.c om...
> > > Hi!
> > >
> > > I have a query regarding the computation of Gardner Timing Error
> > > detector (TED). Dr. Gardner on his article "A BPSK/QPSK Timing Error
> > > Detector for Sampled Receivers" states that Timing Error (for BPSK)
> > > can be computed as:
> > >
> > > U(r) = x(r-1/2)[x(r)-x(r-1)]
> > >
> > > He calls the samples x(r) as "STROBE" values and those like x(r-1/2)
> > > as "MIDWAY"
> > > between strobes [like x(r-1/2) is midway between x(r) and x(r-1)].
> > >
> > > Its also mentioned that this algorithm uses only 2 samples/symbol to
> > > compute the error. I need to ask that:
> > >
> > > 1)What he means by these STROBE and MIDWAY values ,x(r) and x(r-1/2) ,
> > > and how to compute them from the available 2 samples/symbol?
> > >
> > > 2) For the complete timing recovery loop, there is also an
> > > interpolator. If the interpolator is fed with 2 samples/symbol, then
> > > what is it computing? is it computing the midway value x(r-1/2) or
> > > what?
> > >
> > > Regards
> > > Rider
> >
> > You are sampling two samples per symbol, so if the loop is locked and
you
> > are sampling at the symbol instants and midway between the samples, then
> > x(r-1), x(r-1/2), and x(r) will be three successive samples. These
three
> > would comprise two strobes and one midway or one strobe and two midway
> > samples.
> >
> > The TED can be used to adjust a PLL that controls the sample clock.
Then no
> > interpolation is required; the sample clock is just adjusted so that the
ADC
> > samples at the correct points. That's mostly an older method. The
other
> > type of loop uses an ADC that samples at a constant rate, and then a
> > resampling filter controlled by the loop interpolates the value at the
> > sampling instant, which will be between the ADC samples. Before the
loop is
> > locked, the interpolator will not be sampling at the symbol instants,
but
> > the error value will be large and will push the loop until the
interpolated
> > values occur at the symbol instants and the midway points.
> > -Kevin
>
>
> Thanks Kevin!
>
> Yes i m trying to follow the new method in which the sampling clock of
> the receiver is fixed at Fs and an all-digital interpolation/TED loop
> is use to recover timing. Like in the fig below:
>
>
> (~) Fixed Clk --/--Fs= Nfb (fb=symbol clock , Fs=sampling clock)
>
> Fs=1/Ts|-------------| Fi=1/Ti |---|
> x(mTs)----|Interpolator |--------->|TED|
> ----^-------- ---
> | |
> |-------| |---------|
> |Control|<-------------|Loop Fil |
> --------- -----------
>
>
> However, i still cant understand a point . When the Gardner algorithm
> says it uses 2 samples/symbol, then does it mean that Fs in the above
> figure SHOULD exactly be 2fb? In my work, i have chosen fb=1khz and
> Fs=8khz. Do i have to decimate this 8khz signal to 2 Khz before
> applying it to the interpolator or i can feed this 8Khz signal
> directly?
>
> Thanks

For the Gardner algorithm I don't think the extra samples help at all. There
are other algorithms that use more samples per symbol but I think you can
throw out three of every four in your case.

Then you would be left with a sample rate of about 2kHz and the control
block would continuously adjust the phase so that the sampling instants are
at the strobes and midpoints. The rate at which this phase "rolls" would be
the difference in frequencies between the symbol rate and 1KHz (half the
sample rate). This phase is often called mu in the texts. If the symbol
rate is exactly 1kHz, for example, mu will remain constant. The error value
would only be calculated every other sample.

I think you may still operate the loop with an 8kHz sample rate and no
decimation. But in that case, mu would have to have a large derivative and
you would have to design the tracking loop especially for this mode of
operation. Normally you want to design the loop so the derivative of mu is
small. So it's probably best to get the effective sample rate as close to
2X the symbol rate as possible. If it's helpful, you can still use a
lowpass-filter-based decimator instead of just throwing out samples; this
increases precision. Decimation by 4 can give you two extra bits of
precision, I think.

In case you're not clear on the interpolator: it is fed with the 2kHz
sample rate, and then it interpolates new samples that are ahead by a phase
determined by mu. The Farrow filter is often used for this.
-Kevin

"Steve Underwood" <steveu@coppice.org> wrote in message
news:80bddbd7.0404011640.df1e222@posting.google.co m...
> "Kevin Neilson" <kevin_neilson@removethiscomcast.net> wrote in message
news:<8oOac.52137$K91.127528@attbi_s02>...
> > > You are sampling two samples per symbol, so if the loop is locked and
you
> > are sampling at the symbol instants and midway between the samples, then
> > x(r-1), x(r-1/2), and x(r) will be three successive samples. These
three
> > would comprise two strobes and one midway or one strobe and two midway
> > samples.
> >
> > The TED can be used to adjust a PLL that controls the sample clock.
Then no
> > interpolation is required; the sample clock is just adjusted so that the
ADC
> > samples at the correct points. That's mostly an older method. The
other
> > type of loop uses an ADC that samples at a constant rate, and then a
> > resampling filter controlled by the loop interpolates the value at the
> > sampling instant, which will be between the ADC samples. Before the
loop is
> > locked, the interpolator will not be sampling at the symbol instants,
but
> > the error value will be large and will push the loop until the
interpolated
> > values occur at the symbol instants and the midway points.
> > -Kevin
>
> Why do you call tuned sampling an older method? It is really a matter
> of circumstances, rather than age. If you are, say, building modems to
> run in the PCM telephone network you have no control over the
> sampling, and have to fiddle things around by filtering. If you have
> control over the sampler, especially for very high speed channels,
> tuning the sampling time to match the symbols simplies things quite a
> bit.
>
> Regards,
> Steve

That was just my observation from what I've seen. In the past, the PLL was
usually external, because it took too many gates to do interpolation of new
samples. I think the trend is to get rid of the analog loop, not because it
is more complex, but because everybody wants to get rid of external analog
parts and suck all that logic into the FPGA/ASIC. It's complex to do the
loop and the Farrow filter interpolation in logic, but gates are cheaper
than VCOs and analaog loop filters. People like to see reference designs
that don't involve a lot of tweaking analog parts.
-Kevin

"Kevin Neilson" <kevin_neilson@removethiscomcast.net> wrote in message news:<EOibc.161285$_w.1784112@attbi_s53>...
> "rider" <shabana_rizvi@yahoo.com> wrote in message
> news:ca3a68c8.0404012151.576d0986@posting.google.c om...
> > "Kevin Neilson" <kevin_neilson@removethiscomcast.net> wrote in message
> news:<8oOac.52137$K91.127528@attbi_s02>...
> > > "rider" <shabana_rizvi@yahoo.com> wrote in message
> > > news:ca3a68c8.0403312046.7db78add@posting.google.c om...
> > > > Hi!
> > > >
> > > > I have a query regarding the computation of Gardner Timing Error
> > > > detector (TED). Dr. Gardner on his article "A BPSK/QPSK Timing Error
> > > > Detector for Sampled Receivers" states that Timing Error (for BPSK)
> > > > can be computed as:
> > > >
> > > > U(r) = x(r-1/2)[x(r)-x(r-1)]
> > > >
> > > > He calls the samples x(r) as "STROBE" values and those like x(r-1/2)
> > > > as "MIDWAY"
> > > > between strobes [like x(r-1/2) is midway between x(r) and x(r-1)].
> > > >
> > > > Its also mentioned that this algorithm uses only 2 samples/symbol to
> > > > compute the error. I need to ask that:
> > > >
> > > > 1)What he means by these STROBE and MIDWAY values ,x(r) and x(r-1/2) ,
> > > > and how to compute them from the available 2 samples/symbol?
> > > >
> > > > 2) For the complete timing recovery loop, there is also an
> > > > interpolator. If the interpolator is fed with 2 samples/symbol, then
> > > > what is it computing? is it computing the midway value x(r-1/2) or
> > > > what?
> > > >
> > > > Regards
> > > > Rider
> > >
> > > You are sampling two samples per symbol, so if the loop is locked and
> you
> > > are sampling at the symbol instants and midway between the samples, then
> > > x(r-1), x(r-1/2), and x(r) will be three successive samples. These
> three
> > > would comprise two strobes and one midway or one strobe and two midway
> > > samples.
> > >
> > > The TED can be used to adjust a PLL that controls the sample clock.
> Then no
> > > interpolation is required; the sample clock is just adjusted so that the
> ADC
> > > samples at the correct points. That's mostly an older method. The
> other
> > > type of loop uses an ADC that samples at a constant rate, and then a
> > > resampling filter controlled by the loop interpolates the value at the
> > > sampling instant, which will be between the ADC samples. Before the
> loop is
> > > locked, the interpolator will not be sampling at the symbol instants,
> but
> > > the error value will be large and will push the loop until the
> interpolated
> > > values occur at the symbol instants and the midway points.
> > > -Kevin
> >
> >
> > Thanks Kevin!
> >
> > Yes i m trying to follow the new method in which the sampling clock of
> > the receiver is fixed at Fs and an all-digital interpolation/TED loop
> > is use to recover timing. Like in the fig below:
> >
> >
> > (~) Fixed Clk --/--Fs= Nfb (fb=symbol clock , Fs=sampling clock)
> >
> > Fs=1/Ts|-------------| Fi=1/Ti |---|
> > x(mTs)----|Interpolator |--------->|TED|
> > ----^-------- ---
> > | |
> > |-------| |---------|
> > |Control|<-------------|Loop Fil |
> > --------- -----------
> >
> >
> > However, i still cant understand a point . When the Gardner algorithm
> > says it uses 2 samples/symbol, then does it mean that Fs in the above
> > figure SHOULD exactly be 2fb? In my work, i have chosen fb=1khz and
> > Fs=8khz. Do i have to decimate this 8khz signal to 2 Khz before
> > applying it to the interpolator or i can feed this 8Khz signal
> > directly?
> >
> > Thanks
>
> For the Gardner algorithm I don't think the extra samples help at all. There
> are other algorithms that use more samples per symbol but I think you can
> throw out three of every four in your case.
>
> Then you would be left with a sample rate of about 2kHz and the control
> block would continuously adjust the phase so that the sampling instants are
> at the strobes and midpoints. The rate at which this phase "rolls" would be
> the difference in frequencies between the symbol rate and 1KHz (half the
> sample rate). This phase is often called mu in the texts. If the symbol
> rate is exactly 1kHz, for example, mu will remain constant. The error value
> would only be calculated every other sample.
>
> I think you may still operate the loop with an 8kHz sample rate and no
> decimation. But in that case, mu would have to have a large derivative and
> you would have to design the tracking loop especially for this mode of
> operation. Normally you want to design the loop so the derivative of mu is
> small. So it's probably best to get the effective sample rate as close to
> 2X the symbol rate as possible. If it's helpful, you can still use a
> lowpass-filter-based decimator instead of just throwing out samples; this
> increases precision. Decimation by 4 can give you two extra bits of
> precision, I think.
>
> In case you're not clear on the interpolator: it is fed with the 2kHz
> sample rate, and then it interpolates new samples that are ahead by a phase
> determined by mu. The Farrow filter is often used for this.
> -Kevin


Kevin!

Thanks for the reply again. For sometimes i have been confused by the
term Interpolator itself. Interpolator (opposite of decimator) is
normally used to mention an upsampling of the signal sample rate (zero
insertion and then an anti-imaging filter). But here in case of timing
recovery, they sometimes mention that Ti > Ts (means fi<fs , with
reference to fig above), so it means the "interpolator" is actually
decimating? So we may ask:

1) Is Ti always > Ts?

2) Whatever filter structure we may employ for interpolator (like
farrow) , wouldn't it be running at a fix clock (like in hardware, the
clock signal driving the output register of interpolator would be
fix?) . So how does it adjust the phase, if it is outputting samples
after a fixed time?

Regards
Rider

shabana_rizvi@yahoo.com (rider) wrote in message news:<ca3a68c8.0404040722.6c9882c5@posting.google. com>...
> Kevin!
>
> Thanks for the reply again. For sometimes i have been confused by the
> term Interpolator itself. Interpolator (opposite of decimator) is
> normally used to mention an upsampling of the signal sample rate (zero
> insertion and then an anti-imaging filter). But here in case of timing
> recovery, they sometimes mention that Ti > Ts (means fi<fs , with
> reference to fig above), so it means the "interpolator" is actually
> decimating? So we may ask:
>
> 1) Is Ti always > Ts?
>
> 2) Whatever filter structure we may employ for interpolator (like
> farrow) , wouldn't it be running at a fix clock (like in hardware, the
> clock signal driving the output register of interpolator would be
> fix?) . So how does it adjust the phase, if it is outputting samples
> after a fixed time?
>
> Regards
> Rider

If the term interpolator is confusing you, try saying phase shifter.
That is what the filter is doing here. For example. Lets say you want
a root raised cosine filter at you receiver. I think your incoming
samples are at 8kHz, and your symbol rate is 1kHz. A +-4 symbol RRC
filter would be 65 samples long (since you usually add one and make it
odd). So, we could apply this 65 sample FIR every 4 incoming samples,
and feed the results into the equalizer at 2k/s, and let the Gardner
TED work out how well those samples are placed. However we have no way
to apply a fine timing correction when the TED says we need one.
Correcting by shifting in 8kHz samples is very coarse, and leads to
problems.

Now consider what happens if you ask the RRC filter designer to treat
the input sample rate as 16*8kHz, and the filter length as 16*65
samples. Now, if you have a very long set of coefficients. If you use
every 16th coeff. from this filter in the original RRC calculation,
you will end up with much the same result as before. However, if you
step along 1 coefficient, and take every 16th coeff. the RRC output
will be phase shifted by 1/16th of an 8kHz sample. Why? because those
coeffs were designed to be multiplied by samples 1/16th of an 8kHz
sample away from the samples you are actually using. You have 16 of
these phase shifting RRC filter coefficient sets spanning 0/16th of a
sample to 15/16th of a sample of phase shift.

Now, you have something fine enough to work with the TED. As the TED
keeps telling you to nudge the timing in a particular direction, you
step along the coefficient sets until you reach the end. Then you jump
back to the other end, and hop one whole 8kHz sample. Voila, smooth
TED operation.

For very higher order QAM 16 is probably not high enough. For simple
16 state QAM is may be overkill. You need to tune that figure to the
needs of your system, to stabilise the equalizer adaption and the
constellation sufficiently for you actual system.

Regards,
Steve

Hi!

Thanks for detailed reply Steve and Kevin. I am now trying to
implement the Interpolator . The simplest form is the Linear
Interpolator defined as:

y[k] = x[m(k)] + u[k]*{ x[m(k)+1] - x[m(k)] }

I have a confusion in understanding the "base-point" index m(k). This
is defined in Gardner's Article (Interpolation in digital Modems
P-I)as:

m= m(k) - i

1) Is x[m(k)+1] the "recent/current" sample and x[m(k)] is
previous/delayed sample of x or vice-versa?

2)Its written in paper that correct set of signals is identified by
m(k) and the current set of filter samples (coefficients) is
identified by u[k]. However, the Farrow structure doesnt show any
detail as to how m(k) is involved in computation. Its just a
combination of tapped delay lines and their sums controleld by u[k],
which is supplied by controller. About m(k) the article says: "The NCO
is operated such that its avarage period is Ti. Recycling the NCO
register indicates that a new interpolant is to be computed, using the
signal samples currently residing in the interpolator's shift
register. Thus, base-point index is identified by flagging the correct
set of signal samples, rather than explicitly computing m(k)". What
does this mean? We have to compute the interpolant on every clock
cycle Ts? If yes, what is the significance of
m(k)?

Regards
Rider