> It seems to me quite interesting but I have a doubt: If we don't
> synchronize the transmiter/Receiver clock adjusting the symbol clock, is
> not possible to loose a packet of bits?
>
> The scenario I'm figuring out is a Tx working with its clock, and a Rx
> with a slightly different one because of component tolerance, since we
> don't modyfy the clock (just reinvent the sample by interpolation)
>
> Then, if the Rx clock freq is slightly lower than the Tx, and we don't
> correct this, in the end we'll overrun a sample...Won't we?
No, because like you said the receiver samples above the bandwidth
of the transmitted signal. Think about it: if you are sampling above
the
Nyquist rate of the transmit signal, then you can reconstruct the
signal
anyway, and then re-sample at the correct instances. Thankfully, the
same thing can be done by using a timing estimator, timing adjustor,
and using this adjusted estimate to feed into an interpolator -- all in
the
discrete-time domain.
Hope that helps,
Julius
Reply by Calabi_yau●January 17, 20072007-01-17
Hi,
A basic question about QAM. I've noticed that there are timing recovery
systems that calculate the 'correct' sample by means of an interpolation
guessing wich one of this intermediate samples is the most suitable
instead of closing a loop and recovering and using the recovered clock as
a receiver clock..
It seems to me quite interesting but I have a doubt: If we don't
synchronize the transmiter/Receiver clock adjusting the symbol clock, is
not possible to loose a packet of bits?
The scenario I'm figuring out is a Tx working with its clock, and a Rx
with a slightly different one because of component tolerance, since we
don't modyfy the clock (just reinvent the sample by interpolation)
Then, if the Rx clock freq is slightly lower than the Tx, and we don't
correct this, in the end we'll overrun a sample...Won't we?
Am I missing some point about this method?
Any paper/review about this issue will be welcomed.
Thanks.