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Blind Channel phase offset recovery

Started by Ali A Nasir August 13, 2009
> I have the information about the constellation that I am using but not >allowed to use any training or preamble ( since its blind recovery ).
Let
>us say I am using pi/4 QPSK >
pi/4 QPSK constellation is the same as 8PSK. Possible to use blind phase/frequency offset estimator based on 8-th power. Or you need only the removal of phase ambiguity?

Alexander Petrov wrote:

> pi/4 QPSK constellation is the same as 8PSK. Possible to use blind > phase/frequency offset estimator based on 8-th power.
You would be better off by considering the two subsequent symbols and the 4th power. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
>> I have the information about the constellation that I am using but
not
>>allowed to use any training or preamble ( since its blind recovery ). >Let >>us say I am using pi/4 QPSK >> > >pi/4 QPSK constellation is the same as 8PSK. Possible to use blind >phase/frequency offset estimator based on 8-th power. > >Or you need only the removal of phase ambiguity? > >
I need the removal of small carrier frequency offset in addition to constant phase offset. Can you please explain the offset estimator based on 8th power ? Any paper or book reference ? Will that be blind estimator ? Regards, Ali
> > >Alexander Petrov wrote: > >> pi/4 QPSK constellation is the same as 8PSK. Possible to use blind >> phase/frequency offset estimator based on 8-th power. > >You would be better off by considering the two subsequent symbols and >the 4th power. > > >Vladimir Vassilevsky >DSP and Mixed Signal Design Consultant >http://www.abvolt.com >
Thanks Vladimir for your response. Can you please give me some reference so that I could better understand how to implement the blind phase and frequcency offset estimation considering two subsequent symbols and fourth power ? Regards Regards, Ali
> I need the removal of small carrier frequency offset in addition to >constant phase offset. Can you please explain the offset estimator based
on
>8th power ? Any paper or book reference ? Will that be blind estimator ? > >Regards, >Ali >
Look at these simulink models: http://electronix.ru/forum/index.php?showtopic=23652
"Ali A Nasir" <aliarshad46@hotmail.com> writes:

>>> I have the information about the constellation that I am using but > not >>>allowed to use any training or preamble ( since its blind recovery ). >>Let >>>us say I am using pi/4 QPSK >>> >> >>pi/4 QPSK constellation is the same as 8PSK. Possible to use blind >>phase/frequency offset estimator based on 8-th power. >> >>Or you need only the removal of phase ambiguity? >> >> > > > I need the removal of small carrier frequency offset in addition to > constant phase offset. Can you please explain the offset estimator based on > 8th power ?
Ali, If you have an M-ary PSK signal, the possible phases are r[n] = exp(i * m[n] * 2 * pi / M + phi), m[n] \in {0, 1, ..., M-1}, where phi is some unknown phase. Thus (r[n])^M = exp(i * m[n] * 2 * pi + M * phi) = exp(i * M * phi) This is the basis of the Mth-power technique for carrier recovery. However, note that this introduces a 2 * pi / M phase ambiguity, since if phi = m * 2 * pi / M, then M * phi = m * 2 * pi == 0.
> Any paper or book reference ? Will that be blind estimator ?
Yes, it's a blind estimator in that it doesn't require any special source symbol sequence. -- Randy Yates % "Midnight, on the water... Digital Signal Labs % I saw... the ocean's daughter." mailto://yates@ieee.org % 'Can't Get It Out Of My Head' http://www.digitalsignallabs.com % *El Dorado*, Electric Light Orchestra
Randy Yates <yates@ieee.org> writes:
> [...] > However, note that this introduces a 2 * pi / M phase ambiguity, since > if > > phi = m * 2 * pi / M, > > then > > M * phi = m * 2 * pi > == 0.
Let me rewrite this: if phi = m * 2 * pi / M + theta, then M * phi = m * 2 * pi + M * theta == M * theta. -- Randy Yates % "Bird, on the wing, Digital Signal Labs % goes floating by mailto://yates@ieee.org % but there's a teardrop in his eye..." http://www.digitalsignallabs.com % 'One Summer Dream', *Face The Music*, ELO
>Randy Yates <yates@ieee.org> writes: >> [...] >> However, note that this introduces a 2 * pi / M phase ambiguity, since >> if >> >> phi = m * 2 * pi / M, >> >> then >> >> M * phi = m * 2 * pi >> == 0. > >Let me rewrite this: > >if > > phi = m * 2 * pi / M + theta, > >then > > M * phi = m * 2 * pi + M * theta > == M * theta. >-- >Randy Yates % "Bird, on the wing, >Digital Signal Labs % goes floating by >mailto://yates@ieee.org % but there's a teardrop in his
eye..."
>http://www.digitalsignallabs.com % 'One Summer Dream', *Face The Music*,
ELO
>
Thanks a lot Randy for such detailed and good explanation. I am clear now.