> 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?

# Blind Channel phase offset recovery

Started by ●August 13, 2009

Reply by ●August 18, 20092009-08-18

Reply by ●August 18, 20092009-08-18

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

Reply by ●August 20, 20092009-08-20

>> I have the information about the constellation that I am using butnot>>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

Reply by ●August 20, 20092009-08-20

> > >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

Reply by ●August 20, 20092009-08-20

> I need the removal of small carrier frequency offset in addition to >constant phase offset. Can you please explain the offset estimator basedon>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

Reply by ●August 20, 20092009-08-20

"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

Reply by ●August 20, 20092009-08-20

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

Reply by ●August 21, 20092009-08-21

>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 hiseye...">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.