# Blind Channel phase offset recovery

Started by 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.

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.
>
>
>DSP and Mixed Signal Design Consultant
>http://www.abvolt.com
>

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
>

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.