Gentle People,
If we model a low-pass, binary PAM signal in zero-mean noise as
z(t) = +/- a * e^{j\phi} + n(t),
where \phi is unknown carrier phase, then one way to estimate this
carrier phase is as follows.
We square the signal to get
z^2(t) = a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t).
We then average the signal, so that, analytically,
E[z^2(t)] = E[a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t)]
= E[a^2 * e^{j2*\phi}] + E[2 * a * e^{j\phi} * n(t)] +
E[n^2(t)]
= a^2 * e^{j2*\phi} + \sigma^2_n, [1]
where \sigma^2_n is the variance of the noise, and where the middle
term cancels due to the assumption of zero-mean noise.
If we ignore the noise variance term in [1], then we can find the
carrier phase by taking the arctan of E[z^2(t)]:
arctan(E[z^2(t)] = 2*\phi.
However, the truth is, the noise variance term is not 0 and will
instroduce some error into the phase estimate.
So my question is this: Is there any way to remove the effect of the
noise variance from the carrier phase estimate?
--Randy Yates
Estimation of Carrier Phase Offset
Started by ●June 24, 2006
Reply by ●June 24, 20062006-06-24
Randy Yates wrote:> Gentle People, > > If we model a low-pass, binary PAM signal in zero-mean noise as > > z(t) = +/- a * e^{j\phi} + n(t), > > where \phi is unknown carrier phase, then one way to estimate this > carrier phase is as follows. > > We square the signal to get > > z^2(t) = a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t). > > We then average the signal, so that, analytically, > > E[z^2(t)] = E[a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t)] > = E[a^2 * e^{j2*\phi}] + E[2 * a * e^{j\phi} * n(t)] + > E[n^2(t)] > = a^2 * e^{j2*\phi} + \sigma^2_n, [1] > > where \sigma^2_n is the variance of the noise, and where the middle > term cancels due to the assumption of zero-mean noise. > > If we ignore the noise variance term in [1], then we can find the > carrier phase by taking the arctan of E[z^2(t)]: > > arctan(E[z^2(t)] = 2*\phi. > > However, the truth is, the noise variance term is not 0 and will > instroduce some error into the phase estimate. > > So my question is this: Is there any way to remove the effect of the > noise variance from the carrier phase estimate? > > --Randy YatesIf the noise is complex, it will add vectorially (is that a word?) to the signal vector and move it off of +/- 180 degrees randomly, resulting in a noisy phase. There is no way I know of to obtain a noiseless phase estimate for a noisy phase. If the noise is purely real or imaginary, then maybe so. John
Reply by ●June 24, 20062006-06-24
john wrote:> Randy Yates wrote: >> Gentle People, >> >> If we model a low-pass, binary PAM signal in zero-mean noise as >> >> z(t) = +/- a * e^{j\phi} + n(t), >> >> where \phi is unknown carrier phase, then one way to estimate this >> carrier phase is as follows. >> >> We square the signal to get >> >> z^2(t) = a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t). >> >> We then average the signal, so that, analytically, >> >> E[z^2(t)] = E[a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t)] >> = E[a^2 * e^{j2*\phi}] + E[2 * a * e^{j\phi} * n(t)] + >> E[n^2(t)] >> = a^2 * e^{j2*\phi} + \sigma^2_n, [1] >> >> where \sigma^2_n is the variance of the noise, and where the middle >> term cancels due to the assumption of zero-mean noise. >> >> If we ignore the noise variance term in [1], then we can find the >> carrier phase by taking the arctan of E[z^2(t)]: >> >> arctan(E[z^2(t)] = 2*\phi. >> >> However, the truth is, the noise variance term is not 0 and will >> instroduce some error into the phase estimate. >> >> So my question is this: Is there any way to remove the effect of the >> noise variance from the carrier phase estimate? >> >> --Randy Yates > > If the noise is complex, it will add vectorially (is that a word?) to > the signal vector and move it off of +/- 180 degrees randomly, > resulting in a noisy phase. There is no way I know of to obtain a > noiseless phase estimate for a noisy phase. If the noise is purely real > or imaginary, then maybe so.One nice thing about carriers is that they are steady for long periods. That makes a long-term average meaningful, and useful for suppressing noise. The variance of the noise determines the needed averaging time for a given S/N and degree of necessary accuracy. It seems to me that Randy's procedure is good, depending only on the averaging time. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 24, 20062006-06-24
"Jerry Avins" <jya@ieee.org> wrote in message news:RN-dnZcBOInR-gDZnZ2dnUVZ_oqdnZ2d@rcn.net...> john wrote: > > Randy Yates wrote: > >> Gentle People, > >> > >> If we model a low-pass, binary PAM signal in zero-mean noise as > >> > >> z(t) = +/- a * e^{j\phi} + n(t), > >> > >> where \phi is unknown carrier phase, then one way to estimate this > >> carrier phase is as follows. > >> > >> We square the signal to get > >> > >> z^2(t) = a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t). > >> > >> We then average the signal, so that, analytically, > >> > >> E[z^2(t)] = E[a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t)] > >> = E[a^2 * e^{j2*\phi}] + E[2 * a * e^{j\phi} * n(t)] + > >> E[n^2(t)] > >> = a^2 * e^{j2*\phi} + \sigma^2_n, [1] > >> > >> where \sigma^2_n is the variance of the noise, and where the middle > >> term cancels due to the assumption of zero-mean noise. > >> > >> If we ignore the noise variance term in [1], then we can find the > >> carrier phase by taking the arctan of E[z^2(t)]: > >> > >> arctan(E[z^2(t)] = 2*\phi. > >> > >> However, the truth is, the noise variance term is not 0 and will > >> instroduce some error into the phase estimate. > >> > >> So my question is this: Is there any way to remove the effect of the > >> noise variance from the carrier phase estimate? > >> > >> --Randy Yates > > > > If the noise is complex, it will add vectorially (is that a word?) to > > the signal vector and move it off of +/- 180 degrees randomly, > > resulting in a noisy phase. There is no way I know of to obtain a > > noiseless phase estimate for a noisy phase. If the noise is purely real > > or imaginary, then maybe so. > > One nice thing about carriers is that they are steady for long periods. > That makes a long-term average meaningful, and useful for suppressing > noise. The variance of the noise determines the needed averaging time > for a given S/N and degree of necessary accuracy. It seems to me that > Randy's procedure is good, depending only on the averaging time. >Averaging won't get rid of phase noise. Suppose we have cos(wt + phi(t)) where phi(t) is random noise. Averaging is identical to integration -if we integrate the above we make things worse. M.P -- Posted via a free Usenet account from http://www.teranews.com
Reply by ●June 24, 20062006-06-24
Mad Prof wrote:> "Jerry Avins" <jya@ieee.org> wrote in message > news:RN-dnZcBOInR-gDZnZ2dnUVZ_oqdnZ2d@rcn.net... >> john wrote: >>> Randy Yates wrote: >>>> Gentle People, >>>> >>>> If we model a low-pass, binary PAM signal in zero-mean noise as >>>> >>>> z(t) = +/- a * e^{j\phi} + n(t), >>>> >>>> where \phi is unknown carrier phase, then one way to estimate this >>>> carrier phase is as follows. >>>> >>>> We square the signal to get >>>> >>>> z^2(t) = a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t). >>>> >>>> We then average the signal, so that, analytically, >>>> >>>> E[z^2(t)] = E[a^2 * e^{j2*\phi} + 2 * a * e^{j\phi} * n(t) + n^2(t)] >>>> = E[a^2 * e^{j2*\phi}] + E[2 * a * e^{j\phi} * n(t)] + >>>> E[n^2(t)] >>>> = a^2 * e^{j2*\phi} + \sigma^2_n, [1] >>>> >>>> where \sigma^2_n is the variance of the noise, and where the middle >>>> term cancels due to the assumption of zero-mean noise. >>>> >>>> If we ignore the noise variance term in [1], then we can find the >>>> carrier phase by taking the arctan of E[z^2(t)]: >>>> >>>> arctan(E[z^2(t)] = 2*\phi. >>>> >>>> However, the truth is, the noise variance term is not 0 and will >>>> instroduce some error into the phase estimate. >>>> >>>> So my question is this: Is there any way to remove the effect of the >>>> noise variance from the carrier phase estimate? >>>> >>>> --Randy Yates >>> If the noise is complex, it will add vectorially (is that a word?) to >>> the signal vector and move it off of +/- 180 degrees randomly, >>> resulting in a noisy phase. There is no way I know of to obtain a >>> noiseless phase estimate for a noisy phase. If the noise is purely real >>> or imaginary, then maybe so. >> One nice thing about carriers is that they are steady for long periods. >> That makes a long-term average meaningful, and useful for suppressing >> noise. The variance of the noise determines the needed averaging time >> for a given S/N and degree of necessary accuracy. It seems to me that >> Randy's procedure is good, depending only on the averaging time. >> > > Averaging won't get rid of phase noise. > > Suppose we have > > cos(wt + phi(t)) > > where phi(t) is random noise. > > Averaging is identical to integration -if we integrate the above we make > things worse.Remember: zero mean. With 4 times the number of averaged points, the mean deviation from the mean is halved. The sum is a random walk, but id divided by the number of measurements, it converges. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●June 24, 20062006-06-24
Randy Yates queried:> So my question is this: Is there any way to remove the effect of the > noise variance from the carrier phase estimate?Randy, The straightforward way to accomplish this is to do decision-directed phase estimation. If you can make a decision about each value of a[n], then you can multiplity z[n] by this. Then the noise will average to zero, but the phase estimate will be more accurate. But maybe in your application, you can't get a reliable decision. Hopefully, someone else has a better answer for this. I would like to see one.
Reply by ●June 24, 20062006-06-24
Randy: [snip]> So my question is this: Is there any way to remove the effect of the > noise variance from the carrier phase estimate? > > --Randy Yates[snip] Don't all of the "usual" techniques for "estimating" a signal in noise work here? i.e. Weiner filtering, or maximum likelihood detection, or whatever... This problem is actually easier than when the signal to be extracted from the noise carries information and is actually the result of a stochastic process, since usually such a "carrier" is itself a very well determined signal. i.e. it carries exactly zero information because it's "message" never changes. At the transmitter it's frequency and phase are constants! Am I missing something? -- Pete Indialantic By-the-Sea, FL
Reply by ●June 24, 20062006-06-24
"Peter O. Brackett" <none@no-such-domain.nul> wrote in message news:ejkng.1859$ii.45@newsread3.news.pas.earthlink.net...> Randy: > > [snip] > > So my question is this: Is there any way to remove the effect of the > > noise variance from the carrier phase estimate? > > > > --Randy Yates > [snip] > > Don't all of the "usual" techniques for "estimating" a signal in noisework> here? > > i.e. Weiner filtering, or maximum likelihood detection, or whatever... > > This problem is actually easier than when the signal to be extracted from > the noise carries information and is actually the result of a stochastic > process, since usually such a "carrier" is itself a very well determined > signal. i.e. it carries exactly zero information because it's "message" > never changes. At the transmitter it's frequency and phase are constants! > > Am I missing something? > > -- > Pete > Indialantic By-the-Sea, FL > >Ordinary filtering doesn't work since the noise processes at work are not additive, they are multiplicative in nature. M.P -- Posted via a free Usenet account from http://www.teranews.com
Reply by ●June 26, 20062006-06-26
john wrote:> [...] > If the noise is complex, it will add vectorially (is that a word?) to > the signal vector and move it off of +/- 180 degrees randomly, > resulting in a noisy phase. There is no way I know of to obtain a > noiseless phase estimate for a noisy phase. If the noise is purely real > or imaginary, then maybe so.John, You caught my main error - that the noise is complex - but then (I think) missed the wonderful conclusion: the noise doesn't matter. Starting with the equation E[z^2] = alpha^2 * e^{j * 2 *\phi} + E[n^2], let's examine E[n^2]: n^2 = (nx + j*ny) * (nx + j*ny) = nx^2 - ny^2 + j * (nx * ny + ny * nx) and therefore E[n^2] = E[nx^2 - ny^2] + j * (E[nx * ny] + E[ny * nx]) = E[nx^2] - E[ny^2] + j * (E[nx * ny] + E[ny * nx]) = varx^2 - vary^2 + j * (E[nx * ny] + E[ny * nx]) If the complex noise is composed of two independent, identically-distributed noise processes, then E[nx * ny] = E[nx]*E[ny] = 0 and varx^2 = vary^2, therefore E[n^2] = 0. --Randy
Reply by ●June 26, 20062006-06-26
