Estimation of Carrier Phase Offset

Randy Yates wrote:
> 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

This proves that the estimator is unbiased. As you average more and
more data, the noise contribution to the estimate tends to zero. 

John

john wrote:
> [...]
> This proves that the estimator is unbiased. As you average more and
> more data, the noise contribution to the estimate tends to zero.

True, and the previous estimator (assuming real noise) was not. That's
kinda the point, although I omitted the academic label.

--Randy

Randy Yates wrote:
> john wrote:
> > [...]
> > This proves that the estimator is unbiased. As you average more and
> > more data, the noise contribution to the estimate tends to zero.
>
> True, and the previous estimator (assuming real noise) was not. That's
> kinda the point, although I omitted the academic label.
>
> --Randy

I'm not sure what you mean by "noise doesn't matter". In a practical
system the estimator will have a non-infinite signal to noise ratio,
depending on the averaging time, noise power, and signal power. Those
details typically do matter.

John

john wrote:
> Randy Yates wrote:
> > john wrote:
> > > [...]
> > > This proves that the estimator is unbiased. As you average more and
> > > more data, the noise contribution to the estimate tends to zero.
> >
> > True, and the previous estimator (assuming real noise) was not. That's
> > kinda the point, although I omitted the academic label.
> >
> > --Randy
>
> I'm not sure what you mean by "noise doesn't matter".

Hi John,

What I mean is that, analytically, the noise doesn't matter. The more
erudite way to put it is "the estimate is unbiased." Or more
verbosely, (but equivalently) the difference between the actual phase
and the expected value (expected value in the analytic sense, not in a
practical, averaging sort of sense) of the phase estimate is zero.

> In a practical system the estimator will have a non-infinite signal
> to noise ratio, depending on the averaging time, noise power, and
> signal power. Those details typically do matter.

A final approach must work both in theory and practice. However, until
you can get from here to there in the analytical, theoretical world,
then practical considerations are irrelevent.

--Randy

>>>>> "Randy" == Randy Yates <yates@ieee.org> writes:

    Randy> 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.

    Randy> John,

    Randy> You caught my main error - that the noise is complex - but then (I
    Randy> think) missed the wonderful conclusion: the noise doesn't matter.

    Randy> Starting with the equation

    Randy>   E[z^2] = alpha^2 * e^{j * 2 *\phi} + E[n^2],

    Randy> let's examine E[n^2]:

    Randy> n^2 = (nx + j*ny) * (nx + j*ny)
    Randy>     = nx^2 - ny^2 + j * (nx * ny + ny * nx)

If n is complex, don't you really want E[|n|^2] instead of E[n^2]?

Ray

Raymond Toy wrote:
> >>>>> "Randy" == Randy Yates <yates@ieee.org> writes:
>
>     Randy> 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.
>
>     Randy> John,
>
>     Randy> You caught my main error - that the noise is complex - but then (I
>     Randy> think) missed the wonderful conclusion: the noise doesn't matter.
>
>     Randy> Starting with the equation
>
>     Randy>   E[z^2] = alpha^2 * e^{j * 2 *\phi} + E[n^2],
>
>     Randy> let's examine E[n^2]:
>
>     Randy> n^2 = (nx + j*ny) * (nx + j*ny)
>     Randy>     = nx^2 - ny^2 + j * (nx * ny + ny * nx)
>
> If n is complex, don't you really want E[|n|^2] instead of E[n^2]?

Hi Ray,

I don't think so. If, instead of straight-forward squaring, we multiply
by the complex conjugate so that the noise term becomes |n|^2, then
the term with the phase information cancels as well
(e^{j*\phi} * e^{j*-\phi} = 1). 

--Randy

>>>>> "Randy" == Randy Yates <yates@ieee.org> writes:

    Randy> Raymond Toy wrote:
[snip]
    >> If n is complex, don't you really want E[|n|^2] instead of E[n^2]?

    Randy> Hi Ray,

    Randy> I don't think so. If, instead of straight-forward squaring, we multiply
    Randy> by the complex conjugate so that the noise term becomes |n|^2, then
    Randy> the term with the phase information cancels as well
    Randy> (e^{j*\phi} * e^{j*-\phi} = 1). 

Right.  I was thinking of the wrong thing. 

Ray