Question For Craig Markwardt re Pioneer 10 Data

Thanks Craig for the clarification that the transmitter
frequency for all transmissions to Pioneer 10 was the same
and produced by a circuit that multiplied 96 in the early
days;48, later, times a very precise 24 MHz local oscillator
frequency And that this then could produce a very reliable
difference in
the received frequency and the transmitted frequency.
My question: What is ratio of error sum of squares around the
selected frequency to the sum of squares around the mean?
Does item 101 Average Doppler Residual in the NASA data
tape have something to do with the numerator of this ratio
eg 3 times the sq rt of numerator would lead to a 99percent
confidence interval for the true received doppler shift?
Ralph Sansbury


The process of digging the weak received signal out of noise
as I understand it involved the
representation of nanosecond voltage variations as a Fourier
series with the largest weighted sine component of frequency
around 2292MHz and the other sine components much smaller.
The specific phase and frequency is detected using filters,
Fast Fourier Transforms and Phase Locked Loops. And if you
subtracted
the received voltage values at each nanosecond or fraction of a
nanosecond from those predicted by the detected frequency and
phase, you would get a set of numbers that was normally
distributed around zero indicating that these differences were
noise.
Of course if the component of frequency in the expected range
has the same weight as those in other ranges then this would
indicate that it too was noise also.
If the sum of squares of the observed around a predicted set of
values is as great as the sum of squares about the mean of the
set of values then the predicted set of values is worthless and I
suppose some sort of criteria is the basis for saying that the
receptions from Pioneer 10 are now lost in noise.
It would be nice to get a little more clarification on this
point eg What is ratio of error sum of squares around the
selected frequency to the sum of squares around the mean?
Does item 101 Average Doppler Residual have something
to do with the numerator of this ratio eg 3 times the sq rt of
numerator would lead to a 99percent confidence interval for
the true received doppler shift?
Ralph Sansbury





"Craig Markwardt" wrote in
message news





"ralph sansbury" writes:

Hi Craig,
Re the transmitter frequency subtracted from the received
frequency
to the get the doppler shift and motion of Pioneer 10
relative to
the earth
at any specific time.
If the multiplier is exactly 48 for the DCOcase but 96
earlier
and this corresponds to something
specific in the phyical circuit, that would be ok.
(What does it correspond to?)


The hardware has a fixed integer multiplier between the
reference
oscillator and the transmitted frequency. For the VCO the
multiplier
was 96, for the DCO it was 48. This is not a tunable
parameter,
i.e. it is fixed exactly by the electronics and microwave
components
of the oscillator and amplifier.

The "choice" of 48 vs. 96 comes in the modeling software. The
multiplier in the software must match the multiplier used in
the
hardware. There is no subjective choice involved.


Exactly and that is my question???? If the milliHz terms
supposedly
used to show a small anomalous acceleration would have been
changed
by using a different multiplier and there is no independent
reason for choosing
48 or 48.1 etc, then there is a problem!!!!



And, to reiterate, there is no fitting or tuning involved in
the DCO
multiplier.

Another problem is how do we know the transmitter frequency
was
always
exactly the same as the frequency produced by the DCO times
48?

Because that is how the system was designed, tested and
productively
used for more than a decade.