Pioneer 10 rx error and tx frequencies?

In message , Craig Markwardt
writes

"ralph sansbury" writes:
You and George have not clearly answered the question as to
the possibility and probability of sine functions with other
frequencies near the one frequency detected using the FFT
procedure and phase locked loops.

The probability of such an occurrence is essentially zero. Only a
spacecraft moving on Pioneer 10's trajectory, or one very near it
(within a few kilometers) and with very neary the same motion (within
1 mm/s). A different trajectory is ruled out at extremely high
confidence.

I wonder if this is another variation of Ralph's idea that the Venus
radar maps are produced by only choosing the data they want to see.
--
Rabbit arithmetic - 1 plus 1 equals 10
Remove spam and invalid from address to reply.

Jonathan Silverlight writes:
In message , Craig Markwardt
writes

"ralph sansbury" writes:
You and George have not clearly answered the question as to
the possibility and probability of sine functions with other
frequencies near the one frequency detected using the FFT
procedure and phase locked loops.

The probability of such an occurrence is essentially zero. Only a
spacecraft moving on Pioneer 10's trajectory, or one very near it
(within a few kilometers) and with very neary the same motion (within
1 mm/s). A different trajectory is ruled out at extremely high
confidence.

I wonder if this is another variation of Ralph's idea that the Venus
radar maps are produced by only choosing the data they want to see.

He has so many "ideas" and he jumps so randomly between them, that
it's hard to keep track. The simple fact is that if one selects the
data with his "idea" in mind (some fixed, small, amount of light
travel time), then the the tracking solution is destroyed.

Craig

The noisy variation (error bars) around a mean or linear or
polynomial regression line is analagous to a sine curve fitting
tens of thousands of sample voltage values per second for many
seconds. Of course there is no least squares estimation procedure
to calculate the sine curve. Instead there is a FFT(FT) procedure
of selecting as often as necessary, a possible frequency and
multiplying the sine function values of this frequency times the
observed sequence of voltage samples and computing the average of
these products.
1)If the average is half the rms value of the selected sine
function(and observed sequence of voltage samples) then the
difference frequency is zero.
(from the equation sinA*sinB =(1/2)(cos(A-B)-cos(A+B) and the
fact that the av value of any cosine function except cos(0)=1 is
zero)
2)If the average is taken over .5sec(or .0005sec) for example
and the difference frequency is (1.000001-1)MHz=1
Hz(or(1.001-1)MHZ= 1kHz) etc then the average will be 1/4 the
rms value of the orginal.
3)So if the average is taken over a few seconds and it is
less than 1/4 the rms value then one can conclude the frequencies
are not equal. and you have to try another frequency.
But the FFT procedure here assumes that the unknown sequence
of voltage samples is a clear sine curve or periodic sequence of
voltage samples as in the analysis of the whine of a defective
gear or bearing a machine.
In the context of analysing a radar signal with clutter of
various sorts or the Pioneer 10 transmission billions of miles
away there is noise to contend with.
How does the picture you refer to, explain how this is handled?
Please explain how the frequencies in the Gaussian power curve
around the chosen frequency in the FFT graph can be ignored
when the S/N ratio of .01 is asserted.
Ralph

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...

"ralph sansbury" wrote in message
...

"George G. Dishman" wrote in
message
om...
"ralph sansbury" wrote in message
...

Or it could mean that the frequency received had
more
or
less Doppler shift than predicted.

No, that would result in a single peak somewhere other
than the expected position. In fact that is the nature
of the anomaly reported by Anderson et al, the signal
at the end of 1994 was about 3Hz away from where it
was expected.

A small shift over any small time interval might be
inside
the error bars but if it is sustained over many such small
time
this is like the error bars of a sample mean being 1/sqrt(n)
where n is the size of the sample.

There are no "error bars". The signal just needs to be
within the band being examined. Page 10 of DSN document
209, which I keep suggesting you look at, shows that
the smallest bandwidth is 1kHz. As long as the signal
is in that or an adjacent band, it will be found.

You say that you get a normal curve with the peak at
this
frequency and that you integrate under the curve to get
the power
and that would seem to imply your summands or integrands
include
power associated with greater and lesser frequencies
around the
central frequency.

Jitter turns a high narrow peak into a smaller, broader
peak

Jitter connotes interference of parts of the circuitry
on one
another

No, the effect is produced by the noise included with the
signal.

and a small back and forth movement of a distinct wave
form on the scope ie small eg .1 cycle symmetric changes in
phase
of a distinct wave.

That's right, it describes the effect, not the cause.

Here the wave form on the scope is not distinct since the
true
waveform is embedded in noise

For the example you were discussing of a signal to noise
voltage ratio of 100:1, the noise amplitude is ~1% of the
signal so the phase jitter would be around 1 degree rms.
What you would see would be completely indistinguishable
from a pure sine wave but moving slightly back and forth
as you describe.

The total power is just that fraction of what was
transmitted
that impinges on the receive anntenna.

plus all sorts of other noisy radiation and noise within
the
receiver circuitry.

Only that part of the received noise that falls within the
width of the peak and receiver noise is negligible due to
the LNA.

This says to me that the peak frequency is the most
likely
frequency in this particular "sample" but that a .99
confidence
interval for the "population" frequency would be plus or
minus 3
standard deviations around this sample frequency.
The SAMPLING of the population here could be regarded
as many
hypothetical repetitions of the receiving of radiation
procedure
over the same time interval.

It is more complex.
What is complex is the way you are jumping to
another
but related viewpoint:
A decision procedure that will give for the long term a
certain number
of rejections of frequency estimates when they are true and
acceptance
of frequency estimates when they are false


For random noise you have a distibution
of component amplitudes and the probability of getting a
false detection depends on how far above the mean level you
set the threshold. There are two factors, the noise has to
be much higher than average and the signal has to be much
lower than the average, both rare events anway, before the
noise can exceed the signal.


Again though, such a false detection is incredibly unlikely
to be repeated at the same frequency on the repeat test
done
some time later on a new set of samples, the PLL would not
lock on, the sub-carrier would not be present and the data
correction would indicate an unusable Bit Error Rate.


Maybe but what are the reasons?

Reasons for what? Each of the aspects I listed needs a
completely different answer. What you say next doesn't
seem related to any of the above. Can you deal with them
separately please.

I sense that over billions
or millions of repetitions of zero crossings at the same
interval
or on average at the same interval with small symmetric
jitter
like
deviations at each interval imposed on the observed sequence
of voltage values that such a specific frequency is analagous
to
a specific sample mean of billions or millions of individual
samples and
so the true frequency confidence interval of plus or minus 3
standard
deviations divided by the sqrt(a billion).
What would this be in Hz?

George

"Craig Markwardt" wrote in
message news

"ralph sansbury" writes:
You and George have not clearly answered the question as
to
the possibility and probability of sine functions with other
frequencies near the one frequency detected using the FFT
procedure and phase locked loops.

The probability of such an occurrence is essentially zero.
Only a
spacecraft moving on Pioneer 10's trajectory, or one very near
it
(within a few kilometers) and with very neary the same motion
(within
1 mm/s). A different trajectory is ruled out at extremely high
confidence.

That is what you are trying to find out. You observe a sequence
of voltages with a pattern that suggests zero crossings at
regular intervals but what exactly are the regular intervals is
the question. When the signal is very weak there are a lot of
such intervals in which noise greater than zero makes you think
there is no zero crossing where there should be one. Similarly
noise added to a non zero signal value will give zero when there
should not be a zero.


Your supposition of an unaccounted-for Doppler shift is
irrelevant. A
Doppler shift would shift the whole peak. Since, by
construction the
tracking hardware can detect any carrier signal within the
bandpass,
the spacecraft signal would still be detected. That is, after
all,
the purpose of the tracking system: to detect unaccounted-for
changes
in the spacecraft motion, and based on that, apply corrections
to the
spacecraft navigation.


We are not talking about navigation at this point but just
about Doppler shifts within the bandpass but not as predicted.

Your supposition of a harmonic is completely unsupported. The
first
harmonic of the carrier is at 4.5 GHz, which is not even in the
S-band.
I am not talking about harmonics of the carrier but of
harmonics of any other frequency in a Fourier representation of
the observed periodic pattern.

And, your speculation of a light travel time of a few seconds
is
utterly unfounded. As I already pointed out, there are many
cases
(about 30% of the data set) where the uplink transmitter was
off, and
yet at the same time, high quality downlink signal and
telemetry were
still received.
If you are saying that the uplink transmitter was of at the
site where downlink doppler signals were being received
then I agree.
But if you are saying merely that computers said telemetry
reception occurred that must have been sent hours before from
another site then I disagree

There is no way your supposed scenario can function
in those cases.


And furthermore, assuming that the light travel time is
different than
d/c completely destroys the Doppler tracking solution.
Assuming that
the light travel time to a few seconds causes residuals of
thousands
of Hertz. Based on the expected rms of a few mHz, that
assumption is
ruled out with essentially 100% confidence. Even a change of
the
speed of light by one part in one million is ruled out.

CM

"Craig Markwardt" wrote in
message news

"ralph sansbury" writes:

And furthermore, assuming that the light travel time is
different than
d/c completely destroys the Doppler tracking solution.
I say it is about 1 second and that though the speed of light
as
a wave front or whatever implies the Doppler formula this is
not an if and only if sort of implication. The Doppler formula
could be simply due to the movement of the receiver over the
one second that the received signal rises above threshold in the
receiver.

Assuming that
the light travel time to a few seconds causes residuals of
thousands
of Hertz. Based on the expected rms of a few mHz, that
assumption is
ruled out with essentially 100% confidence. Even a change of
the
speed of light by one part in one million is ruled out.


No it is not ruled out because (1+v/c)f is different in all of
these cases
where c is still needed in the formula but has nothing to do with
the
speed of massless photons or wave fronts but rather with the
"elasticity"
of the oscillating charge inside electrons and protons in the
receiving material.
Ralph

"Craig Markwardt" wrote in
message news

"ralph sansbury" writes:
2)Am concerned now about what to do when a physical eof is
encountered so as to continue reading after it and changing
the
subsequent fields as before?

There is no data beyond the end of the file. Do not change the
original data fields.

I assume the named file eg87037t071 with
9209088 bytes has a physical "eof" at the end of this file and
that this
is the only eof that is responded to by the io system, but does
not respond to
what TRK2-25 calls a "software eof".
And does not respond to what you call "markers
which comes at the end of each
"physical record". ?
In your notes you describe a function that counts the number
of bytes in a "file"
which in this example gives you 9209088 and that you divide this
number
by (28 times 288) +1 =8065 and this is the number of "physical
records"
in the "file". Thus the first byte in the 29th logical record is
byte number 8066
etc..
I hope I can use the gcount() function in the c++ ifstream
function to do this and maybe you
know of a good reference on how to use the other functions in
ifstream to
go to selected bytes in the file without reading each byte and
ignoring or
using depending on a programmed condition?
Ralph

"ralph sansbury" wrote in message
...
The noisy variation (error bars) around a mean or linear or
polynomial regression line is analagous to a sine curve fitting
tens of thousands of sample voltage values per second for many
seconds.

Yes, but that is not relevant to an FFT.

Of course there is no least squares estimation procedure
to calculate the sine curve.

Right.

Instead there is a FFT(FT) procedure
of selecting as often as necessary, a possible frequency and
multiplying the sine function values of this frequency times the
observed sequence of voltage samples and computing the average of
these products.

Not quite. An FFT involves calculating the amplitude
of _every_ possible frequency in the set of samples.
It is the combination of _all_ the componenets that
reproduces the original set.

1)If the average is half the rms value of the selected sine
function(and observed sequence of voltage samples) then the
difference frequency is zero.

and it is exactly that vaule that is calculated for
each component therefore there are no errors.

(from the equation sinA*sinB =(1/2)(cos(A-B)-cos(A+B) and the
fact that the av value of any cosine function except cos(0)=1 is
zero)

Right.

2)If the average is taken over .5sec(or .0005sec) for example
and the difference frequency is (1.000001-1)MHz=1
Hz(or(1.001-1)MHZ= 1kHz) etc then the average will be 1/4 the
rms value of the orginal.
3)So if the average is taken over a few seconds and it is
less than 1/4 the rms value then one can conclude the frequencies
are not equal. and you have to try another frequency.

Wrong. An FFT does not try to look for individual frequencies,
it calculates _all_ of them.

But the FFT procedure here assumes that the unknown sequence
of voltage samples is a clear sine curve or periodic sequence of
voltage samples as in the analysis of the whine of a defective
gear or bearing a machine.
In the context of analysing a radar signal with clutter of
various sorts or the Pioneer 10 transmission billions of miles
away there is noise to contend with.

Wrong again. In a sample of a given length, the series of
entirely random voltage samples is exactly reproduced by
the combination of _all_ possible frequencies. There is
no assumption made and there is no error regardless of
the type of signal.

How does the picture you refer to, explain how this is handled?

If you look at the description of the "baseband channel",
it states "Variable width, 1 kHz – 16 MHz". This tells
you that the smallest width is 1kHz and you can see the
shape has a flat top that says all frequencies are treated
with equal gain so there is no bias.

Please explain how the frequencies in the Gaussian power curve
around the chosen frequency in the FFT graph can be ignored
when the S/N ratio of .01 is asserted.

We were talking of signal to noise of 100:1, not 0.01
but remember the ratio depends on the bandwidth. If I
represent the output from the FFT in a similar fashion
plotting power versus frequency it might look like this
for a very broad, low peak:

| |
| |
| /\ |
|_________/ \__________________|
| |
+-------------------------------+

In reality it would look more like this:

| |
| | |
| | |
|__________|____________________|
+-------------------------------+
0 1kHz

because the gaussian would be too narrow to see on this
scale. The one we looked at some posts back was over a
very small frequency range.

If the signal was not where it was predicted, it might
look like this:

| |
| | |
| | |
|_____________|_________________|
+-------------------------------+
0 1kHz

The noise produces the broad flat line while the signal
is all concentrated in the single narrow peak. Since the
power ratio is the area under these curves, the FFT can
locate the signal where other techniques could not. The
diagrams above could represent 1:1 signal to noise ratio
in the 1kHz bandwidth of the sub-channel.

However, this only says where the signal is, it is then
up to the PLL to actually lock on to it. It is the PLL
that actually rejects the noise by using a very narrow
bandwidth once the signal has been found.

George

"George Dishman" writes:
Wrong again. In a sample of a given length, the series of
entirely random voltage samples is exactly reproduced by
the combination of _all_ possible frequencies. There is
no assumption made and there is no error regardless of
the type of signal.

George, to clarify even more, noise is the combination of all possible
frequencies, with *equal* expected amplitudes. Thus, no one frequency
is expected to be any stronger than the other. [ There will be some
fluctuations about this amplitude level of course. ]

Craig

In message , Craig Markwardt
writes

"George Dishman" writes:
Wrong again. In a sample of a given length, the series of
entirely random voltage samples is exactly reproduced by
the combination of _all_ possible frequencies. There is
no assumption made and there is no error regardless of
the type of signal.

George, to clarify even more, noise is the combination of all possible
frequencies, with *equal* expected amplitudes. Thus, no one frequency
is expected to be any stronger than the other. [ There will be some
fluctuations about this amplitude level of course. ]

Just to nitpick a bit and educate myself, aren't there different types
of noise? "Pink" noise and "white" noise, for instance, with different
spectra?
--
Rabbit arithmetic - 1 plus 1 equals 10
Remove spam and invalid from address to reply.

"Jonathan Silverlight" wrote
in message ...
In message , Craig Markwardt
writes

"George Dishman" writes:
Wrong again. In a sample of a given length, the series of
entirely random voltage samples is exactly reproduced by
the combination of _all_ possible frequencies. There is
no assumption made and there is no error regardless of
the type of signal.

George, to clarify even more, noise is the combination of all possible
frequencies, with *equal* expected amplitudes. Thus, no one frequency
is expected to be any stronger than the other. [ There will be some
fluctuations about this amplitude level of course. ]

Just to nitpick a bit and educate myself, aren't there different types
of noise? "Pink" noise and "white" noise, for instance, with different
spectra?

That's right. 'Pink' implies there is more power at lower
frequencies while 'white' means equal power across the band.
The terms describe a trend so are gradual changes and we are
talking about a carrier at 2.295GHz. If the spectral power
density varied by 2:1 from 1.7GHz to 2.7GHz, it would vary
by only 1.000001:1 over the 1kHz baseband filter which of
course is far less than the intrinsic statistical variation
from component to component that Craig mentions. In other
words, pink noise is white when considering a sufficiently
small bandwidth. To show any structure in the FFT, you would
need a background noise source with a linewidth less than 1kHz.

George