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Question For Craig Markwardt re Pioneer 10 Data



 
 
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  #1  
Old November 11th 03, 06:16 PM
ralph sansbury
external usenet poster
 
Posts: n/a
Default 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.






  #2  
Old November 13th 03, 07:20 AM
Craig Markwardt
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Posts: n/a
Default Question For Craig Markwardt re Pioneer 10 Data


"ralph sansbury" writes:
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?


You appear to be asking about the signal to noise ratio. The ATDF
Doppler tracking records do not appear to contain a measure of the
carrier signal to noise ratio. However, data quality measures for the
Doppler data *are* present, including Doppler noise and estimate cycle
slips, which are of course the most relevant quantities for Doppler
tracking.

However, there are published measures of the carrier signal to noise
ratio. For example, Watola (1992) shows a plot from Pioneer 10 on 19
December, 1991, where the signal to noise ratio is 20 dB for a
bandwidth of ~0.2 Hz, which means the mean signal is stronger than the
noise by a factor of 100. The epoch of the measurement is in the
middle of the data arc of published papers, such as Anderson et al or
my own paper. Of course, signal to noise ratio depends on many
environmental and technical factors, not always constant.

The "average Doppler residual" field of ATDF records is not a signal
to noise measure. Rather, it is the difference between the measured
and predicted Doppler frequencies. I.e., the navigation group
prepared a prediction of the Doppler frequency before the track, and
the "residual" field represents the difference between the measured
frequency and the prediction.

CM

References
Watola, D. A. 1992, TDA Progress Report 42-111
  #3  
Old November 15th 03, 02:14 PM
ralph sansbury
external usenet poster
 
Posts: n/a
Default Question For Craig Markwardt re Pioneer 10 Data


"Craig Markwardt" wrote in
message news

"ralph sansbury" writes:
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?


You appear to be asking about the signal to noise ratio. The

ATDF
Doppler tracking records do not appear to contain a measure of

the
carrier signal to noise ratio. However, data quality measures

for the
Doppler data *are* present, including Doppler noise and

estimate cycle
slips, which are of course the most relevant quantities for

Doppler
tracking.


Yes I was afraid of that from looking at some of the papers
you
suggested in your notes. And I dont see how to relate "Doppler
Noise"
and "cycle slips" in the data. Re 100 times
stronger signal over noise to this clearer statistical accuracy
measure.
Maybe obs(i)=sig(i)+error(i) where error(i) is approximately
sig(i)/100
That is when you subtract the voltage changes observed over
time,obs(i), from|
one or a sequence of systematic sine voltages
obtained using Fast Fourier Transform techniques etc, sig(i), the
error(i) would
have this average magnitude, sometimes it would be 1/99 sometimes
1/101 etc.




However, there are published measures of the carrier signal to

noise
ratio. For example, Watola (1992) shows a plot from Pioneer 10

on 19
December, 1991, where the signal to noise ratio is 20 dB for a
bandwidth of ~0.2 Hz, which means the mean signal is stronger

than the
noise by a factor of 100. The epoch of the measurement is in

the
middle of the data arc of published papers, such as Anderson et

al or
my own paper. Of course, signal to noise ratio depends on many
environmental and technical factors, not always constant.

The "average Doppler residual" field of ATDF records is not a

signal
to noise measure. Rather, it is the difference between the

measured
and predicted Doppler frequencies. I.e., the navigation group
prepared a prediction of the Doppler frequency before the

track, and
the "residual" field represents the difference between the

measured
frequency and the prediction.


I was afraid of that also, that I couldn't use the average
doppler residual
But it is good to know one can use the papal Watola measure of
signal to noise at one time
as representative of the signal to noise for years around 1991.
I gather that the Anomalous Doppler Shift measurments are 100
times
greater than noise and that they are systematic over time?.



CM

References
Watola, D. A. 1992, TDA Progress Report 42-111



  #4  
Old November 17th 03, 08:49 AM
Craig Markwardt
external usenet poster
 
Posts: n/a
Default Question For Craig Markwardt re Pioneer 10 Data


"ralph sansbury" writes:
I gather that the Anomalous Doppler Shift measurments are 100
times greater than noise and that they are systematic over time?.


Question 1: from a statistical standpoint, that is approximately
correct; the statistical errors in the anomalous acceleration are a
few percent of the measured value. Question 2: yes, it is systematic
over time.

CM

  #5  
Old November 17th 03, 04:41 PM
ralph sansbury
external usenet poster
 
Posts: n/a
Default Question For Craig Markwardt re Pioneer 10 Data


"Craig Markwardt" wrote in
message news

"ralph sansbury" writes:
I gather that the Anomalous Doppler Shift measurments are

100
times greater than noise and that they are systematic over

time?.

Question 1: from a statistical standpoint, that is

approximately
correct; the statistical errors in the anomalous acceleration

are a
few percent of the measured value.


Craig, ..
Yes I see from the papers you refer to that the difference
between the received sky frequency(oscillating voltage vlaues)
and the known transmitted frequency(voltage values) is obtained
digitally as follows: a sequence of oscillating voltages is
received from the sky where the spacecraft is, through a filter
that suppresses frequencies outside the expected range and this
is input to one gate of a dual gate transistor while an
oscillation at a preset frequency is input to the other gate so
that the output of the transistor controlled by both of these
inputs contains the sum ,difference, and both input frequencies.
A resonance tuner picks out the difference frequency and a a
sequence of voltages at this difference frequency.(mixer and
repeated heterodyne up and down conversion etc is the jargon and
the engineering details I am trying to avoid).
This sequence of oscillations is digitized into a set of 1s
and 0s (1 if the analogue voltage is above a certain amount etc)
and an Fast Fourier Transform procedure is used to find the
underlying "sine" pattern of 1s and 0s that most closely fits
this. Since the incoming frequency is constantly changing
slightly because of the motions of the earth etc, the detected
underlying sine patterns will change.
The 100 times greater noise figure in the Watola paper means
then that the differences between the FFT sequence value and the
corresponding observed sequence value is zero, 99 times out of a
100.
Is this what Watola says? I should hope this is quoted
somewhere in the Anderson papers or yours?
Ralph



Question 2: yes, it is systematic
over time.

CM



  #6  
Old November 17th 03, 11:03 PM
George Dishman
external usenet poster
 
Posts: n/a
Default Question For Craig Markwardt re Pioneer 10 Data

Hi Ralph,

"ralph sansbury" wrote in message
...

Craig, ..
Yes I see from the papers you refer to that the difference
between the received sky frequency(oscillating voltage vlaues)
and the known transmitted frequency(voltage values) is obtained
digitally as follows: a sequence of oscillating voltages is
received from the sky where the spacecraft is, through a filter
that suppresses frequencies outside the expected range and this
is input to one gate of a dual gate transistor while an


Where did you read that it was a dual gate transistor?

oscillation at a preset frequency is input to the other gate so
that the output of the transistor controlled by both of these
inputs contains the sum ,difference, and both input frequencies.
A resonance tuner


Wrong, we have been over this dozens of times. It is
digitally filtered, there is no "resonance tuner"
involved. The characteristics are fundamentally
different.

picks out the difference frequency and a a


Wrong, it doesn't pick out one frequency, it passes a
complete band of frequencies to the FFT.

http://eis.jpl.nasa.gov/deepspace/ds...05/209/209.pdf

See page 10 (yet again).

sequence of voltages at this difference frequency.(mixer and
repeated heterodyne up and down conversion etc is the jargon and
the engineering details I am trying to avoid).


Instead you are inventing a process that doesn't exist
and describing it in far more (and incorrect) detail
than exists in the published documentation.

This sequence of oscillations is digitized into a set of 1s


The entire band is digitised.

and 0s (1 if the analogue voltage is above a certain amount etc)
and an Fast Fourier Transform procedure is used to find the
underlying "sine" pattern of 1s and 0s that most closely fits
this.


Nope, the amplitude of _all_ frequencies in the band is
calculated and passed on to the next stage without any
judgement. Ralph, it's as if all the weeks I spent
talking you through the DSN documentation by email had
never happened. You seemed to grasp it at the time, why
have you reverted to this grossly inaccurate description
of the process?

George


  #7  
Old November 18th 03, 04:14 PM
ralph sansbury
external usenet poster
 
Posts: n/a
Default Question For Craig Markwardt re Pioneer 10 Data


"George Dishman" wrote in message
...
Hi Ralph,

"ralph sansbury" wrote in message
...

Craig, ..
Yes I see from the papers you refer to that the difference
between the received sky frequency(oscillating voltage

vlaues)
and the known transmitted frequency(voltage values) is

obtained
digitally as follows: a sequence of oscillating voltages is
received from the sky where the spacecraft is, through a

filter
that suppresses frequencies outside the expected range and

this
is input to one gate of a dual gate transistor while an


Where did you read that it was a dual gate transistor?


It is surely more complex than this but this is the principle
used in obtaining an intermediate frequency.


oscillation at a preset frequency is input to the other gate

so
that the output of the transistor controlled by both of these
inputs contains the sum ,difference, and both input

frequencies.
A resonance tuner


Wrong, we have been over this dozens of times. It is
digitally filtered, there is no "resonance tuner"
involved. The characteristics are fundamentally
different.

I appreciate your pointing out to me how bandpass filters
made up of low pass and high pass can do the same thing as
resonant filters. But I am not talking about such filters now.
I am talking about the way a typical mixer that produces the
intermdiate frequency is tuned typically and if there is a
digital version here, then feel free to explain it. I did not
understand your email explanations.


picks out the difference frequency and a a


Wrong, it doesn't pick out one frequency, it passes a
complete band of frequencies to the FFT.

OK relax. It is a small band of frequencies around the single
difference frequency. This is always understood.

http://eis.jpl.nasa.gov/deepspace/ds...05/209/209.pdf


See page 10 (yet again).



sequence of voltages at this difference frequency.(mixer and
repeated heterodyne up and down conversion etc is the jargon

and
the engineering details I am trying to avoid).


Instead you are inventing a process that doesn't exist
and describing it in far more (and incorrect) detail
than exists in the published documentation.



If you understand by difference frequency a small band around
a specific difference frequency then there is no problem. This is
obviously the meaning of what I have said.

This sequence of oscillations is digitized into a set of

1s

The entire band is digitised.


It is clearer to say that the observed sequence of
oscillations is digitized.


and 0s (1 if the analogue voltage is above a certain amount

etc)
and an Fast Fourier Transform procedure is used to find the
underlying "sine" pattern of 1s and 0s that most closely fits
this.


Nope, the amplitude of _all_ frequencies in the band is
calculated and passed on to the next stage without any
judgement.


I am talking about the final stage and I mentioned that the
movement of the Earth etc requires different patterns to be
obtained successively but the point is that the FFT procedure
finds the underlying pattern and it is this that is used to
compare to the given sequence of 1s and 0s.
This is the procedure I understood from your comments and
various books and links.

Ralph, it's as if all the weeks I spent
talking you through the DSN documentation by email had
never happened. You seemed to grasp it at the time, why
have you reverted to this grossly inaccurate description
of the process?

Again I think you have misunderstood what I have said. I
dont think it is inaccurate if you replace single intermediate
frequency by small range of frequencies around the single
intermediate frequency.
And if you want to try and describe the digital version of the
mixer please do so. It was not clear from your emails.
I was simply trying to summarize what I understood from your
sometimes helpful comments and the various links and texts I have
been reading.

Ralph


  #8  
Old November 18th 03, 06:50 PM
George Dishman
external usenet poster
 
Posts: n/a
Default Question For Craig Markwardt re Pioneer 10 Data


"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...
Hi Ralph,

"ralph sansbury" wrote in message
...

Craig, ..
Yes I see from the papers you refer to that the difference
between the received sky frequency(oscillating voltage

vlaues)
and the known transmitted frequency(voltage values) is

obtained
digitally as follows: a sequence of oscillating voltages is
received from the sky where the spacecraft is, through a

filter
that suppresses frequencies outside the expected range and

this
is input to one gate of a dual gate transistor while an


Where did you read that it was a dual gate transistor?


It is surely more complex than this but this is the principle
used in obtaining an intermediate frequency.


OK, the principe used is multiplication. A dual gate
device is one way of producing multiplication but is
very specific and quite possibly incorrect.

oscillation at a preset frequency is input to the other gate so
that the output of the transistor controlled by both of these
inputs contains the sum ,difference, and both input frequencies.
A resonance tuner


Wrong, we have been over this dozens of times. It is
digitally filtered, there is no "resonance tuner"
involved. The characteristics are fundamentally
different.

I appreciate your pointing out to me how bandpass filters
made up of low pass and high pass can do the same thing as
resonant filters. But I am not talking about such filters now.
I am talking about the way a typical mixer that produces the
intermdiate frequency is tuned typically


That is my point, the mixer is not tuned at all.

and if there is a
digital version here, then feel free to explain it. I did not
understand your email explanations.


picks out the difference frequency and a a


Wrong, it doesn't pick out one frequency, it passes a
complete band of frequencies to the FFT.

OK relax. It is a small band of frequencies around the single
difference frequency. This is always understood.


No, it is a very _wide_ band of frequencies selected out of
an even wider band as shown in the diagrams:

http://eis.jpl.nasa.gov/deepspace/ds...05/209/209.pdf


See page 10 (yet again).



sequence of voltages at this difference frequency.(mixer and
repeated heterodyne up and down conversion etc is the jargon and
the engineering details I am trying to avoid).


Instead you are inventing a process that doesn't exist
and describing it in far more (and incorrect) detail
than exists in the published documentation.



If you understand by difference frequency a small band around
a specific difference frequency then there is no problem. This is
obviously the meaning of what I have said.


I know, but it is wrong. It is not a small band, it is
a wide band.

This sequence of oscillations is digitized into a set of 1s


The entire band is digitised.


It is clearer to say that the observed sequence of
oscillations is digitized.


Again it may be clearer but it is wrong. It is not just
the carrier oscillations that are digitised, it is the
whole signal, oscillations plus random thermal noise
and any other sources such as the galactic background.

and 0s (1 if the analogue voltage is above a certain amount etc)
and an Fast Fourier Transform procedure is used to find the
underlying "sine" pattern of 1s and 0s that most closely fits
this.


Nope, the amplitude of _all_ frequencies in the band is
calculated and passed on to the next stage without any
judgement.


I am talking about the final stage


The final stage is the carrier PLL, not the FFT. All
the FFTs are removed from the chain once the PLL locks
on and they play no further part in the process. It is
the PLL that tracks the drifting signal and gives us
the accurate measurement.

and I mentioned that the
movement of the Earth etc requires different patterns to be
obtained successively but the point is that the FFT procedure
finds the underlying pattern and it is this that is used to
compare to the given sequence of 1s and 0s.


No it isn't. The final FFT is only used to set initial
frequency for the carrier PLL. If that locks, the
bandwidth is reduced to improve the signal/noise ratio.
The output of that is fed to the sub-carrier PLL. Again
that has to lock before the signal can be decoded using
a phase detector. Then it gets decoded through the error
correction scheme. There are many critical steps after
the FFT, and in fact the FFT plays no part in the decoding
process whatsoever.

This is the procedure I understood from your comments and
various books and links.


Then you need to read my mails again. I have gone over
this at least a dozen times with you and asked repeatedly
to look at page 10 of handbook 209 where the bandwidths
and IF processing are laid out in simple charts.

Ralph, it's as if all the weeks I spent
talking you through the DSN documentation by email had
never happened. You seemed to grasp it at the time, why
have you reverted to this grossly inaccurate description
of the process?

Again I think you have misunderstood what I have said. I
dont think it is inaccurate if you replace single intermediate
frequency by small range of frequencies around the single
intermediate frequency.


It is very inaccurate when the DSN document tells you
the analog band is the digitised band is 110MHz wide and
the signals of interest are of the order of 1Hz wide.

And if you want to try and describe the digital version of the
mixer please do so. It was not clear from your emails.


The mixer is analog. The output is digitised and a baseband
extracted as shown on page 10. The details of the method
of mixing are not given but the principle is simply
multiplication of the incoming signal (including noise) by
the reference sine wave.

V_out = V_in * V_ref

where

V_ref = A * sin(wt)

I was simply trying to summarize what I understood from your
sometimes helpful comments and the various links and texts I have
been reading.


OK, but somehow you have gone back to using all the
descriptions I told you were wrong. We went over and
over this stuff many times and at the end you seemed
to understand it. Now a few weeks later you have
reverted to your original inaccurate descriptions.

George


  #9  
Old November 19th 03, 07:52 PM
ralph sansbury
external usenet poster
 
Posts: n/a
Default Question For Craig Markwardt re Pioneer 10 Data

George, I know you are a superior EE and that
you understand or at least can imitate the jargon in the nasa
documentation.
But you and nasa are talking to EEs and not to an audience
that has some elementary understanding of electrical circuits
and basic physics.
I am trying to represent this in a way which is intelligible
without going into details that obscure the basic procedure
and why it is reliable.
I am surprised at your lack of understanding of this
and general lack of good will.

----- Original Message -----
From: "George Dishman"
Newsgroups: sci.astro
Sent: Tuesday, November 18, 2003 1:50 PM
Subject: Question For Craig Markwardt re Pioneer 10 Data



"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...
Hi Ralph,

"ralph sansbury" wrote in message
...

Craig, ..
Yes I see from the papers you refer to that the

difference
between the received sky frequency(oscillating voltage

vlaues)
and the known transmitted frequency(voltage values) is

obtained
digitally as follows: a sequence of oscillating

voltages is
received from the sky where the spacecraft is, through a

filter
that suppresses frequencies outside the expected range

and
this
is input to one gate of a dual gate transistor while an

Where did you read that it was a dual gate transistor?


It is surely more complex than this but this is the

principle
used in obtaining an intermediate frequency.


OK, the principe used is multiplication.

No the principle is what I said. The law of superposition
suggests
electrical fields are added, not multiplied. When you say
something
like this you should also say that the sum of sine functions etc
can
be represented as a product of related sine functions etc.

A dual gate
device is one way of producing multiplication but is
very specific and quite possibly incorrect.


Watch your language. Your are contradicting yourself.
You mean inexact, it is correct in principle as you just said.


oscillation at a preset frequency is input to the other

gate so
that the output of the transistor controlled by both of

these
inputs contains the sum ,difference, and both input

frequencies.
A resonance tuner

Wrong, we have been over this dozens of times. It is
digitally filtered, there is no "resonance tuner"
involved. The characteristics are fundamentally
different.

I appreciate your pointing out to me how bandpass

filters
made up of low pass and high pass can do the same thing as
resonant filters. But I am not talking about such filters

now.
I am talking about the way a typical mixer that produces

the
intermediate frequency is tuned typically


That is my point, the mixer is not tuned at all.

Please explain how a mixer works if it does not contain one
sort
of tuning to get the difference frequency.
out of the superposition or combination of the input frequencies,
another
sort to get the sum frequency, another sort to get the single
frequencies.
By tuning I mean a resonant inductance and capacitance to tune or
filter
out all but the chosen frequency or perhaps striclty capacitive
high and low pass filters.


and if there is a
digital version here, then feel free to explain it. I did not
understand your email explanations.

Please answer this question.




picks out the difference frequency and a a

Wrong, it doesn't pick out one frequency, it passes a
complete band of frequencies to the FFT.

OK relax. It is a small band of frequencies around the

single
difference frequency. This is always understood.


No, it is a very _wide_ band of frequencies selected out of
an even wider band as shown in the diagrams:


We are talking about the size of the intermediate frequency range
relative
to the original range 1MHz is small relative to 200MHz but not
to 1Hz



sequence of voltages at this difference frequency.(mixer

and
repeated heterodyne up and down conversion etc is the

jargon and
the engineering details I am trying to avoid).

Instead you are inventing a process that doesn't exist
and describing it in far more (and incorrect) detail
than exists in the published documentation.


As I have detailed above you are misunderstanding what I
am saying




If you understand by difference frequency a small band

around
a specific difference frequency then there is no problem.

This is
obviously the meaning of what I have said.


I know, but it is wrong. It is not a small band, it is
a wide band.

1MHz is small relative 200MHz but not to 1Hz. Obviously
that
is what is meant here when we are talking about the reduction of
the sky frequency to something that is more amenable to analysis
And the reason this is possible is that the difference between
the lower
frequencies is the same as that between the higher.


This sequence of oscillations is digitized into a set

of 1s

The entire band is digitised.


It is clearer to say that the observed sequence of
oscillations is digitized.


Again it may be clearer but it is wrong. It is not just
the carrier oscillations that are digitised, it is the
whole signal, oscillations plus random thermal noise
and any other sources such as the galactic background.

Your understanding is wrong. I did not say CARRIER
oscillations
You can't change the meaning of
'oscillations' to mean only the
part due to the spacecraft transmitter
I accept your apology but maybe you are similarly misreading
the nasa documents and that is why you are missing the essence
of the procedure. You cant see the forest from the trees.

and 0s (1 if the analogue voltage is above a certain

amount etc)
and an Fast Fourier Transform procedure is used to find

the
underlying "sine" pattern of 1s and 0s that most closely

fits
this.

Nope, the amplitude of _all_ frequencies in the band is
calculated and passed on to the next stage without any
judgement.


I am talking about the final stage


The final stage is the carrier PLL, not the FFT. All
the FFTs are removed from the chain once the PLL locks
on and they play no further part in the process. It is
the PLL that tracks the drifting signal and gives us
the accurate measurement.


The FFT as I was using the term includes the PLL. The point
which you insist on obscuring is that this technique
gets at the right sine frequency starting at the right time from
the
sum of sine functions of various frequencies equivalent as
Fourier showed to
the noisy oscillations observed.


and I mentioned that the
movement of the Earth etc requires different patterns to be
obtained successively but the point is that the FFT procedure
finds the underlying pattern and it is this that is used to
compare to the given sequence of 1s and 0s.


No it isn't. The final FFT is only used to set initial
frequency for the carrier PLL. If that locks, the
bandwidth is reduced to improve the signal/noise ratio.


You are saying the same thing that I was saying. I think
it is clearer to say it without the jargon.

The output of that is fed to the sub-carrier PLL.

Whatever the details a sequence of 1s and 0s
is obtained that is a digitised intermediate version of the
sky frequency.



Again
that has to lock before the signal can be decoded using
a phase detector. Then it gets decoded through the error
correction scheme. There are many critical steps after
the FFT, and in fact the FFT plays no part in the decoding
process whatsoever.


The bottom line is a sine representation of a sum of sine
frequency represention
of an oscillating pattern made possible by the FFT procedure
essentially
and this includes the phase locked loop procedure perhaps
involving
the recognition of some code modulation of the carrier to insure
that
the fitted frequency starts at the right time.


The fact that this representation is a much smaller frequency
than
the GHz sky frequency is ok because when you look at the
difference
between this and a small frequency representation of the
transmitted
frequency the difference is the same as the difference between
the
original frequencies. And it is this difference that is used to
get the
Doppler shift.




This is the procedure I understood from your comments and
various books and links.



Ralph, it's as if all the weeks I spent
talking you through the DSN documentation by email had
never happened. You seemed to grasp it at the time, why
have you reverted to this grossly inaccurate description
of the process?

Again I think you have misunderstood what I have said.

I
dont think it is inaccurate if you replace single

intermediate
frequency by small range of frequencies around the single
intermediate frequency where small is relative the original

frequency.


It is very inaccurate when the DSN document tells you
the analog band is the digitised band is 110MHz wide and
the signals of interest are of the order of 1Hz wide.


You are quoting the wrong document. We are talking about the
intermediate frequency
being smaller that the original frequency. Is that so hard for
you to understand.
I am continually amazed that a person of your knowledge and
intelligence has so many
blindspots.

And if you want to try and describe the digital version of

the
mixer please do so. It was not clear from your emails.


The mixer is analog. The output is digitised and a baseband
extracted as shown on page 10. The details of the method
of mixing are not given but the principle is simply
multiplication of the incoming signal (including noise) by
the reference sine wave.

V_out = V_in * V_ref

where

V_ref = A * sin(wt)


This makes no sense. Electrical oscillations add by the law of
superposition;
Tthey dont multiply. The mathematical fact that a sum of sine
and cosine functions can
be represented as a product of related sine and cosine functions
has to be mentioned dont you think?
Ralph



  #10  
Old November 19th 03, 10:29 PM
George Dishman
external usenet poster
 
Posts: n/a
Default Question For Craig Markwardt re Pioneer 10 Data


"ralph sansbury" wrote in message
...
George, I know you are a superior EE and that


I like the word "adequate" in that respect.

you understand or at least can imitate the jargon in the nasa
documentation.


I understand most extremely well, I tell you when there
is something I don't understand.

But you and nasa are talking to EEs and not to an audience
that has some elementary understanding of electrical circuits
and basic physics.
I am trying to represent this in a way which is intelligible
without going into details that obscure the basic procedure
and why it is reliable.


Understood.

I am surprised at your lack of understanding of this
and general lack of good will.


Let me put it this way. Suppose I tried to explain America
to you and said "it is a land where all are equal" and you
said "that's too difficult to understand, I'm going to call
it a dictatorship, it's close enough because both are
political systems". Would you think that reasonable? It is
what you are doing.

"ralph sansbury" wrote in message
...

"George Dishman" wrote in message
...
Hi Ralph,


Where did you read that it was a dual gate transistor?

It is surely more complex than this but this is the principle
used in obtaining an intermediate frequency.


OK, the principe used is multiplication.

No the principle is what I said. The law of superposition suggests
electrical fields are added, not multiplied.


Lets write down some voltages for a mixer with two inputs
and an output. This is "in principle", actuals signal
levels are much smaller:

V_a V_b V_out

0.00 0.00 0.00
0.50 0.50 +0.25
-0.50 0.50 -0.25
0.50 -0.50 -0.25
-0.50 -0.50 +0.25
0.50 1.00 +0.50
1.00 1.00 +1.00
-1.00 0.50 -0.50
-1.00 1.00 -1.00
-2.00 2.00 -4.00
-3.00 3.00 -9.00

This is a multiplication table. Superposition doesn't mix
the signals, you have to build something special to do it.

When you say something
like this you should also say that the sum of sine functions etc can
be represented as a product of related sine functions etc.


You already know that so I don't need to say it again. Anyway
it is not relevant to the fact that a mixer takes voltage V_a
and voltage V_b and produces:

V_out = V_a * V_b

A dual gate
device is one way of producing multiplication but is
very specific and quite possibly incorrect.


Watch your language. Your are contradicting yourself.
You mean inexact, it is correct in principle as you just said.


I mean 'incorrect' in that they may be achieving multiplication
but not using a dual gate FET. If you say "multiplication" it
is non-specific and correct, saying "dual gate transistor" is
specific and may be wrong because they may be using a set of
matched diodes or something completely different.

oscillation at a preset frequency is input to the other gate so
that the output of the transistor controlled by both of these
inputs contains the sum ,difference, and both input frequencies.
A resonance tuner

Wrong, we have been over this dozens of times. It is
digitally filtered, there is no "resonance tuner"
involved. The characteristics are fundamentally
different.
I appreciate your pointing out to me how bandpass filters
made up of low pass and high pass can do the same thing as
resonant filters. But I am not talking about such filters now.
I am talking about the way a typical mixer that produces the
intermediate frequency is tuned typically


That is my point, the mixer is not tuned at all.

Please explain how a mixer works if it does not contain one sort
of tuning to get the difference frequency.


Tuning means selecting one frequency from many. The mixer works
by multiplication of two voltages, one is the reference and the
other is a wide band signals with many frequencies. The result
contains sum and difference frequencies as two wide bands but
these are also separated by a gap. The result is filtered, not
tuned, to discard one entire wide band but retain the other
entire wide band.

Suppose the signal is a band from 2265MHz to 2375MHz and it
is multiplied by a pure sine wave of 2000MHz. The sum is a
band from 4265MHz to 4375MHz while the difference is a band
from 265MHz to 375MHz. To get just the difference, all you
need is a simple filter that rejects anything above say
1500MHz but lets through everything below 500MHz with equal
gain (between 500MHz and 1500Mhz, the gain can fall gently).

out of the superposition or combination of the input frequencies, another
sort to get the sum frequency, another sort to get the single

frequencies.
By tuning I mean a resonant inductance and capacitance to tune or filter
out all but the chosen frequency


That is the correct meaning of 'tuning' in this context. The
point is that this is not done in the system, they work with
wide bands of frequencies and treat them with equal gain, so
cannot use tuning.

or perhaps striclty capacitive
high and low pass filters.


That is correct but very different. One filter might pass
everything above 250MHz equally but have a rapidly diminishing
gain below that. The other would pass everything below say
400MHz equally and have a rapidly diminishing gain above that.
In the region of interest, the gain is constant over all of the
band of frequencies.

and if there is a
digital version here, then feel free to explain it. I did not
understand your email explanations.

Please answer this question.


Sorry, I thought I had replied to that. The mixer is
analogue, not digital. Maybe it was in another reply.
My email explanations related to the later stages after
the signal is digitised.

picks out the difference frequency and a a

Wrong, it doesn't pick out one frequency, it passes a
complete band of frequencies to the FFT.

OK relax. It is a small band of frequencies around the single
difference frequency. This is always understood.


No, it is a very _wide_ band of frequencies selected out of
an even wider band as shown in the diagrams:


We are talking about the size of the intermediate frequency range
relative
to the original range 1MHz is small relative to 200MHz but not
to 1Hz


The terms "narrow band" and "wide band" compare the width
of the equipment to the width of the signal being processed.
Wide band in this context means sufficiently wide that it
does not exclude any frequency of interest or produce any
modification of the signal such as emphasising one frequency
more than another.

sequence of voltages at this difference frequency.(mixer

and
repeated heterodyne up and down conversion etc is the

jargon and
the engineering details I am trying to avoid).

Instead you are inventing a process that doesn't exist
and describing it in far more (and incorrect) detail
than exists in the published documentation.


As I have detailed above you are misunderstanding what I
am saying


What you are saying is very different to what is being
done. It may be that this is because you are using terms
in an unconventional manner but you will then hit problems
in referring to your text books.

If you understand by difference frequency a small band around
a specific difference frequency then there is no problem.

This is
obviously the meaning of what I have said.


I know, but it is wrong. It is not a small band, it is
a wide band.

1MHz is small relative 200MHz but not to 1Hz. Obviously
that
is what is meant here when we are talking about the reduction of
the sky frequency to something that is more amenable to analysis
And the reason this is possible is that the difference between
the lower
frequencies is the same as that between the higher.


I don't quite follow that. The examples frequency bands
I gave above may clarify the situation.

This sequence of oscillations is digitized into a set

of 1s

The entire band s digitised.

It is clearer to say that the observed sequence of
oscillations is digitized.


Again it may be clearer but it is wrong. It is not just
the carrier oscillations that are digitised, it is the
whole signal, oscillations plus random thermal noise
and any other sources such as the galactic background.

Your understanding is wrong. I did not say CARRIER
oscillations
You can't change the meaning of
'oscillations' to mean only the
part due to the spacecraft transmitter


"oscillations" means something regular, the noise is
random so not regular in any way and that is critically
important to extracting the signal. Your use of the word
is inappropriate and confusing.

I accept your apology but maybe you are similarly misreading
the nasa documents and that is why you are missing the essence
of the procedure. You cant see the forest from the trees.


NASA don't talk of 'oscillations', they correctly talk of
the signal.

and 0s (1 if the analogue voltage is above a certain

amount etc)
and an Fast Fourier Transform procedure is used to find

the
underlying "sine" pattern of 1s and 0s that most closely

fits
this.

Nope, the amplitude of _all_ frequencies in the band is
calculated and passed on to the next stage without any
judgement.

I am talking about the final stage


The final stage is the carrier PLL, not the FFT. All
the FFTs are removed from the chain once the PLL locks
on and they play no further part in the process. It is
the PLL that tracks the drifting signal and gives us
the accurate measurement.


The FFT as I was using the term includes the PLL.


The two are entirely diffeent and separate.

The point
which you insist on obscuring is that this technique
gets at the right sine frequency starting at the right time from
the
sum of sine functions of various frequencies equivalent as
Fourier showed to
the noisy oscillations observed.


You said "an Fast Fourier Transform procedure is used to find
the underlying "sine" pattern of 1s and 0s that most closely fits"

The FFT is not applied to "1s and 0s", it is applied to voltage
samples. The frequency is found and the PLL commanded to start at
that frequency. The PLL locks on and tracks the carrier and it
uses a digital phase comparator that probably treats the signal
as 1s and 0s.

There are several levels of processing that you are skipping over
which are very important in establishing that the signal is
genuine and from the right craft. Ultimately that is your main
concern, isn't it?

and I mentioned that the
movement of the Earth etc requires different patterns to be
obtained successively but the point is that the FFT procedure
finds the underlying pattern and it is this that is used to
compare to the given sequence of 1s and 0s.


No it isn't. The final FFT is only used to set initial
frequency for the carrier PLL. If that locks, the
bandwidth is reduced to improve the signal/noise ratio.


You are saying the same thing that I was saying. I think
it is clearer to say it without the jargon.


Clearer but completely wrong. The FFT does not compare
anything to a pattern of 1s and 0s. It does not compare
anything to anything else and in this case it does not
work on 1s and 0s.

The output of that is fed to the sub-carrier PLL.

Whatever the details a sequence of 1s and 0s
is obtained that is a digitised intermediate version of the
sky frequency.


No, a series of voltages samples like +0.25, -0.375, +0.112 etc.
is the result of digitising the IF.

Again
that has to lock before the signal can be decoded using
a phase detector. Then it gets decoded through the error
correction scheme. There are many critical steps after
the FFT, and in fact the FFT plays no part in the decoding
process whatsoever.


The bottom line is a sine representation of a sum of sine frequency

represention
of an oscillating pattern made possible by the FFT procedure essentially
and this includes the phase locked loop procedure perhaps involving
the recognition of some code modulation of the carrier to insure that
the fitted frequency starts at the right time.


The fact that this representation is a much smaller frequency than
the GHz sky frequency is ok because when you look at the difference
between this and a small frequency representation of the transmitted
frequency the difference is the same as the difference between the
original frequencies. And it is this difference that is used to
get the
Doppler shift.




This is the procedure I understood from your comments and
various books and links.



Ralph, it's as if all the weeks I spent
talking you through the DSN documentation by email had
never happened. You seemed to grasp it at the time, why
have you reverted to this grossly inaccurate description
of the process?
Again I think you have misunderstood what I have said.

I
dont think it is inaccurate if you replace single

intermediate
frequency by small range of frequencies around the single
intermediate frequency where small is relative the original

frequency.


It is very inaccurate when the DSN document tells you
the analog band is the digitised band is 110MHz wide and
the signals of interest are of the order of 1Hz wide.


You are quoting the wrong document. We are talking about the
intermediate frequency
being smaller that the original frequency. Is that so hard for
you to understand.


What matters is how wide the frequency range is compared
to what you are looking at. If the equipment only handles
a band that is small in comparison to the signal, the edges
will be chopped off, or if the Doppler shift was more than
expected the signal might be lost entirely. If th system
is 'wide band' then there is no such risk. How wide it is
compared to the original is completely irrelevant. Now this
matters because I know you are rferring to text boks and
those will use "wide band" and "narrow band" as terms relating
the width of the channle to the width of the signal, so if
you look up the text for "narrow band", you are going to get
entirely misleading information.

I am continually amazed that a person of your knowledge and
intelligence has so many
blindspots.


On the contrary, I can see potential mistakes you are
about to make through your unfamiliarity with the jargon
and I am trying to educate you in these terms to avoid
those pitfalls before you reach them.

And if you want to try and describe the digital version of

the
mixer please do so. It was not clear from your emails.


The mixer is analog. The output is digitised and a baseband
extracted as shown on page 10. The details of the method
of mixing are not given but the principle is simply
multiplication of the incoming signal (including noise) by
the reference sine wave.

V_out = V_in * V_ref

where

V_ref = A * sin(wt)


This makes no sense. Electrical oscillations add by the law of
superposition;


Yes, which is why it takes a special ciruit to get around that.

Tthey dont multiply.


Dual gate fets and other methods of implementing mixers do
because that is their intended function and we poor designers
have to make them do it well. It's what engineers get paid
for (though I personally work on the digital side).

The mathematical fact that a sum of sine
and cosine functions can
be represented as a product of related sine and cosine functions
has to be mentioned dont you think?


Only if you don't already know it. The circuit multiplies
the two voltages together and since the product is the
same as a combination of sum and difference, you can then
discard all of (say) the sum components and keep all of the
difference components by a simple filter. Tuning is not
required, highly undesirable, and is definitely not included
in that part of the DSN system, it uses filtering instead
and to remove the jargon, that means it doesn't select a
single frequency from a range, it accepts the whole range,
treats it all equally, and only rejects a mirror image of
the range very far away.

George


 




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