GPS Quantization Levels

Hi,
  I have been pondering about this all day and I think I need some
clarification.
  In current GPS systems the post-IF A/D uses 1-2 bits.  Since in GPS
signals the SNR is -19 how can this be enough resolution to recover the
signal?  I am sure it has to do with DS-SS but I have not been able to
find a good source that can explain the choice of levels.  If anyone has
any clues please let me know.  Thanks.

--Ryan

They also have an AGC designed to hold the percentage of 0's and 1's at
about 50%

effimofunk wrote:
> Hi,
>   I have been pondering about this all day and I think I need some
> clarification.
>   In current GPS systems the post-IF A/D uses 1-2 bits.  Since in GPS
> signals the SNR is -19 how can this be enough resolution to recover the
> signal?  I am sure it has to do with DS-SS but I have not been able to
> find a good source that can explain the choice of levels.  If anyone has
> any clues please let me know.  Thanks.
> 
> --Ryan

effimofunk wrote:
> Hi,
>   I have been pondering about this all day and I think I need some
> clarification.
>   In current GPS systems the post-IF A/D uses 1-2 bits.  Since in GPS
> signals the SNR is -19 how can this be enough resolution to recover the
> signal?  I am sure it has to do with DS-SS but I have not been able to
> find a good source that can explain the choice of levels.  If anyone has
> any clues please let me know.  Thanks.
> 
> --Ryan

Sorry that you haven't received much comment on this.  It's related to 
the spread spectrum signaling utilized.  The actual data bit rate is 
really very slow, and yet the transmitted bit rate is many many many 
times higher.  It uses a PRBS (Psuedo-random bit stream) to 'encode' the 
actual data to spread it across a much larger frequency range.

This means that when you correlate the apparent white noise signal with 
the same synched up PRBS at the receiver you do actually get a large 
amplitude correlation result.  Hence the signal isn't white noise at 
all, it has some structure (which the receiver knows).

 From shannon's capacity theory you can see this can hold true
C = BW*log(SNR)
A negative SNR will result in a very small value for the log(SNR) 
portion, however if applied over a very large bandwidth it will still be 
capable of some measure of bit capacity.  This is enough for the GPS data.

Bevan Weiss wrote:

>  From shannon's capacity theory you can see this can hold true
> C = BW*log(SNR)
> A negative SNR will result in a very small value for the log(SNR)
> portion, however if applied over a very large bandwidth it will still be
> capable of some measure of bit capacity.

Something fishy with that statement.  The log of a negative quantity is
undefined.
If you mean a quantity smaller than unity, the log of which is
negative, wouldnt' that imply a negative capacity by the above formula?

cs_posting@hotmail.com wrote:
> Bevan Weiss wrote:
> 
>>  From shannon's capacity theory you can see this can hold true
>> C = BW*log(SNR)
>> A negative SNR will result in a very small value for the log(SNR)
>> portion, however if applied over a very large bandwidth it will still be
>> capable of some measure of bit capacity.
> 
> Something fishy with that statement.  The log of a negative quantity is
> undefined.
> If you mean a quantity smaller than unity, the log of which is
> negative, wouldnt' that imply a negative capacity by the above formula?

Crap... I believe I forgot a 1... lol, haven't had coffee yet

C = BW*log(1+SNR)
where SNR is simply Signal Power/ Noise Power (and isn't dB scale).

That's better :)  simple equation, but I can still forget the details :s

I understand the spread spectrum part of it.  The Gold Code correlation
used in the acquisition process is clear but what is not clear to me is
how the signal is not lost after the path loss from the satellite and such
a low quantization level.  If we have 2 bits we can only get about 12 dB of
SNR and the signal is -19 dB below the noise.  The processing gain of the
DSP section only helps if you retain the structure of the signal itself. 
If you lose the signal due to low quantization all the correlation in the
world wouldn't help you.  How bout does it really have to do with the zero
crossings due to the phase modulated signal?  Thanks for your responses so
far. 

--Ryan

>cs_posting@hotmail.com wrote:
>> Bevan Weiss wrote:
>> 
>>>  From shannon's capacity theory you can see this can hold true
>>> C = BW*log(SNR)
>>> A negative SNR will result in a very small value for the log(SNR)
>>> portion, however if applied over a very large bandwidth it will still
be
>>> capable of some measure of bit capacity.
>> 
>> Something fishy with that statement.  The log of a negative quantity
is
>> undefined.
>> If you mean a quantity smaller than unity, the log of which is
>> negative, wouldnt' that imply a negative capacity by the above
formula?
>
>Crap... I believe I forgot a 1... lol, haven't had coffee yet
>
>C = BW*log(1+SNR)
>where SNR is simply Signal Power/ Noise Power (and isn't dB scale).
>
>That's better :)  simple equation, but I can still forget the details :s
>

effimofunk wrote:
> I understand the spread spectrum part of it.  The Gold Code correlation
> used in the acquisition process is clear but what is not clear to me is
> how the signal is not lost after the path loss from the satellite and such
> a low quantization level.  If we have 2 bits we can only get about 12 dB of
> SNR and the signal is -19 dB below the noise.  The processing gain of the
> DSP section only helps if you retain the structure of the signal itself. 
> If you lose the signal due to low quantization all the correlation in the
> world wouldn't help you.  How bout does it really have to do with the zero
> crossings due to the phase modulated signal?  Thanks for your responses so
> far. 

Well, what kind of noise is the channel noise represented as?
And what kind of noise is the quantization noise?
What's the mean of both these noises?

So if I apply 0.0001V to this noise at the input to the receiver and 
average over billions of samples, what is the measured output?  (The 
receiver is just the basic integrate and dump matched filter, ie 
baseband transmission)

Now imagine that instead of just applying that 0.0001V signal you now 
make it toggle every second sample (thus it has 2 samples at 0.0001V and 
2 samples at -0.0001V).  You apply this to the input of the receiver, 
but now the receiver matched filter also inverts it again, so that you 
again have 0.001V as the input to the integrate and dump.  Will the 
output be the same as in the first situation?

The next logical step to reduce the effect of narrow band interferer's 
is to provide an additional level of 'randomness' to the toggling of the 
signal.  This is done by using a known PRBS.

There is still the issue of AGC and offsets, AGC isn't such a big issue, 
as large amounts of clipping can occur, the energy isn't the big swings, 
it's simply in the small offsets.  Offset correction is aided by the 
PRBS having a very uniform distribution.  On average it has as many 1's 
as it has 0's, so that over a large period of time it can be tuned to 
eliminate the offset pre-decoding.

I see what are saying but I am still hung up on something.  If you have a
signal that is three to four times smaller than the noise and then you
digitize the full-scale swing to lets say one-bit isn't the resultant
digital signal just the noise?  It seems just by intuition on my end that
the signal will be lost and all you will see is the noise.  Granted the
mean will be zero over time but you can no longer recover your signal. 
That is where I am getting confused.  Thanks for your help.  I am sure I
am asking the wrong questions or misunderstanding your responses.
--Ryan


>Well, what kind of noise is the channel noise represented as?
>And what kind of noise is the quantization noise?
>What's the mean of both these noises?
>
>So if I apply 0.0001V to this noise at the input to the receiver and 
>average over billions of samples, what is the measured output?  (The 
>receiver is just the basic integrate and dump matched filter, ie 
>baseband transmission)
>
>Now imagine that instead of just applying that 0.0001V signal you now 
>make it toggle every second sample (thus it has 2 samples at 0.0001V and

>2 samples at -0.0001V).  You apply this to the input of the receiver, 
>but now the receiver matched filter also inverts it again, so that you 
>again have 0.001V as the input to the integrate and dump.  Will the 
>output be the same as in the first situation?
>
>The next logical step to reduce the effect of narrow band interferer's 
>is to provide an additional level of 'randomness' to the toggling of the

>signal.  This is done by using a known PRBS.
>
>There is still the issue of AGC and offsets, AGC isn't such a big issue,

>as large amounts of clipping can occur, the energy isn't the big swings,

>it's simply in the small offsets.  Offset correction is aided by the 
>PRBS having a very uniform distribution.  On average it has as many 1's 
>as it has 0's, so that over a large period of time it can be tuned to 
>eliminate the offset pre-decoding.
>

effimofunk wrote:
> I see what are saying but I am still hung up on something.  If you have a
> signal that is three to four times smaller than the noise and then you
> digitize the full-scale swing to lets say one-bit isn't the resultant
> digital signal just the noise?  It seems just by intuition on my end that
> the signal will be lost and all you will see is the noise.  Granted the
> mean will be zero over time but you can no longer recover your signal. 
> That is where I am getting confused.  Thanks for your help.  I am sure I
> am asking the wrong questions or misunderstanding your responses.
> --Ryan

Thinking about the first question...
You seem to be implying that the signal will have no influence over the 
quantized signal.  Essentially you're writing the signal off simply 
because it's lower in amplitude to the noise.

At some points in time the noise will have lower amplitude than the 
signal, and hence only the signal will be received.  If you average the 
noise over a large time, then the average noise amplitude will be zero, 
however the signal amplitude will be some non zero finite value.

So again, I ask you...
What is the mean of the channel noise?
What is the mean of the quantization noise?

"effimofunk" <rfrankel@ufl.edu> wrote in message 
news:xs2dnQCvMsY41OzenZ2dnUVZ_smdnZ2d@giganews.com...
>I see what are saying but I am still hung up on something.  If you have a
> signal that is three to four times smaller than the noise and then you
> digitize the full-scale swing to lets say one-bit isn't the resultant
> digital signal just the noise?

No.  If the noise is random, then the digitized noise+signal will be 
slightly non-random.  When the amplitude of the noise just happens to be 
very small in a given sample instant, the sample will reflect the signal. 
Over a long enough time, the correlation between the digitized signal and 
the SS code will be significant enough to detect reliably.  It's like a 
casino making reliable profits over time by skewing the odds of each bet a 
tiny bit in their favour.

--
Matt