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

# GPS Quantization Levels

Started by ●November 4, 2005

Reply by ●November 7, 20052005-11-07

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

Reply by ●November 7, 20052005-11-07

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. > > --RyanSorry 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.

Reply by ●November 7, 20052005-11-07

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?

Reply by ●November 7, 20052005-11-07

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

Reply by ●November 8, 20052005-11-08

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 stillbe>>> capable of some measure of bit capacity. >> >> Something fishy with that statement. The log of a negative quantityis>> undefined. >> If you mean a quantity smaller than unity, the log of which is >> negative, wouldnt' that imply a negative capacity by the aboveformula?> >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 >

Reply by ●November 8, 20052005-11-08

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.

Reply by ●November 8, 20052005-11-08

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. >

Reply by ●November 8, 20052005-11-08

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. > --RyanThinking 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?

Reply by ●November 8, 20052005-11-08

"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