I'm trying to characterize the ADC of my board and have a question related to SNR. My first question is related to process gain. I understand that you get 10log(fs/2*BW) process gain added to your ADCs SNR while filtering for a single rate filter. By using a decimating filter or even decimating at the output of the filter, am I losing this process gain as the sampling frequency changes? I am assuming that I will only see this process gain outside of the passband of the filter. Or in other words, the noise floor of the ADC will remain the noise floor within the passband of the filter. Am I correct to think that if I am only interested in the passband frequencies, I will not see any benefits from the process gain? I know my wording is a bit confusing and can try to put it in different words. Hope someone can shed some light on this for me. Thanks
ADC SNR and process gain.
Started by ●February 24, 2007
Reply by ●February 24, 20072007-02-24
"tbrim20" <tjbrimeyer@gmail.com> writes:> I'm trying to characterize the ADC of my board and have a question related > to SNR. > > My first question is related to process gain. I understand that you get > 10log(fs/2*BW) process gain added to your ADCs SNR while filtering for a > single rate filter. By using a decimating filter or even decimating at the > output of the filter, am I losing this process gain as the sampling > frequency changes? > > I am assuming that I will only see this process gain outside of the > passband of the filter. Or in other words, the noise floor of the ADC will > remain the noise floor within the passband of the filter. Am I correct to > think that if I am only interested in the passband frequencies, I will not > see any benefits from the process gain? > > I know my wording is a bit confusing and can try to put it in different > words. Hope someone can shed some light on this for me.Hi, I sense your confusion, and I think if you can understand a couple of things about quantization everything will become clear. Quantization (A/D conversion) at a sample rate Fs and a certain number of bits N produces a total noise power Pn that is a function of N but not a function of Fs. So, for any given N, you get Pn watts of noise power no matter if the sample rate is 1 sample per second or 1E9 samples per second. The next thing to understand is that the quantization noise power Pn is spread evenly across all frequencies in the digital signal. So if you're sampling at Fs samples per second, the noise will be Pn/(Fs/2) watts per Hertz. For example, if Pn is 1 watt and Fs is 2000 samples/second. Then there will be that Pn/(Fs/2) = 1 / 1000 = 0.001 watt/Hz of noise power in the frequency domain. So in a 1 Hz bandwidth you'd have 1 milliwatt; in a 10 Hz bandwidth you'd have 10 milliwatts; in the total bandwidth of 1000 Hz you'd have 1 watt. So, now we get to the point of your question. Let's say your total signal power in the bandwidth of interest is Ps, the total noise power due to quantization is Pn, and the signal bandwidth is B Hz. If you sample at 2*B samples/second, then the amount of quantization noise power Pnb in the signal bandwidth B is Pnb = B * (Pn / (Fs/2)) = B * (Pn / B) = Pn and the SNR is then SNR = Ps / Pnb = Ps / Pn So there's nothing you can do to improve the SNR. However, if you oversample by some amount, say, Fs = M*2*B samples per second, where M > 1, where B is the bandwidth of interest, then the amount of quantization noise power Pnb in the signal bandwidth B is Pnb = B * (Pn / (Fs/2)) = B * (Pn / M*B) = Pn / M and the SNR is then SNR = Ps / Pnb = M * Ps / Pn. Thus you have improved your SNR by a factor of M. So, if M = 2, you improve the SNR by 3 dB; if M = 4, you improve the SNR by 6 dB (or 1 bit); etc. So, to answer your questions note the following: 1. what is important is the SNR in the bandwidth of interest. 2. oversampling improves the SNR in the bandwidth of interest 3. decimation (assuming you prefilter with the appropriate lowpass filter) does not destroy the SNR improvement in the bandwidth of interest. --Randy -- % Randy Yates % "Bird, on the wing, %% Fuquay-Varina, NC % goes floating by %%% 919-577-9882 % but there's a teardrop in his eye..." %%%% <yates@ieee.org> % 'One Summer Dream', *Face The Music*, ELO http://home.earthlink.net/~yatescr
Reply by ●February 24, 20072007-02-24
Randy Yates wrote: ...> Hi, > > I sense your confusion, and I think if you can understand a couple > of things about quantization everything will become clear.... Randy, That is the clearest explanation I can imagine, far more cogent that what I was beginning to formulate. My hat is off to you. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●February 24, 20072007-02-24
Jerry Avins <jya@ieee.org> writes:> Randy Yates wrote: > > ... > >> Hi, >> I sense your confusion, and I think if you can understand a couple >> of things about quantization everything will become clear. > > ... > > Randy, > > That is the clearest explanation I can imagine, far more cogent that > what I was beginning to formulate. My hat is off to you.Thank you, Jerry. That is very kind of you. -- % Randy Yates % "Midnight, on the water... %% Fuquay-Varina, NC % I saw... the ocean's daughter." %%% 919-577-9882 % 'Can't Get It Out Of My Head' %%%% <yates@ieee.org> % *El Dorado*, Electric Light Orchestra http://home.earthlink.net/~yatescr
Reply by ●February 26, 20072007-02-26
Thank you Randy for the great explanation. The theory is more clear to me, however I'm trying to explain what I am seeing in reality. I have a setup where I am filtering the incoming ADC data and then producing the FFT. As I continue to reduce the bandwidth of the filter by a factor of 2, I would expect to see a 3dB drop in the noise everywhere(in band and out of band). However, I am only seeing the drop in noise in the out of band. The pass band noise level doesn't appear to budge. What am I missing here? Also, how would I compute process gain if I have 2 separate filters? I'm using a polyphase decimate by 4 FIR windowing filter(larger bandwidth) followed by a single rate FIR(narrow bandwidth). I'm guessing that I just get process gain from the narrower filter. If I am correct, then would I use my decimated sample rate for the process gain calculation since this is the rate at which the filter is running? Thanks again.>Jerry Avins <jya@ieee.org> writes: > >> Randy Yates wrote: >> >> ... >> >>> Hi, >>> I sense your confusion, and I think if you can understand a couple >>> of things about quantization everything will become clear. >> >> ... >> >> Randy, >> >> That is the clearest explanation I can imagine, far more cogent that >> what I was beginning to formulate. My hat is off to you. > >Thank you, Jerry. That is very kind of you. >-- >% Randy Yates % "Midnight, on the water... >%% Fuquay-Varina, NC % I saw... the ocean's daughter." >%%% 919-577-9882 % 'Can't Get It Out Of My Head' >%%%% <yates@ieee.org> % *El Dorado*, Electric Light Orchestra >http://home.earthlink.net/~yatescr >
Reply by ●February 26, 20072007-02-26
On Feb 26, 10:01 am, "tbrim20" <tjbrime...@gmail.com> wrote:> T I would expect to see a 3dB drop in the noise everywhere(in > band and out of band). However, I am only seeing the drop in noise in the > out of band. The pass band noise level doesn't appear to budge. ...The in band power spectral -density- does not change, as you are seeing. It is the broadband noise power that has dropped 3 dB. If there is a single in band tone, the broadband SNR has improved 3 dB. 'Broadband SNR' meaning that the 'N' is the all the noise represented by the sampled data stream. If you perform the same sized FFT after filtering and desampling by 2, the noise -per bin- will have dropped 3 dB, because the power spectral -density- has stayed the same, but the analysis band width (bin size) has been reduced by a factor of 2. If this is your definition of 'process' then the process has a 3 db gain in SNR for the single in band tone. Dale B. Dalrymple
Reply by ●February 26, 20072007-02-26
tbrim20 wrote:> Thank you Randy for the great explanation. The theory is more clear to me, > however I'm trying to explain what I am seeing in reality. > > I have a setup where I am filtering the incoming ADC data and then > producing the FFT. As I continue to reduce the bandwidth of the filter by > a factor of 2, I would expect to see a 3dB drop in the noise everywhere(in > band and out of band). However, I am only seeing the drop in noise in the > out of band. The pass band noise level doesn't appear to budge. What am I > missing here?I don't see Randy's response, so I'll try to pinch hit. How do you measure noise? If on a per-Hz basis, why do you expect the noise in the passband to be reduced by a filter that passes everything else? If on a total power basis, of course the noise and signal are suppressed together in the stopband and passed together in the passband. ... Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●February 26, 20072007-02-26
"tbrim20" <tjbrimeyer@gmail.com> writes:> Thank you Randy for the great explanation. The theory is more clear to me, > however I'm trying to explain what I am seeing in reality. > > I have a setup where I am filtering the incoming ADC data and then > producing the FFT. As I continue to reduce the bandwidth of the filter by > a factor of 2, I would expect to see a 3dB drop in the noise everywhere(in > band and out of band).Why? Why would you expect that? Let's stop thinking about noise for the moment. Consider a signal that contains two frequencies: 1 kHz and 2 kHz. Let's say, just for example, that you're sampling the signal with an ADC at 10,000 samples/second. If you look at the FFT of the output of the ADC without any filtering, you'll see both sine waves. If you filter the output of the ADC with a lowpass filter having a cutoff frequency of 1500 Hz, then observe the output of the filter with an FFT, you wouldn't expect the signal component at 1000 Hz to be gone, would you? Noise is exactly the same way - in frequency, the level varies depending on the filtering. The noise that was present before a filter is going to remain if it's within the passband of the filter. When one says the noise drops 3 dB, one necessarily means "3 dB in the entire bandwith."> Also, [...]Let's get an understanding of the basics first, then we can elaborate on more complex schemes. -- % Randy Yates % "Bird, on the wing, %% Fuquay-Varina, NC % goes floating by %%% 919-577-9882 % but there's a teardrop in his eye..." %%%% <yates@ieee.org> % 'One Summer Dream', *Face The Music*, ELO http://home.earthlink.net/~yatescr
Reply by ●February 27, 20072007-02-27
On Feb 26, 3:49 pm, Randy Yates <y...@ieee.org> wrote:> > When one says the noise drops 3 dB, one necessarily means "3 dB > in the entire bandwith."[...]> Let's get an understanding of the basics first, then we can elaborate > on more complex schemes.When one says the noise drops 3 dB, one can certainly mean "3 dB in the entire bandwidth" but one needn't mean that. Let's get clear statement of the basics first, then we can elaborate on more complex schemes. Dale B. Dalrymple
Reply by ●February 27, 20072007-02-27
dbd wrote:> On Feb 26, 3:49 pm, Randy Yates <y...@ieee.org> wrote: >> When one says the noise drops 3 dB, one necessarily means "3 dB >> in the entire bandwith." > [...] >> Let's get an understanding of the basics first, then we can elaborate >> on more complex schemes. > > When one says the noise drops 3 dB, one can certainly mean "3 dB > in the entire bandwidth" but one needn't mean that. > > Let's get clear statement of the basics first, then we can elaborate > on more complex schemes.Consider the process under discussion and Randy's answer in light of that. Would you please elaborate on the processes by which a bandpass filter can reduce noise in the passband? Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
