# SNR improvement from placing analog filter ahead of digital filter

We have designed a PCB containing a data collection circuit for a customer, and their external expert presents this "helpful suggestion":

Question about signal averaging vs. longer integration times: Averaging $N=20$ samples provide us with a digital SNR improvement of $\sqrt{N} = 4.47$. However, the signal of interest is slowly varying. So, instead we could integrate the signal longer (say, $N$ times longer) and reduce the filter bandwidth by $N$, resulting in an SNR improvement of $\sqrt{N}\sqrt{N} = N$.

Is reducing the sampling rate and corner frequency of the anti-aliasing filter that much more effective than decimation via a digital filter?

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I've posted my reasoning about this question as an answer, but I'm very interested in whether other users see something I'm missing here. –  Ben Voigt Apr 29 at 15:50

No, you can't double-dip like that.

While it is true that a lowpass integrating stage that reduces noise bandwidth by a factor of $N$ does improve linear SNR by $\sqrt{N}$, you cannot simply cascade two stages to get an SNR improvement of $N$. Here's why:

The SNR improvement from averaging $N$ consecutive samples is conditional on the characteristics of sampled additive white Gaussian noise, specifically that they are independently distributed. The following relationship holds for the sum of independent Gaussian random variables:

If

$$Y = \sum_i X_i$$

with

$$X_i \sim \mathcal{N}(\mu_i, \sigma_i^2) \quad \forall i$$

then

$$Y \sim \mathcal{N}(\mu_y, \sigma_y^2)$$

where

$$\mu_y = \sum_i \mu_i$$

and

$$\sigma_y^2 = \sum_i \sigma_i^2$$

If the variables are i.i.d, which is to say that $\mu_i = \mu_0 \; \forall i$ and $\sigma_i = \sigma_x \; \forall i$, then averaging gives

$$\frac{1}{N} \sum_{i=1}^N X_i \sim \mathcal{N}\left(\mu_x,\left(\frac{\sigma_x}{\sqrt{N}}\right)^2\right)$$

As the acquisition system DC offset is properly zeroed, $\mu_x = 0$. In this case the noise is reduced by a factor of $\sqrt{N}$ by averaging. The same result holds for averaging in the analog domain via integration.

However, this result only applies when the Gaussian random variables are independent (white noise). As soon as a lowpass filter is added in front of the averaging block, the noise contributions become highly correlated, and integration of a signal which is already band-limited results in no additional reduction.

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The original question in the highlighted box talks about integrating N times longer, which would indeed give you the additional factor of sqrt(N) at the reduced bandwidth, so I don't exactly understand which question you are answering. –  dustincarr Apr 29 at 16:31
@dustincarr: Clearly if $N$ is increased, then the ratio is improved. But that's not a way of increasing the ratio from $\sqrt{N}$ to $N$. "Integrate the signal longer" means lowering the ADC sampling rate, don't you think? –  Ben Voigt Apr 29 at 16:41
I see. If the ADC sampling rate stays the same, then changing the anti-aliasing filter indeed does nothing (except reduce aliasing, which is presumably not a problem). In fact, analog filtering could hurt you with some types of ADC if you don't have enough high bandwidth signal to dither the bits. –  dustincarr Apr 29 at 16:48

However you do it — analog domain or digital domain — reducing the bandwdith of the system by a factor of N reduces the noise by a factor of N.

Note that we're talking about noise power here, which means that the noise voltage is reduced by a factor of $\sqrt{N}$.

But you can't reduce the bandwidth to the same value twice; there's no additional benefit to doing this, except for the tiny contribution of having the transition band being steeper.

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There could be a difference if the ADC has low resolution and itself lowers the SNR. Then, filtering in the digital domain (after ADC) will reduce the quantization noise. –  Juancho Apr 29 at 16:42
@Juancho: I agree with that. But it didn't seem as important as the issue of filtering already-colored noise, which is why I didn't try to cover that in my explanation. –  Ben Voigt Apr 29 at 16:43
@Juancho: That's a completely different question. Quantization noise is injected at the boundary between the analog and digital domains, and is not part of the original signal. –  Dave Tweed Apr 29 at 16:44
@DaveTweed: Well, it is part of output SNR. –  Ben Voigt Apr 29 at 16:46
Yes, but the way your question is worded, it seems that the level of noise in the analog signal is significantly higher than the quantization noise. Otherwise, the "outside expert" would never have made that suggestion. –  Dave Tweed Apr 29 at 16:49

It should be a comment. However I do not have enough reputation to comment, so here is a tiny answer made of questions..

@Ben Voigt I think you are right saying that averaging has effect when the samples are i.i.d. However, are you sure that the samples after an analog low pass filter are not i.i.d any more? IMHO, they are also i.i.d. Just like the results of averaging Gaussian samples are still i.i.d.

And it also seems that there is a subtle difference between averaging and analog lowpass. Am I correct to say that averaging reduce the noise spectral density at a same level on the whole frequency domain? While lowpass only reduce the higher frequency part.

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I.i.d. means independent, identically distributed. After filtering they are no longer independent. Instead they are mutually Gaussian.. A moving average is a lowpass filter (it is the same as convolution by a rectangular pulse, try calculating its fourier transform) –  Ben Voigt May 8 at 14:13
Are you performing a moving average? It seems that you take N new samples and average them every time. –  richieqianle May 8 at 14:16
That's the same as moving average followed by decimation. So it has the same passband characteristics as convolution by rectangular pulse. –  Ben Voigt May 8 at 14:41
1) Why are they samples from analog low pass filter become dependent? As far as I know, pink noises can still be i.i.d. 2) Gaussian assumes flat noise density with infinite bandwidth. I think averaging samples will only lead to reduce of noise density, not the bandwidth. –  richieqianle May 9 at 2:53
No, it can't. No, it doesn't. No, it won't. Where did you get these wrong definitions? –  Ben Voigt May 9 at 3:36