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I have 8 digital complex input channels digitized with 8 bits each and I want to combine them with certain complex weights. In the easiest case where all weights would be 1 the combination would be just a sum of the 8 channels. The resulting number would be an 10 bit number. Is it correct to say that the resolution has improved? Would the summed 10-bit number be equivalent to a case in where the summation is done analogically and then digitized with a 10-bit ADC?

This is interesting because it means I could decrease intentionally the number of bits of individual channels to decrease the throughput with no loss.

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TL;DR: You're trying to invent oversampling (just not in time direction). That works, when the necessary conditions hold.


The resulting number would be an 10 bit number. Is it correct to say that the resolution has improved?

Well, certainly, before, you could have 256 possible 8 bit numbers, now you have 1024 possible 10 bit numbers. Why I don't think this is a good way to look at it, see below (you don't get 10 significant digits).

Looking at the input/output relation of the system, however, your input are four 8-bit vectors, so $(2^8)^4=2^{32}$ possible inputs, and you reduce them to $4\cdot 2^8 = 2^{10}$ possible outputs. So, to me that looks like a reduction in data. (That's really not surprising: aside from the min and the max, there's always more than one input combination mapping to one output.)

Now,

Would the summed 10-bit number be equivalent to a case in where the summation is done analogically and then digitized with a 10-bit ADC

No! And different quantization approaches will be better or worse under the metric of the analog sum quantized being the same as the sum of the quantized numbers.

Simple example: you quantize analog values to their nearest integer from $\{0,1,\ldots, 254,255\}$. So, the analog values $(1.4, 1.3, 1.45, 1.49)$ would be quantized to $(1,1,1,1)$, sum $4$. If you first sum in analog, the sum would be $5.64$, so would get quantized to $6$.

So, analog addition would always seem preferable! Aside from: it isn't, sometimes.

In reality, your ADC input and the digitization process is noisy – and you typically select an ADC whose number of bits exceeds the information that you can actually get about the observed phenomenon (you'll notice that for all ADCs you buy, ENOB < number of bits).

Then, (assuming your summation does not add more noise, which it usually does) analog summation of signals with independent noise would have the sum of variances as total variance. In your example, assuming the four channels had the same noise power, but independent noise realizations, that means you'd get four times the noise power; power is the square of amplitude, so you'd get twice the noise amplitude. In other words, if before, your least significant bit of each channel were within the standard deviation of the noise amplitude, then now the two least significant bits are. This implies that instead of now getting 10 bits, you only get 9 bits of useful sample information.

This is interesting because it means I could decrease intentionally the number of bits of individual channels to decrease the throughput with no loss.

Yes; you could arrive at the same idea without taking ADCs into account at all, as I sketched above:

If what you care about is the sum of $n$ numbers with $b$ bits, then that sum will have (at most) $\log_2(n)\cdot b$ (significant) bits. To transport the $n$ numbers, you would have needed $n\cdot b$ bits. Since $\log_2 n \le n\quad\forall \mathbb R \ni n > 2$, that is always going to be fewer bits to transport.

Information-theoretically, the uncertainty about your inputs $X_i, \quad i=1,\ldots n$ isn't zero when knowing the sum: $H(X_1, X_2,\ldots X_n | \sum_{i=1}^n X_i)>0$, at least for practically relevant distributions of independent $X$. So, the mutual information between input ($n$ continuous values) and output (one quantized sum) must be lower than the entropy in the input, and there can be a quantization it with fewer than $n$ times the input entropy that preserves the information of the sum.

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  • $\begingroup$ I don't get the log2(n)⋅b relation.With n=8, b=8 then 24. The analogue summation typically ocurred in my system, done with 12 bit. The point is if I digitize each of the 8 channels and then do the sum, can I avoid having to digitize each of the 8 channels with 12 bit? or can I relax the number of bits of each channel? Intuitively I would say the more channels I have the more I can relax their number of bits. However, I reckon this conclusion may not always apply because it depends on the nature of the input signals (in my case typically a chirp signal surrounded by noise) $\endgroup$
    – Albert
    Commented Mar 18 at 12:26
  • $\begingroup$ I think I answered that! It depends on how many of your bits contain useful information in the sum; but in general, yes, you get $\log_2(n)$ as bit gain in width. This is, however, really something that depends on your application. Can't tell you without knowing what these signals are, and what their sum represents, and why you're digitizing it! $\endgroup$ Commented Mar 18 at 12:28
  • $\begingroup$ I agree it depends a lot on the statistical nature of the input signals, but I now understand there is some underlying gain in general terms. I was going to ask about a case where input signals are Gaussian, but it will also depend on how correlated they are. I'm trying to construct a digital beam-former in fact...Thanks a lot !! $\endgroup$
    – Albert
    Commented Mar 18 at 12:56
  • $\begingroup$ if you're doing a digital beam former, well, your addition will have to happen digitally, otherwise it's not going to be a digital beam former (you could use multiplying DACs to generate the voltgaes you sum up, but that's going to just introduce the same quantization problems, but at a different end). $\endgroup$ Commented Mar 18 at 14:58
  • $\begingroup$ Note that in different RF receiver chains, we can typically very well model noise to be uncorrelated. $\endgroup$ Commented Mar 18 at 14:59

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