# Is it theoretically possible to perfectly quantize a continuous signal?

So, I'm completely new to digital signal processing, but while reading a piece this morning about quantization it got me daydreaming: could a machine ever be fast enough to sample the position and amplitude of each particle that makes up a wave?

• Why would you want to? All you're measuring is thermal noise. – endolith Nov 18 '14 at 14:35
• I hope information is ultimately quantized too – Laurent Duval Aug 30 '17 at 20:29
• @LaurentDuval, ironically it's quantum theory that leaves us very little hope that information is quantised. Even a single qubit contains (uncountable) infinite classical information. And unlike classical information, quantum information grows exponentially with system size. While the information stored in classical systems of size N and M is simply the sum of the information in N and M, quantum theory requires us to encode the product of the individual information with the extra information residing in the non-local and non-classical correlations, or "entangled" degrees of freedom. – Jazzmaniac Sep 2 '17 at 11:50
• So even if the classical information in one qubit turned out to be finite, the whole system would pretty quickly grow into practically infinite much information you'd have to deal with. – Jazzmaniac Sep 2 '17 at 11:53
• I like this argument – Laurent Duval Sep 2 '17 at 12:57

Depends on assumptions you are willing to make and what type of signals are you trying to sample, but in theory I think that sampling rate equal to the Planck time would be a gold standard for anything...

This translates to sampling frequency of $1.855 \times 10 ^ {43} \mathtt{Hz}$ ($18.55$ tredecillion hertz). Personally I believe that machines will never be so fast. Obviously we can limit our assumptions to more realistic ones.

Obviously there is more constraints, such as: what sensor is capable of measuring displacement/velocity with such accuracy? Obviously this assumption is applicable even for subatomic scale, and most likely we don't need such accuracy in measurements of sound signals. Additionally when frequency is increasing, sound attenuation in media (air in this case) is increasing to enormous amounts and ultra-sonic sound waves are not propagating on large distances. At some point you will start measuring the Brownian motion. You also mentioned about possibility of measuring each particle separately and it is another, very difficult problem to tackle.

• Wow, so considering a computer running a 1 GHz processor is $1 \times 10^{9} \mathtt{Hz}$ then we've got some ways to go! – armadadrive Aug 12 '14 at 13:00
• That was time sampling. What about amplitude quantization now :) – Laurent Duval Aug 30 '17 at 20:14
• @LaurentDuval I guess it would require infinite amount of bits ;) – jojek Aug 30 '17 at 20:24
• @LaurentDuval Actually the amplitude will be bounded by the Planck's distance? We are getting really crazy here... It's a cool question to ask this guy. – jojek Aug 31 '17 at 12:05

I'd like to point out Heisenberg Uncertainty principle, based on which theoretical achievable precision is limited. It states that one can not measure two complementary qualities (e.t. here time and charge) concurrently and there is a trade off between amount of precision you can get from one or another. In ADCs, for example theoretical limit for resolution of an Analog to Digital Converter (quantizer), that is sampling a wave with rate of 1 Gs/s, is limited to 22 Bits (~132dB)[ 1 ]. Based on the figure below, even if there is no implementation error due to the Heisenberg (the pink curve) we would go beyond a certain level of accuracy.

But why we do not encounter it and no body talk about in real applications? Because the trade off appears in applications which are far from current real world applications.

• The article defines the conditions as 50 ohm impedance and 1 V peak-to-peak input signal. – Olli Niemitalo Sep 2 '17 at 9:25
• I quite strongly disagree with this argument. First of all, Heisenberg's (not "hisenburg") uncertainty only applies to very specific observables that are canonically conjugate. Time and voltage are not of that kind. There is a different uncertainty principle that is much weaker and less general relation between time and energy. Using this to argue about the entropy of Energy as a function of time would already be a far stretch, but the idea is even more invalidated by the fact that electric potential is not energy but energy relative to the charge of a probe. – Jazzmaniac Sep 2 '17 at 12:31
• So which probing charge should one assume? A single electron? A significant number of electrons? Interestingly the uncertainty gets smaller with increasing probe charge, so that there is no lower bound due to charge quantisation. That leaves the whole argument about entropy bounds meaningless, even on a very superficial level. If you do it properly, looking at the physics of electromagnetic field quantisation and the actual process of measuring voltage it becomes even clearer that this is nonsense. – Jazzmaniac Sep 2 '17 at 12:34
• You might be right and I am not giving very exact explanation, it is my idea and you are free to edit the post :) In case of voltage and time relation you are right it is not exactly voltage or charge. – MimSaad Sep 2 '17 at 14:27
• @OlliNiemitalo , yeah, I guess I need to re read the article. – MimSaad Oct 16 '17 at 20:23

# Is it theoretically possible to perfectly quantize a continuous signal?

No. A quantization has an information content obviously countable as bits.

Now, if you have a continuously distributed 1D random variable $$X$$, then the event that any of these real numbers $$x$$ occurred is unbounded ("infinite"):

$$I(x) = -\log_2\left(P(X=x)\right)$$

So, for any (non-degenerate) continuous distribution, there's always uncountably many $$x$$ for which $$P(X=x) = 0$$ (that's why we work on PDFs with continuous distributions). So, the information content of the event "hey, I measured the value $$x$$!" would be infinite.

That means you can't use a finite number of bits to represent that sample.

That means there's no existing quantization that can be used to perfectly represent a continuously valued signal.

No, and the reason is not so much a question of how fast one can sample a continuous-time signal (as the accepted answer and another one says) but rather the impossibility of representing a real number with perfect accuracy via a quantized representation of the real number (as noted in the answer by Marcus Muller). At best, even if we assume an infinite number of quantization levels, the number of levels is countably infinite whereas there are uncountably infinitely many real numbers that we need to represent perfectly. Ergo, we cannot represent even one real-valued sample perfectly in terms of a quantized representation; that the next sample cannot be closer than $$\left(1.855 \times 10 ^{43}\right)^{-1}$$ seconds away from the first is more of a minor concern at this point. Let's learn to walk before trying to run.