# Under what conditions is there a one-to-one mapping between continuous-time and discrete-time signals?

As the sampling theorem dictates that the uniform sampling frequency must be at least twice the maximum frequency present in the bandlimited signal (Nyquist rate), a question arises about the uniqueness of this representation of the analog signal to digital one. Apart from practical errors related to the truncation of the infinitely-long energy signal when discretized to digital, can we find a UNIQUE mapping between the two signal forms? If yes, then what are the conditions that limit or control this unique representation?

Note: I am not assuming by default that the sampling topology should be uniform. Random sampling also could be considered as well.

Apart from practical errors related to the truncation of the infinitely-long energy signal when discretized to digital, can we find a UNIQUE mapping between the two signal forms?

No.

I think about this differently than Tim Wescott's in his excellent answer and I hope we can make a fun discussion out of this. Tim accurately describes that any bandlimited signal can be sampled without any loss of information and can be reconstructed into the original analog signal again.

So where is the rub? Turns out bandlimited signals do NOT exist. In order to be bandlimited a signal must be infinitely long in time. That's not possible. The same goes for the reconstruction filter: not only does it require infinite time, it's also infinitely non-causal: you have to go backwards in time past the big band to "correctly" apply.

My point here is that dealing with finite sequences isn't just a "practical" problem, it's a fundamental one. A unique mapping does exist in theory but only under assumptions which are ALWAYS guaranteed to be violated. Whether that qualifies are not, really depends on what exactly do you mean by "exist".

Here is an interesting example: let's say you have an signal that's sampled at 44.1kHz. You ought be to be able to sample rate convert this to 48 kHz and then back to 44.1kHz getting back the original samples. That's possible in "theory" but it's impossible to actually build a contraption that can do this.

• What you are describing is correct for band-limited signals, which are important because their structure is time-shift invariant. However, there are many other ways to map continuous-time to discrete-time signals, and they don't come with the restrictions you mention. In particular, a bijection exists if the dimension of the function space spanning the continuous-time signal is finite. That's a sufficient condition, not a necessary one. A necessary condition is harder to establish and subject of the mathematical subject of "harmonic analysis". There's no short answer for that. – Jazzmaniac Jan 20 at 9:51
• But you can broaden the sufficient condition somewhat by requiring the finite number of dimensions for every finite duration part of the continuous-time signal space. That's rather general and practical. Identifying the number of dimensions and finding a proper basis is a different and entirely non-trivial story. – Jazzmaniac Jan 20 at 9:52

Apart from practical errors related to the truncation of the infinitely-long energy signal when discretized to digital, can we find a UNIQUE mapping between the two signal forms?

Yes.

Say you're starting with $$x(t)$$, which is perfectly bandlimited to less than $$f_B = \frac{1}{2 T_s}$$, and you take samples at every $$t = T_s k$$: $$x_k = x(T_s k)$$.

All you need to do is to treat the as a train of Dirac impulses (i.e., let $$x_p(t) = \sum x_k \delta(t - T_s k)$$), and then filter $$x_p(t)$$ with a brick-wall low-pass filter. I.e., let $$x_f(t) = h(x_p(t))$$, where $$h$$ is that brick-wall filter. $$h$$ will have an infinitely long impulse response, so you cannot realize this system in the real world - but not only will the mapping from $$x_k$$ to $$x_f$$ be exact, but if $$h$$ has zero phase delay then you'll find that $$x_f(t) = x(t)$$.

IMHO the question has and obvious and boring answer.

Given no additional constrains, no continuous signal can be equivalent to their information-destroying discretized representation.

Let's introduce some constraints:

- the signal level is perfectly constant between samplings
- level change is perfectly synchronized with sampling by a KNOWN function


Then we could perhaps have isomorphism.

A trivial example: constant signal.

• Nice and simple one. However, it still needs more to add. Apart from constant signals, this means that for any unknown analog signal there is NO way to have UNIQUE mapping to "discrete" form unless there is an INFINITE number of samples, as @Hilmar pointed out above. Now this also means, from practical perspective, we can't sample a signal and then be able to PERFECTLY reconstruct it. This is only imaginable theoretically, as the classical WKS sampling theory dictates, but not practically. – HYMD Jan 28 at 15:33
• @HYMD indeed all of that. But if you want a 'practical' instead of a 'theoretical' answer do indicate that in your question. Also define what 'practical' is. – Vorac Jan 28 at 15:50
• My question considers both theory and practice at once. Meaning, with the theoretical infinitely-long sampling, the sinc interpolation function becomes a dirac delta, and the "discrete" signal is no more discrete, actually, it is the continuous signal itself! Practically, sampling a bandlimited signal truncated to an observation window not only introduces a truncation-related errors, but also errors related to the FINITE sampling rate! Since the truncation itself convert the signal into a non-bandlimited signal. Hence, we need an infinite sampling rate again to get rid of the aliasing errors. – HYMD Jan 28 at 16:44