As the title says, I'm looking for a way to implement the point-wise multiplication of 2 signals somehow as a convolution.

Here's why:

In my system, I have an input signal $x[n]$ which has been sampled uniformly, however with periodically missing samples. Best way to model this is as follows:

  1. I defined a finite (Kronecker's) delta train of M samples as : $$ v[n] = \sum_{l=0}^{M-1} \delta[n-l] $$
  2. I defined the periodically missing samples as follows: Denote the point-wise multiplication: $$ r[n]=\sum_{l=0}^{P-1} v[n-lK] $$ Where $K>M$ is the period of the periodically missing sample function. enter image description here
  3. Then, the resulting function is actually my input signal $x[n]$ multiplied by $r[n]$: $$p[n]=x[n]\cdot r[n]$$

This is related to a previous question I asked which was focused on a specific input signal and in continuous time.

So to summarize: I'm given $p[n]$ and I know $M,P,K$. What is a good method of recovering $x[n]$ from $p[n]$?

Initially, I thought of coming up with another model that uses the convolution operator in the time-domain between $x[n]$ and some other function $h[n]$ to get to the same resulting signal $p[n]$, and then filter it out in the frequency domain.

  • $\begingroup$ Hi Sammy! I think your problem statement is self-contradictory: You start with "Suppose I have two …, $x[n]$ and $r[n]$", then go on to ask "Is there a way to come up with … $x[n]$ and $r[n]$"; which makes no sense, because they're already given. I think you might have just mixed up symbols there, or should use symbols you've not used before. $\endgroup$ Nov 16, 2023 at 14:43
  • $\begingroup$ Hi, I've restated the problem in order to be a bit more clear. $\endgroup$ Nov 16, 2023 at 15:13
  • $\begingroup$ Why would you want to do this? Convolution requires point-wise multiplication (and lots of it), so what possible benefit would be in turning a simple operation into a much more complicated one ? $\endgroup$
    – Hilmar
    Nov 16, 2023 at 15:31
  • $\begingroup$ Good question, the motivation behind this is that I'm actually analyzing a way to recover x[n] by filtering in the frequency domain. Meaning, I'm trying to model my incoming signal x[n] which undergoes multiplication by r[n] as a convolution system so that I'll be able to express it as multiplication of dtft's in the frequency domain. $\endgroup$ Nov 16, 2023 at 15:42
  • 2
    $\begingroup$ Then this is an XY Problem. The answer to your full question ("How do I recover $x(t)$ when I have $x(t)r(t)$") is different from your question as asked, and isn't closely related. You should either edit your question to say what you're really trying to do, or you should walk away from this question and ask another one. I'm not sure what the rest of the community would prefer, but I think that even if I'd already answered this (incomplete) question I'd want you to edit what you have. $\endgroup$
    – TimWescott
    Nov 16, 2023 at 16:12

1 Answer 1


No, that's impossible.

Simple consideration based on the neutral elements of "multiplying with"!

  1. The neutral element of right-hand multiplication is the constant 1-function, i.e. to make $p=x$, for all possible $x$, it's necessary that $r[n]=1\forall n$. Since $\beta$ is applied to each element of $r$ independently, it follows that $u[n]= \text{const.}\ne 0$ ($\ne0$ due to linearity)
  2. The neutral element of convolution is the Kronecker delta, where decidedly $\delta[0] \ne \delta[1]$.

Hence, no such function $\beta$ can exist. By commutivity, neither can $\alpha$.


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