Implementing point-wise multiplication as a convolution

Question

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.
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.

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 $\alpha$.