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I have two signal s1(t) and s2(t) that I want to extract but I can only measure:

y1(t) = s1(t) + s2(t-d1)

y2(t) = s1(t-d2) + s2(t-d3)

The time delays d1, d2 and d3 are unknown, though I have a rough estimate for d2.

Maybe (but only if necessary) I could approximate this by shifting y2 so that

y1(t) = s1(t) + s2(t-d1)

y2(t) = s1(t) + s2(t-d3)

Is there any way to separate the signals? What happens if I have three measurements and three signals?

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  • $\begingroup$ I would use cross-correlation between $y1$ and $y2$ to try to estimate the offset $d1-d3$ (whether this works depends a lot on the structure of $s1$ and $s2$ - do they look similar or different; are they equally strong or can one be much stronger?). Once you have that, subtracting $y1$ and a suitably delayed version of $y2$ should eliminate $s2$, so that you are left with something like $s1(t)-s1(t-\Delta)$. From this you can get back $s1$ except for its DC, which is lost in the process. $\endgroup$
    – Florian
    Jul 12 '19 at 8:43
  • $\begingroup$ How can I get back s1 from s1(t)−s1(t−Δ)? $\endgroup$
    – torpedo
    Jul 12 '19 at 10:19
  • $\begingroup$ Are the time delays constant for the duration you are interested in? $\endgroup$
    – A_A
    Jul 12 '19 at 10:20
  • $\begingroup$ Yes, the time delays are constant $\endgroup$
    – torpedo
    Jul 12 '19 at 10:21
  • $\begingroup$ Well, $s_1(t)-s_1(t-\Delta)$ is nothing but $s_1(t)$ convolved with $\delta(t)-\delta(t-\Delta)$. Many ways to deconvolve, e.g., dividing in frequency domain. Depends on your signals, I guess we're talking about discrete/sampled signals? Then it's a difference equation which you can solve sample by sample algebraically. $\endgroup$
    – Florian
    Jul 12 '19 at 11:01
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Elaborating a little bit on what I discussed in the comments: Let's assume the delays $d_1$, $d_2$ and $d_3$ are integer multiples of the sampling time interval $t_0$ so that we can treat it as an integer problem.

Then, as a first step, I would try to use the cross correlation between $y_1$ and $y_2$ to determine the delays. Ideally, it should have a peak at an offset of $d_2$ (when the shifted copies of $s_1(t)$ and $s_1(t-d_2)$ align) and another one at $d_3-d_1$ (when the shifted copies of $s_2(t-d_1)$ and $s_2(t-d_3)$ align).

Now compute $z[n] = y_1[n] - y_2[n-(d_3-d_1)]$ (using $[n]$ as a discrete time / sample index for clarity). If your estimate was correct, this cancels $s_2(t)$ and we are left with $z[n] = s_1[n] - s_1[n-\Delta]$ where $\Delta = d_3-d_1-d_2$.

Now, your signals ought to start somewhere (causality) so that we can assume $s_1[n]=0$ for $n<0$ (if you have different initial conditions, you need to work them in here). Then, we have $$z[n] = s_1[n]-s_1[n-\Delta] = s_1[n]\quad \forall n<\Delta.$$ Therefore $$s_1[n] = \begin{cases} 0 & n<0 \\ z[n] & n<\Delta \\ z[n]+s_1[n-\Delta] & \mbox{otherwise}\end{cases}.$$ This allows you to reconstruct $s_1[n]$ sample by sample.

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