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Suppose you have one real-valued signal $s(t)$ and another the real-valued signal $r(t)$ related to $s(t)$ by a linear integral transformation with a real kernel $k$ such that $r(t) = \int dt'\, k(t-t') s(t')$. The kernel depends on particulars of the problem, e.g. boundary conditions or the wave equation governing the physical system we're modeling.

Instead of working with real-valued signals, we can work with complex-valued signals, and then at the end of the day take the real part.

The analytic signals $s_a(t)$ and $r_a(t)$ related to the real-valued signals are given by $x_a(t)=x(t)+i\cal H[x](t)$ where $\cal H$ is the Hilbert transform and $x$ stands for $s$ or $r$.

Given $s(t)$, we can calculate $r(t)$ in two ways:

  1. We could use the integral transformation with kernel $k$.

  2. We could calculate $s_a(t)$ from $s(t)$, then perform an (so far unknown) integral transformation $r_a(t) = \int dt'\, h(t-t') s_a(t')$ with kernel $h$, and then take the real part of $r_a(t)$.

Obviously, the results must be the same, but what must $h$ be? In particular, how is $h$ related to $k$?

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  • $\begingroup$ Would Signal Processing or Mathematics be a better home for this question? $\endgroup$ – Qmechanic Aug 8 '16 at 8:48
  • $\begingroup$ I don't really think so. When in physics we solve wave equations for real wave functions, we usually don't use real wave functions but complex ones. Instead of sin(x), we operate with exp(x) and take the real part at the end of the calculation. I was wondering how the Green's functions in these two cases are related. $\endgroup$ – ThomasS Aug 8 '16 at 8:59
  • $\begingroup$ This looks like Signal Processing to me. It is entirely maths. I don't see any physics in it. It is clearer now you've edited, but it is still maths/Signal Processing. $\endgroup$ – sammy gerbil Aug 8 '16 at 20:47
  • $\begingroup$ Did you try working this out yourself? In the end this problem requires nothing beyond calculus. What did you try and where did you get stuck? $\endgroup$ – DanielSank Aug 9 '16 at 3:05
  • $\begingroup$ You're right. There's no physics in it but it is essential for doing physics :-) How can I withdraw the question? $\endgroup$ – ThomasS Aug 9 '16 at 6:30

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