Suppose there are two independent signals, $s(t-\tau)$ and $n(t)$, and they are doubtly uncorrelated so that $\mathbb{E}${$s(t-\tau)\times n(t)$}=0. I wonder if the equation $\mathbb{E}${$\frac{\partial s^*(t-\tau)}{\partial \tau}\times n(t)$} still equal to zero, which are uncorrelated?

I read the expression from a radar paper: enter image description here

Here the "Fj" is the wrong print and the "$r(t)-as(t-\tau)$" represnts the noise received at radar. My calculation finally leads to the (19) equation puls a $n^*(t) \times \frac{\partial^2 s(t-\tau)}{\partial \tau}$ and a $n(t) \times \frac{\partial^2 s'(t-\tau)}{\partial \tau}$. (Sorry I have no idea why the conjugate symbol can not be used in the latter expressions)

It seems the uncorrelated relation between the noise and the 2nd derivative of the signal conjugate can derive the (19), but I have no idea how to proof or understand the equation.

  • $\begingroup$ are you looking for a proof of "if signals uncorrelated, then derivatives uncorrelated", or, as you state in your question, based on the stronger requirement "if signals independent, then derivatives uncorrelated"? $\endgroup$ May 17 at 11:11
  • $\begingroup$ The latter one...Thank you for telling me the uncertain description in my question. I'll amend my title.@MarcusMüller $\endgroup$ May 17 at 13:31


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