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I have a signal of the form $f(\omega,x) = g(\omega) e^{i \omega (x +c)},$ for $g: [\omega_1 , \omega_2] \rightarrow C, \, x\in R$ and some constant $c.$ I want to get the function $$ F (\omega,x) = \mathcal{H} ( \Re \{f(\cdot,x)\}) (\omega), $$ where $\mathcal{H}$ denotes the Hilbert transform with respect to frequency $\omega,$ and $\Re$ the real part. I need the function $F$ to be again symbolic on $x$ such that I can differentiate it later.

I am trying to implement it in Matlab using:

L = 1e-2; K = 100; c = 1;
omega = (1/L)*(-K:K-1);

g = exp(-omega.^2/4);

syms x; f(x) = real(g.*exp(1i*omega*(x+c)));

but then the hilbert build-in function is not working. Do you have any idea how to do it?

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1 Answer 1

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The real part, the Hilbert transform and the derivation are all operators that commute. So you can move the derivative just close to the function.

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    $\begingroup$ "commune"? or "commute"? $\endgroup$ May 21, 2019 at 23:14
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    $\begingroup$ I don't even have the excuse that letters are close on the keyboard $\endgroup$ May 23, 2019 at 7:15
  • $\begingroup$ Thanks for your comment, but I think this does answer my question. The derivative is with respect to x and the Hilbert with respect to omega. My question is if the hilbert transform can be applied to f, that is symbolic at x, and result to a symbolic function at x. $\endgroup$
    – Nikolas
    May 23, 2019 at 8:58

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