$\hat{x}(t)$ is the Hilbert Transform of $x(t)$. Can anybody help me proving the above.


closed as off-topic by Tendero, MBaz, lennon310, A_A, Stanley Pawlukiewicz Sep 15 '18 at 19:25

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  • $\begingroup$ Yes, with pleasure, if you provide us with the point at where you are stuck or have doubts $\endgroup$ – Laurent Duval Sep 13 '18 at 12:47
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    $\begingroup$ I'm voting to close this question as off-topic because the OP hasn't showed any effort to solve the problem by himself. $\endgroup$ – Tendero Sep 13 '18 at 13:13

The Hilbert transform of $x(t)$ denoted by $\hat{x}(t)$ has the following Fourier transform \begin{equation} \mathcal{F}(\hat{x}) = H(w)X(w) = -j \operatorname{sgn}(w)X(w) \end{equation} because \begin{equation} \hat{x}(t) = \int\limits_{-\infty}^{\infty} \frac{x(u)}{\pi(t-u)} \ du \end{equation} Notice that \begin{equation} \vert \mathcal{F}(\hat{x}) \vert^2 = \vert -j \operatorname{sgn}(w)X(w) \vert^2 = \vert X(w) \vert^2 \end{equation} i.e. \begin{equation} \int\limits_{-\infty}^{\infty} \vert \mathcal{F}(\hat{x}) \vert^2 \ dw = \int\limits_{-\infty}^{\infty} \vert X(w) \vert^2 \ dw \end{equation} Use Parsevals equation which gives you \begin{equation} \int\limits_{-\infty}^{\infty} \vert \hat{x}(t) \vert^2 \ dt = \int\limits_{-\infty}^{\infty} \vert x(t) \vert^2 \ dt \end{equation} P.S: For real signals, you could indeed remove the magnitudes (absolute values).

  • $\begingroup$ Thanks, I am not sure whether $x(t)$ is real or not. $\endgroup$ – RAKESH GANDHI Sep 14 '18 at 5:51

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