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Consider this exercise in stochastic processes:

Consider a sine signal with random phase, $X(t) = A sin(\omega_0 t + B)$, with $A$ and $\omega_0 $ constants and $B$ a random phase shift with PDF (probability density function) uniform in the interval of $[-\pi,\pi]$.

How does one calculate the average and autocorrelation functions for that signal?

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    $\begingroup$ What are you thoughts about that? What have you tried so far? Where have you got stuck? $\endgroup$ – Taroccoesbrocco Mar 19 '18 at 22:12
  • $\begingroup$ Hi Taroccoesbrocco. Thanks for your reply. I have many question about this exercise, I have also other 3 related to same subject that I have questions about. Ideally, I'd like to chat with someone on this forum and ask a couple of question via email or phone. Do you think is there any way to get that type of help from here? Thank you. $\endgroup$ – Sergio Mar 20 '18 at 1:29
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$$\mu\left[X(t)\right]=\mu\left[A\sin(\omega_0t+B)\right]=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}A\sin(\omega_0t+B)dB=0\\R\left[X(t+\tau)X^{*}(t)\right]=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}A^2\sin(\omega_0t+B)\sin\left(\omega_0(t+\tau)+B\right)dB\\=\dfrac{1}{4\pi}\int_{-\pi}^{\pi}A^2\cos(\omega_0\tau)dB-\dfrac{1}{4\pi}\int_{-\pi}^{\pi}A^2\cos(2\omega_0t+2B+\omega_0\tau)dB\\=\dfrac{A^2}{2}\cos(\omega_0\tau)$$

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  • $\begingroup$ Hi. Thanks for the reply. Isn't the integral of sin(x) -cos(x)? Maybe I missed something but apparently, you considered the Integral o sin(x) as cos(x). Thank you. $\endgroup$ – Sergio Mar 20 '18 at 1:24
  • $\begingroup$ Yes that's right but also mind the variable the integrals are getting calculated respect to. In the auto correlation function integrals, the first contains a variable-free integrand and the second one is zero since the integral is over one period. $\endgroup$ – Mostafa Ayaz Mar 20 '18 at 7:32

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