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I have a pulse train of a fixed period, but a random amplitude following a Gaussian distribution $(\mu=0,\sigma=1)$. I know how to work with a random period or delay, but how do I proceed to calculate the autocorrelation with a random amplitude?

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Suppose that your pulse train is denoted $X(t) = Y\cdot g(t)$ where where $g(t)$ is the basic pulse train and $Y$ is the random amplitude of the pulse train. Then, $X(t)$ is not a wide-sense-stationary process and its autocorrelation function is $$R_X(t_1,t_2) = E[X(t_1)X(t_2)] = E[Y^2]\cdot g(t_1)g(t_2).$$ If $Y$ has mean $\mu$ and variance $\sigma_2$, then $E[Y^2] = \mu^2+\sigma^2$. If $g(t)$ is a function of period $T$, that is, $g(t) = g(t+T)$ for all $t$, then $R_X(t_1,t_2)$ is periodic in both arguments: \begin{align} R_X(t_1+T,t_2) &= (\mu^2+\sigma^2)g(t_1+T)g(t_2) = (\mu^2+\sigma^2)g(t_1)g(t_2) = R_X(t_1,t_2)\\ R_X(t_1,t_2+T) &= (\mu^2+\sigma^2)g(t_1)g(t_2+T) = (\mu^2+\sigma^2)g(t_1)g(t_2) = R_X(t_1,t_2) \end{align} I am quite sure that this is not what you hoped to hear and so you must revise your question appropriately, giving more details and describing more clearly what it is you want to find out.

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  • $\begingroup$ Actually, this was the answer I was looking for. I was making a silly mistake before. Thanks! $\endgroup$ – Marco Jan 27 '16 at 21:09

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