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Consider a signal $v(t)$ given by $$v(t)=\sum_{n=-\infty}^{\infty} b[n]p(t-nT).$$ Assume that $b[n]$ is uncorrelated with zero mean, i.e. $\mathbb{E}[b[n]b^*[m]]=\mathbb{E}[|b[n]|^2]\delta[n-m]$ and $\mathbb{E}[b[n]]=0$.

How can one show, using the autocorrelation of $v(t)$, that the PSD of $v(t)$ is given by the following?

$$S_{v}(f)=\frac{\mathbb{E}[|b[n]|^2]}{T}|P(f)|^2$$

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    $\begingroup$ The derivation of this result is given in almost any textbook on digital communications. $\endgroup$ – Matt L. Nov 4 '20 at 17:12

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