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Suppose we have a power signal $n(t)$ whose power- auto-correlation function is: $R_{n}(\tau)$.

Now we multiply the signal by a cosine as $$\cos(2\pi f_ct) \cdot n(t)$$ whose auto-correlation is $R_{n_i}(\tau)$.

How is $R_{n}(\tau)$ mathematically related to $R_{n_i}(\tau)$ ?

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$$R_{n}(\tau) = E(n(t)n(t-\tau))$$ $$R_{n_i}(\tau) = \cos(2\pi f_ct)\cos(2\pi f_c(t-\tau))E(n(t)n(t-\tau))=\cos(2\pi f_ct)\cos(2\pi f_c(t-\tau))R_{n}(\tau) $$ Using $\cos(a-b) = \cos a \cos b + \sin a \sin b$, we get $$R_{n_i}(\tau) = \cos(2\pi f_ct)[ \cos(2\pi f_ct ) \cos(2\pi f_c\tau)+ \sin(2\pi f_ct) \sin(2\pi f_c\tau)]R_{n}(\tau) $$ or $$R_n(\tau) = \frac{R_{n_i}(\tau)}{\cos(2\pi f_ct)[ \cos(2\pi f_ct ) \cos(2\pi f_c\tau)+ \sin(2\pi f_ct) \sin(2\pi f_c\tau)]}$$

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  • $\begingroup$ How can we take out the cosine terms out of the expectation operator? It contains terms dependent on t, right? $\endgroup$ – helloworld1e. Jul 8 at 10:24
  • $\begingroup$ but $\cos(f(t))$ is not a random process. Hence seen as constant with respect to $E(.)$ $\endgroup$ – Ahmad Bazzi Jul 8 at 10:25
  • $\begingroup$ A power signal need not be a random signal. $\endgroup$ – Dilip Sarwate Jul 8 at 21:37

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