# Expectation and autocorrelation for modulated sinusoid

Given

$$Y(t) = A X(t) \cos(\omega t + \phi)$$

with $$X(t)$$ is zero-mean WSS (wide-sense stationary) process, $$\phi$$ ~ Unif$$(0,2\pi)$$. Suppose $$X(t)$$ and $$\phi$$ are independent random variables. I want to compute the mean and autocorrelation functions of $$Y(t)$$.

My attempt is (using independence of $$X(t)$$ and $$\phi$$, and $$E(X(t)) = 0$$)

$$E(Y(t))= E(A X(t) \cos(\omega t + \phi)) = A \cdot E(X(t)) \cdot E(\cos(\omega t + \phi)) = 0$$

Similarly we have

$$R_Y(t,t+\tau)= E(A X(t) \cos(\omega t + \phi) \cdot A X(t+\tau) \cos(\omega (t + \tau) + \phi)) = 0$$

Is this a correct use of independence? If not, why?

• Thanks. But this also means that the autocorrelation function is zero too which I don't think is right. Am I missing something? Mar 27 at 18:49

## 1 Answer

Your use of independence for computing $$E[Y(t)]$$ is correct. However, for the autocorrelation you get

\begin{align}E[Y(t)Y(t+\tau)]&=A^2E[X(t)X(t+\tau)\cos(\omega t+\phi)\cos(\omega(t+\tau)+\phi)]\\&=A^2E[X(t)X(t+\tau)]\cdot E[\cos(\omega t+\phi)\cos(\omega(t+\tau)+\phi)]\\&=A^2R_X(\tau)E[\cos(\omega t+\phi)\cos(\omega(t+\tau)+\phi)]\tag{1}\end{align}

Note that in general $$X(t)$$ and $$X(t+\tau)$$ are not independent, so $$E[X(t)X(t+\tau)]\neq E[X(t)]E[X(t+\tau)]$$.

You can compute the expectation in $$(1)$$ by using a trigonometric identity to write the product of the two cosines as a sum of two cosines. If you do it right, it should turn out that $$E[Y(t)Y(t+\tau)]$$ only depends on $$\tau$$ but not on $$t$$.

• I see thank you! Mar 27 at 21:26