A sample of a random process is given as:
$$ x(t) = Acos(2\pi f_0t) + Bw(t) $$
where w(t) is a white noise process with 0 mean and a power spectral density of N0/2, and f, A and B are constants. Find the auto-correlation function.
Here's my attempt at a solution: $$ let\\a = 2\pi f_0t\\ b = 2\pi f_0(t+\tau) $$ $$ \begin{align} Autocorrelation & = E\{x(t)x(t + \tau)\}\\ & = E\{(Acos(a) + Bw(t))(Acos(b) + Bw(t+\tau))\}\\ & = E\{A^2cos(a)cos(b) + ABcos(a)w(t+\tau) + ABcosb(wt) + B^2w(t)w(t+\tau)\}\\ & = E\{A^2cos(a)cos(b)\} + E\{ABcos(a)w(t+\tau)\} + E\{ABcosb(wt)\} + E\{B^2w(t)w(t+\tau)\}\\ & = E\{A^2cos(a)cos(b)\} + E\{B^2w(t)w(t+\tau)\}\\ & = E\{A^2cos(a)cos(b)\} + B^2(R_w(\tau))\\ & = E\{A^2cos(a)cos(b)\} + B^2(N_0/2)(\delta(\tau))\\ \end{align} $$
The expectation terms with the noise in them all equal 0 (the last is just the auto correlation of white noise ... hence the simplification above. Using trig identities: $$ cos(a)cos(b) = (1/2)[cos(a + b) + cos(a - b)] $$
we have: $$ \begin{align} Autocorrelation & = E\{A^2cos(a)cos(b)\} + B^2(N_0/2)(\delta(\tau))\\ & = E\{(A^2)(1/2)[cos(a+b)+cos(a-b)]\} + B^2(N_0/2)(\delta(\tau))\\ & = (A^2/2)[E\{cos(a+b)\} + E\{cos(a-b)\}] + B^2(N_0/2)(\delta(\tau))\\ \end{align} $$ We're dealing with constant terms, so expectation term goes away and subbing in our initial conditions we get: $$ \frac {A^2}2 [cos(2\pi f_o(2t + \tau) + cos(2\pi f_o\tau)] + B^2(N_0/2)(\delta(\tau)) $$
For some reason I can't help but feel I did something incorrectly calculating that autocorrelation ... it's supposed to be a function of Tau, but has a t in there ... I would very much appreciate it if someone could point me in the right direction, or explain what I messed up. I don't know whether it matters, but in this class we're dealing with only wide sense stationary processes.