I'm working through my textbook for a communications course I'm taking, and this problem is confusing me big time. Like always, the math questions give me the most problems. Maybe I should take the hint. Anyways, I'll just get to it I guess.
Given a carrier phase synchronization error $x$, it can be shown that the probability of bit error for coherent BPSK is:
$\displaystyle P(e) = Q \left( \sqrt{\frac{2E_b}{N_0}cos^2(x)} \right)$
If $x$ is a random variable uniformly distributed over $-a \leq x \leq a$, where $0 < a \ll \pi$, determine the average probability of bit error as a function of $a$.
The question itself seems pretty straightforward. Bit error is a function of a random variable. I need to find the average bit error. That means I need to figure out an expression for the expected value of bit error. Here's my brief attempt of a solution:
$\displaystyle E[P(e)]=\int_{-\infty}^{\infty}P(e)f_x(x)dx = \frac{1}{2a}\int_{-a}^{a}Q \left( \sqrt{\frac{2E_b}{N_0}cos^2(x)} \right)dx$
Beyond this, I really don't know how to approach the problem. I don't think I can really do anything with the Q-function as-is, so I was thinking to express it as an integral and hoping that I could do something with that.
$\displaystyle E[P(e)]=\frac{1}{(\sqrt{2\pi})(2a)}\int_{-a}^{a} \int_{\sqrt{\frac{2E_b}{N_0}cos^2(x)}}^{\infty} exp \left( -\frac{u^2}{2}\right)du \thinspace dx$
This is obviously pretty messy and I'm not sure if this is even the right way to be going with it. I haven't used the information that $0<a \ll \pi$ yet, so if I had to guess, there might be some sort of trickery I could apply to that double integral and work from there, but I'm not seeing it at this point. If $0<a \ll \pi$ is implying $a < \frac{\pi}{10}$, I could make some (crazy) assumption that $a = 0$, but that would completely break the question and make it immediately solvable from the start with a solution not in terms of $a$, not to mention that it's a pretty broad assumption to make. If it's just suggesting that $a$ is closer to $0$ than to $\pi$, I have no idea how I can use that information to make anything easier. The only possible thing I can think of is that $cos^2(x)$ is decreasing on that interval, but again, no idea how that might help. Maybe I can go back to my previous expression (with the Q-function still intact) and say something about the integral being over a small region of the Q-function and make some assumption there? Again, my knowledge isn't really up to par here and I don't know if any of this is possible.
So that's about it. I'm honestly completely stumped at this point. Any help would be much appreciated.
All you really need to do here is use the fact that $a$ is very small. You have everything else written down OK:
$$E[P(X)] = \frac{1}{2 a} \int_{-a}^a dx \: Q\left(\sqrt{\frac{2 E_b}{N_0}} |\cos{x}|\right)$$
When $a$ is small, you can approximate the integral by $2 a$ times the integrand evaluated at $a$. Thus
$$E[P(X)] \approx Q\left(\sqrt{\frac{2 E_b}{N_0}} \cos{a}\right)$$
You can make further approximations on $\cos{a}$, but I am not sure how much that will further simplify things.