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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.

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What is the $Q$ function? $Q=\rm{erf}$? – Mario Carneiro Nov 1 '13 at 2:53

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.

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This approach finds the worst-case error probability (when $x$ has minimum value $a$) as an approximation to the average error probability, and so approximations for $\cos(a)$ might be useful. – Dilip Sarwate Mar 18 '13 at 1:43
I fail to see why this approximation should be any better than $$E[P(X)] \approx Q\left(\sqrt{\frac{2 E_b}{N_0}}\right).$$ – Did Feb 22 '15 at 13:06

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