So I'm asked to find the time-averaged power of the following continuous signal: $$z(t) = \sqrt{E} \sum_{k=-\infty}^{+\infty} A_k g(t-kT)$$ where $A_k =1$ if $|k|$ is a prime number and $A_k=-1$ otherwise, and where $g(t)=1$ for $0 \leq t \leq T$ and $g(t) = 0$ otherwise. $T$ is the fundamental period.
I have no idea how I can do this. For now, all I did was sketching $z(t)$ for $0 \leq t \leq 4T$, but I don't see how it could help me.
If you could give me a hint, that would be nice, thank you.
Just a quick reminder for those who are interested, the power of a continuous signal is defined by: $$\lim_{T\to\infty} \frac{1}{2T} \cdot \int_{-T}^{T} |x(t)|^2 dt$$
I think you can just compute it directly:
$$P=\lim_{T\rightarrow\infty}\frac{1}{2T}\int_{-T}^T|z(t)|^2dt.$$
Squaring the sum gives terms $A_iA_jg(t-iT)g(t-jT)$. Next, $g(t-kT)=1$ for $0\leq t-kT\leq T$, which gives $kT\leq t\leq T(k+1)$ when $g=1$. Thus the integral of each such term of the sum gives
$$EA_iA_j[\min(T,T(i+1),T(j+1))-\max(T,Ti,Tj)].$$
So for example when $i>j>1$ this becomes $EA_iA_j[T(i+1)-Tj]=EA_iA_jT(i-j+1)$.
For any fixed $t$, the sum is finite (e.g. $g(t-kT)=0$ for large $k$), so you can see how it behaves for each $t$. Can you finish it from here?