0

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: $$\lim_{T\to\infty} \frac{1}{2T} \cdot \int_{-T}^{T} |x(t)|^2 dt$$

Skyris
  • 376

1 Answers1

2

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?

Alex R.
  • 32,771
  • I lost you when you say "the integral of each term of the sum gives $EA_iA_j[\mathrm{min}(T,T(i+1),T(j+1)) - \mathrm{max}(T,Ti,Tj)]$" Could you explain this please ? – Skyris Feb 28 '18 at 20:44
  • $g(t-iT)g(t-jT)$ is nonzero on the interval which is the intersection of $[iT,T(i+1)]$ and $[jT,T(j+1)]$, so the integral will be the length of this intersection. – Alex R. Feb 28 '18 at 22:38
  • I think I got it, thank you – Skyris Mar 01 '18 at 01:45