I have started learning digital signal processing and came across with a question about energy and power of a digial signal.
I have calculated the correct answers but i'm not quite sure about that i have done the calculations correctly, in other words, i may have solved them by luck. I tried finding similar questions on Youtube or on other websites but i couldn't. Could someone who has experience with solving that kind of problems have a quick look at my solution and suggest me a book ?
$$ x[n] = \Bigg\{ (-1)^n.n, for \ n=1,2,3 \\ 0 \ otherwise $$
We also have its periodic version $$ y\left[n\right]=\sum _{n=-\infty \:}^{\infty }\:x\left[n+7k\right] $$
I'm supposed to calculate the energy and power of $x[n]$ and $y[n]$.
For of $x[n]$ $$ E=\sum _{n=-\infty \:}^{\infty }\:\left|x\left[n\right]\right|^2=\left(\left|-1\right|+\left|2\right|+\left|-3\right|\right)^2=14 $$ $$ \lim \:_{n\to \:\:\infty }\left(\frac{1}{2N+1}\:\sum _{n=-N}^N\:\left|x\left[n\right]\right|^2\:\right)=\lim \:\:_{n\to \infty \:}\frac{1}{2N+1}\left[\sum _{n=-N}^0\left(0\right)\:+\sum _{n=1}^3\left(14\right)\:\sum _{n=4}^{\infty }\left(0\right)\:\right]=\lim \:\:\:_{n\to \:\infty \:\:}\frac{14}{2N+1}=0 $$
I think i have calculated $x[n]$ correctly yet the problem comes with periodic $y[n]$ function. Since it is periodic the energy will be equal to infinity because we will sum 14 to the last.
For power of $y[n]$ we have the formula below because it is periodic. $$ P_x=\frac{1}{N}\sum _{n=0}^{N-1}\:\left|x\left[n\right]\right|^2 $$ Since $y[n]$ is periodic we know that it is circulating, we also know that we will get the same values after every 7th iteration. Since energy of $x[n]=14$ and the sample size is 7. I thought we can get the answer as:
$$ P_x=\frac{1}{N}\sum _{n=0}^{N-1}\:\left|x\left[n\right]\right|^2=\frac{14}{7}=2 $$
The answers are correct but since i couldn't understand how to solve periodic discrete function (because of shifting). I'm not sure about the solution. I would appreciate that if someone enlighten me and provide me some sources.