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A doubt came up to me on an Oppenheim's Signals and Systems 2ed exercise:

1.3) (f) Determine $P_{\infty}$ and $E_{\infty}$ for the following signal: $x\left[n\right]=\cos\left(\frac{\pi}{4}n\right)$

On theory, $P_{\infty}$ and $E_{\infty}$ are defined as:

$$ \begin{align} E_{\infty}&\triangleq\lim\limits_{N\to \infty}{\sum_{n=-N}^{+N}{|x\left[n\right]|^{2}}}\\ P_{\infty}&\triangleq\lim\limits_{N\to \infty}{\frac{E_{\infty}}{2N+1}} \end{align} $$

Starting by $E_{\infty}$:

$$ \begin{align} E_{\infty}&=\lim\limits_{N\to \infty}{\sum_{n=-N}^{+N}{\left|\cos\left(\frac{\pi}{4}n\right)\right|^{2}}}\\ &=\lim\limits_{N\to \infty}{\sum_{n=-N}^{+N}{\cos^{2}\left(\frac{\pi}{4}n\right)}}\\ &=\lim\limits_{N\to \infty}{\sum_{n=-N}^{+N}{\left(\frac{e^{j\frac{\pi}{4}n}+e^{-j\frac{\pi}{4}n}}{2}\right)^{2}}}\\ &=\lim\limits_{N\to \infty}{\frac{1}{4}\sum_{n=-N}^{+N}{\left(e^{j\frac{\pi}{2}n}+2+e^{-j\frac{\pi}{2}n}\right)}}\\ &=\lim\limits_{N\to \infty}{\frac{1}{4}\sum_{n=-N}^{+N}{\left(j^{n}+2+j^{-n}\right)}}\\ &=\lim\limits_{N\to \infty}{\frac{1}{4}\left[\sum_{n=-N}^{+N}{2}+\sum_{n=-N}^{+N}{j^{n}}+\sum_{n=-N}^{+N}{\left(-j\right)^{n}}\right]} \end{align} $$

So far, so good... Now here's what I imagined:

$$ \begin{align} \sum_{n=-N}^{+N}{j^{n}}&=\color{red}{\sum_{n=-N}^{-1}{j^{n}}}+1+\sum_{n=1}^{+N}{j^{n}} \\ &=\color{red}{\sum_{n=1}^{+N}{\left(-j\right)^{n}}}+1+\sum_{n=1}^{+N}{j^{n}} \end{align} $$

and

$$ \begin{align} \sum_{n=-N}^{+N}{\left(-j\right)^{n}}&=\color{red}{\sum_{n=-N}^{-1}{\left(-j\right)^{n}}}+1+\sum_{n=1}^{+N}{\left(-j\right)^{n}} \\ &=\color{red}{\sum_{n=1}^{+N}{j^{n}}}+1+\sum_{n=1}^{+N}{\left(-j\right)^{n}} \end{align} $$

Which gives us:

$$\sum_{n=-N}^{+N}{j^{n}}+\sum_{n=-N}^{+N}{\left(-j\right)^{n}}=2+2\sum_{n=1}^{+N}{j^{n}}+2\sum_{n=1}^{+N}{\left(-j\right)^{n}}$$

Putting this all back into $E_{\infty}$:

$$ \begin{align} E_{\infty}&=\lim\limits_{N\to \infty}{\frac{1}{4}\left[2+2\sum_{n=-N}^{+N}{1}+2\sum_{n=1}^{+N}{j^{n}}+2\sum_{n=1}^{+N}{\left(-j\right)^{n}}\right]}\\ &=\lim\limits_{N\to \infty}{\frac{1}{2}\left[\left(2N+2\right)+\sum_{n=1}^{+N}{j^{n}}+\sum_{n=1}^{+N}{\left(-j\right)^{n}}\right]} \end{align} $$

That's where I got stuck, because this should sum up to:

$$E_{\infty}=\lim\limits_{N\to \infty}{\frac{1}{2}\left[2N+1\right]}=\infty$$

So that:

$$P_{\infty}=\lim\limits_{N\to \infty}{\frac{E_{\infty}}{2N+1}}=\frac{1}{2}\left(\lim\limits_{N\to \infty}{\frac{2N+1}{2N+1}}\right)=\frac{1}{2}$$

Which are the books answers to this exercise...

What am I doing wrong or missing here???

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  • $\begingroup$ The sums for $j$ are zero. The only problem is the $2$ $\endgroup$ – Rafa Budría May 30 '17 at 21:29
  • $\begingroup$ Yeah, there's always the possibility that the book's answers are mistaken... $\endgroup$ – bertozzijr May 31 '17 at 0:55
  • $\begingroup$ No, I don't think so. I meant I see the two sums cancel each other, but I cannot see how the $2$ can be $1$. $\endgroup$ – Rafa Budría May 31 '17 at 4:40
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I got to a solution!

All the previous assumtions about the sums were correct... Now:

$$ \begin{align} E_{\infty}&=\lim\limits_{N\to \infty}{\frac{1}{4}\left[2+2\sum_{n=-N}^{+N}{1}+2\sum_{n=1}^{+N}{j^{n}}+2\sum_{n=1}^{+N}{\left(-j\right)^{n}}\right]}\\ &=\lim\limits_{N\to \infty}{\frac{1}{2}\left[\left(2N+2\right)+\underbrace{\sum_{n=1}^{+N}{j^{n}}+\sum_{n=1}^{+N}{\left(-j\right)^{n}}}_{0}\right]}\\ &=\lim\limits_{N\to \infty}{\frac{1}{2}\left(2N+2\right)}\\ &=\lim\limits_{N\to \infty}{\left(N+1\right)}= \infty \end{align} $$

Then: $$ \begin{align} P_{\infty}&=\lim\limits_{N\to \infty}{\frac{E_{\infty}}{2N+1}}\\ &=\lim\limits_{N\to \infty}{\frac{N+1}{2N+1}}\\ &=\underbrace{\lim\limits_{N\to \infty}\frac{\frac{d}{dN}\left(N+1\right)}{\frac{d}{dN}\left(2N+1\right)}}_{L'Hopital}=\frac{1}{2} \end{align} $$

Actually, I was doing nothing wrong... My mistake was trying to adapt my soluton induced by the answers I already had from the book.

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