I am trying to find the way to reduce the standard expression to compute the power of a generic sequence $x(n)$:

$$P_{\text{x}}= \lim\limits_{N \to \infty}\frac{1}{2N + 1}\sum\limits_{n=-N}^{N}|x(n)|^2\text,$$

when $x(n) = x(n + N_0)$ ($N_0$-periodic sequence), into the simplified formula

$$P_{\text{x}}= \frac{1}{N_0}\sum\limits_{n=0}^{N_0-1}|x(n)|^2\text.$$

How can I demonstrate that second one could be obtained from the first one, using the hypothesis of periodicity? I've tried something like this, but I get stuck in after $\text{(iii)}$:

$$\begin{align} N &= N_0\tag{i}\\ P_{\text{x}}&= \lim\limits_{N_0 \to \infty}\frac{1}{2N_0 + 1}\sum\limits_{n=-N_0}^{N_0}|x(n)|^2\\ &= \lim\limits_{N_0 \to \infty}\frac{1}{2N_0 + 1}\sum\limits_{n=-N_0}^{N_0}|x(n + N_0)|^2\tag{ii}\\ &m = n + N_0\text{, so }\\ P_{\text{x}}&= \lim\limits_{N_0 \to \infty}\frac{1}{2N_0 + 1}\sum_{m=0}^{2N_0}|x(m)|^2\tag{iii} \end{align} $$

  • 1
    $\begingroup$ Your approach with $N_0=N$ is not fully general, it should be valid for all $N$. This only shows the convergence of a subsequence $\endgroup$ – Laurent Duval Nov 27 '16 at 3:26

The basic trick is to bound the series above and below. Let us do it on one side, for positive indices.

For any $N> 0$, you can write $N=kN_0+r_N$, with $0\le r_N< N_0$. Then if $a_n$ (here $a_n = |x_n|^2$) is positive, $\sum_{n=0}^{N-1} a_n$ is increasing. Now $kN_0 \le N< (k+1)N_0$, hence you have:

$$ \sum_{n=0}^{kN_0-1} a_n \le \sum_{n=0}^{N-1} a_n \le \sum_{n=0}^{(k+1)N_0-1} a_n $$ or

$$ k\sum_{n=0}^{N_0-1} a_n \le \sum_{n=0}^{N-1} a_n \le (k+1)\sum_{n=0}^{N_0-1} a_n $$

since $a_n$ is $N_0$-periodic. Now since

$$\frac{1}{(k+1)N_0} \le \frac{1}{N} \le \frac{1}{kN_0}$$ by a simple point-wise product of the threefold inequality (valid because positive):

$$ \frac{k}{(k+1)N_0} \sum_{n=0}^{N_0-1} a_n \le \frac{1}{N}\sum_{n=0}^{N-1} a_n \le \frac{(k+1)}{kN_0}\sum_{n=0}^{N_0-1} a_n \,. $$

Since $k/(k+1)\to 1$ and $(k+1)/k\to 1$ as $N\to\infty$, you have your result, mostly. You can split the original series in $\sum_{-N}^{N}a_n = \sum_{0}^{N} a_n+ \sum_{-N}^{0}a_n - a_0$, take care of the positive and negative indices separately, and leave $a_0/(2N+1)$ tend to $0$.


Another intuitive way:

$P_{\text{x}}= \lim\limits_{N \to \infty}\frac{1}{2N + 1}\sum_{n=-N}^{N}|x(n)|^2$

Now the period is $N_0$ and it extends from -N to N, so total number of periods between $-N$ to $N$ are $2N/N_0$.

$P_{\text{x}}= \lim\limits_{N \to \infty}\frac{2N}{N_0(2N + 1)}\sum_{n=0}^{N_0-1}|x(n)|^2$

$P_{\text{x}}= \lim\limits_{N \to \infty}\frac{1}{N_0(1 + 1/2N)}\sum_{n=0}^{N_0-1}|x(n)|^2$

As ${N\to \infty}$, ${1/2N\to 0}$

$P_{\text{x}}= \frac{1}{N_0}\sum_{n=0}^{N_0-1}|x(n)|^2$

  • $\begingroup$ This intuitive has the limit it is valid only for multiples of the period, and strictly speaking does not show the convergence. $\endgroup$ – Laurent Duval Nov 27 '16 at 3:16
  • $\begingroup$ Even if it's not the multiples of the period, the factor will be $(2N + 1)/N_0$ instead of $2N/N_0$ yielding the same result. I guess since I am just decomposing the term, the initial conditions still hold for convergence. Correct me if I am wrong.. $\endgroup$ – Navin Prashath Nov 27 '16 at 3:49
  • $\begingroup$ Not exactly I think. You will have a reminder on the numerator, a part of a truncated sum over a period $\endgroup$ – Laurent Duval Nov 27 '16 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.