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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 $\text{(iii)}$:

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

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    $\begingroup$ 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
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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 $0$.

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

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

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  • $\begingroup$ 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

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