0
$\begingroup$

I am calculating the average power of a vector. I would like to compare the final expression with the simulation. However, they are not equal. Please help me to point out which steps are wrong. Thank you very much.

Calculate the average power of a vector:

$x(n) = \sum\limits_{i = 1}^{L - 1} {\sum\limits_{l = 0}^{L - 1} {\frac{1}{L}{e^{ - j2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right){r_l}}}} } {e^{ - j2\pi \frac{i}{L}l}}s\left( {n- \frac{i}{L}N} \right),n = 0,...,N - 1$

where $s$ is independent and identically distributed data symbols with zero mean and variance $\sigma _s^2$, $r = ({r_0},...,{r_{L - 1}})$ has Gaussian distribution with mean ${\mu _r}$ and variance $\sigma _r^2$, and $\frac{N}{L}$ is integer.

The average power of $x$ is:

$\bar P = \frac{1}{N}\sum\limits_{n = 0}^{N - 1} {E\left\{ {{{\left| {x\{ n\} } \right|}^2}} \right\}}$ (1)

where

$\begin{array}{l} E\left\{ {{{\left| {x\{ n\} } \right|}^2}} \right\} = \sum\limits_{i = 1}^{L - 1} {\frac{{\sigma _s^2}}{{{L^2}}}} E\left\{ {{{\left| {\sum\limits_{l = 0}^{L - 1} {{e^{ - j2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right){r_l}}}{e^{ - j2\pi \frac{i}{L}l}}} } \right|}^2}} \right\}\\ = \sum\limits_{i = 1}^{L - 1} {\frac{{\sigma _s^2}}{{{L^2}}}} \sum\limits_{l = 0}^{L - 1} {\sum\limits_{m = 0}^{L - 1} E } \left\{ {{e^{ - j2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right)({r_l} - {r_m})}}{e^{ - j2\pi \frac{i}{L}(l - m)}}} \right\} \end{array}$ (2)

When $l=m$, then $E\left\{ {{{\left| {x\{ n\} } \right|}^2}} \right\} = \sum\limits_{i = 1}^{L - 1} {\frac{{\sigma _s^2}}{{{L^2}}}} \sum\limits_{l = 0}^{L - 1} E \left\{ {{e^{ - j2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right)({r_l} - {r_l})}}{e^{ - j2\pi \frac{i}{L}(l - l)}}} \right\} = \frac{{\sigma _s^2}}{{{L^2}}}(L - 1)L = \frac{{\sigma _s^2}}{L}(L - 1)$ (3)

When $l \ne m$, then $\begin{array}{l} E\left\{ {{{\left| {x\{ n\} } \right|}^2}} \right\} = \sum\limits_{i = 1}^{L - 1} {\frac{{\sigma _s^2}}{{{L^2}}}} \sum\limits_{l = 0}^{L - 1} {\sum\limits_{m = 0\atop m \ne l}^{L - 1} E } \left\{ {{e^{ - j2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right){r_l}}}{e^{ - j2\pi \frac{i}{L}l}}} \right\}\left\{ {{e^{j2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right){r_m}}}{e^{ - j2\pi \frac{i}{L}m}}} \right\}\\ = \sum\limits_{i = 1}^{L - 1} {\frac{{\sigma _s^2}}{{{L^2}}}} ({L^2} - L){\left| {{\Phi _r}\left( {2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right)} \right)} \right|^2} = \frac{{\sigma _s^2}}{L}(L - 1)\sum\limits_{i = 1}^{L - 1} {{{\left| {{\Phi _r}\left( {2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right)} \right)} \right|}^2}} \end{array}$ (4)

where ${{\Phi _r}\left( {2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right)} \right)}$ is the characteristic function of the random variable $r$, which is defined as (for Gaussian distribution):

${\left| {{\Phi _r}\left( {2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right)} \right)} \right|^2} = {e^{ - {{\left( {2\pi \left( {\frac{n}{N} - \frac{i}{L}} \right)} \right)}^2}\sigma _r^2}} \approx 1 - 4{\pi ^2}{\left( {\frac{n}{N} - \frac{i}{L}} \right)^2}\sigma _r^2$ (5)

Therefore, from (2), (3), (4), and (5), we have:

$\begin{array}{l} E\left\{ {{{\left| {x\{ n\} } \right|}^2}} \right\} = \frac{{\sigma _s^2}}{L}(L - 1) + \frac{{\sigma _s^2}}{L}(L - 1)\sum\limits_{i = 1}^{L - 1} {\left( {1 - 4{\pi ^2}{{\left( {\frac{n}{N} - \frac{i}{L}} \right)}^2}\sigma _r^2} \right)} \\ = \frac{{\sigma _s^2}}{L}(L - 1) + \frac{{\sigma _s^2}}{L}{(L - 1)^2} - \frac{{4{\pi ^2}\sigma _s^2\sigma _r^2}}{L}(L - 1)\sum\limits_{i = 1}^{L - 1} {\left( {\frac{{{n^2}}}{{{N^2}}} - 2\frac{n}{N}\frac{i}{L} + \frac{{{i^2}}}{{{L^2}}}} \right)} \\ = \frac{{\sigma _s^2}}{L}(L - 1) + \frac{{\sigma _s^2}}{L}{(L - 1)^2} - \frac{{4{\pi ^2}\sigma _s^2\sigma _r^2}}{L}(L - 1)\left( {\frac{{{n^2}}}{{{N^2}}}(L - 1) - 2\frac{n}{N}\frac{{L(L - 1)}}{{2L}} + \frac{{(L - 1)L(2L - 1)}}{{6{L^2}}}} \right)\\ = \frac{{\sigma _s^2}}{L}(L - 1) + \frac{{\sigma _s^2}}{L}{(L - 1)^2} - \frac{{4{\pi ^2}\sigma _s^2\sigma _r^2}}{L}{(L - 1)^2}\left( {\frac{{{n^2}}}{{{N^2}}} - \frac{n}{N} + \frac{{2L - 1}}{{6L}}} \right) \end{array}$ (6)

From (1) and (6), the average power of $x$ becomes:

$\begin{array}{l} \bar P = \frac{{\sigma _s^2}}{L}(L - 1) + \frac{{\sigma _s^2}}{L}{(L - 1)^2} - \frac{{4{\pi ^2}\sigma _s^2\sigma _r^2}}{{NL}}{(L - 1)^2}\sum\limits_{n = 0}^{N - 1} {\left( {\frac{{{n^2}}}{{{N^2}}} - \frac{n}{N} + \frac{{2L - 1}}{{6L}}} \right)} \\ = \frac{{\sigma _s^2}}{L}(L - 1) + \frac{{\sigma _s^2}}{L}{(L - 1)^2} - \frac{{4{\pi ^2}\sigma _s^2\sigma _r^2}}{{NL}}{(L - 1)^2}\left( {\frac{{(N - 1)(2N - 1)}}{{6N}} - \frac{{N - 1}}{2} + \frac{{N(2L - 1)}}{{6L}}} \right) \end{array}$

Which steps are not correct?

$\endgroup$
  • $\begingroup$ What's wrong with RMS? $\endgroup$ – David K Jul 16 '14 at 20:29
  • $\begingroup$ Hi @DavidK, what do you mean about RMS? $\endgroup$ – Steven Huynh Jul 17 '14 at 8:47
  • $\begingroup$ From your formulas I suppose that $s(t)$ a continuous random process. When computing $E\{|x(n)|^2\}$ why don't you get values of the autocorrelation of $s(t)$? You should have terms like $E\{s(n/N-i/L)s(n/N-j/L)\}$, and I think you just considered $i=j$. $\endgroup$ – Matt L. Jul 17 '14 at 9:22
  • $\begingroup$ Hi @MattL., indeed, $s(t)$ is not a continuous random process, I assume it as independent and identically distributed data symbols. So, there is still autocorrelation of $s(t)$? P/s: I will edit $s(t)$ in question to avoid misunderstanding. Thank you for your help. $\endgroup$ – Steven Huynh Jul 17 '14 at 9:55
  • $\begingroup$ But obviously the argument of $s$ is not integer, is it? At least I thought that $(n/N-i/L)$ is the argument of $s$. $\endgroup$ – Matt L. Jul 17 '14 at 10:00

Your Answer

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

Browse other questions tagged or ask your own question.