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I was reading something on communication, then I came across the following equation:

$Power_{rx}=Power_{tx}*|R|^2/(1+d^2)$ where $Power_{tx}$ and $d$ can be assume to be constant, and R is the Rayleigh random variable.

Then it says that the $Power_{rx}$ have a distribution of

$f_{power_{rx}}(Power_{rx})=1/P_{mean}*exp(-Power_{rx}/P_{mean})$,

where $P_{mean}$ is the expectation, $E[Power_{rx}]$.

Confusion:

$|R^2|$ alone having the distribution as above I have no problem, however, I don't understand why the entire $Power_{rx}$ also have the same distribution. At least, I see they multiple a bunch of constant to R, shouldn't that will having some effect on the distribution as well?

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  • $\begingroup$ What is the difference between $Power_{tx}$ and $Power_{rx}$? $\endgroup$
    – Marc
    Jun 15 '14 at 12:15
  • $\begingroup$ Power_{tx} (transmitting power) is just a constant, but Power_{rx} (received power) is a function of a random variable, so Power_{rx} is a random variable. $\endgroup$
    – kou
    Jun 15 '14 at 16:22
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$|R^2|$ alone having the distribution as above I have no problem, however, I don't understand why the entire $Power_{rx}$ also have the same distribution. At least, I see they multiple a bunch of constant to R, shouldn't that will having some effect on the distribution as well?

Yes, it does have an effect on the distribution -- the expectations are different, although the type of distribution is the same (i.e. Exponential).

You seem okay with the fact that if $R$ has a Rayleigh distribution, then $X := R^2$ has an Exponential distribution; i.e., the distribution of $X$ has the density function $$f_{X}(x)=(1/(EX)) e^{-x/(EX)}(x \ge 0) $$ and the cumulative distribution function is (by integration) $$pr(X \le x) = (1 - e^{-x/(EX)})(x \ge 0) $$ where $EX$ denotes the expectation of $X$.

But then $Y := c\cdot X$, where $c > 0$ is any constant (e.g., $c=Power_{tx}/(1+d^2)$), also has an Exponential distribution: $$pr(Y \le y) = pr(c\cdot X \le y) = pr(X \le y/c) = (1 - e^{-(y/c)/(EX)})(y/c \ge 0) = (1 - e^{-y/(c\cdot EX)})(y \ge 0). $$ That is, $c\cdot X$ also has an Exponential distribution, but with expectation $c \cdot EX$.

In more of your original symbols, the densities are

$f_{R^2}(x)=(1/X_{mean}) exp(-x/X_{mean})$, where $X_{mean}$ is $E(R^2)$, and

$f_{Power_{rx}}(p)=(1/P_{mean}) exp(-p/P_{mean})$, where $P_{mean}$ is $E(Power_{rx}) = E(R^2)\cdot Power_{tx}/(1+d^2)$.

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