Re: rician random number

*To*: mathgroup at smc.vnet.net*Subject*: [mg15142] Re: [mg15117] rician random number*From*: "Tomas Garza" <tgarza at mail.internet.com.mx>*Date*: Wed, 16 Dec 1998 03:11:17 -0500*Sender*: owner-wri-mathgroup at wolfram.com

"gcseng at yahoo.com" wrote: >Anyone knows how to generate random number base on the rician >probability density function. >The rician pdf is given by: >f(r)=(r/sigma^2)*exp(1)^(-1*(r^2+A^2)/(2*sigma^2))*Io(r*A/sigma^2) >where 2*sigma^2 is the average power in the scatter component of the >fade, and Io(.) is the modified Bessel function of the first kind and >zeroth order. I suggest you use an approximate method based on an idea by Paul Abbott: First, give some constant value to your A and sigma, say, A=sigma=1 (no loss of generality). f[r_]=\!\(E\^\(1\/2\ \((\(-1\) - r\^2)\)\)\ r\ BesselI[0, r]\) and construct a table of values of the corresponding cumulative distribution function, with as many points as you wish. A plot of the density function indicates that it is positive on the range 0 to 7, at most, so that you can take 70 points, i.e. one every tenth of a unit. The table, say t, is calculated by means of t=Table[{j/10,NIntegrate[f[r],{r,0,j/10}]},{j,0,70}] and you get an approximation to the true distribution function through h=Interpolation[t] You can check on the goodness of the approximation by plotting a graph of the derivative of h and comparing it with the original f. You'll be surprised (anyway, I was!). Construct a plot of h: s=Plot[h[x],{x,0,7}] Extract the plotpoints (which takes advantage of the adaptive code built into Plot): w=Cases[s,Line[{x__}]->x,\[Infinity]]; and then numerically compute the inverse function using interpolation: c=Interpolation[Reverse/@w] Then if you take a random number uniformly distributed in (0, 1) and transform it through c, you get a random number which follows the distribution f, i.e. the rician distribution. A sample of ten rician random numbers: Table[c[Random[]],{10}] {2.38887,2.36907,1.5057,2.28679,1.29854,1.692,0.69509,2.32733,2.09194,2.0422 5} Good luck, Tomas Garza Mexico City