floating point - Handling large numbers and precision of Ricean Fading PDF python -


i'm trying calculate ricean fading pdf using following equation. ricean fading pdf. 'y' normalized envelope , 'gamma' snr

if k value large, 

math.exp(-((1.+_gamma)*pow(_y,2.) + _gamma))

exp results in big floating point (e.q. 1.01e-5088). in python shows '0.0' value

mpmath.besseli(0,2. * _y * np.sqrt(_gamma * (1. + _gamma)))

the value of bessel function shows big int value (e.q. 7.78e+5092). in python shows '**inf**' value

how can store big integer , floating point value in python , calculate pdf?

def rice_pdf(self, _y, _gamma):    return 2. * _y * (1. + _gamma) * math.exp(-((1.+_gamma)*pow(_y,2.) + _gamma)) * special.i0(2. * _y * np.sqrt(_gamma * (1. + _gamma)))

thanks.

if have way compute logarithm of bessel function, can avoid multiplication of large , small numbers transforming summation subsequent exonentiation, should solve numerical problems (exploit fact exp(a) * exp(b) == exp(a + b)).

def rice_pdf(_y, _gamma):     = np.log(2. * _y * (1. + _gamma))     b = -((1.+_gamma)*pow(_y,2.) + _gamma)     c = lni(2. * _y * np.sqrt(_gamma * (1. + _gamma)))      return np.exp(a + b + c)

this function assumes there exist implementation of lni computes log(i0(z)). however, aware of no existing implementation of such function. can work around using mpmath intermediate results:

def lni(z):     i0 = mpmath.besseli(0, z)  # may become big number     logi0 = mpmath.log(i0)  # logarithm brings big number sensible range     return float(logi0)  # convert normal floating point

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