>Steve Pope wrote: >> Oli Charlesworth <catch@olifilth.co.uk> wrote: >> >>> One very important point is that the following is true (ignoringscaling):>>> >>> p(r|b) = exp{-|b-r|^2 / sigma^2} >> >> Good thing you said ignoring scaling, since p(r|b) = 0. >> Assuming p() is supposed to indicate probability. > >Perhaps it's late, and I'm missing something?... > > >-- >OliHi Oli, You wrote "One very important point is that the following is true (ignoring scaling): p(r|b) = exp{-|b-r|^2 / sigma^2} " and Steve just give his comment on that . I'm not sure on what you think when You said "(ignoring scaling)" Can You explain this. Thanks
Soft demmaper - LLRs calculations
Started by ●February 28, 2009
Reply by ●March 1, 20092009-03-01
Reply by ●March 1, 20092009-03-01
Melinda wrote:>> Steve Pope wrote: >>> Oli Charlesworth <catch@olifilth.co.uk> wrote: >>> >>>> One very important point is that the following is true (ignoring > scaling): >>>> p(r|b) = exp{-|b-r|^2 / sigma^2} >>> Good thing you said ignoring scaling, since p(r|b) = 0. >>> Assuming p() is supposed to indicate probability. >> Perhaps it's late, and I'm missing something?... >> >> >> -- >> Oli > > Hi Oli, > You wrote > "One very important point is that the following is true (ignoring > scaling): > > p(r|b) = exp{-|b-r|^2 / sigma^2} " > > and Steve just give his comment on that . > > I'm not sure on what you think when You said "(ignoring scaling)" > Can You explain this.I don't understand Steve's comment, but what I was talking about is that the Gaussian distribution always has a normalising constant on the front, and I can't remember the precise arrangement. It's something along the lines of 1/(sigma.sqrt(2.pi)). -- Oli
Reply by ●March 2, 20092009-03-02
>Melinda wrote: >>> Steve Pope wrote: >>>> Oli Charlesworth <catch@olifilth.co.uk> wrote: >>>> >>>>> One very important point is that the following is true (ignoring >> scaling): >>>>> p(r|b) = exp{-|b-r|^2 / sigma^2} >>>> Good thing you said ignoring scaling, since p(r|b) = 0. >>>> Assuming p() is supposed to indicate probability. >>> Perhaps it's late, and I'm missing something?... >>> >>> >>> -- >>> Oli >> >> Hi Oli, >> You wrote >> "One very important point is that the following is true (ignoring >> scaling): >> >> p(r|b) = exp{-|b-r|^2 / sigma^2} " >> >> and Steve just give his comment on that . >> >> I'm not sure on what you think when You said "(ignoring scaling)" >> Can You explain this. > >I don't understand Steve's comment, but what I was talking about is that>the Gaussian distribution always has a normalising constant on the >front, and I can't remember the precise arrangement. It's something >along the lines of 1/(sigma.sqrt(2.pi)). > >-- >Oli >Hi Oli, Thanks very much and let me sumarize this: The Gaussian function(G) is (wikipedia site): G(x) = 1/(sigma*sqrt(2*pi)) * exp{-(b-r)^2 / "2"*(sigma^2)} so lets me comment each section of this formula: the "1/[(sigma*sqrt(2*pi))]" part can be note as : sigma = sqrt(sigma^2); -> 1/[sqrt(sigma^2) * sqrt(2*pi)] = 1/[sqrt((sigma^2)* 2*pi)] = 1/[sqrt((No/2)*2*pi)] = 1/[sqrt(No*pi)]. --> so: "1/[(sigma*sqrt(2*pi))]" is 1/[sqrt(No*pi)] the "exp{-(b-r)^2 / "2"*(sigma^2)}" section can be note as : exp(-(Di^2)/2*(No/2)) = exp(-(Di^2)/No) where Di is distance/s between received and ideal constellation points. As You see I mark "2" when I wrote G(x)=.... In Your previous post You didn't write this 2 factor beside sigma^2. ("p(r|b) = exp{-|b-r|^2 / "here" sigma^2} " ) and in Gausian function there is a factor 2 beside sigma^2. This is the way I calculate my p(r|b=0) i.e. p(r|b=1), i.e. LLRs(b) = log[ {1/[sqrt(No*pi)] * sum exp(-(Di^2)/No)} --> Di(b=0) -------------------------------------- {1/[sqrt(No*pi)] * sum exp(-(Di^2)/No)} ] --> Di(b=1) Is my calculation good? And one more question - on dependence of No(noise power level) the upper and lower 'probabilities'(measures like Steve said) {...}/{...} can be as 0.01345 or 0 or 12.42 or 47.456 ... and alike. { for example: LLR = log(44.56 / 0.008) } This is the reason for my questions and doubts and can You one more time confirm or not my calculations. Thanks and best regards
Reply by ●March 2, 20092009-03-02
Melinda wrote:>> Melinda wrote: >>>> Steve Pope wrote: >>>>> Oli Charlesworth <catch@olifilth.co.uk> wrote: >>>>> >>>>>> One very important point is that the following is true (ignoring >>> scaling): >>>>>> p(r|b) = exp{-|b-r|^2 / sigma^2} >>>>> Good thing you said ignoring scaling, since p(r|b) = 0. >>>>> Assuming p() is supposed to indicate probability. >>>> Perhaps it's late, and I'm missing something?... >>>> >>>> >>>> -- >>>> Oli >>> Hi Oli, >>> You wrote >>> "One very important point is that the following is true (ignoring >>> scaling): >>> >>> p(r|b) = exp{-|b-r|^2 / sigma^2} " >>> >>> and Steve just give his comment on that . >>> >>> I'm not sure on what you think when You said "(ignoring scaling)" >>> Can You explain this. >> I don't understand Steve's comment, but what I was talking about is that > >> the Gaussian distribution always has a normalising constant on the >> front, and I can't remember the precise arrangement. It's something >> along the lines of 1/(sigma.sqrt(2.pi)). >> >> -- >> Oli >> > > Hi Oli, > Thanks very much and let me sumarize this: > The Gaussian function(G) is (wikipedia site): > G(x) = 1/(sigma*sqrt(2*pi)) * exp{-(b-r)^2 / "2"*(sigma^2)} so lets me > comment each section of this formula: the "1/[(sigma*sqrt(2*pi))]" part can > be note as : > sigma = sqrt(sigma^2); -> > 1/[sqrt(sigma^2) * sqrt(2*pi)] = 1/[sqrt((sigma^2)* 2*pi)] = > 1/[sqrt((No/2)*2*pi)] = 1/[sqrt(No*pi)]. > > --> so: "1/[(sigma*sqrt(2*pi))]" is 1/[sqrt(No*pi)] > > the "exp{-(b-r)^2 / "2"*(sigma^2)}" section can be note as : > exp(-(Di^2)/2*(No/2)) = exp(-(Di^2)/No) where Di is distance/s between > received and ideal constellation points. As You see I mark "2" when I wrote > G(x)=.... In Your previous post You didn't write this 2 factor beside > sigma^2. ("p(r|b) = exp{-|b-r|^2 / "here" sigma^2} " ) and in Gausian > function there is a factor 2 beside sigma^2. > > This is the way I calculate my p(r|b=0) i.e. p(r|b=1), i.e. > LLRs(b) = log[ {1/[sqrt(No*pi)] * sum exp(-(Di^2)/No)} --> Di(b=0) > -------------------------------------- > {1/[sqrt(No*pi)] * sum exp(-(Di^2)/No)} ] --> Di(b=1) > > Is my calculation good? > And one more question - on dependence of No(noise power level) the upper > and lower 'probabilities'(measures like Steve said) {...}/{...} can be as > 0.01345 or 0 or 12.42 or 47.456 ... and alike. > { for example: LLR = log(44.56 / 0.008) } > > This is the reason for my questions and doubts and can You one more time > confirm or not my calculations.The only thing I'd say is that you can safely eliminate the constant scaling factor (1/sqrt(No.pi)), as it's the same in the numerator and denominator! -- Oli
