Hi All On AWGN channel (y=x+n), the LLR, L_y, corresponding to the channel output can be written as L=2/sigma^2*y, where x is BPSK modulated symbols. However, when the channel is frequency selective, and at the receiver end, a filter based equalizer (LE or DFE) is used to mitigate the ISI, do we still calculate the LLR the same way i.e. L=2/sigma^2*x_hat, where x_hat is the output of the equalizer. Your opinion matters a lot. Best Regards Chintan

# Gaussian assumption and Calculating the loglikelihood ratio

Started by ●May 25, 2010

Reply by ●May 25, 20102010-05-25

On May 25, 7:54�am, "cpshah99" <cpshah99@n_o_s_p_a_m.rediffmail.com> wrote:> > L=2/sigma^2*y, where x is BPSK modulated symbols. >PEMDAS says that this means 2/(sigma^2*y) since multiplications are done before divisions. Is this what you mean? Or is it (2/sigma^2)*y ??> However, when the channel is frequency selective, and at the receiver end, > a filter based equalizer (LE or DFE) is used to mitigate the ISI, do we > still calculate the LLR the same way i.e. > > L=2/sigma^2*x_hat, where x_hat is the output of the equalizer.It depends on whether the noise variance at the output of the equalizer is still sigma^2 (not always true), and whether the noise can still be regarded as Gaussian (not always true either, but often explained away by invocations of central limit theorems and the like). --Dilip Sarwate

Reply by ●May 25, 20102010-05-25

Hello Prof. Sarwate Thanks for your reply.>PEMDAS says that this means 2/(sigma^2*y) since multiplications >are done before divisions. Is this what you mean? Or is it >(2/sigma^2)*y ?? >Sorry for this typo. It is Indeed (2/sigma^2)*y. I have no issues with that.>It depends on whether the noise variance at the output >of the equalizer is still sigma^2 (not always true), and >whether the noise can still be regarded as Gaussian >(not always true either, but often explained away by >invocations of central limit theorems and the like).It is also true that the noise variance at the o/p of the equalizer may not be sigma^2. However, is it valid to *assume* that the variance is sigma^2? Best Regards Chintan

Reply by ●May 25, 20102010-05-25

>On May 25, 7:54=A0am, "cpshah99" <cpshah99@n_o_s_p_a_m.rediffmail.com> >wrote: > >> >> L=3D2/sigma^2*y, where x is BPSK modulated symbols. >> > >PEMDAS says that this means 2/(sigma^2*y) since multiplications >are done before divisions. Is this what you mean? Or is it >(2/sigma^2)*y ?? > >> However, when the channel is frequency selective, and at the receiverend=>, >> a filter based equalizer (LE or DFE) is used to mitigate the ISI, do we >> still calculate the LLR the same way i.e. >> >> L=3D2/sigma^2*x_hat, where x_hat is the output of the equalizer. > >It depends on whether the noise variance at the output >of the equalizer is still sigma^2 (not always true), and >whether the noise can still be regarded as Gaussian >(not always true either, but often explained away by >invocations of central limit theorems and the like). > >--Dilip Sarwate > >The reason I am asking is that, I am using LDPC code with linear equalizer. I am assuming perfect channel knowledge at the receiver and using MMSE criteria. Now, when I use this 2/sigma^2*y or (2/sigma^2)*y (I dont see whats the difference......) I see some kind of strange behavior in the BER plot. The BER plot follows the waterfall shape up to some SNR and then the BER goes up at higher SNR. Has anybody come across this problem before?? Best Regards Chintan

Reply by ●June 10, 20102010-06-10