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From: cpshah99 on 25 May 2010 08:54

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
From: dvsarwate on 25 May 2010 10:55
On May 25, 7:54 am, "cpshah99" <cpshah99(a)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

From: cpshah99 on 25 May 2010 13:03

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

From: cpshah99 on 25 May 2010 14:08
>On May 25, 7:54=A0am, "cpshah99" <cpshah99(a)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 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=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
From: Maba Mani on 10 Jun 2010 11:23
Hello,

I want to simulate ldpc code over AWGN channel.
can you help me please?

Best,
Mahba
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