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SISO decoder for repetition (supplementary Chapter 4)

2022-06-11 19:13:00 Bai Xiaosheng of Ming Dynasty

Preface :

        stay LDPC,Polar Code will be involved repetion code, Here we will focus on the corresponding principle ,

Why can we add directly


One     Overall process

Repetition code(3,1):

Sending process :

news bit: m Range [0,1],    1bit

After the coding : M=[m,m,m],3bit

BPSK:     1 Modulation into -1,

                0 Modulation into +1

Receiver decoding process

r=[r_1,r_2,r_3]

L_i:

        beliefs that C_i is 0

Here we mainly focus on the decoder :

Practical application factor Fixed for 1( Because it doesn't affect the final hardout This value is greater than 0, Namely 0, Less than 0 Namely 1)

  Example :

For example, I received it r[3,2,4]

3:      beliefs about first bit       base on r1 alone

2:      beliefs about second bit base on r2 alone

4:      beliefs about third bit      base on r3 alone

as follows r[0.1,-2.-0.5]


Two   intrinsic LLR

      We also call it channel LLR perhaps input LLR

      Here's a detailed derivation of its principle

    Let's start with a priori probability (Prior), front BEC,BSC It's all like this

     P(c_1=0)=P(c_1=1)=\frac{1}{2}

     c_1=0,Symbol=+1,r_1 \sim N(1,\sigma^2)

     c_1=1, Symbol=-1,r_1 \sim N(-1,\sigma^2)

be Using Bayesian principles :

    p(c_1=0|r_1)=\frac{f(r_1|c_1=0)p(c_1=0)}{f(r_1)}

   p(c_1=1|r_1)=\frac{f(r_1|c_1=1)p(c_1=1)}{f(r_1)}

     ratio =\frac{p(c_1=0|r_1)}{p(c_1=1|r_1)}=\frac{f(r_1|c_1=0)}{f(r_1|c_1=1)}

            among

            f(r_1|c_1=0)=\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{-(r_1-1)^2}{2\sigma^2}}

            f(r_1|c_1=1)=\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{-(r_1+1)^2}{2\sigma^2}}

              ratio = e^{\frac{1}{\sigma^2}2r_1}

    be

       L_1= log (ratio)= r_1\frac{2}{\sigma^2}

              = r_1 *factor

      Simplify writing

       L_1=r_1


      3、 ... and   extrinsic LLR

          Now it's given r_1,r_2,r_3  How to find L_i

         L_i= log \frac{P(c_i=0|r_1,r_2,r_3)}{P(c_i=1|r_1,r_2,r_3)}

          Here we use L_1 For example :

          p(c_1=0|r_1,r_2,r_3)=\frac{f(r_1,r_2,r_3|c_1=0)p(c_1=0)}{f(r_1,r_2,r_3)}

          p(c_1=1|r_1,r_2,r_3)=\frac{f(r_1,r_2,r_3|c_1=1)p(c_1=1)}{f(r_1,r_2,r_3)}

    

          ratio=\frac{p(c_1=0|r_1,r_2,r_3)}{p(c_1=1|r_1,r_2,r_3)}

                    =\frac{f(r_1,r_2,r_3|c_1=0)}{f(r_1,r_2,r_3|c_1=1)}

         log(ratio)=log \frac{f(r_1,r_2,r_3|c_1=0)}{f(r_1,r_2,r_3|c_1=1)}

           

    When c_1=0,QPSK-> [+1,+1,+1]

   r_1=1+N_1(0,\sigma^2)

   r_2=1+N_2(0,\sigma^2)

   r_3=1+N_3(0,\sigma^2)

    because r_1,r_2,r_3  Conditions are independent

  therefore   f(r_1,r_2,r_3|c_1=0)

 =e^{\frac{-(r_1-1)^2}{2\sigma^2}}*e^{\frac{-(r_2-1)^2}{2\sigma^2}}*e^{\frac{-(r_3-1)^2}{2\sigma^2}}

  Empathy f(r_1,r_2,r_3|c_1=1)

  =e^{\frac{-(r_1+1)^2}{2\sigma^2}}*e^{\frac{-(r_2+1)^2}{2\sigma^2}}*e^{\frac{-(r_3+1)^2}{2\sigma^2}}

        

     

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