Difference between revisions of "Channel polarization"

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It turns out that the channel <math>W^{-}: U_1 \rightarrow (Y_1, Y_2)</math> reduces to another BSC. To see this, consider the case where <math>(Y_1, Y_2) = (0,0)</math>. To minimize the error probability, we must decide the value of <math>U_1</math> that has the greater likelihood. For <math>0 < p < 0.5</math>, <math> 0.5(1-2p+2p^2) > p(1-p)</math> so that the maximum-likelihood (ML) decision is <math>U_1 = 0</math>. Using the same argument, we see that the ML decision is <math>U_1 = 0</math> for <math>(Y_1,Y_2)=(1,1)</math>. More generally, the receiver decision is to set <math>\hat{U_1} = Y_1 \oplus Y_2</math>. Indeed, if the crossover probability is low, it is very likely that <math>Y_1 = U_2 + U_1</math> and <math>Y_2 = U_2</math>. Solving for <math>U_2</math> and plugging it back produces the desired result.
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It turns out that the channel <math>W^{-}: U_1 \rightarrow (Y_1, Y_2)</math> reduces to another BSC. To see this, consider the case where <math>(Y_1, Y_2) = (0,0)</math>. To minimize the error probability, we must decide the value of <math>U_1</math> that has the greater likelihood. For <math>0 < p < 0.5</math>, <math> 0.5(1-2p+2p^2) > p(1-p)</math> so that the maximum-likelihood (ML) decision is <math>U_1 = 0</math>. Using the same argument, we see that the ML decision is <math>U_1 = 0</math> for <math>(Y_1,Y_2)=(1,1)</math>. More generally, the receiver decision is to set <math>\hat{U_1} = Y_1 \oplus Y_2</math>. Indeed, if the crossover probability is low, it is very likely that <math>Y_1 = U_2 \oplus U_1</math> and <math>Y_2 = U_2</math>. Solving for <math>U_2</math> and plugging it back produces the desired result.

Revision as of 04:48, 7 May 2021

Synthetic channels

Binary erasure channel

Let be the erasure probability of a binary erasure channel (BEC).

00 0? 01 ?0 ?? ?1 10 1? 11
00
01
10
11
00 0? 01 ?0 ?? ?1 10 1? 11
0
1

Binary symmetric channel

Let be the crossover probability of a binary symmetric channel (BSC).

00 01 10 11
00
01
10
11
00 01 10 11
00
01

It turns out that the channel reduces to another BSC. To see this, consider the case where . To minimize the error probability, we must decide the value of that has the greater likelihood. For , so that the maximum-likelihood (ML) decision is . Using the same argument, we see that the ML decision is for . More generally, the receiver decision is to set . Indeed, if the crossover probability is low, it is very likely that and . Solving for and plugging it back produces the desired result.