for your help: 1/4 (code rate ) turbo code

[fielix]

The code rate (=1/4) can be obtained from mother code (=1/3) by repetiton.

Is there anybody who knows how to implemente the process ?

Thank you in advance!

 

[m_llaa]

Hi
your question is not clear,
but if u work with turbo codes, i think u can add 1 more interleaver and convolutional encoder.
but if your question is general, i think you can repeat some bits to decrease your rate and increase BER. but it is not common.

 

[fielix]

I am sorry, but I am a beginner in turbo code. What I mean is how to get low-rate turbo code like 1/4 code rate in a common way.

Only add 1 more interleaver and convolutional encoder, Is it right?

Thank you very much!

 

[m_llaa]

Hi
yes, u can add 1 more interleaver and encoder.

 

[cwjcwjcwj]

Further to what you all discussing....I guess you can also try repeat the parity code.....so you need not use another interleaver/deinterleaver and encoder/decoder. What I mean is.....let say your ouput for time t1 is I1 P1 P2....instead of just transmitting these three coded bits.....you repeat again one of the parity bit and become.....I1 P1 P2 P2.....therefore, you can obtained the coding rate of 1/4.......however, in your decoder......you make use of these three parity bits ann also information bit....for decoding and exchange extrinsic information between each others............

 

[fielix]

But the question is how to decode I1 P1 P2 P2 compared to the decoding of I1 P1 P2 ?

 

[tina37]

But the question is how to decode I1 P1 P2 P2 compared to the decoding of I1 P1 P2 ?

Hello,
Did you find out how to proceed with the decoding? I am working on a similar project and got stuck with the decoding. Thanks.