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: Yes, reg3 does not affect output 1 in direct way - by the argument you describe. However, reg3 affect reg1 : reg1=input+(reg2+reg3), and hence affect output 1 on the next performance cycle. [[User:Alexander Chervov|Alexander Chervov]] ([[User talk:Alexander Chervov|talk]]) 18:42, 8 October 2009 (UTC)
: I wonder if the author intended the adder above/between reg1/reg2 to add the encoder input to the reg1 output; this is based on the numerator of the first transfer function. Check my work:
* <math>reg_1=(1+reg_2+reg_3)*z^{-1},\,</math>
* <math>reg_2=reg_1*z^{-1},\,</math>
* <math>reg_3=reg_2*z^{-1},\,</math>
* combining these:
** <math>reg_1=z^{-1}+reg_1*z^{-2}+reg_1*z^{-3}=\frac{z^{-1},1-z^{-2}-z^{-3}},\,</math>
== Turbo Code for Planetary Exploration? ==
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