Talk:Convolutional code: Difference between revisions

Revision as of 19:30, 30 November 2009

Revision as of 19:41, 30 November 2009

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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>

Line 23 ⟶ 24:

* <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> <span style="font-size: smaller;" class="autosigned">—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/64.203.138.235|64.203.138.235]] ([[User talk:64.203.138.235|talk]]) 19:29, 30 November 2009 (UTC)</span><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

** <math>reg_2=\frac{z^{-2}}{1-z^{-2}-z^{-3}},\,</math>

** <math>reg_3=\frac{z^{-3}}{1-z^{-2}-z^{-3}},\,</math>

now, I see the first transfer function from the drawing as:

: <math>H_1=input+reg_2+reg_3+reg_1+reg_3=input+reg_1+reg_2=\frac{1+z^{-1}+z^{-2}}{1-z^{-2}-z^{-3}},\,</math>

[[Special:Contributions/64.203.138.235|64.203.138.235]] ([[User talk:64.203.138.235|talk]]) 19:41, 30 November 2009 (UTC)

== Turbo Code for Planetary Exploration? ==