Talk:Convolutional code

Re the recent change to say that code rates could be equal to 1, I don't know of any such codes ever being used in practice, and it certainly doesn't seem very useful...

There *are* codes that when combined with M-ary modulation schemes produce one output symbol per input bit, but the output symbols are non-binary. But I don't think that's what you meant. Can you clarify?? User: Karn

From BrettE: This page doesn't descripte the ENCODING at all. Perhaps I'll add some of that later...

Josh Rubin: I second BrettE. The page for Reed-Solomon codes gives a rough but accurate definition; this page manages to talk about convolutional codes without defining them. A definition of "convolution" as used in mathematics and signal processing is in order too.

Sequential decoding

I notice there is some confusion on sequential decoding and algorithms. The Fano algorithm (which doesn't currently seem to have a page on Wikipedia) is a low-memory sequential decoding algorithm. Shannon-Fano coding is a data compression technique and AFAIK is just related by having the same inventer. Perhaps we need a page on sequential decoding? Edratzer 22:46, 12 February 2007 (UTC)[reply]

Wrong picture?

The Img. 2 is said to have a constraint length of 4, but I can only see a length of 3, since reg3 has no effect on any output (it is used twice for output 1 (which cancels out in xor)). Am I wrong?

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. Alexander Chervov (talk) 18:42, 8 October 2009 (UTC)[reply]

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:

$reg_{1}=(1+reg_{2}+reg_{3})*z^{-1},\,$
$reg_{2}=reg_{1}*z^{-1},\,$
$reg_{3}=reg_{2}*z^{-1},\,$
combining these:
- $reg_{1}=z^{-1}+reg_{1}*z^{-2}+reg_{1}*z^{-3}={\frac {z^{-1}}{1-z^{-2}-z^{-3}}},\,$
- $reg_{2}={\frac {z^{-2}}{1-z^{-2}-z^{-3}}},\,$
- $reg_{3}={\frac {z^{-3}}{1-z^{-2}-z^{-3}}},\,$

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

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}}},\,

64.203.138.235 (talk) 19:41, 30 November 2009 (UTC)[reply]

Turbo Code for Planetary Exploration?

'"However, in the interest of planetary exploration this may someday be done."'?

Can somebody explain this one to me? --74.13.201.210 14:05, 3 July 2007 (UTC)[reply]

I've added a request for citation to the article. However I support of the statement I can say that deep-space channels are mainly characterised by Gaussian noise and are not subject to fading or ISI to the same degree as terrestrial channels. Therefore striving for the theoretical Shannon Limit (traditionally defined in terms of Gaussian noise) is worthwhile for deep-space channels. Edratzer (talk) 19:41, 4 December 2008 (UTC)[reply]

I am curious as well. There is no reason to concatenate a turbo code with a hard decision algebraic code other than to remove a residual error floor. The original rationale for code concatenation was that the convolutional code alone got too high a complexity if its constraint length was too long. If the statement cannot be justified, it should be removed from the article. Nageh (talk) 19:48, 8 October 2009 (UTC)[reply]

Dashed Arrows?

Please can someone explain what the difference is between solid and dashed arrows in the trellis diagram? Thanks. 190.160.224.58 (talk) 22:46, 16 November 2008 (UTC)[reply]

I've now expanded the caption. Edratzer (talk) 19:33, 4 December 2008 (UTC)[reply]