BCH Block Length?

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Article #207

#207

From: alavarre
Date: Mon, 13 Jun 1994 12:03

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Why must a BCH code have as a length the multiplicative order 
of alpha?  

Michelson, p. 121, states, without proof, that "The block 
length of a  BCH code is the order of the element used in 
defining the consecutive roots [of the generator polynomial]."  
Gallager gives the same as a defined property of a BCH code.  
My question is "why?"

I can see how the order (q) of the elements on GF(q) results in 
q distinct roots, hence a generator polynomial of degree q, and 
that for an extension field GF(2^m), q=2^m is always even, 
making the generator polynomial always of odd length.
  
But it seems that the information word length k can be 
arbitrary, and the resulting codeword must then be of arbitrary 
degree (and length).

So how does the choice of a particular generator constrain the 
number of information bits and hence the length of the codeword?  And 
why does the order of alpha matter?

TIA,
Andy La Varre
alavarre@ids.net

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From: alavarre@ids.net
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Subject: BCH Block Length?
Date: Mon, 13 Jun 94 07:03:17 EST
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