Reply by dvsarwate February 22, 20112011-02-22
On Feb 22, 6:11&#4294967295;pm, Tim Wescott <t...@seemywebsite.com> wrote:
> > How about using GF(49) for the 35-tone case, and just never assigning > the missing 14 words?
There are far more than 14 codewords in any reasonable code over GF(49). A (n, k) code will have 49^k codewords and ***very few*** of these will be such that all the symbols in the codeword are in the set of 35 and none among the remaining 14. It would be far better to code over GF(32) which leaves 3 symbols unused. These could be put to other uses such as pilot tones, control of the comm system etc. And if the OP rge11x is curious but crazy, he should look at the newsgroup sci.math which is currently considering the question: "Is there are number that is divisible by 30, but is not divisible by both 5 and 6?" --Dilip Sarwate
Reply by rge11x February 22, 20112011-02-22
curious but crazy
Reply by Tim Wescott February 22, 20112011-02-22
On 02/22/2011 02:29 PM, rge11x wrote:
> Hi > > or 35, or pq symbols in general where p and q are relative prime so that both GF(p) and
> GF(q) exist? What I am asking about is if there is some sort of nontrivial "product"
> code that would let us combine say a Reed-Solomon or BCH code over GF(p) and another over GF(q)? > > What I mean by trivial is this:let us say that we use 35-ary orthogonal modulation, > say FSK; there is no GF(35) but both GF(5) and GF(7) exist. We generate a singly extended > code RS(5,3) over GF(5) and a shortened RS(5,3) over GF(7) and uniquely associate any symbol > pair (a,b) from a<-GF(5) and b<-GF(7) as one of the 35 possible tones. Then write the > information symbols in a 3x3 square. Now we could code columns and rows for both codes and > create a 5x5 square, this way we have a pair of (25,9) product codes but the two constituent > product codes have not much to do with each other although noticing decoding failure for one > may be used as indication of erasures for the other. > > I am asking if there is a better way of combining these codes? If not over fields as there is > no GF(pq), over some polynomial rings, or over some other structures, maybe?
Are you just curious, or are you actually contemplating doing this crazy thing? How about using GF(49) for the 35-tone case, and just never assigning the missing 14 words? -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by rge11x February 22, 20112011-02-22
Hi

or 35, or pq symbols in general where p and q are relative prime so that both GF(p) and GF(q) exist? What I am asking about is if there is some sort of nontrivial "product" code that would let us combine say a Reed-Solomon or BCH code over GF(p) and another over GF(q)? 

What I mean by trivial is this:let us say that we use 35-ary orthogonal modulation, say FSK; there is no GF(35) but both GF(5) and GF(7) exist. We generate a singly extended code RS(5,3) over GF(5) and a shortened RS(5,3) over GF(7) and uniquely associate any symbol pair (a,b) from a<-GF(5) and b<-GF(7) as one of the 35 possible tones. Then write the information symbols in a 3x3 square. Now we could code columns and rows for both codes and create a 5x5 square, this way we have a pair of (25,9) product codes but the two constituent product codes have not much to do with each other although noticing decoding failure for one may be used as indication of erasures for the other.

I am asking if there is a better way of combining these codes? If not over fields as there is no GF(pq), over some polynomial rings, or over some other structures, maybe?

Thanks
Robert