FPGA Central

Does a modulation with q different symbol states absolutely require a
code over X^n, where X = {a_1, a_2, ..., a_q}? Or can you use a
symbol set (X) with a different order (count)?

A slightly different question: is this *typically* done (q symbol states
==> order q symbol set)? I.e., although it's not required, is it
typically done this way?

As a simple example, could a binary (7,4) Hamming code be applied to 16
QAM, such that 4/7 codewords fit into one symbol?
--
Randy Yates % "She tells me that she likes me very much,
Digital Signal Labs % but when I try to touch, she makes it
mailto://[email protected] % all too clear."
http://www.digitalsignallabs.com % 'Yours Truly, 2095', *Time*, ELO

Randy Yates wrote:
> Does a modulation with q different symbol states absolutely require a
> code over X^n, where X = {a_1, a_2, ..., a_q}? Or can you use a
> symbol set (X) with a different order (count)?
> A slightly different question: is this *typically* done (q symbol states
> ==> order q symbol set)? I.e., although it's not required, is it
> typically done this way?

Ideally you want coding to be matched to the modulation, so the Hamming
distance corresponds to the Euclidean distance.

> As a simple example, could a binary (7,4) Hamming code be applied to 16
> QAM, such that 4/7 codewords fit into one symbol?

The result is likely to be worse then the uncoded QAM. Proakis Chapter 8
is exactly about your questions.

Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com

Randy Yates wrote:
> Does a modulation with q different symbol states absolutely require a
> code over X^n, where X = {a_1, a_2, ..., a_q}? Or can you use a
> symbol set (X) with a different order (count)?
>
> A slightly different question: is this *typically* done (q symbol states
> ==> order q symbol set)? I.e., although it's not required, is it
> typically done this way?
>

I *believe* ( but could easily be wrong ) that the map
from symbol states to symbol order sets is opportunistic,
and based on which sorts of maps empirically behave the best.

Mathematically, any function could work. And whether or
not the function is bijective/symmetrical seems
irrelevant, long as you don't need the symmetry for
computational reasons.

ObDisclosure: I'm not a coding theory guy at all, but
I've worked with several on instrumentation.

> As a simple example, could a binary (7,4) Hamming code be applied to 16
> QAM, such that 4/7 codewords fit into one symbol?

I don't see why not - isn't the choice of
16 QAM largely independent of what's being modulated
over it? Once you have conditions necessary to
describe one bit in 16QAM... it's a bit. It is,
form a dataflow perspective, just a by-16 PSK stream....
once you get it into baseband...

16 & 7 modularize together in an interesting, 56-bit
sort of way, too....

--
Les Cargill

Vladimir Vassilevsky <[email protected]> writes:

> Randy Yates wrote:
>> Does a modulation with q different symbol states absolutely require a
>> code over X^n, where X = {a_1, a_2, ..., a_q}? Or can you use a
>> symbol set (X) with a different order (count)?
>> A slightly different question: is this *typically* done (q symbol states
>> ==> order q symbol set)? I.e., although it's not required, is it
>> typically done this way?
>
> Ideally you want coding to be matched to the modulation, so the
> Hamming distance corresponds to the Euclidean distance.

Euclidean distance as in the correlation metrics when you're doing
soft-decision decoding?

>> As a simple example, could a binary (7,4) Hamming code be applied to 16
>> QAM, such that 4/7 codewords fit into one symbol?
>
> The result is likely to be worse then the uncoded QAM. Proakis Chapter
> 8 is exactly about your questions.

That narrows it down to 120 pages or so. Thanks...
--
Randy Yates % "And all you had to say
Digital Signal Labs % was that you were
mailto://[email protected] % gonna stay."
http://www.digitalsignallabs.com % Getting To The Point', *Balance of Power*, ELO

On Aug 19, 7:47*pm, Randy Yates <ya...@ieee.org> asked:
> Does a modulation with q different symbol states absolutely require a
> code over X^n, where X = {a_1, a_2, ..., a_q}? *Or can you use a
> symbol set (X) with a different order (count)?

This kind of stuff is routine. Consider, for example, using
a Reed-Solomon code (say (255,223) code over GF(256))
on the binary symmetric (or additive white Gaussian noise)
channel. As another example, one of the first implementations
of a (2,1) convolutional code was for a QPSK modem in which
the two coded bits were the I and Q inputs to the modem,
and the channel rate in bauds was the same as the input
data rate in bits.

>As a simple example, could a binary (7,4) Hamming code
>be applied to 16-QAM, such that 4/7 codewords fit into one
>symbol?

Yes, though the implementation as well as any analysis would
be messy. The first codeword would have its bits in the first
and second QAM symbols transmitted, but the *second*
codeword would have one bit in the second QAM symbol
transmitted, the next four bits in the third QAM symbol, and
two bits in the fourth QAM symbol. The 7 bits of the third
codeword would have .... ; the whole cycle repeating
after 7 QAM symbols (i.e. 4 codewords). Whether the
increased complexity is worth the benefits reaped remains
to be seen in 120 or so pages. Or the answer exists in some
unpublished master's thesis since negative results tend not
to get published: the number of papers saying "Our scheme
works" is far outweighed by the number of papers saying
"We tried this scheme and it did not work."

--Dilip Sarwate

[email protected] <[email protected]> wrote:

>On Aug 19, 7:47*pm, Randy Yates <ya...@ieee.org> asked:

>>As a simple example, could a binary (7,4) Hamming code
>>be applied to 16-QAM, such that 4/7 codewords fit into one
>>symbol?

>Yes, though the implementation as well as any analysis would
>be messy. The first codeword would have its bits in the first
>and second QAM symbols transmitted, but the *second*
>codeword would have one bit in the second QAM symbol
>transmitted, the next four bits in the third QAM symbol, and
>two bits in the fourth QAM symbol. The 7 bits of the third
>codeword would have .... ; the whole cycle repeating
>after 7 QAM symbols (i.e. 4 codewords). Whether the
>increased complexity is worth the benefits reaped remains
>to be seen in 120 or so pages.

I would expect the Hamming code thus applied to exhibit
slightly less normalized coding gain than the same code
applied to a BPSK-modulated system.

There are two things that make the hit you're taking from
this mismatch smaller than one might think: in AWGN,
the noise vectors on the I and Q dibits are independent;
and if those dibits are grey-coded, as they typically
are, the bit errors within a dibit move closer to looking
independent of each other.

Still it would be worth placing an interleaver between
the code and the channel.

Steve