0
$\begingroup$

I'm having trouble understanding what the right way to think about the presentation of discrete channels and codes in text by Cover and Thomas.

A discrete channel is a system consisting of an alphabet $\mathcal{X}$ and output alphabet $\mathcal{Y}$ and a probability transition matrix $p(y|x)$ expressing the probability of observing output symbol $y$ given that we send symbol $x.$

How do we typically think of the $\mathcal{X}$ and $\mathcal{Y}$ alphabets here? Are we mapping individual symbols from $\mathcal{X}$ to individual symbols of $\mathcal{Y}$? In this case, both alphabets should have the same cardinality? I don't see how we could take $\mathcal{X} = \{a,b,\ldots, z\}$ and $\mathcal{Y} = \{0,1\}$, for example.

Next, after introducing channel extensions, they define:

An $(M,n)$ code for the channel $(\mathcal{X}, p(y|x), \mathcal{Y})$ consists of

  1. An index set $\{1,\ldots, M\}$
  2. Encoding function $X^n:\{1,\ldots, M\} \rightarrow \mathcal{X}^n,$ yielding codewords $x^n(1), x^n(2), \ldots, x^n(M).$ The set of codewords is called the code-book.
  3. Decoding function $g:\mathcal{Y}^n \rightarrow \{1,\ldots, M\},$ deterministic rule assigning guess to each possible received vector.

So we are taking some arbitrary "message" from the index set, encoding it using $n$ symbols from $\mathcal{X}$, sending those $n$ symbols over the channel, receiving a noisy version of $n$ symbols in $\mathcal{Y}$, and then trying to recover the original message based on those $n$ symbols?

Why do the codewords all have to have length $n$? What's the advantage of letting $\mathcal{X}$ be different from $\mathcal{Y}$?

I think I understand the formal stuff going on here, but my intuition as to why it's interesting is a bit lacking. Can someone shed some light on this?

$\endgroup$
0
$\begingroup$

It's not strictly necessary to have $\mathcal{X}$ equal to $\mathcal{Y}$. They can also be different sets. The purpose of a channel based on $(M,n)$-codes is to map a word of symbols from an alphabet to another word of symbols of another alphabet of the same length $n$ (i.e. so blocks of symbols). However, in a channel code you can also have communication errors, that are modeled by a conditional probability table $p(y\mid x)$. The advantage of letting $\mathcal{X}$ be different from $\mathcal{Y}$ is the fact that the communication channel can be defined in order to switch the alphabet from $\mathcal{X}$ to $\mathcal{Y}$, but not necessarily.

$\endgroup$
5
  • $\begingroup$ Something I still don't understand: if both the encoder and decoder both operate on words of length $n$, then doesn't this disallow for any sort of compression of the signal? $\endgroup$
    – theQman
    Jan 30 '21 at 19:30
  • $\begingroup$ Discrete channel codes are not suitable only for compression, but also for reliability. However, the compression task is the encoder's task while the decompression task is the decoder's task. If the main task of the channel is to ensure reliability, then the encoder will add some kind of redundancy in order to have noise prevention. $\endgroup$
    – loreloc
    Jan 30 '21 at 21:18
  • $\begingroup$ But how can the encoder compress anything if it is forced to map messages to strings of length $n$? $\endgroup$
    – theQman
    Feb 1 '21 at 21:09
  • $\begingroup$ Messages can be of whatever length, they are just enumerated from $1$ to $M$, but they can be arbitrarily large. $\endgroup$
    – loreloc
    Feb 1 '21 at 21:26
  • $\begingroup$ I sort of see your point, but I still don't see how this would connect to codes that were previously covered in the book, such as Huffman codes. $\endgroup$
    – theQman
    Feb 2 '21 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.