Please can someone verify these basic results about error correcting codes.

For this example I am using Reed-Solomon because there are functions to calculate the rate $r$ and distance $d$ of the code, which are

$$r = \frac{k}{n}$$ $$d = n - k + 1.$$

1.0 The rate and distance are independent for this code.

1.1 By introducing the parameter $c$, the rate can be changed without affecting the distance

$$r = \frac{k+c}{n+c}$$ $$d = n + c - k - c + 1 = n - k + 1.$$

1.2 The distance can be changed without affecting the rate

$$r = \frac{ck}{cn} = \frac{k}{n}$$ $$d = c(n - k) + 1.$$

I understand that there is nothing in these equations that prevents me from setting the rate higher than the channel capacity, but the proof of Shannon's noisy coding theorem is very complicated for me.

Am I on the right track with these ideas? Can you recommend any references to help me understand the capacity constraint?

  • $\begingroup$ No, you are not on the right track with these ideas. Statement 1.0 is false: $d = n(1-r) +1$ is very much a function of the rate, instead of being independent of the rate (whatever that word independent means to you). $\endgroup$ – Dilip Sarwate Jan 12 at 23:38
  • $\begingroup$ @DilipSarwate do you not agree that the equations 1.1 and 1.2 are correct? $\endgroup$ – user827822 Jan 13 at 9:19
  • $\begingroup$ No, 1.2 is also false. $d$ remains at $n-k+1$ instead of increasing as you think it does. $\endgroup$ – Dilip Sarwate Jan 13 at 17:45
  • $\begingroup$ @DilipSarwate The equation, the proof, says otherwise and I'm not convinced that you know what you're talking about. I could elaborate on 1.2 but I couldn't make it any clearer by putting it into words, except to benefit someone who doesn't understand the formula. $\endgroup$ – user827822 Jan 13 at 19:41
  • $\begingroup$ @DilipSarwate This question is too difficult. Let's look at 1.2 by itself dsp.stackexchange.com/questions/63227/… $\endgroup$ – user827822 Jan 14 at 8:49

Your Answer

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

Browse other questions tagged or ask your own question.