Furthermore the erasure channel is used 3× to transmit the 3-bit code resulting in a coded information rate of 2H(X) × 1 ∕ 3. The erasure channel is represented in directed graph as Y3 erasure channel
Since most possible outputs are "error detected and re-transmit", there are 4 possible outcomes for each of the 4 cases. Therefore, Decodable Outcome = 4 × 4 = 16. Probabilities for these decodable outcomes was Ps = 0.896.
Therefore, for all the remaining outcomes (which the sink requests for a re-transmit), the Probabilities for the re-transmit will be
The question then is, for perfect transmission (i.e., without wrong symbol transmission)What is the information rate penalty?
In other words,How often, on the average must a triplet be transmitted (to successfully deliver the information)?
To answer this let us represent the transmission process graphically. Consider three states: ◯ Transmission state, ◯ Decode state and ◯ Accept state such that, probability to transmit, P = 1. This is shown as . For respective decoding case, the probability to decode Ps = 0.896. Thus . Also, for each case there are four possibilities but except for one the remaining possibilities are requested as re-transmit by the sink. The probability to re-transmit is Pr = 1 − Ps or .This is called a Markov Process.
Let us define
Let us also define Effective Information Rate
[If I(X; Y) is maximized w.r.t p(xi) at the source to channel capacity and . Then, transmission with arbitrary information loss is possible (eg. Turbo Codes).]
Referring to the above directed graph:
Recall that information may be classified in terms of quality as useful and useless information. Using this classification
- Noise information is regarded useless by a communication engineer
- Noise information is regarded useful by a cryptographer. ❸