Any experts in communications/information theory can help me on this?
Consider a fiber optic communication channel with cross over probability of 10^-2 and a wireless mobile channel with erasure probability of 10^-2. Calculate the mutual information assuming equiprobable inputs for both types of communication channels. Which system provides more information for the same bit error rates?
Many thanks
Certainly! I can help you with that.
To calculate the mutual information for each channel, we need to know the channel cross over probability for the fiber optic channel and the erasure probability for the wireless mobile channel.
The mutual information is calculated using the following formula:
I(X;Y) = Σ Σ p(x,y) * log2(p(x,y) / (p(x) * p(y)))
where X and Y are the random variables representing the input and output of the communication channel respectively, and p(x), p(y), and p(x,y) are the probabilities associated with the channel.
In this case, since we assume equiprobable inputs, the input probabilities, p(x), for both channels would be equal. Therefore, we can simplify the formula to:
I(X;Y) = Σ p(y) * log2(1 / p(y))
For the fiber optic channel, the crossover probability is given as 10^-2. This means that the probability of receiving the correct bit is 1 - 10^-2 = 0.99. Therefore, the probability of erasure is p(y) = 10^-2.
For the wireless mobile channel, the erasure probability is given as 10^-2. This means that the probability of receiving the correct bit is 1 - 10^-2 = 0.99. Therefore, the probability of erasure is p(y) = 10^-2.
Now, we can calculate the mutual information for both channels using the simplified formula:
For the fiber optic channel:
I(X;Y) = p(y) * log2(1 / p(y)) = 10^-2 * log2(1 / 10^-2) = 10^-2 * log2(10^2) = 10^-2 * 2 = 0.02 bits
For the wireless mobile channel:
I(X;Y) = p(y) * log2(1 / p(y)) = 10^-2 * log2(1 / 10^-2) = 10^-2 * log2(10^2) = 10^-2 * 2 = 0.02 bits
As you can see, both channels provide the same amount of information, which is 0.02 bits, assuming equiprobable inputs and the given bit error rates.
Therefore, both systems provide the same amount of information for the same bit error rates.