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But the question of the code rate is not without merit either!!! In the Mariner application it would have meant that it takes a longer time to transmit a single image using $R(5,1)$ than it would with $R(4,1)$. The eager astronomers can wait a bit longer to get the image, but a serious concern is that probe uses a fixed amount of battery power per transmitted bit. This would also need to be taken into account, so my figures are not fair to the shorter code. In terrestrial communication systems we carry out extensive simulations before we choose one coding scheme over another, and plot the probability of an error vs. energy per transmitted bit. With Mariner, we could try and estimate the relation between $p$ and energy consumption per bit, but I don't have the time to get into that.

Assume that noise has deviation $\sigma=0.5$. With $R(5,1)$ we then get a bit error, when noise exceeds $+2\sigma$, so this happens with probability $p_5=1-\Phi(2.0)=0.0228$. The corresponding bit error probability when using $R(4,1)$ is then $p_4=1-\Phi(2.0\sqrt{5/3})=1-\Phi(2.58)=0.0049$, because this time we $\sqrt{5/3}$ times as much noise as earlier. The test is then to compare $$ 1-P_5(0.0228)^5=2.35\cdot10^{-6} $$ to $$ 1-P_4(0.0049)^6=6.01\cdot10^{-6}. $$ We see that we do have a better chance of correctly receiving 5 pixels worth of data using the longer code, but the difference is not nearly as dramatic as the earlier figures, disregarding the energy consumption, would have indicated.

But the question of the code rate is not without merit either!!! In the Mariner application it would have meant that it takes a longer time to transmit a single image using $R(5,1)$ than it would with $R(4,1)$. The eager astronomers can wait a bit longer to get the image, but a serious concern is that the probe uses a fixed amount of battery power per transmitted bit. This would also need to be taken into account, so my figures are not fair to the shorter code. In terrestrial communication systems we carry out extensive simulations before we choose one coding scheme over another, and plot the probability of an error vs. energy per transmitted bit. With Mariner, we could try and estimate the relation between $p$ and energy consumption per bit, but I don't have the time to get into that.

Assume that noise has deviation $\sigma=0.5$. With $R(5,1)$ we then get a bit error, when noise exceeds $+2\sigma$, so this happens with probability $p_5=1-\Phi(2.0)=0.0228$. The corresponding bit error probability when using $R(4,1)$ is then $p_4=1-\Phi(2.0\sqrt{5/3})=1-\Phi(2.58)=0.0049$, because this time we need $\sqrt{5/3}$ times as much noise as earlier to receive a bit incorrectly. The test is then to compare $$ 1-P_5(0.0228)^5=2.35\cdot10^{-6} $$ to $$ 1-P_4(0.0049)^6=6.01\cdot10^{-6}. $$ We see that we do have a better chance of correctly receiving 5 pixels worth of data using the longer code, but the difference is not nearly as dramatic as the earlier figures, disregarding the energy consumption, would have indicated.

The 5-dimensional Reed-Muller code of length 16 and minimum Hamming distance is capable of correcting 3 errorneous bits. The 6-dimensional Reed-Muller code of length 32 and minimum Hamming distance 16 is capable of correcting 7 errorneous bits. Let us assume a very simplistic model in which all the bits are received errorneously at the same probability $p\in(0,1)$, independently from each other. The probability of correctly decoding a received word of $R(4,1)$ is thus

The 5-dimensional Reed-Muller code of length 16 and minimum Hamming distance 8 is capable of correcting 3 errorneous bits. The 6-dimensional Reed-Muller code of length 32 and minimum Hamming distance 16 is capable of correcting 7 errorneous bits. Let us assume a very simplistic model in which all the bits are received errorneously at the same probability $p\in(0,1)$, independently from each other. The probability of correctly decoding a received word of $R(4,1)$ is thus

Assume that noise has deviation $\sigma=0.5$. With $R(5,1)$ we then get a bit error, when noise exceeds $+2\sigma$, so this happens with probability $p_5=1-\Phi(2.0)=0.0228$. The corresponding bit error probability when using $R(4,19$ is then $p_4=1-\Phi(2.0\sqrt{5/3})=1-\Phi(2.58)=0.0049$, because this time we $\sqrt{5/3}$ times as much noise as earlier. The test is then to compare $$ 1-P_5(0.0228)^5=2.35\cdot10^{-6} $$ to $$ 1-P_4(0.0049)^6=6.01\cdot10^{-6}. $$ We see that we do have a better chance of correctly receiving 5 pixels worth of data using the longer code, but the difference is not nearly as dramatic as the earlier figures, disregarding the energy consumption, would have indicated.

Assume that noise has deviation $\sigma=0.5$. With $R(5,1)$ we then get a bit error, when noise exceeds $+2\sigma$, so this happens with probability $p_5=1-\Phi(2.0)=0.0228$. The corresponding bit error probability when using $R(4,1)$ is then $p_4=1-\Phi(2.0\sqrt{5/3})=1-\Phi(2.58)=0.0049$, because this time we $\sqrt{5/3}$ times as much noise as earlier. The test is then to compare $$ 1-P_5(0.0228)^5=2.35\cdot10^{-6} $$ to $$ 1-P_4(0.0049)^6=6.01\cdot10^{-6}. $$ We see that we do have a better chance of correctly receiving 5 pixels worth of data using the longer code, but the difference is not nearly as dramatic as the earlier figures, disregarding the energy consumption, would have indicated.