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Network Coding and Wireless Physical-layer ... - Jacobs University

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Chapter 5: <strong>Network</strong> <strong>Coding</strong> with LT Codes as Erasure Codes 65<br />

network decoding as<br />

⎧<br />

⎪⎨<br />

ψ(i) ≈<br />

⎪⎩<br />

18<br />

µ(i), i = 1<br />

19<br />

18<br />

µ(i) + 1<br />

19 19<br />

∑ i−1<br />

j=1 µ(j)µ(i − j), otherwise. (5.7)<br />

Figures 5.4 <strong>and</strong> 5.5 compare the plot of the robust soliton distribution µ(i) in the<br />

erasure-free case with the distorted distribution due to erasures at BD obtained by the<br />

analytical approximation ψ(i) in (5.7) <strong>and</strong> that obtained from simulation by counting the<br />

degree of 10 7 symbols received at D around the two peaks at 2 <strong>and</strong> 42, respectively. We<br />

can see that the steepness of the peaks is reduced in the distorted case. Moreover, an<br />

unwanted peak is formed at 44.<br />

Although the distortion of the degree distribution is small, its effect on the performance<br />

is clearly visible. Figure 5.6 shows the histogram of the number of LT-coded symbols<br />

needed to be transmitted until the original symbols can be recovered by D. Figure 5.7<br />

makes a comparison between the number of symbols needed to be transmitted when<br />

the erasure probability of BD is 0.1 in the network in Fig. 5.1 <strong>and</strong> that when the erasure<br />

probability is 0.05 in normal point-to-point communication. Although the average erasure<br />

probabilities in both cases are the same, the network coding case requires more symbols<br />

due to the distortion of the degree distribution.<br />

5.5 The First Solution: Wise Assignment of <strong>Network</strong><br />

Codes<br />

In case erasures occur only along the edge BD with the erasure probability of 0.1, the<br />

solution to the degree distribution distortion problem is very simple.<br />

We change our<br />

network codes from those in Fig. 5.1 into those in Fig. 5.8. Since we now transmit b 1 + b 2<br />

instead of b 1 along BD, when b 1 + b 2 is erased, D only receives b 1 , yielding no distortion,<br />

when b 1 + b 2 is not erased, D can network-decode <strong>and</strong> receive both b 1 <strong>and</strong> b 2 , yielding no<br />

distortion either.

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