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This condition is always satisfied when b ¼ 0:5. This means that if the sourceencoder can make b ¼ 0:5 for a BSC, then1: Channel capacity C c ¼ 1 OðaÞ bit=symbol ð3:87Þ2: Source information rate R r ¼ r b OðbÞ bit=sec ð3:88aÞ3: if b ¼ 0:5; R r ¼ r b bit=sec ð3:88bÞ4: The maximum rate at which information can be transferred acrossa BSC channel isC m ¼ r b C c ¼ ½1OðaÞŠr b bit=sec ð3:89ÞSome books refer to this expression as the channel capacity. Inpractice, Eq. (3.87) is often referred to as the channel capacity.In general, in calculating the channel capacity, whether or not the input datatransfer rate is specified, OðaÞ is replaced with each channel’s self-entropy,HðX Þ.Example 3.5: Consider a digital satellite system to comprise three channels:the uplink, onboard processor, and the downlink. The uplink has an errorprobability of 0.001. The processing unit has an error probability of 0.00025.The downlink error probability is 0.01. CalculateSolution1. Each channel’s capacity2. The system’s overall conditional probability3. The overall system channel capacity1. Define each channel’s error probability pðy 1 Þ. pðy 2 Þ¼1 pðy 1 Þ.Using (3.80) and (3.87), we can define each channel capacity as1C c ¼ 1 pðy 1 Þ log 2pðy 1 Þ þ pðy 12Þ log 2 bit=symbolpðy 2 Þð3:90ÞThus:Channels pðy 1 Þ pðy 2 Þ Capacity, C cUplink 0.001 0.999 0.9886Onboard processor 0.00025 0.99975 0.9966Downlink 0.01 0.99 0.9192Copyright © 2002 by Marcel Dekker, Inc. All Rights Reserved.