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Unformatted text preview: S/N=100 . Then, with Shannons theorem, we get the maximum achievable rate R Shannon max = Hlog 2 (1+S/N)=3* log 2 101 that is larger than 3*log 2 64=18 kbits/sec, but smaller than 3*log 2 128=32 kbits/sec. From this, all 8.1, 8.2 and 8.3 could be correct choices. However, from Nyquist theorem, we know the maximum achievable rate R Nyquist max =2Hlog 2 V, where V=2 since the signal is binary . So we further have R Nyquist max =6 kbits/sec , for which only 8.1 and 8.2 are satisfactory . Taking both into account, we can conclude that only 8.1 and 8.2 are correct choices. Page 1 of 1 1...
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This note was uploaded on 10/14/2009 for the course EECS 333 taught by Professor Guo during the Winter '08 term at Northwestern.
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