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Unformatted text preview: 2 (1 + SNR) We have W = 300 Hz (SNR) dB = 3 Therefore, SNR = 10 0.3 C = 300 log 2 (1 + 10 0.3 ) = 300 log 2 (2.995) = 474 bps 3.16 Using Nyquist's equation: C = 2B log 2 M We have C = 9600 bps a. log 2 M = 4, because a signal element encodes a 4bit word Therefore, C = 9600 = 2B × 4, and B = 1200 Hz b. 9600 = 2B × 8, and B = 600 Hz Additional Problem: a.+ 00 + 000+ + 00 + b.+ 00 +00+ 00...
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This note was uploaded on 11/02/2010 for the course EE 136 taught by Professor Kang xi during the Spring '10 term at NYU Poly.
 Spring '10
 KANG XI

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