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We know from Shannon's Theorem,Maximum data rate of a channel in bps (B) = Hlog2 ( 1+S/N ) _ 1Where H = bandwidth in HzS/N = signal-to-noise ratioWe also know thatDb = 10log10 S/NWhere Db= signal-to-noise ratio in decibel (which is in this case 20)So20 = 10 log10 S/N2 = log10 S/NAnd we get S/N = 100Putting this value of S/N in equation 1 we getB = 6000 log2 (1+100) = 6000 x 6.6582= 39949.2 bps or 39.949 kbpsSo the maximum achievable data rate here is 39.949 kbps
Ribonucleic acid (RNA) and deoxyribonucleic acid (DNA) have ribose and deoxyribose as the pentose sugar, respectively. They also differ, to some extent, in possession of nit
Define the Utilization of Fatty Acids In the presence of oxygen, fatty acids are utilized to provide energy and heat. Waste products produced are carbon dioxide and water. Fatt
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so say if a car was going at 50km in a hour I would not no how to work it out help me:(
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