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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
1.) Referring directly to the locations of igneous activity identified on the below diagram (figure 5), and restricting yourself to any of the following terms and rocks (sedimentar
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