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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
Explain the Digestion of Food Salivary amylase or ptyalin acts on cooked starches (polysaccharides) and changes them into maltose as shown herewith. Boiled starch → soluble
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