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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
with introduction description structures and conclusion
Explain how the uterus supports the development of a baby during gestation.
taken two oil samples A&B ,1st reaction A iodine value will be high and reichert meissl value will be low,and then 2nd reaction B iodine value low but reichert meissl value is high
PROCEDURE : Your Counsellor will be performing the following experiments. You are advised to observe the disposal methods carefully. Experiment 1: Test for studying miscibility
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what is the speed and velocity of a car that travels 32 meters to the east and 12 meters in the same direction? the total time of travel in 3.0 seconds
what are the problems
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