Homework3 - IT 420 Homework #2 Key 1. Given a channel...

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IT 420 Homework #2 Key 1. Given a channel bandwidth of 25 KHz, what is the maximum throughput on the channel if we use 12 signal levels? data rate (D) = 2 B log 2 K where B = bandwidth and K = signal levels D = 2 (25 x 10 3 cycles/sec) log 2 12 D = (50 x 10 3 cycles/sec)(3.58) D = 179,248 bps 4. Given a channel bandwidth of 1 MHz, what is the maximum throughput on the channel if we use 8 signal levels? data rate (D) = 2 B log 2 K where B = bandwidth and K = signal levels D = 2 (1 x 10 6 cycles/sec) log 2 8 D = (2 x 10 6 cycles/sec)(3) D = 6 x 10 6 bps = 6 Mbps 5. How many signal levels would be required to obtain 40,000 bps through put over a channel with a bandwidth of 8000 Hz? data rate (D) = 2 B log 2 K where B = bandwidth and K = signal levels 40,000 bps = 2 (8000 cycles/sec) log 2 K 2.5 = log 2 K K = 2 2.5 K = 5.66 but, since we can’t have a fraction of a signal level we round up to 6 6. Given a channel bandwidth of 25 KHz, what is the maximum throughput on the channel if the signal to noise ration is 40db? SNR = 10 log
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This note was uploaded on 12/09/2009 for the course CS 431,430,48 taught by Professor Scher,statica during the Spring '09 term at NJIT.

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Homework3 - IT 420 Homework #2 Key 1. Given a channel...

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