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Telephone channels have a bandwidth of about 3.1 kHz. Answer the following questions. (If you like, you can use Excel to do the calculations and then cut/paste your analysis into your homework). (15 points) a) If a telephone channel’s signal-to-noise ratio (SNR) is 2,000 (the signal strength is 2,000 times as much as the noise strength), use the Shannon Equation to calculate the maximum data rate of a telephone channel? b) How fast could a telephone channel carry data if the SNR were increased massively, from 2,000 to 20,000? (Note: This would not be realistic in practice.) c) With an SNR of 2,000, how fast could a telephone channel carry data if the bandwidth were increased to 4 kHz? Show your work or no credit. d) What do you think is a better way to increase data rate, and why?

User Bob Ross
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Answer: a) 34 Kb/s b) 44 Kb/s c) 44 Kb/s

Step-by-step explanation:

The Shannon - Hartley equation, relates the channel capacity (which is the maximum data rate achievable) with the channel bandwidth, and the S/N ratio, as follows:

C (bps) = B (Hz) log₂ (1 +( S(w) / N(w)))

a) The problem tells us that the BW is of about 3.1 Khz, and that S/N is 2,000.

Replacing by the values, we get the value of C:

C = 3.1 Khz log₂ (1 + 2,000) = 3.1 Khz . 11 = 34 Kbps.

b) If the S/N were increased to 20,000, the Shannon equation will become like this:

C = 3.1 Khz log₂ (1 + 20000) = 3.1 Khz . 14 = 44 kb/s.

c) In a clear example of which is called bandwidth vs S/N trade-off, if we increase the BW, instead to have a better S/N, we will have the following:

C= 4 Khz. log₂ (1 + 2000) = 4 Khz . 11 = 44 Kb/s

d) The best way to increase the data rate, is increasing the BW, because the S/N is always limited by the thermal noise, which is impossible to remove.

The BW increment is possible, for instance, using more symbols to represent the voice samples, or using digital modulation schemes more efficient, like QPSK or 8PSK.

User Gandjustas
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