06-12-2011, 06:10 PM
Help for Question No
You can solve Q No 1 with the help of this question
We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally
has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually
3162. For this channel the capacity is calculated as
BitRate = 2 × 3000 × log2 2 = 6000 bps
BitRate = 2 × 3000 × log2 4 = 12,000 bps
Capacity = Bandwidth × log2 (1 + SNR)
C = B log2 (1 + SNR) = B log2 (1 + 0) = B log2 (1) = B × 0 = 0
C = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163)
C = 3000 × 11.62 = 34,860 bps
You can solve Q No 1 with the help of this question
We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally
has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually
3162. For this channel the capacity is calculated as
BitRate = 2 × 3000 × log2 2 = 6000 bps
BitRate = 2 × 3000 × log2 4 = 12,000 bps
Capacity = Bandwidth × log2 (1 + SNR)
C = B log2 (1 + SNR) = B log2 (1 + 0) = B log2 (1) = B × 0 = 0
C = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163)
C = 3000 × 11.62 = 34,860 bps
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