Given a channel with intended capacity of 100Mbps the bandwidth of the channel is 20 MHz What signal to noise ratio is required to achieve this capacity?

Answer 1

Using the Shanon Capacity formula,

C = B log2 (1+ SNR)

Where B = 20 X 106 Hz

In order to calculate C (channel capacity) one must be mindful that 1Kbps = 1024 bps, NOT 1000 bps. Having this in mind,

C = 100 X 1.024 * 106 bps = 1.024 * 108

Substituting C, B, and solving for SNR:

1.024 * 108 = 20* 106 * log2 (1+SNR)

1.024 * 102 = 20 * log2 (1+SNR)

5.12 = log2 (1+SNR)

25.12 = 1 + SNR

2 5.12 - 1= SNR

SNR = 33.8

if in decibels = 10 log (33.8) = 15.29 dB

Answer 2

C = Blog2(1+SNR)

20 = 3*log2(1+SNR)

log2(1+SNR) =20/3

SNR = 100.6

SNRdb=10*log(SNR)=20 db