Post Wed Aug 29, 2012 10:09 pm

Shannon's Equation and Channel Capacity Example

Consider a white Gaussian noise with B = 3KHz. Find the maximum value of S/N in dB for reliable information transmission at R = 2400, 4800 and 9600 bps.

Solution:

Using Shannon's Equation and Channel Capacity

For R=2400 bps = C
B=3 KHz = W
C = W log2⁡(1+SNR)
2400 = 3000 log2⁡(1+SNR)
2400/3000 = log2⁡(1+SNR)
0.8= log2⁡(1+SNR)
2^0.8 = 1+ SNR
1.7411-1=SNR
SNR = 0.7411 – Signal Power is 0.7411times the noise power
SNR = 10*log10 (⁡0.7411)
SNR = -1.3012dB


For R=4800 bps = C
B=3 KHz = W
C = W log2(1+SNR)
4800 = 3000 log2(1+SNR)
4800/3000 = log2(1+SNR)
1.6= log2(1+SNR)
21.6 = 1+ SNR
3.03−1=SNR
SNR = 2.03 – Signal Power is 2.03 times the noise power
SNR = 10 ∗log10 (2.03)
SNR = 3.075 dB

For R=9600 bps = C
B=3 KHz = W
C = W log2⁡(1+SNR)
9600 = 3000 log2⁡(1+ SNR)
9600/3000 = log2⁡(1+SNR)
3.2= log2⁡(1+SNR)
2^3.2 = 1+ SNR
9.19-1= SNR
SNR = 8.19 – Signal Power is 8.19 times the noise power
SNR = 10*log10⁡(8.19)
SNR = 9.13 dB
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