Answer to Question #140483 in Computer Networks for suale

Question

MTN operating on a frequency of 2.3MHz aim at achieving 15Mbps of data throughput with a successful receiver interpreting an incoming signal to noise (S/N)dB = 12, 18, 25. Calculate the maximum bite rate of each S/N without using calculator explain the tradeoff factor, compare your results with the calculator results, which of
this is enough for 4G which aim at achieving 14
Mbps and why?

Answer 1

Given,

Bandwidth=2.3Mhz=$2.3\times 10^6hz$

(A) When SNR$_{db}=12$

As we know

$SNR_{db}=10log_{10}(SNR)$

$SNR=10^{1.2}$

=15.84

So Maximum rate=Bandwidth$\times log_2(1+SNR)$

=$2.3\times log_2(1+15.84)$

= 2.3 log$_2(16.84)$

=$4.073Mbps$

(b) when $SNR_{db}=18$

$SNR_{db}=10\times log_{10}(SNR)$

$SNR=10^{1.8}$

=63.09

So Maximum rate=Bandwidth$\times log_2(1+SNR)$

=$2.3\times log_2(1+63.09)$

=$2.3\times log_2(64.09)$

=$6mbps$

(c) When SNR$_{db}=25$

$SNR_{db}=10\times log_{10}(SNR)$

$SNR=10^{2.5}$

=316.22

So Maximum rate=Bandwidth$\times log_2(1+SNR)$

=$2.3\times log_2(1+316.22)$

=$2.3\times log_2(317.22)$

=$8.30 mbps$

Tradeoff factor:-For digital communication between two geographical apart stations the quality of received signals at the receiver end is an important factor. To achieve the desired quality of received

signal at receiver end We can have a trade-off between major perimeters .i.e. Bandwidth, bit width and hardware complexity.