Answer to Question #292438 in Computer Networks for viiicccc

Question

CAT-6 twisted pair cable is a commonly used computer networks cable that provides a frequency bandwidth of up to 250 MHz. It is suitable for most varieties of Ethernet over twisted pair up to 1000 BASE-T (Gigabit Ethernet). Determine whether a signal-to-noise ratio (SNR) of 20 dB is adequate to transmit 0.9 Gbps on this cable?

Answer 1

first find the channel capacity

C=Blog2(1+SNR)

C=channel capacity

B=bandwidth

SNR=signal power/noise power

thus; 20db=10log10(SNR)

2db=log10(SNR)

100=SNR

therefore capacity now is calculated as follows:

C=Blog2(SNR)

B=250MHZ

SNR=100

C=250log2(1+100) ;

log2(101)= 6.6582;

C=250x6.6582

C=1664.55Mbps

1664.55Mbps - this is the channel capacity.

hence when we convert 0.9Gbps to Mbps becomes ; 1Gb = 1000Mb so 0.9Gb becomes 900Mbs. 900Mbps is below the actual channel capacity of 1664.55.

and thus a signal to noise ratio of 2dbs is adequate to transmit 0.9Gbs