Question

If a binary signal is transmitted through a 4kHz noiseless channel, what is the maximum data rate? If the SNR of the channel is 20dB, what is the maximum data rate?

Answer 1

Given H = 4 Khz

= 4000 hz

Usually noise channel equation ( i,e) Maximum data rate = 2H log2V bits/sec

= 2.(4000) log₂2 (V is discrete levels that a signal posses)  

= 8000 .( log₂2)

= 8000(1) [since log₂2 = 1]

= 8000 bits/sec

When SNR =20 db   it implies  10.log(S/N) =20 db

Hence S/N =100

Maximum data rate = H.log2(1 + S/N)

= 4 khz.(log2(1 + S/N)

= 4.1000 .(log2(1 + 100) [ since 1khz= 1000 hz]

= 4000 .log2(101)

= 4000.(6.6582)

= 26632.8

= 26632 bits/sec