Question
(c) Consider a communication channel with 500 kHz bandwidth.
(i) If one wishes to transmit at a data rate of 8.3 Mbps, what is the minimum signal-to-noise ratio in dB required to accomplish this?
(5 marks)
(ii) If the effective data rate is 80% of the maximum channel data rate.
What is ratio of signal energy per bit to noise power density (i.e. Eb/N0)?

(c)(i) Answer
SNRdB = 10log10 (SNR)
C =B log2(1 + SNR)
8.3 Mbps = 500 000 log2 (1 + SNR)
SNR = ?
SNRdB = 10log10 (SNR)
SNRdB = 10log10 (SNR ??)
SNRdB = ?

(ii)
Effective data rate (80%) = 8.3 * 0.8 = 6.64
Eb/N0 = (S/N) * (B/R)
= SNR / (500*103 / 6.64*106)
= ?? W/Hz

Anyone able to help me? Appreciate your help. Thanks.

This post was edited by evgeric on Mar 2 2018 04:53pm