Question

Consider a point-to-point channel. Assume the bandwidth of the channel is 5 MHz and the signal-to-noise ratio of this channel is 24 dB.

a) What is the capacity of this channel?

b) How many signal levels are required to achieve the above maximum transmission rate?

Answer 1

Answer:

Given Bandwidth = B = 5MHz,

SNR = 24 dB.

SNRdB = 10log10SNR,

hence SNR = 10(SNR(dB)/10)

SNR = 10(24/10)

SNR = 102.4

SNR = 251.18,

Now we calculate maximum capacity of network, by using Noisy Channel : Shannon Capacity

C = B*log2(1+SNR)

C= 5 * 106 * log2(1+251.18)

C= 5 * 106 * log2(252.18)

C = 5 * 106 * (ln(252.18) ÷ ln(2))

C = 5 * 106 * 7.97

C = 39.85 MHz.

Now we calculate number of signal levels by using formula of Noiseless Channel : Nyquist Bit Rate.

BitRate = 2 * Bandwidth * log2(L)

39.85 * 106 = 2 * 5 * 106 * log2(L)

log2(L) = (39.85 * 106 ) / (2 * 5 * 106)

log2(L) = 3.985

L = 23.985

L = 15.83

Hence number of levels required = 15.83 (approx 16)