Question

Suppose that the bandwidth is 1MHz and SNR = 24dB. Find

1. Maximum data rate that can be transmitted over a communication channel
2. How many signaling levels (i.e., M) are required to transmit the data we found in part (a).

Answer 1

A) Maximum data rate:

Given that bandwidth = 1 MHz, SNR = 24 dB.

According to shannon-hartley theorem the maximum data rate is given by the formula,

Data rate (C) = Bandwidth(B) * log₂(1+SNR), Where bandwidth in Hz, and SNR- Signal to Noise Ratio.

Therefore Data rate = 1MHz * log₂(1+24) = 1*10⁶ Hz * log₂(25) = 4.644 * 10⁶ bits/sec 4644 Kbits/sec.

B) Signaling levels:

Nyquist channel capacity theorem gives us the formula about the data rate. i.e, 2 * bandwidth * log₂ (M) where M is modulation level or signaling levels.

We have already found out the data rate for the given problem, so equate both the formula and the value and we will get the M value.

Therefore, 2 * bandwidth * log₂ (M) = 4.644 * 10⁶ bits/sec   2* 10⁶ Hz * log₂(M) = 4.644 * 10⁶ bits/sec

log₂(M) = 4.644 / 2 = 2.322 log₂(M) =2.322 (apply reverse of logarithms)

Therefore M = 2^2.322    5 levels.

Therefore 5 signaling levels are required to transmit the data we found in part (a).