Question

Suppose that the passband of a channel is between 3 MHz and 6 MHz. The signal-to-noise ratio of the channel is 39 dB. What is the capacity of the channel? What is the minimum number of signaling levels required to achieve a data rate close to the channel capacity?

Answer 1

The capacity of a channel can be find out using Shannon capacity equation,

Capacity = Bandwidth × log2(1 + SNR)

Given that,

Bandwidth=6MHz-3MHz=3MHz

SNR=39dB

SNR = 10^(SNR(dB)/10)
SNR = 10^3.9 = 7943.282

Then,

Capacity=3×10⁶ ×log₂(7943.282)

Capacity=38866559 Hz

To find the signaling level required for this capacity

Consider the equation

Capacity = 2 × Bandwidth × log2(L)

Log₂(L)=(38866559) ÷(2×3×10⁶)

Log2(L)=6.47776

Then signaling level ,L=89.125

So the signaling level required to data rate which is close to capacity will be

L=89