Question

n channels of bandwidth Bc are multiplexed together with FDM on a link. The guard band between two adjacent channels has bandwidth Bg. What is the minimum required bandwidth for the link ? (Write the formula for the total bandwidth in terms of the symbols given)

Answer 1

Answer: The answer to the question is as follows:

Minimum BW= n*Bc + (n-1) Bg

Step-by-step explanation:

FDM is a multiplexing technique that takes several baseband signals of n Khz wide (let's assume that all have the same bandwidth), and translates them in frequency so they can be transmitted together, at a higher frequency.

If we assume no guard bands between the different signals, the mimum bandwidth needed would be n times the bandwidth of a single signal.

In order to avoid crosstalk between signals, as the communication channel is not perfect, it usually leaves some room between any 2 signals, which it is called a band guard, and because there is a band guard for any 2 contiguous signals in the spectrum, the total number of bandguards (Bg) will be equal to (n-1) signals, so the total bandwidth to be used will be as follows:

BW needed: n*Bc + (n-1) Bg.