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: The answer to the question is as follows:

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

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.