A signal of $f_m=10\text{ kHz}$ is modulated by a carrier of $f_c=1\text{ MHz}$. If the bandwidth of the channel is $B=50\text{ kHz}$, how many such signals can be accommodated in the channel without interference?

Solution

1. The bandwidth required for one amplitude-modulated signal is $$\begin{array}{rl}{B}_{signal}& =2\times f_m\\ & =2\times 10\text{ kHz}\\ & =20\text{ kHz}\end{array}$$.
2. The number of signals that can fit in the channel is calculated by dividing the total channel bandwidth by the bandwidth per signal: $$\begin{array}{rl}N& =\frac{B}{B_{signal}}\\ & =\frac{50\text{ kHz}}{20\text{ kHz}}\\ & =2.5\end{array}$$.
3. Since only complete signals can be accommodated without interference, we round down to the nearest integer, which is 2.
4. Hence the answer is (A).