$f_1$ = 30 kHZ and $f_2$= 75 kHZ

Therefore, Bandwidth B = $f2 – f1$ = 75-30 = 45 kHZ…………..(1)

Let us assume that $f_S= 2B$ =2*45 = 90 kHZ………………..(2)

From equation (1) and (2) we observe that neither $f_1$ nor $f_2$ is harmonically related to $f_S$. Hence we have to use the general bandpass sampling theorem stated as follows:

If a bandpass signal x (t) has a bandwidth “B” and the highest frequency “$f_M$” then x (t)

can be recovered from its sampled version if $f_S= \frac{(2f_M)}{k}$ where k is the largest integer not

exceeding $\frac{f_M}{B}$ .All higher sampling rates are not necessarily usable unless they exceed 2 $f_M$.