Q160 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Physics
A message signal is super-imposed with a carrier signal. The resulting modulating signal $C_m(t)$ is given by
$C_m(t)=A_1 \sin \left(\omega_1 t\right)+A_2 \sin \left(\omega_2 t\right)-A_2 \sin \left(\omega_3 t\right) \text {, }$
where $\omega_2 < \omega_1 < \omega_3$. The modulation index and the angular frequency fo the message signal respectively, are
- A$\frac{A_2}{A_1}, \frac{\omega_3-\omega_2}{2}$
- B$\frac{2 A_2}{A_1}, \omega_3-\omega_2$
- C$\frac{A_1}{2 A_2}, \frac{\omega_3-\omega_2}{2}$
- Correct answerD$\frac{2 A_2}{A_1}, \frac{\omega_3-\omega_2}{2}$
Solution
Modulating signal is given as
$C_m(t)=A_1 \sin \left(\omega_1 t\right)+A_2 \sin \left(\omega_2 t\right)-A_3 \sin \left(\omega_3 t\right)$ $\ldots(\mathrm{i})$
where, $\quad \omega_1>\omega_2$ and $\omega_3>\omega_1$
Standard equation of modulating signal is given as
$C_m(t)=A_c \sin \left(\omega_c t\right)+\frac{\mu A_c}{2} \cos \left(\omega_c-\omega_m\right) t-\frac{\mu A_c}{2}$
$\cos \left(\omega_c+\omega_m\right) t$ ...(ii)
Clearly, $\omega_c>\omega_c-\omega_m$ and $\omega_c+\omega_m>\omega_c$
Eq. (i) can also be written as,
$C_m(t)=A_1 \sin \left(\omega_1 t\right)+A_2 \cos \left(\frac{\pi}{2}+\omega_2 t\right)-A_3 \cos$ $\left(\frac{\pi}{2}+\omega_3 t\right) \ldots$ (iii)
Comparing Eq. (ii) and Eq. (iii), we get
$A_1=A_c$
$\omega_c=\omega_1$
$\omega_c-\omega_m=\omega_2$ $\ldots(\mathrm{iv})$
and $\quad \omega_c+\omega_m=\omega_3$ $\ldots(\mathrm{v})$
Also, $\quad \frac{\mu A_c}{2}=A_2 \Rightarrow \mu=\frac{2 A_2}{A_1}$ $\ldots(\mathrm{vi})$
Substracting Eq. (iv) from Eq. (v), we have $\omega_3-\omega_2=\left(\omega_c+\omega_m\right)-\left(\omega_c-\omega_m\right)$
$\Rightarrow \quad 2 \omega_m=\omega_3-\omega_2$
$\Rightarrow \quad \omega_m=\frac{\omega_3-\omega_2}{2}$ $\ldots(\mathrm{vii})$
From Eq. (vi) and (vii), option (d) is correct.
Hence, the correct answer is option: 'D'.