Question:9
Identify the mathematical expression for amplitude modulated wave:
(a) $\quad A_c \sin \left[\left\{\omega_c+k_1 v_m(t)\right\} t+\phi\right]$.
(b) $A_c \sin \left\{\omega_c t+\phi+k_2 v_m(\mathrm{t})\right\}$.
(c) $\quad\left\{A_c+k_2 v_m(\mathrm{t})\right\} \sin \left(\omega_c t+\phi\right)$.
(d) $A_c v_m(\mathrm{t}) \sin \left(\omega_c t+\phi\right)$.
Answer:
The correct answer is the option (c)
Consider a sinusoidal modulating signal represented by
$
m(t)=A_m \sin \omega_m t ------(I)
$
where, $A_m=$ Amplitude of modulating signal $\omega_m=$ Angular frequency $=2 \pi \nu_m=\phi \nu_m$
Also consider a sinusoidal carrier wave represented by $C(t)=A_c \sin \omega_c t -------(ii)$
Thus, modulated wave is given by
$
\begin{aligned}
C_m(t) & =\left(A_c+A_m \sin \omega_m t\right) \sin \omega_c t \\
& =A_c\left[1+\frac{A_m}{A_c} \sin \omega_m t\right] \sin \omega_c t
\end{aligned}
$
$
\begin{array}{ll}
\text { Here, } \frac{A_m}{A_c}=M \\
\Rightarrow C_m(t)=\left(A_c+A_c \times \mu \sin \omega_m t\right) \sin \omega_c t
\end{array}
$
Now, we know that $A_c \times \mu=K$ [wave constant]
and $\quad \sin \omega_m t=v_m \quad$ [wave velocity]
Thus, Eq. (iii) becomes
$
C_m(t)=\left(A_c+K \times v_m\right) \sin \omega_c t
$
Now, consider a change in phase angle by $\phi$ then $\sin \omega_c t \rightarrow \sin \left(\omega_c t+\phi\right)$
Thus,
$
C_m(t)=\left(A_c+K v_m\right)\left(\sin \omega_c+\phi\right)
$
