A sinusoidal wave y ( t )=40 (10 10^ 6 t ) is amplitude modulated… - Vidyadip

MCQ

A sinusoidal wave $y ( t )=40 \sin \left(10 \times 10^{6} \pi t \right)$ is amplitude modulated by another sinusoidal wave $x ( t )=20 \sin (1000 \pi t )$. The amplitude of minimum frequency component of modulated signal is ...........

A
$0.5$
B
$0.25$
C
$20$
✓
$10$

✓

Answer

Correct option: D.

$10$

d
$y ( t )=40 \sin \left(10 \times 10^{6} \; \pi t \right)$

$x(t)=20 \sin (1000 \; \pi t)$

$\Rightarrow \omega_{ c }=10^{7} \; \pi$

$\omega_{ m }=10^{3} \; \pi$

$A _{ C }=40$

$A_{m}=20$

Equation of modulated wave $=\left( A _{ C }+ A _{ m } \sin \omega_{ m } t \right)$

$\sin \omega_{ c } t$

$= A _{c}\left(1+\frac{ A _{ m }}{ A _{ c }} \sin \omega_{ m } t \right) \sin \omega_{ c } t$

$= A _{ c }\left(1+\mu \sin \omega_{ m } t \right) \sin \omega_{ c } t , \quad \mu=\frac{ A _{ m }}{ A _{ c }}$

$=A_{c} \sin \omega_{c} t+\frac{\mu A_{c}}{2}\left[\cos \left(\omega_{c}-\omega_{m}\right) t-\cos \left(\omega_{c}+\omega_{m}\right) t\right]$

Amplitude of minimum frequency = $\frac{\mu A _{ c }}{2}=\frac{ A _{ m }}{ A _{ c }} \times \frac{ A _{ c }}{2}=\frac{ A _{ m }}{2}=10$

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