A sinusoidal wave y(t) = 40 sin(10 times 10^6 | vedclass

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(D) Given carrier wave: $y(t) = 40 \sin(10^7 \pi t)$,so $A_c = 40$ and $\omega_c = 10^7 \pi$.Given modulating wave: $x(t) = 20 \sin(1000 \pi t)$,so $A

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$10$

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(D) Given carrier wave: $y(t) = 40 \sin(10^7 \pi t)$,so $A_c = 40$ and $\omega_c = 10^7 \pi$.Given modulating wave: $x(t) = 20 \sin(1000 \pi t)$,so $A_m = 20$ and $\omega_m = 10^3 \pi$.The modulated wave equation is given by $E = A_c(1 + \mu \sin \omega_m t) \sin \omega_c t$,where $\mu = \frac{A_m}{A_c} = \frac{20}{40} = 0.5$.Expanding the equation: $E = A_c \sin \omega_c t + \frac{\mu A_c}{2} \cos(\omega_c - \omega_m)t - \frac{\mu A_c}{2} \cos(\omega_c + \omega_m)t$.The frequency components are $\omega_c$,$(\omega_c - \omega_m)$,and $(\omega_c + \omega_m)$.The minimum frequency component is $(\omega_c - \omega_m)$.The amplitude of this component is $\frac{\mu A_c}{2} = \frac{A_m}{2} = \frac{20}{2} = 10$.

List $I$	List $II$
$A$. Attenuation	$I$. Combination of a receiver and transmitter.
$B$. Transducer	$II$. Process of retrieval of information from the carrier wave at receiver.
$C$. Demodulation	$III$. Converts one form of energy into another.
$D$. Repeater	$IV$. Loss of strength of a signal while propagating through a medium.

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

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