A carrier signal C(t)=25 sin(2.512 times 10^{ | vedclass

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(D) The bandwidth of an amplitude modulated signal is given by $BW = 2f_m$,where $f_m$ is the frequency of the message signal.Given the message signal

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$50 	ext{ MHz}$

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(D) The bandwidth of an amplitude modulated signal is given by $BW = 2f_m$,where $f_m$ is the frequency of the message signal.Given the message signal $m(t) = 5 \sin(1.57 	imes 10^8 t)$,the angular frequency is $\omega_m = 1.57 	imes 10^8 	ext{ rad/s}$.Since $\omega_m = 2\pi f_m$,we have $f_m = \frac{\omega_m}{2\pi}$.Substituting the values,$f_m = \frac{1.57 	imes 10^8}{2 	imes 3.14} = \frac{1.57 	imes 10^8}{6.28} = 0.25 	imes 10^8 	ext{ Hz} = 25 	imes 10^6 	ext{ Hz} = 25 	ext{ MHz}$.Therefore,the bandwidth $BW = 2f_m = 2 	imes 25 	ext{ MHz} = 50 	ext{ MHz}$.

$A$ carrier signal $C(t)=25 \sin(2.512 \times 10^{10} t)$ is amplitude modulated by a message signal $m(t)=5 \sin(1.57 \times 10^{8} t)$ and transmitted through an antenna. What will be the bandwidth of the modulated signal?

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