Answer:
Option B
Explanation:
Given ,
Frequency of message signal $(f_{m})$ =1 KHz= $1 \times 10^{3} Hz$
Peak voltage of messages signal ($E_{m}$)=5V
Carrier frequency $(f_{c})$ =1 MHz= $1 \times 10^{6} Hz$
Peak voltage of carrier $(E_{c})$=15 V
We know that, The equation of amplitude modulated wave is given by
$e(t)= E_{c}\left[1+\frac{E_{m}}{E_{c}}\sin \omega_{m}t\right]\sin \omega_{c}t$
$= E_{c}\left[1+\frac{E_{m}}{E_{c}}\sin (2 \pi f_{m})t\right]\sin 2\pi f_{c}t$
$= 15\left[1+\frac{5}{15}\sin (2 \pi \times10^{3})t\right]\sin 2\pi \times 10^{6}t$
$= 15\left[1+\frac{1}{3}\sin (2 \pi 10^{3}t)\right]\sin ( 2\pi 10^{6})t$