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(A) The given amplitude modulated wave is $C_m = 10[1+0.6 \sin(40 	imes 10^3 t)] \sin(4 	imes 10^6 t) 	ext{ V}$.Comparing this with the standard eq

Vedclass · Accepted Answer

$101: 99$

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(A) The given amplitude modulated wave is $C_m = 10[1+0.6 \sin(40 	imes 10^3 t)] \sin(4 	imes 10^6 t) 	ext{ V}$.Comparing this with the standard equation $C_m = A_c[1 + \mu \sin(\omega_m t)] \sin(\omega_c t)$,we get:$\omega_m = 40 	imes 10^3 	ext{ rad/s}$ and $\omega_c = 4 	imes 10^6 	ext{ rad/s}$.The frequencies are $f_m = \frac{\omega_m}{2\pi}$ and $f_c = \frac{\omega_c}{2\pi}$.The upper sideband frequency is $f_{USB} = f_c + f_m$ and the lower sideband frequency is $f_{LSB} = f_c - f_m$.The ratio is $\frac{f_c + f_m}{f_c - f_m} = \frac{\omega_c + \omega_m}{\omega_c - \omega_m}$.Substituting the values: $\frac{4 	imes 10^6 + 40 	imes 10^3}{4 	imes 10^6 - 40 	imes 10^3} = \frac{4000 	imes 10^3 + 40 	imes 10^3}{4000 	imes 10^3 - 40 	imes 10^3} = \frac{4040}{3960} = \frac{101}{99}$.

List-$I$ (Layer of atmosphere)	List-$II$ (Approximate height over earth's surface)
$(A)$ $F_1$-Layer	$(I)$ $10\,km$
$(B)$ $D$-Layer	$(II)$ $170-190\,km$
$(C)$ Troposphere	$(III)$ $100\,km$
$(D)$ $E$-Layer	$(IV)$ $65-75\,km$

An amplitude modulated wave is represented by $10[1+0.6 \sin(40 \times 10^3 t)] \sin(4 \times 10^6 t)$ volt,where $t$ is in seconds. The ratio of the upper sideband frequency to the lower sideband frequency is:

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