An amplitude modulated wave is represented by | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$101: 99$

Vedclass · Answer

(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: $\frac{4 	imes 10^6 + 40 	imes 10^3}{4 	imes 10^6 - 40 	imes 10^3} = \frac{4000 	imes 10^3 + 40 	imes 10^3}{4000 	imes 10^3 - 40 	imes 10^3} = \frac{4040}{3960} = \frac{101}{99}$.

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:

Similar Questions

Each television channel is allocated a bandwidth of approximately ........?

For transmitting a signal of frequency $1 \text{ kHz}$,the wavelength of the wave should be:

For sky wave propagation,the radio waves must have a frequency range in between

$A$ carrier wave of peak voltage $12 \,V$ is used to transmit a signal. If the modulation index is $75 \%$, the peak voltage of the modulating signal is (in $\,V$)

The maximum amplitude of an $AM$ wave is found to be $20 \,V$ while its minimum amplitude is $4 \,V$. The modulation index is

Vedclass Test Series

Exam Paper Generator

Online Exam Module