Anamplitude modulated signal is given by `V(t)=10[1+0.3 cos (2.2 xx 10 ^4) sin (5.5 xx 10^5t)` Here t is in seconds .The sideband frequencies (in KHz) are `[Given pi = 22//7]`

Options

A. 1785 and 1715
B. 89.25 and 85.75
C. 178.5 and 171.5
D. 892.5 and 857.5

Answer

B

Explanation

The standard equation of AM wave is :
`y(t)=[1+m cos (omega_m t+phi)]A. sin (omega_c t)`
Where `omega_m` : Angular freq. of mod. Wave
`omega_c` : Angular freq. of carrier wave.
if the above equation is expanded as follows :
`y(t) =A sin (omega_ct ) +1/2 Am.sin((omega_c+omega_m)t+phi)+1/2Amsin((omega_c-omega_m)t)`
The first term is the original carrier wave, and the other two are known as sidebands. Comparing the given equation with above equation, sideband frequencies are:
(i)`f_c-f_m="(55-2.2)"/(2pi)xx10^4 approx 84.1 KH_3 `
(ii)`f_c+f_m="(55+2.2)"/(2pi) xx10^4 approx 91KH_3`.