Question

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. Then the ratio of the upper to the lower side band frequencies is

• A. \( 101 : 99 \)
• B. \( 100 : 99 \)
• C. \( 100 : 1 \)
• D. \( 10 : 1 \)

Hint:

For an AM wave given by \( A[1 + m \sin(\omega_m t)] \sin(\omega_c t) \), the frequencies present are the carrier frequency \( f_c = \omega_c / 2\pi \), the upper sideband frequency \( f_{USB} = f_c + f_m \), and the lower sideband frequency \( f_{LSB} = f_c - f_m \), where \( f_m = \omega_m / 2\pi \) is the modulating frequency. Calculate these frequencies and then find the required ratio.

Answer 1

The Correct Option is A

The given amplitude modulated (AM) wave equation is: $$ y(t) = 10[1 + 0.
6 \sin(40 \times 10^3 t)] \sin(4 \times 10^6 t) $$ This can be expanded as: $$ y(t) = 10 \sin(4 \times 10^6 t) + 6 \sin(40 \times 10^3 t) \sin(4 \times 10^6 t) $$ Using the trigonometric identity \( \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \), we can rewrite the second term: $$ \sin(40 \times 10^3 t) \sin(4 \times 10^6 t) = \frac{1}{2} [\cos((4 \times 10^6 - 40 \times 10^3) t) - \cos((4 \times 10^6 + 40 \times 10^3) t)] $$ $$ = \frac{1}{2} [\cos(3960 \times 10^3 t) - \cos(4040 \times 10^3 t)] $$ So, the AM wave can be written as: $$ y(t) = 10 \sin(4 \times 10^6 t) + 3 \cos(3.
96 \times 10^6 t) - 3 \cos(4.
04 \times 10^6 t) $$ The frequencies present in the AM wave are the carrier frequency \( f_c \) and the upper and lower sideband frequencies \( f_{USB} \) and \( f_{LSB} \).
These are related to the angular frequencies by \( \omega = 2\pi f \).
The carrier angular frequency is \( \omega_c = 4 \times 10^6 \) rad/s, so the carrier frequency is \( f_c = \frac{4 \times 10^6}{2\pi} \).
The modulating angular frequency is \( \omega_m = 40 \times 10^3 \) rad/s, so the modulating frequency is \( f_m = \frac{40 \times 10^3}{2\pi} \).
The upper sideband angular frequency is \( \omega_{USB} = \omega_c + \omega_m = 4 \times 10^6 + 40 \times 10^3 = 4.
04 \times 10^6 \) rad/s.
The upper sideband frequency is \( f_{USB} = \frac{4.
04 \times 10^6}{2\pi} \).
The lower sideband angular frequency is \( \omega_{LSB} = \omega_c - \omega_m = 4 \times 10^6 - 40 \times 10^3 = 3.
96 \times 10^6 \) rad/s.
The lower sideband frequency is \( f_{LSB} = \frac{3.
96 \times 10^6}{2\pi} \).
The ratio of the upper to the lower sideband frequencies is: $$ \frac{f_{USB}}{f_{LSB}} = \frac{4.
04 \times 10^6}{3.
96 \times 10^6} = \frac{404}{396} $$ Divide both numerator and denominator by 4: $$ \frac{404}{396} = \frac{101}{99} $$ The ratio of the upper to the lower sideband frequencies is \( 101 : 99 \).