A speech signal given by 11 sin(2200 \( \pi \)t) V is used for amplitude modulation with a carrier signal given by 44 sin(6600 \( \pi \)t) V. The minimum amplitude of modulated wave will be :

Question

The percentage of the total power carried by the sidebands of the AM wave for tone modulation when the modulation index is 0.3 is:

Answer 1

Step 1: The AM wave is y = (A_c + A_m sin \(\omega\)_mt) sin \(\omega\)_ct = A_c(1 + \(\mu\) sin \(\omega\)_mt) sin \(\omega\)_ct with \(\mu\) = A_m/A_c = 11/44 = 1/4.
Step 2: The envelope is smallest when sin \(\omega\)_mt = -1: A_min = A_c(1 - \(\mu\)) = 44 × (1 - 0.25) = 44 × 0.75.
Step 3: A_min = 33 V.
Answer: (A) 33 V

List-I (Modulation Schemes)	List-II (Wave Expressions)
(A) Amplitude Modulation	(I) \( x(t) = A\cos(\omega_c t + k m(t)) \)
(B) Phase Modulation	(II) \( x(t) = A\cos(\omega_c t + k \int m(t)dt) \)
(C) Frequency Modulation	(III) \( x(t) = A + m(t)\cos(\omega_c t) \)
(D) DSB-SC Modulation	(IV) \( x(t) = m(t)\cos(\omega_c t) \)

A speech signal given by 11 sin(2200 \( \pi \)t) V is used for amplitude modulation with a carrier signal given by 44 sin(6600 \( \pi \)t) V. The minimum amplitude of modulated wave will be :

Show Hint

The Correct Option is A

Solution and Explanation

Top Questions on communication systems

Questions Asked in JEE Main exam