Download Aptitude Crack
img not found img not found

Exam Questions Papers - Engineering

Q1:

An FM transmitter delivers 80W to a load of 30 W when no modulation is present. The carrier is now frequency modulated by a single sinusoidal signal and the peak frequency deviation is so adjusted to make the amplitude of the second sideband is zero in the given output.
J0(0, 0) = 1,
J0(2, 4) = 0,
J0(3.8) = - 0.4,
J0(5.1)= - 0.1
J1(2, 4) = 0.52,
J1((3.8) = 0,
J1(5.1) = - 0.33,
J2(2.4) = 0.1
J2(3.8)= 0.41,
J2(5.1) = 0
The power in all the remaining sidebands is :

A 84 W

B 97.44 W

C 100 W

D 77.44 W

ANS:D - 77.44 W

Total power of the FM transmitter = 80 W to a load of 50 W second side band amplitude = J2(β) = 0 ∴ β = 5.1 Average Power = (J0(β))2 x 100 = (0.16)2 x 100 = 2.56 W Power in all remaining = 80 - 2.56 = 77.44 .