Download Aptitude Crack
img not found img not found

Exam Questions Papers

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
Power in the first order sidebands is :

A 5.12 W

B 21.78 W

C 2.56 W

D 16.81 W

ANS:B - 21.78 W

Power in first order side bands = 2(J1(β))2 x 100 = 21.78 W.



img not found
img

For help Students Orientation
Mcqs Questions

One stop destination for examination, preparation, recruitment, and more. Specially designed online test to solve all your preparation worries. Go wherever you want to and practice whenever you want, using the online test platform.