This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Amplitude Modulation”.
1. An AM signal is represented by x(t) = (20 + 4sin(500πt)) cos(2πt x 105)V. What is the modulation index?
a) 20
b) 4
c) 0.2
d) 10
View Answer
Explanation: 20 + 4sin(500πt) = 20(1 + 0.2sin(500πt)), hence the modulation index is 0.2.
2. An AM signal is represented by x(t) = (20 + 4sin(500πt)) cos(2πt x 105)V. What is the total signal power?
a) 208 W
b) 204 W
c) 408 W
d) 416 W
View Answer
Explanation: Pc = 20×20/2 or 200 W. Pt = Pc(1 + 0.2×0.2/4) or 204 W.
3. An AM signal is represented by x(t) = (20 + 4sin(500πt)) cos(2πt x 105)V. What is the total sideband power?
a) 4 W
b) 8 W
c) 16 W
d) 2 W
View Answer
Explanation: Pt – Pc = 204 – 200 = 4W.
4. An AM broadcast station operates at its maximum allowed total output of 50 kW with 80% modulation. The power in the intelligence part is
a) 12.12 kW
b) 31.12 kW
c) 6.42 kW
d) none of the mentioned
View Answer
Explanation: Pi = Pt – Pc = 50 – 37.88 kW or 12.12 kW.
5. The aerial current of an AM transmitter is 18 A when unmodulated but increases to 20 A when modulated. The modulation index is
a) 0.68
b) 0.73
c) 0.89
d) 0.95
View Answer
Explanation: 400/326 = 1 + (α2)/2, therefore α is 0.68.
6. A modulating signal is amplified by a 80% efficiency amplifier before being combined with a 20 kW carrier to generate an AM signal. The required DC input power to the amplifier, for the system to operate at 100% modulation, would be
a) 5 kW
b) 8.46 kW
c) 12.5 kW
d) 6.25 kW
View Answer
Explanation: Pi = Pt -Pc = 30 – 20 = 10 kW. DC input = 10/0.8 or 12.5 kW.
7. A 2 MHz carrier is amplitude modulated by a 500 Hz modulating signal to a depth of 70%. If the unmodulated carrier power is 2 kW, the power of the modulated signal is
a) 2.23 kW
b) 2.36 Kw
c) 1.18 kW
d) 1.26 kW
View Answer
Explanation:Pt = Pc (1 + 0.49/2).
8. A carrier is simultaneously modulated by two sine waves with modulation indices of 0.4 and 0.3. The resultant modulation index will be
a) 1.0
b) 0.7
c) 0.5
d) 0.35
View Answer
Explanation: α2 = 0.32 + 0.42 = 0.52 or α = 0.5.
9. In a DSB-SC system with 100% modulation, the power saving is
a) 100%
b) 55%
c) 75%
d) 100%
View Answer
Explanation: This is so because the power is suppressed by two thirds of the total. hence the power saving is 66%.
10. A 10 kW carrier is sinusoidally modulated by two carriers corresponding to a modulation index of 30% and 40% respectively. The total radiated power is
a) 11.25 kW
b) 12.5 kW
c) 15 kW
d) 17 kW
View Answer
Explanation: The required answer is 1 (1 + 0.42 + 0.32) or 11.25 kW.
Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.
To practice all areas of Electronic Devices and Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.
