# Electronic Devices and Circuits Questions and Answers – Amplitude Modulation

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

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

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

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

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

Explanation: 400/326 = 1 + (α2)/2, therefore α is 0.68.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

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

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

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

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%

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

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.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]