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# Test: Amplitude Modulation

## 10 Questions MCQ Test Electronic Devices | Test: Amplitude Modulation

Description
This mock test of Test: Amplitude Modulation for Electronics and Communication Engineering (ECE) helps you for every Electronics and Communication Engineering (ECE) entrance exam. This contains 10 Multiple Choice Questions for Electronics and Communication Engineering (ECE) Test: Amplitude Modulation (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Amplitude Modulation quiz give you a good mix of easy questions and tough questions. Electronics and Communication Engineering (ECE) students definitely take this Test: Amplitude Modulation exercise for a better result in the exam. You can find other Test: Amplitude Modulation extra questions, long questions & short questions for Electronics and Communication Engineering (ECE) on EduRev as well by searching above.
QUESTION: 1

### (Q.1-Q.3) An AM signal is represented by x(t) = (20 + 4sin(500πt)) cos(2πt x 105)V. Q. The modulation index is

Solution:

20 + 4sin(500πt) = 20(1 + 0.2sin(500πt)), hence the modulation index is 0.2.

QUESTION: 2

### Total signal power is

Solution:

Pc = 20×20/2 or 200 W. Pt = Pc(1 + 0.2×0.2/4) or 204 W.

QUESTION: 3

### Total sideband power is

Solution:

Pt – Pc = 204 – 200 = 4W.

QUESTION: 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

Solution:

Pi = Pt – Pc = 50 – 37.88 kW or 12.12 kW.

QUESTION: 5

The aerial current of an AM transmitter is 18 A when unmodulated but increases to 20 A when modulated.The modulation index is

Solution:

400/326 = 1 + (α2)/2, therefore α is 0.68.

QUESTION: 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

Solution:

Pi = Pt -Pc = 30 – 20 = 10 kW. DC input = 10/0.8 or 12.5 kW.

QUESTION: 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

Solution:

Pt = Pc (1 + 0.49/2).

QUESTION: 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

Solution:

α2 = 0.32 + 0.42 = 0.52 or α = 0.5.

QUESTION: 9

In a DSB-SC system with 100% modulation, the power saving is

Solution:

This is so because the power is suppressed by two thirds of the total. hence the power saving is 66%.

QUESTION: 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

Solution:

The required answer is 1 (1 + 0.42 + 0.32) or 11.25 kW.