Test: Amplitude Modulation

# Test: Amplitude Modulation

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## 10 Questions MCQ Test Analog Electronics | Test: Amplitude Modulation

Test: Amplitude Modulation for Electrical Engineering (EE) 2023 is part of Analog Electronics preparation. The Test: Amplitude Modulation questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Amplitude Modulation MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Amplitude Modulation below.
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Test: Amplitude Modulation - 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

Detailed Solution for Test: Amplitude Modulation - Question 1

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

Test: Amplitude Modulation - Question 2

### Total signal power is

Detailed Solution for Test: Amplitude Modulation - Question 2

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

Test: Amplitude Modulation - Question 3

### Total sideband power is

Detailed Solution for Test: Amplitude Modulation - Question 3

Pt – Pc = 204 – 200 = 4W.

Test: Amplitude Modulation - 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

Detailed Solution for Test: Amplitude Modulation - Question 4

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

Test: Amplitude Modulation - 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

Detailed Solution for Test: Amplitude Modulation - Question 5

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

Test: Amplitude Modulation - 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

Detailed Solution for Test: Amplitude Modulation - Question 6

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

Test: Amplitude Modulation - 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

Detailed Solution for Test: Amplitude Modulation - Question 7

Pt = Pc (1 + 0.49/2).

Test: Amplitude Modulation - 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

Detailed Solution for Test: Amplitude Modulation - Question 8

α2 = 0.32 + 0.42 = 0.52 or α = 0.5.

Test: Amplitude Modulation - Question 9

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

Detailed Solution for Test: Amplitude Modulation - Question 9

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

Test: Amplitude Modulation - 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

Detailed Solution for Test: Amplitude Modulation - Question 10

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

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