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Test: Pulse Width Modulation Inverter


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10 Questions MCQ Test Power Electronics | Test: Pulse Width Modulation Inverter

Test: Pulse Width Modulation Inverter for Electrical Engineering (EE) 2023 is part of Power Electronics preparation. The Test: Pulse Width Modulation Inverter questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Pulse Width Modulation Inverter MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Pulse Width Modulation Inverter below.
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Test: Pulse Width Modulation Inverter - Question 1

In the single-pulse width modulation method, the output voltage waveform is symmetrical about __________

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 1

The waveform is a positive in the first half cycle and symmetrical about π/2 in the first half.

Test: Pulse Width Modulation Inverter - Question 2

The shape of the output voltage waveform in a single PWM is

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 2

Positive and the negative half cycles of the output voltage are symmetrical about π/2 and 3π/2 respectively. The shape of the waveform obtained is called as quasi-square wave.

Test: Pulse Width Modulation Inverter - Question 3

In the single-pulse width modulation method, the Fourier coefficient an is given by

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 3

As the positive and the negative half cycles are identical the coefficient an = 0.

Test: Pulse Width Modulation Inverter - Question 4

In case of a single-pulse width modulation with the pulse width = 2d, the peak value of the fundamental component of voltage is given by the expression

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 4

For the fundamental component put n = 1.

Vo = (4Vs/π) sin (d) sin (ωt)
Hence the peak value is (4Vs/π) sin d.

Test: Pulse Width Modulation Inverter - Question 5

Find the peak value of the fundamental component of voltage with a pulse width of 2d = 90 and Vs = 240 V for single-pulse modulation in a full wave bridge inverter.

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 5

The peak value of the fundamental component of voltage is given by (4Vs/π) sin d.

Test: Pulse Width Modulation Inverter - Question 6

In the single-pulse width modulation method, the output voltage waveform is symmetrical about ____________ in the negative half cycle.

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 6

In the negative half the wave is symmetrical about 3π/2.

Test: Pulse Width Modulation Inverter - Question 7

In the single-pulse width modulation method, the Fourier coefficient bn is given by

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 7

The Fourier analysis is as under:
bn = (2/π) ∫ Vs sin⁡ nωt .d(ωt) , Where the integration would run from (π/2 + d) to (π/2 – d)
2d is the width of the pulse.

Test: Pulse Width Modulation Inverter - Question 8

In the single-pulse width modulation method, when the pulse width of 2d is equal to its maximum value of π radians, then the fundamental component of output voltage is given by

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 8

The Fourier representation of the output voltage is given by

Put 2d = π & n = 1.

Test: Pulse Width Modulation Inverter - Question 9

In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the nth harmonic from the output voltage

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 9

To eliminate, the nth harmonic, nd is made equal to π radians, or d = π/n.
From the below expression,

when nd = π. sin nd = 0 hence, that output voltage harmonic is eliminated.

Test: Pulse Width Modulation Inverter - Question 10

In case of a single-pulse width modulation with the pulse width = 2d, to eliminate the 3rd harmonic from the output voltage waveform, the value of the pulse width (2d) must be

Detailed Solution for Test: Pulse Width Modulation Inverter - Question 10

To eliminate the nth harmonic, nd = π.
Therefore, d = π/n = π/3 = 60°
Hence, 2d = 120°.

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