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

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Answer & Explanation
Correct AnswerC. nd = π

Quick Explanation

For a single pulse of half-width d, the nth harmonic term contains sin(nd). To eliminate the nth harmonic, set sin(nd)=0; the non-trivial condition used in the options is nd = π.

Formula / Key Relation

Vn ∝ sin(nd)/n
For elimination: sin(nd)=0
⇒ nd = mπ
Smallest non-zero choice: nd = π

Detailed Explanation

Concept & Reasoning

Fourier analysis of the single-pulse waveform shows that the amplitude of an nth harmonic is proportional to sin(nd)/n (with additional symmetry restrictions on even/odd harmonics). Therefore harmonic elimination is achieved by choosing a pulse width such that the sine factor vanishes.

How to Solve It in the Exam

See sin(nd) in harmonic coefficient → set nd to π for first non-zero root.

Important Exam Point

Check whether d is half pulse width; the question defines full pulse width as 2d.

Related Revision Path

This question sits in the revision path Power Electronics → Inverters → Pulse Width Modulation .

Selective Harmonic Elimination in Single-Pulse PWM explanatory diagram

Quick Trick

See sin(nd) in harmonic coefficient → set nd to π for first non-zero root.

Common Mistake

Setting 2d=π irrespective of harmonic number.

Exam Tip

Check whether d is half pulse width; the question defines full pulse width as 2d.

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