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 .
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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