A periodic square wave is formed by rectangular pulses ranging from −1 to +1 and period = 2 units. The ratio of the power in the 7th harmonic to the power in the 5th harmonic for this waveform is equal to ____________
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For an ideal symmetric square wave, only odd harmonics are present and the nth harmonic amplitude is proportional to 1/n. Harmonic power is proportional to amplitude squared, so P7/P5=(5/7)²≈0.51, nearest to 0.5.
Formula / Key Relation
Vn ∝ 1/n (n odd)
P₇/P₅ = (V₇/V₅)²
= (5/7)²
= 25/49 ≈ 0.510
Detailed Explanation
Concept & Reasoning
The Fourier series of a bipolar 50% duty square wave contains odd terms 1, 1/3, 1/5, 1/7, …. If the load resistance is the same for every harmonic, power contributed by one sinusoidal harmonic is proportional to V_n². Thus the ratio is the square of the amplitude ratio.
How to Solve It in the Exam
Square-wave harmonic power falls as 1/n².
Important Exam Point
Use amplitude ∝1/n, but power ∝1/n².
Related Revision Path
This question sits in the revision path Signals and Systems → Fourier Series → Harmonics .
Quick Trick
Square-wave harmonic power falls as 1/n².
Common Mistake
Using 5/7 directly as the power ratio instead of squaring it.
Exam Tip
Use amplitude ∝1/n, but power ∝1/n².
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