Half-bridge single-phase inverter: If a DC battery of voltage V feeds an ideal half-bridge that produces a two-level square wave of +V/2 and −V/2 at the output, what is the rms value of the fundamental component of the output voltage (express your choice in terms of V)?

Introduction / Context:
Fourier analysis of inverter waveforms is central to designing filters and estimating fundamental RMS values. In a half-bridge, the output toggles between +V/2 and −V/2, producing a square wave whose fundamental must be extracted for RMS calculations.

Given Data / Assumptions:

• Ideal half-bridge with amplitude levels ±V/2.
• Pure square wave, 50% duty, no dead-time or device drop.
• We seek RMS of the fundamental component only (not total RMS).

Concept / Approach:
A symmetrical square wave of peak amplitude A has a fundamental (sine) component with peak value V1_peak = 4A/π. The RMS of the fundamental is V1_rms = V1_peak/√2 = (4A/π)/√2 = (2√2/π) * A.

Step-by-Step Solution:
1) Here A = V/2.2) V1_peak = 4*(V/2)/π = 2V/π.3) V1_rms = (2V/π)/√2 = (√2/π)V ≈ 0.450V.
Verification / Alternative check:
Compare with full-bridge where amplitude is ±V, yielding fundamental RMS (2√2/π) * (V/2) * 2? For half-bridge, halving the level halves the fundamental magnitude relative to a ±V square wave, so 0.450V is consistent.

Why Other Options Are Wrong:

• 2V: Dimensionally incorrect for RMS of a component.
• V/√2: That is the RMS of a sinusoid with peak V, not applicable here.
• V/2: Not the fundamental RMS; it is just the DC magnitude half, irrelevant here.
• (2/π) * V ≈ 0.637V: That is the average of a ±V square wave magnitude, not the fundamental RMS.

Common Pitfalls:

• Confusing total RMS of square wave with fundamental RMS.
• Dropping the factor of 1/2 for half-bridge amplitude.

Final Answer:
(√2/π) * V (≈ 0.450 V) (Option B).