Fundamental RMS of a full-bridge square-wave inverter A full-bridge inverter is fed by a DC battery of voltage V (producing a square-wave output of +V and −V). What is the RMS value of the fundamental component of the output voltage?

Introduction / Context:
Square-wave inverters produce rich harmonic spectra. Designers often need the magnitude of the fundamental component for calculating output RMS, filter design, and machine voltage rating. This question asks for the RMS of the fundamental of an ideal H-bridge square wave with amplitude ±V.

Given Data / Assumptions:

• Ideal full-bridge; no dead-time or device drops.
• Square-wave output swings between +V and −V.
• We need only the fundamental (first harmonic) RMS value.

Concept / Approach:

The Fourier series of a bipolar square wave of amplitude ±V has a fundamental (first harmonic) peak value given by V1,peak = 4V/π. The corresponding RMS of the fundamental is V1,rms = V1,peak/√2 = 4V/(π√2). This is lower than the total square-wave RMS (which equals V), because higher-order harmonics carry the remaining energy.

Step-by-Step Solution:

Verification / Alternative check:

Alternatively compute total RMS of square wave (V) and note that harmonics are present; only the fundamental portion is 0.9003… of V for bipolar modulation (since 4/(π√2) ≈ 0.9003).

Why Other Options Are Wrong:

2V/π is average of a half-sine; V/√2 is the RMS of a sine with peak V, not 4V/π; 2V/√2 and 4V/π are not RMS of the fundamental for this waveform.

Common Pitfalls:

Confusing total RMS with the fundamental RMS; forgetting to divide the peak by √2.

Final Answer:

4V/(π√2)