Test: Pulse Width Modulation Inverter

The waveform is a positive in the first half cycle and symmetrical about π/2 in the first half.

Positive and the negative half cycles of the output voltage are symmetrical about π/2 and 3π/2 respectively. The shape of the waveform obtained is called as quasi-square wave.

As the positive and the negative half cycles are identical the coefficient a_n = 0.

For the fundamental component put n = 1.

V_o = (4Vs/π) sin (d) sin (ωt)
Hence the peak value is (4Vs/π) sin d.

The peak value of the fundamental component of voltage is given by (4Vs/π) sin d.

In the negative half the wave is symmetrical about 3π/2.

The Fourier analysis is as under:
b_n = (2/π) ∫ Vs sin⁡ nωt .d(ωt) , Where the integration would run from (π/2 + d) to (π/2 – d)
2d is the width of the pulse.

The Fourier representation of the output voltage is given by

Put 2d = π & n = 1.

To eliminate, the nth harmonic, nd is made equal to π radians, or d = π/n.
From the below expression,

when nd = π. sin nd = 0 hence, that output voltage harmonic is eliminated.

To eliminate the nth harmonic, nd = π.
Therefore, d = π/n = π/3 = 60°
Hence, 2d = 120°.