Question

A single-phase full-bridge bipolar PWM inverter employs selective harmonic elimination method.

The output voltage waveform of the inverter is shown in the figure below. For ${\alpha }_1={23.62}^o$ and ${\alpha }_2={33.30}^o$ , the ratio of amplitude of $7^{th}$ harmonic to fundamental component of unmodulated voltage wave is

• A. 0.247
• B. 0.349
• C. 0.302
• D. 0.494

Answer 1

The correct option is A 0.247

As this voltage waveform has quarter wave symmetry,

$a_n=0$

$\therefore b_n=\frac{4V_0}{\pi }\left[{\int }_0^{{\alpha }_1}sin\right(n\omega t\left)d\right(\omega t)-{\int }_{a_1}^{a_2}sin(n\omega t\left)d\right(\omega t)+{\int }_{a_2}^{\pi /2}sin(n\omega t\left)d\right(\omega t\left)\right]$

$b_n=\frac{4V_s}{\pi }\left[\frac{1-2Cos\left(n{\alpha }_1\right)+2cos\left(n{\alpha }_2\right)}{n}\right]$

$b_7=\frac{4V_s}{\pi }\left[\frac{1-2cos(7\times 23.62)+2cos(7\times 33.3)}{7}\right]$

$b_1=\frac{4V_s}{\pi }$

$\frac{b_7}{b_1}=0.2477$