Question

A single-phase inverter is operated in PWM mode generating a single-pulse of width '2d' in the centre of each half as shown below.

If the pulse width is ${130}^{\circ }$, then ratio of maximum value of third harmonic output voltage to maximum value of fundamental output voltage will be _______ (up to 3 decimal places)

• A. 0.095

Answer 1

The correct option is A 0.095
For single-phase PWM inverter,

Given pulse width is, $2d={130}^{\circ }\Rightarrow d={65}^{\circ }$

Maximum value of third harmonic output voltge,

$V_{m3}=\frac{4V_s}{3\pi }sin(3\times 65)sin\left(\frac{3\pi }{2}\right)$

Similarly, maximum value of fundamental output voltage,

$V_{m1}=\frac{4V_s}{\pi }sin\left({65}^{\circ }\right)sin\left(\frac{\pi }{2}\right)$

$\therefore Ratio,\frac{V_{m3}}{V_{m1}}=\frac{1}{3}\frac{(-0.2588)(-1)}{\left(0.90630\right)\left(1\right)}=0.0951$