A single-phase inverter is fed from a 100 V dc source and is controlled using a quasi square wave modulation scheme to produce an output waveform, v (t) as shown. The angle σ is adjusted to entirely eliminate the 3rd harmonic component from the output voltage. Under this condition, for v (t), the magnitude of the 5th harmonic component as a percentage of magnitude of the fundamental component is _______ (rounded off to two decimal places).

The correct answer is 20 (or 20%).

Step 1: Understand the Output Voltage Waveform

From the given quasi-square wave modulation waveform $v\left(t\right)$, the Fourier series expansion of a quasi-square wave voltage output is given by:

$v\left(t\right)=\sum _{n=1,3,5,\dots }^{\infty }{V}_n\sin \left(n\omega t\right)$

where the peak amplitude of the $n$-th harmonic component $V_n$ is expressed as:

$V_n=\frac{4{\mathrm{V;}}_s}{n\pi }\cos \left(n\sigma \right)$

Here, $V_s=100\text{ V}$ is the DC source voltage and $\sigma$ is the pulse width control angle shown in the image.

Step 2: Determine the Angle $\sigma$ to Eliminate the 3rd Harmonic

To eliminate the 3rd harmonic component (${\mathrm{V;}}_3=0$):

${\mathrm{V;}}_3=\frac{4{\mathrm{V;}}_s}{3\pi }\cos \left(3\sigma \right)=0$

This implies:

$\cos \left(3\sigma \right)=0$

$3\sigma =\frac{\pi }{2}\Rightarrow \sigma =\frac{\pi }{6}\text{ rad }(30°)$

Step 3: Calculate Fundamental and 5th Harmonic Magnitudes

For $n=1$ (Fundamental Component):

${\mathrm{V;}}_1=\frac{4{\mathrm{V;}}_s}{\pi }\cos \left(\frac{\pi }{6}\right)$

For $n=5$ (5th Harmonic Component):

${\mathrm{V;}}_5=\frac{4{\mathrm{V;}}_s}{5\pi }\cos \left(\frac{5\pi }{6}\right)$

Step 4: Compute the Percentage Ratio

Taking the ratio of the magnitude of the 5th harmonic to the magnitude of the fundamental component:

$\frac{\left|{\mathrm{V;}}_5\right|}{\left|{\mathrm{V;}}_1\right|}=\frac{\frac{4{\mathrm{V;}}_s}{5\pi }\left|\cos \left(\frac{5\pi }{6}\right)\right|}{\frac{4{\mathrm{V;}}_s}{\pi }\left|\cos \left(\frac{\pi }{6}\right)\right|}$

Since $\left|\cos \left(\frac{5\pi }{6}\right)\right|=|-\frac{\sqrt{3}}{2}|=\frac{\sqrt{3}}{2}$ and $\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$:

$\frac{\left|{\mathrm{V;}}_5\right|}{\left|{\mathrm{V;}}_1\right|}=\frac{1}{5}=0.20$

In percentage terms:

$\text{Percentage}=0.20\times 100\%=20\%$

Thus, the magnitude of the 5th harmonic component as a percentage of the magnitude of the fundamental component is 20.