GATE 2020 EE Q.22: solution and answer

Official answer: 19.90 to 20.20Source: Official answer key, GATE 2020 EE (IIT Delhi)

1. The output is $v=+V_{dc}$ for $\sigma <\omega t<\pi -\sigma$, $-V_{dc}$ for $\pi +\sigma <\omega t<2\pi -\sigma$, and $0$ in the intervals of width $\sigma$ on either side of $\omega t=0,\pi ,2\pi ,\dots$. Here $V_{dc}=100\text{V}$. The wave has odd symmetry and half-wave symmetry, and it is symmetric about $\omega t=\pi /2$ (quarter-wave symmetry). So only odd-order sine terms are present.

2. Fourier coefficient, using quarter-wave symmetry: $$b_n=\frac{4}{\pi }{\int }_{\sigma }^{\pi /2}{V}_{dc}\sin n\omega td(\omega t)=\frac{4V_{dc}}{n\pi }\cos n\sigma ,n=1,3,5,\dots$$

3. To remove the 3rd harmonic, set $\cos 3\sigma =0$. The smallest non-trivial solution is $3\sigma =90^{\circ }$, so $\sigma =30^{\circ }$. The pulse width is then $\pi -2\sigma =120^{\circ }$.

4. Fundamental: $$V_1=\frac{4\times 100}{\pi }\cos 30^{\circ }=\frac{400}{\pi }(0.8660)\approx 110.27\text{V}$$

5. 5th harmonic: $$V_5=\frac{4\times 100}{5\pi }\cos 150^{\circ },|V_5|=\frac{400}{5\pi }(0.8660)\approx 22.05\text{V}$$

6. Ratio: $$\frac{|V_5|}{V_1}\times 100=\frac{1}{5}\cdot \frac{|\cos 150^{\circ }|}{\cos 30^{\circ }}\times 100=\frac{1}{5}\times 100=20.00\%$$