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I'm really struck in solving the following question

The single-phase full-bridge voltage source inverter (VSI), shown in figure, has an output frequency of 50 Hz. It uses unipolar pulse width modulation with switching frequency of 50 kHz and modulation index of 0.7. For \$V_{in}=100\$ V DC, \$L = 9.55 \$ mH, \$C = 63.66 \mu \$F and \$R = 5\; \Omega \$, the amplitude of the fundamental component in the output \$V_o \$(in Volt) under steady-state is ?
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Here is my approach. I'm using the following formula of the fourier series of the output voltage $$V_{O} = \sum_{n=1,3,5}^{\infty} \frac{8V_{in}}{n\pi} \sin(n\gamma) \sin \left( \frac{nd}{2} \right) \sin(n\omega t)$$ where, \$ 2d = \$ total pulse width per half cycle. \$ \gamma \$ is defined as enter image description here

I'm getting \$2d=1.5 ms \$ which is giving me answer way much less than the actual. The answer is in the range of \$ 60 - 64\$ V. What is the procedure for solving such type of problems ?

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  • \$\begingroup\$ What do you understand is the modulation index - is it related to duty cycle? \$\endgroup\$ – Andy aka Feb 9 '17 at 8:56
  • \$\begingroup\$ It is the ratio of reference voltage to the carrier voltage. I don't think it is related to duty cycle. \$\endgroup\$ – Ansh Kumar Feb 9 '17 at 8:57
  • \$\begingroup\$ I think to work this out you need duty cycle and modulation index. I'm assuming that "the fundamental component" referred to is the 50 Hz component and that the problem boils down to the transfer function of a low pass filter. This doesn't appear to be what you are doing but i could be wrong. \$\endgroup\$ – Andy aka Feb 9 '17 at 9:04
  • \$\begingroup\$ LC network does not seems like a low pass filter, its Laplace transfer function is coming out to be \$ 1/(1 + LC s^2) \$. And for multiple pulse width modulation how would you define duty cycle ? \$\endgroup\$ – Ansh Kumar Feb 9 '17 at 9:07
  • \$\begingroup\$ @Andyaka Yes you were right it was a second order low pass filter. I was previously neglecting resistance. nptel.ac.in/courses/108105066/PDF/… says that modulation index relates the inverter’s dc-link voltage and the magnitude of pole voltage (fundamental component) output by the inverter. So if we multiply modulation index by input dc we must get fundamental output peak by the inverter. I really don't understand much about modulation index though. \$\endgroup\$ – Ansh Kumar Feb 9 '17 at 10:27

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