(i) RMS Output Voltage : The average value of the output voltage is given by

$$E_{0(\mathrm{av})}=\frac{1}{2 \pi} \int_{0}^{2 \pi} e_{0}(\omega t) \mathrm{d} \omega t$$

Now, rms value of the output voltage is given by

$$E_{0(\mathrm{rms})}=\sqrt{\frac{1}{2 \pi} \int_{0}^{\pi / 2} e_{0}^{2}(\omega t) \mathrm{d} \omega t}=\sqrt{\frac{4}{2 \pi} \int_{0}^{\pi / 2} e_{0}^{2}(\omega t) \mathrm{d} \omega t} -----\text{(due to quarter-wave symmetry)}$$

$$=\sqrt{\frac{2}{\pi} \int_{0}^{\pi / 2}\left(\frac{E_{\mathrm{dc}}}{2}\right)^{2}} \mathrm{d} \omega t=\frac{E_{\mathrm{dc}}}{2}-----(1)$$

RMS value of a square-wave is equal to its peak-value.

(ii) Instantaneous Output-Voltage : The fourier-series can be found out by using the following equation, $$e_{0}(\omega t)=\sum_{n=1,2,3, \ldots}^{\infty} C_{n} \sin \left(n \omega t+\phi_{n}\right)$$

where, $$\quad C_{n}=\sqrt{a_{n}^{2}+b_{n}^{2}}, \ and \ \phi_{n}=\tan ^{-1}\left(a_{n} / b_{n}\right)$$

and $$\qquad a_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} e_{0}(\omega t) \cdot \cos (n \omega t) \mathrm{d} \omega t=0,$$

and $$\qquad b_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} e_{0}(\omega t) \cdot \sin (n \omega t) \mathrm{d} \omega t$$

Due to quarter-wave symmetry, $b_{n}=0,$ for all even n

$$\therefore \qquad b_{n}=\frac{4}{\pi} \int_{0}^{\pi / 2} \frac{E_{\mathrm{dc}}}{2} \sin (n \omega t) \mathrm{d} \omega t, \quad \text {for all odd n}$$

$$\therefore \quad b_{n}=\frac{2 E_{\mathrm{dc}}}{n \pi}, \quad \text {for odd value of n}$$

$$\therefore \quad C_{n}=\sqrt{a_{n}^{2}+b_{n}^{2}}=\frac{2 E_{\mathrm{dc}}}{n \pi} \quad \ and \ \quad \phi_{n}=\tan ^{-1}\left(\frac{a_{n}}{b_{n}}\right)=0$$

Therefore, the instantaneous output voltage of a half-bridge inverter can be expressed in fourier-series form as,

$$\begin{aligned} e_{0}(\omega t) &=\sum_{n=1,3,5, \ldots}^{\infty} \frac{2}{\pi} \frac{E_{\mathrm{dc}}}{n} \sin (n \omega t)------(2) \\ &=0, \quad \text { for } n=2,4, \ldots \text { (even values of } n ) \end{aligned}$$

The $n^{\text { th }}$ harmonic-component is given by,

$$e_{0}(n)=\frac{c_{n}}{\sqrt{2}}=\frac{2 E_{\mathrm{dc}}}{n \pi \cdot \sqrt{2}}=\frac{\sqrt{2}}{n} \frac{E_{\mathrm{dc}}}{n} \quad \text{ for n=1,3,5}-----(3)$$

RMS value of fundamental components is obtained by substituting $n=1$ in Equation (3)

$$\therefore \quad E_{1_{(\operatorname{rms})}}=\frac{\sqrt{2}}{\pi} E_{\mathrm{dc}}=0.45 E_{\mathrm{dc}}-----(4)$$

(iii) Switch (Device) Voltage and Current Ratings : From the waveform,

$$V_{\text { CEO(transistor) }} \geq 2 \frac{E_{\mathrm{dc}}}{2} \geq E_{\mathrm{dc}}-----(5)$$

The current waveform for switch is a square-wave with a peak value of $E_{\mathrm{dc}} / 2 R .$

$$\therefore \quad I_{\text { Tavg }}=\frac{1}{T} \int_{0}^{T / 2} \frac{E_{\mathrm{dc}}}{2 R} \mathrm{d} t=\frac{E_{\mathrm{dc}}}{4 R}-----(6)$$

$$\therefore \qquad I_{T_{\operatorname{rms}}}=\sqrt{\frac{1}{T} \cdot \int_{0}^{T / 2}\left(\frac{E_{\mathrm{dc}}}{2 R}\right)^{2}} \mathrm{d} t=\frac{E_{\mathrm{dc}}}{2 \sqrt{2} R}-----(7)$$

and $$\qquad I_{T_{\text { peak }}}=\frac{E_{\mathrm{dc}}}{2 R}-----(8)$$