A single-phase full bridge inverter fed by a 325 V DC produces a symmetric quasi-square waveform across “ab” as shown. To achieve a modulation index of 0.8, the angle θ expressed in degrees should be _______. (Round off to 2 decimal places)

To find the angle $\theta$ in degrees to achieve a modulation index of $0.8$, we analyze the Fourier series representation of the quasi-square output voltage waveform $V_{ab}$.

Step 1: Fourier Series Analysis of the Output Waveform
The output voltage waveform $V_{ab}\left(\omega t\right)$ is symmetric and has quarter-wave symmetry. The Fourier series expansion of such a symmetric quasi-square wave is given by:
$V_{ab}\left(\omega t\right)=\sum _{n=1,3,5,...}^{\infty }{V}_{on}sin\left(n\omega t\right)$
where the peak amplitude of the $n$-th harmonic component $V_{on}$ is calculated as:
$V_{on}=\frac{4}{\pi }{\int }_{\theta }^{\pi /2}{V}_{dc}sin\left(n\omega t\right)d\left(\omega t\right)$
Evaluating this integral for the fundamental component ($n=1$):
$V_{o1}=\frac{4V_{dc}}{\pi }{\left[-cos\left(\omega t\right)\right]}_{\theta }^{\pi /2}=\frac{4V_{dc}}{\pi }cos\theta$

Step 2: Relation with Modulation Index
The modulation index $m$ for a single-phase full bridge inverter is defined as the ratio of the peak of the fundamental output voltage to the input DC voltage:
$m=\frac{V_{o1}}{V_{dc}}$
Substituting the expression for $V_{o1}$ into the modulation index formula:
$m=\frac{4}{\pi }cos\theta$

Step 3: Calculating the Angle $\theta$
Given the modulation index $m=0.8$, we can solve for $cos\theta$:
$0.8=\frac{4}{\pi }cos\theta$
$cos\theta =\frac{0.8\pi }{4}=0.2\pi \approx 0.6283185$
Now, taking the inverse cosine to find $\theta$:
$\theta ={cos}^{-1}\left(0.6283185\right)\approx 0.891397\text{radians}$
Converting the angle $\theta$ from radians to degrees:
$\theta =0.891397\times \frac{180}{\pi }\approx 51.0724°$

Rounding to 2 decimal places, the angle $\theta$ is 51.07 degrees.