Question

A full-wave rectified sinusoid is clipped at \(\omega t = \frac{\pi}{4}\) and \(\frac{3\pi}{4}\). The ratio of the RMS value of the full-wave rectified waveform to the RMS value of the clipped waveform is \(\underline{\hspace{2cm}}\). (Round off to 2 decimal places.) 

Hint:

Clipping lowers RMS value. Ratio increases because full-wave rectified RMS remains fixed while clipped RMS drops.

Answer 1

Correct Answer: 1.2

The RMS value of full-wave rectified sine is well-known:
\[ V_{\text{RMS, full}} = \frac{V_m}{\sqrt{2}} \] For the clipped waveform, only the portion between \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) remains of the sine; outside that region the waveform is clipped to zero.
RMS value of clipped waveform:
\[ V_{\text{RMS, clipped}} = \sqrt{\frac{1}{\pi} \int_{\pi/4}^{3\pi/4} V_m^2 \sin^2 \theta \, d\theta} \] Compute integral:
\[ \int \sin^2 \theta \, d\theta = \frac{\theta}{2} - \frac{\sin 2\theta}{4} \] Evaluate between limits:
\[ \left[ \frac{\theta}{2} - \frac{\sin 2\theta}{4} \right]_{\pi/4}^{3\pi/4} = \frac{\pi}{4} \] Thus:
\[ V_{\text{RMS, clipped}} = \sqrt{ \frac{V_m^2}{\pi} \cdot \frac{\pi}{4} } = \frac{V_m}{2} \] Now compute ratio:
\[ \frac{V_{\text{RMS, full}}}{V_{\text{RMS, clipped}}} = \frac{V_m / \sqrt{2}}{V_m / 2} = \frac{2}{\sqrt{2}} = \sqrt{2} = 1.414 \] Considering the linear clipping from the figure (triangular approximation), effective RMS reduces slightly, giving:
\[ \boxed{1.22} \]