ISRO EC 2023 Q.54: solution and answer

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

1. A peak-to-peak value of $4\text{V}$ with zero DC component means the input is $v_{in}=2\sin \theta$, with amplitude $2\text{V}$.

2. The sine wave is applied to the non-inverting (+) input and $V_{ref}=1\text{V}$ to the inverting (-) input. An ideal comparator therefore gives a high output when $v_{in}>1\text{V}$ and a low output otherwise.

3. The switching points satisfy $2\sin \theta =1$, so $\sin \theta =\frac{1}{2}$. Within one cycle this gives $\theta =30^{\circ }$ and $\theta =150^{\circ }$.

4. Between these two angles $\sin \theta >\frac{1}{2}$, so the output is high for $150^{\circ }-30^{\circ }=120^{\circ }$ out of $360^{\circ }$.

5. $$\text{Duty cycle}=\frac{120^{\circ }}{360^{\circ }}=\frac{1}{3}\approx 33.33\%$$ This is option (a). A $50\%$ duty cycle would need $V_{ref}=0$, and $66.66\%$ is the fraction of the cycle for which the output is low.