One positive pulse with tw = 75 µs is applied to one of the inputs of an exclusive-OR circuit. A second positive pulse with tw = 15 µs is applied to the other input beginning 20 µs after the leading edge of the first pulse. Which statement describes the output’s relation with the inputs?(a) The exclusive-OR output is a 20 s pulse followed by a 40 s pulse, with a separation of 15 s between the pulses(b) The exclusive-OR output is a 20 s pulse followed by a 15 s pulse, with a separation of 40 s between the pulses(c) The exclusive-OR output is a 15 s pulse followed by a 40 s pulse(d) The exclusive-OR output is a 20 s pulse followed by a 15 s pulse, followed by a 40 s pulseThe question was posed to me in an interview.My enquiry is from Fast Adder & Serial Adder topic in section Arithmetic Circuits of Digital Circuits

One positive pulse with tw = 75 µs is applied to one of the inputs of

1.	One positive pulse with tw = 75 µs is applied to one of the inputs of an exclusive-OR circuit. A second positive pulse with tw = 15 µs is applied to the other input beginning 20 µs after the leading edge of the first pulse. Which statement describes the output’s relation with the inputs?(a) The exclusive-OR output is a 20 s pulse followed by a 40 s pulse, with a separation of 15 s between the pulses(b) The exclusive-OR output is a 20 s pulse followed by a 15 s pulse, with a separation of 40 s between the pulses(c) The exclusive-OR output is a 15 s pulse followed by a 40 s pulse(d) The exclusive-OR output is a 20 s pulse followed by a 15 s pulse, followed by a 40 s pulseThe question was posed to me in an interview.My enquiry is from Fast Adder & Serial Adder topic in section Arithmetic Circuits of Digital Circuits
Answer» Right option is (d) The exclusive-OR OUTPUT is a 20 s pulse followed by a 15 s pulse, followed by a 40 s pulse Best explanation: When both the INPUT pulses are high or low X-OR output is low. But when one of the input is high and another is low or vice-versa, output is high. In this PROBLEM for the first 20uS one input is high and another is low. So, OBVIOUSLY output is a high. for next 15uS both the input is high so output is low and for remaining 40uS(75-20-15) first input is still high and SECOND one is low so output is high.

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