19 votes 19 votes In serial communication employing $8$ data bits, a parity bit and $2$ stop bits, the minimum band rate required to sustain a transfer rate of $300$ characters per second is $2400$ band $19200$ band $4800$ band $1200$ band Computer Networks gate1998 computer-networks communication serial-communication normal out-of-gate-syllabus + – Kathleen asked Sep 25, 2014 edited Jun 16, 2018 by Pooja Khatri Kathleen 8.6k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 18 votes 18 votes Since stop bit is given it is asynchronous communication and $1$ start bit is implied. So, $\text{(8 + 2 + 1 + 1) * 300 = 3600 bps}$ Minimum band rate required would be $4800$ here. Correct Answer: $C$ Arjun answered Mar 5, 2016 edited May 18, 2019 by Naveen Kumar 3 Arjun comment Share Follow See all 16 Comments See all 16 16 Comments reply Show 13 previous comments Saikumar_Baalu commented Jan 25, 2020 reply Follow Share If we solve this question using synchronous serial we will get accurate answer. In synchronous mode, baud rate = 2*bit rate baud rate = 2 * (300*8) baud rate = 2*2400 baud rate = 4800 Note : In synchronous mode we won't consider about start, stop and parity bits. Q : If asynchronous/synchronous mode is not given in the question then which one(asynchronous/synchronous) we need to consider to solve problem ? 0 votes 0 votes vaibhavkedia968 commented Dec 23, 2020 reply Follow Share If nothing is mentioned then I think we have to consider both the cases and see which option is matching. If its a NAT then such ambiguity is usually not there, even if it is there then we will have to go with any one method (they might make the answer key as “3600 or 4800” but not really sure about it). Anyway such questions of serial communication are not really in syllabus right now, but this situation is applicable to a lot of other questions as well. 0 votes 0 votes mani312 commented Feb 1, 2021 reply Follow Share Is Serial Communication part of the 2021 GATE syllabus? Please let me know 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes Transfer rate = Baud rate * (data bits/total bits) 300 * 8 = B * (8/12) B = 3600 2400 < 3600 < 4800 4800 baud. Barney Ross answered Jul 4, 2018 Barney Ross comment Share Follow See all 0 reply Please log in or register to add a comment.