20 20 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-gatecse-syllabus + – Kathleen 13.4k views answer comment Share Follow Print See all 2 Comments 2 2 Comments reply Sherlock_Holmes commented May 20, 2025 reply Follow flag NEVER HEARD OF THIS CONCEPT IS THIS STILL IN SYLLABUS 0 0 replyShare P0535_Yedidyah_Sagar commented Oct 7, 2025 reply Follow flag Out of cse syllabus 0 0 replyShare Please log in or register to add a comment.
Best answer 19 19 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 16 16 Comments reply Show 13 previous comments Saikumar_Baalu commented Jan 25, 2020 reply Follow flag 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 0 replyShare vaibhavkedia968 commented Dec 23, 2020 reply Follow flag 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 0 replyShare mani312 commented Feb 1, 2021 reply Follow flag Is Serial Communication part of the 2021 GATE syllabus? Please let me know 0 0 replyShare Please log in or register to add a comment.
4 4 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 1 comment 1 1 comment reply Sherlock_Holmes commented May 20, 2025 reply Follow flag IS THIS TOPIC INCLUDED IN SYLLABUS? 0 0 replyShare Please log in or register to add a comment.
