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 8.8k 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 Prashant Sharma 1 commented Jun 27, 2016 reply Follow Share What is Band Rate? 2 votes 2 votes Arjun commented Jun 27, 2016 reply Follow Share Data is traveling through a medium. You can say "band rate" is the width of the medium or the amount of bits per second the medium can carry. 6 votes 6 votes Pranabesh Ghosh 1 commented Jul 27, 2016 reply Follow Share How 4800 is coming? 7 votes 7 votes Brij Mohan Gupta commented May 18, 2017 reply Follow Share Because they asking for minimum but 3600 is not the option show that we will take lowest one which is 4800. 7 votes 7 votes dhruvkc123 commented Sep 1, 2017 reply Follow Share why we are multiplying (300 characters/sec) with the bits..i am not getting dimensionally..? 0 votes 0 votes saxena0612 commented Sep 1, 2017 reply Follow Share @dhruvkc123 : Because for one character you need 12 bits then how many bits you will need for 300 characters? Its 12*300=3600bits right. 6 votes 6 votes Tuhin Dutta commented Oct 5, 2017 reply Follow Share Is start bit present in synchronous communication? 0 votes 0 votes Tuhin Dutta commented Oct 5, 2017 reply Follow Share So here bandwidth is 3600? 0 votes 0 votes it_snow commented Dec 20, 2017 reply Follow Share NO, For synchronous transmission, data is not transferred byte-wise so there are no start- or stop bits indicating the beginning or end of a character. Instead, there is a continuous stream of bits which have to be split up into bytes. 3 votes 3 votes rajoramanoj commented Jan 30, 2018 reply Follow Share plz explain how you got 4800 ....??? 0 votes 0 votes it_snow commented Feb 2, 2018 reply Follow Share 3600 is the actual answer but it is not given in the options, hence we took the next bigger option. 0 votes 0 votes zeeshanmohnavi commented Dec 23, 2018 reply Follow Share @Arjun Please highlight, in your answer, why $4800$ is chosen when the actual answer comes out to be $3600$. This point is evident when going through the comments above, but in GO PDF, it is really very confusing to comprehend. 0 votes 0 votes VIDYADHAR SHELKE 1 commented Jan 16, 2019 reply Follow Share please anyone mention in ans Baud rate in Synchronous and Asynchronous communication. 0 votes 0 votes 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.