32 32 votes How many bytes of data can be sent in $15$ seconds over a serial link with baud rate of $9600$ in asynchronous mode with odd parity and two stop bits in the frame? $10,000$ bytes $12,000$ bytes $15,000$ bytes $27,000$ bytes Computer Networks gateit-2008 computer-networks communication serial-communication normal out-of-gatecse-syllabus + – Ishrat Jahan 25.8k views answer comment Share Follow Print See all 5 Comments 5 5 Comments reply Show 2 previous comments JashanArora commented Mar 5, 2020 reply Follow flag A_i_$_h baud rate = 2 * bit rate This is very incorrect. Baud rate = Number of symbols sent per unit time. When by "symbol" we mean a "bit", then we say Bit rate = Number of bits sent per unit time. 0 0 replyShare ronak.ladhar commented Nov 27, 2020 reply Follow flag Baud Rate = 2 * Bit Rate, Only in the case where we are using manchestor and differential manchestor encoding. 2 2 replyShare Venky8 commented Apr 26, 2021 reply Follow flag For those confused between bit rate and baud rate for serial communications see: https://gateoverflow.in/1019/gate-cse-2004-question-22?show=10129#a10129 0 0 replyShare Please log in or register to add a comment.
0 0 votes . akshay_123 answered Apr 4 akshay_123 comment Share Follow 0 reply Please log in or register to add a comment.
–1 –1 vote it should be 13090 bytes= 9600x15/11 Marv Patel answered Dec 26, 2014 Marv Patel comment Share Follow See 1 comment 1 1 comment reply pushkar dk commented Oct 9, 2016 reply Follow flag Start and stop bits for a byte of data are a must in asynchronous transmission. Therefore, even if start bit isn't mentioned in the question, we must add 1 start bit while calculating. 1(start) + 8(data byte in bits) + 1(parity) + 2(stop bits as per question) = 12 bits per character(i.e 12 bits per byte of 'data') 3 3 replyShare Please log in or register to add a comment.
