GATE | Gate IT 2008 | Question 17
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?
(A) 10,000 bytes
(B) 12,000 bytes
(C) 15,000 bytes
(D) 27,000 bytes
Background required – Physical Layer in OSI Stack
In serial Communications, information is transferred in or out one bit at a time.
The baud rate specifies how fast data is sent over a serial line. It’s usually expressed in units of bits-per-second (bps). Each block (usually a byte) of data transmitted is actually sent in a packet or frame of bits. Frames are created by appending synchronization and parity bits to our data.
Attention reader! Don’t stop learning now. Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.
Learn all GATE CS concepts with Free Live Classes on our youtube channel.
"9600 baud" means that the serial port is capable of transferring a maximum of 9600 bits per second. 1 sec--------> 9600 bits 15 sec------->9600*15 bits Total Data To send in 1 frame = 1 bit(start) + 8 bits(char size) + 1 bit(Parity) + 2 bits(Stop) = 12 bits. Number of 8-bit characters that can be transmitted per second = (9600 * 15)/12 = 12000 bytes.
This explanation is contributed by Pranjul Ahuja.