Related Articles

GATE | GATE-IT-2004 | Question 45

• Difficulty Level : Basic
• Last Updated : 25 Sep, 2017

A serial transmission Ti uses 8 information bits, 2 start bits, 1 stop bit and 1 parity bit for each character. A synchronous transmission T2 uses 3 eight bit sync characters followed by 30 eight bit information characters. If the bit rate is 1200 bits/second in both cases, what are the transfer rates of Ti and T2?
(A) 100 characters/sec, 153 characters/sec
(B) 80 characters/sec, 136 characters/sec
(C) 100 characters/sec, 136 characters/sec
(D) 80 characters/sec, 153 characters/sec

Explanation:
Serial communication :

Total number of bits transmitted = 8 + 2 + 1 + 1 = 12 bits
Bit rate = 1200 / second
Transfer Rate = 1200 * (8/12) = 800 bits/sec = 100 bytes/sec = 100 characters/sec

Synchronous transmission :

Total number of bits transmitted = 3 + 30 = 33 bits
Transfer Rate = 1200 * (30/33) = 1090.90 bits/sec
Given size of one character is 8 bits.
So, Transfer Rate = 1090.90 bits/sec = 1090/8 = 136 characters/sec.

So,

Thus, option (C) is correct.

Please comment below if you find anything wrong in the above post.

Quiz of this Question

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.

My Personal Notes arrow_drop_up