Open In App

GATE | GATE-IT-2004 | Question 45

Like Article
Like
Save
Share
Report

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


Answer: (C)

Explanation:

Transfer Rate = Bit rate * ( Information bits / Total number of bits transmitted)

Serial communication : 
Total number of bits transmitted = 8 + 2 + 1 + 1 = 12 bits

Information bits = 8 bits

Transfer Rate = 1200 * (8/12) = 800 bits/sec

Given size of one character is 8 bits. 

So, Transfer Rate = 800/8 = 100 characters/sec 

Synchronous transmission : 
Total number of bits transmitted = 3 + 30 = 33 bits 

Information bits = 30 bits

Transfer Rate = 1200 * (30/33) = 1090.90 bits/sec 

Given size of one character is 8 bits. 

So, Transfer Rate = 1090.90/8 ≈ 136 characters/sec.
 
Thus, option (C) is correct. 
 
Please comment below if you find anything wrong in the above post. 

Quiz of this Question


Quiz of this Question
Please comment below if you find anything wrong in the above post


Last Updated : 25 Sep, 2017
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads