GATE | GATE-IT-2004 | Question 45

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. 

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