GATE CSE 1998 | Question: 1.16

Question

In serial communication employing $8$ data bits, a parity bit and $2$ stop bits, the minimum band rate required to sustain a transfer rate of $300$ characters per second is

• A. $2400$ band
• B. $19200$ band
• C. $4800$ band
• D. $1200$ band

Answer 1

Since stop bit is given it is asynchronous communication and $1$ start bit is implied. So, 

$\text{(8 + 2 + 1 + 1) * 300 = 3600 bps}$

Minimum band rate required would be $4800$ here. 

Correct Answer: $C$

Comments

If we solve this question using synchronous serial we will get accurate answer.

In synchronous mode,  baud rate = 2*bit rate

baud rate = 2 * (300*8)

baud rate = 2*2400

baud rate = 4800

Note : In synchronous mode we won't consider about start, stop and parity bits.

Q : If asynchronous/synchronous mode is not given in the question then which one(asynchronous/synchronous) we need to consider to solve problem ?

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If nothing is mentioned then I think we have to consider both the cases and see which option is matching. If its a NAT then such ambiguity is usually not there, even if it is there then we will have to go with any one method (they might make the answer key as “3600 or 4800” but not really sure about it). Anyway such questions of serial communication are not really in syllabus right now, but this situation is applicable to a lot of other questions as well.

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Is Serial Communication part of the 2021 GATE syllabus?

Please let me know

Answer 2

Transfer rate = Baud rate * (data bits/total bits)

300 * 8 = B * (8/12)

B = 3600

2400 < 3600 < 4800

4800 baud.