GATE1998-1.16

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. 

Comments

What is Band Rate?

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Data is traveling through a medium. You can say "band rate" is the width of the medium or the amount of bits per second the medium can carry.

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How 4800 is coming?

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Because they asking for minimum but 3600 is not the option show that we will take lowest one which is 4800.

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why we are multiplying (300 characters/sec) with the bits..i am not getting dimensionally..?

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@dhruvkc123 : Because for one character you need 12 bits then how many bits you will need for 300 characters?

Its 12*300=3600bits right.

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fine..got it..thnx for your explanation

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Is start bit present in synchronous communication?

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So here bandwidth is 3600?

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NO, For synchronous transmission, data is not transferred byte-wise so there are no start-
or stop bits indicating the beginning or end of a character. Instead, there is a continuous
stream of bits which have to be split up into bytes.

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plz explain how you got 4800 ....???

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3600 is the actual answer but it is not given in the options, hence we took the next bigger option.

Answer 2

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

300 * 8 = B * (8/12)

B = 3600

2400 < 3600 < 4800

4800 baud.