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+11 votes

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

  1. $2400$ band
  2. $19200$ band
  3. $4800$ band
  4. $1200$ band
asked in Computer Networks by Veteran (59.7k points)
edited by | 2k views

2 Answers

+12 votes
Best answer

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. 

answered by Veteran (367k points)
edited by
What is Band Rate?
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.

How 4800 is coming?

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

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

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

Its 12*300=3600bits right.
0 it..thnx for your explanation
Is start bit present in synchronous communication?
So here bandwidth is 3600?
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.
plz explain how you got 4800 ....???
3600 is the actual answer but it is not given in the options, hence we took the next bigger option.
0 votes

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

300 * 8 = B * (8/12)

B = 3600

2400 < 3600 < 4800

4800 baud.

answered by (137 points)

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