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+10 votes
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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 (59.5k points)
edited by | 1.7k views

2 Answers

+11 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 (352k points)
edited by
0
What is Band Rate?
+1
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.
+2

How 4800 is coming?

+1
Because they asking for minimum but 3600 is not the option show that we will take lowest one which is 4800.
0

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

+3
@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
fine..got it..thnx for your explanation
0
Is start bit present in synchronous communication?
0
So here bandwidth is 3600?
+1
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.
0
plz explain how you got 4800 ....???
0
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 (43 points)


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