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A serial transmission $T1$ 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  $T1$ and $T2$?

  1. $100$ characters/sec, $153$ characters/sec
  2. $80$ characters/sec, $136$ characters/sec
  3. $100$ characters/sec, $136$ characters/sec
  4. $80$ characters/sec, $153$ characters/sec
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3 Answers

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66 votes
  1. T1: 1 char.$= ( 8 + 2 + 1 + 1) = 12\ bit$

          Transfer Rate $=\dfrac{1200}{12}=100\ char/sec$.
     
  2. T2: Transfer character in bits $= 24 + 240=264\ bits$

                 In  $264 = 30$ character

       Then  $1200 = ?$

       $\dfrac{264}{30}=\dfrac{1200}{X}$

       $X =136.3\ char/sec.$

  So, correct option is (C).

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15 votes
15 votes

Prerequisite

Bit rate is a specific case of baud rate.

Baud rate means the rate at which symbols(or elements or characters) get sent across a channel.

$\text{Baud rate} = \text{Number of symbols being sent per unit time} / \text{Size of a symbol}$

Very clearly, baud rate is nothing but the transfer rate.

 

When we say bit rate, we mean that 1 symbol = 1 bit.


Now, coming to the question

The bit rate given in the question is 1200 bits/second.

We have to get the data transfer rates.

 

For serial transmission, as per the question, Size of a character = $12\ bits$.

Baud rate/Transfer rate = $\frac{1200\ bits/sec}{1 \ character}$

=> $\frac{1200\ bits/sec}{12 \ bits}=100\ characters/sec$

 

For synchronous transmission, as per the question, Size of 30 characters = $30(8)+3(8)$.

Hence, size of a character = $8.8\ bits$

Baud rate/Transfer rate = $\frac{1200\ bits/sec}{1 \ character}$

=> $\frac{1200\ bits/sec}{8.8 \ bits}\approx136\ characters/sec$

 

Option C

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Serial communication :

Total number of bits transmitted = 8 + 2 + 1 + 1 = 12 bits
Bit rate = 1200 / second
Transfer Rate = 1200 * (8/12) = 800 bits/sec = 100 bytes/sec = 100 characters/sec

Synchronous transmission :

Total number of bits transmitted = 3 + 30 = 33 bits
Transfer Rate = 1200 * (30/33) =1090.90 bits/sec= 136.36 bytes/sec=136.36 characters/sec

 
Thus, option (C) is correct.
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