GATE IT 2004 | Question: 45

Question

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 $T2$?

• A. $100$ characters/sec, $153$ characters/sec
• B. $80$ characters/sec, $136$ characters/sec
• C. $100$ characters/sec, $136$ characters/sec
• D. $80$ characters/sec, $153$ characters/sec

Comments

C should be ans..

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I don't think we can use more than 1 start bit in serial communication. Correct me if I am wrong?

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For transmission T2

the bit rate is given as 1200 bits per second.

we can imagine it as………….

I---------------------------------------1200 ------------------------------------I

I--------------------------1056------------------------I--------144------------I

264 bits	264 bits	264 bits	264 bits	144 bits	Total 1200 bits
30 char	30 char	30 char	30 char	24 sync bits + 120 data bits(15 chars)	120 + 12 = 135 chars

@Bikram sir is this the correct procedure.

Answer 1

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).

Comments

Yes.

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I have not yet understand the T2 part precisely. Could anybody please explain using very Layman approach?

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could you pls explain t2 in more detail

Answer 2

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.

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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

Comments

Is this question in current Gate syllabus?

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Jaya06 I don't think so, but better have an overview.

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Finally understood. Thanks :)

Answer 3

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.

Comments

Why for T2 only 30 characters are considered?

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First of all (1200*30)/33 is not 136.What you've considered is 33 bytes not bits.

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First of all (1200*30)/33 is not 136.

He didn't mentioned that (1200*30)/33 is 136. What he mentioned is 1090.90 which is further divided by 8 as the and is given in char/sec.

I think he has to write (1200*240)/264 = 1090.90 after dividing by 8 we will get 136.36