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How many $8-bi$t characters can be transmitted per second over a $9600$ baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits and one parity bit?

  1. $600$
  2. $800$
  3. $876$
  4. $1200$
in Computer Networks edited by
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2 Comments

I am not able to find this question in GO BOOK
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3 Answers

47 votes
47 votes
Best answer

The baud rate is the rate at which information is transferred in a communication channel.
Serial ports use two-level (binary) signaling, so the data rate in bits per second is equal to the symbol rate in bauds.
Ref: https://en.wikipedia.org/wiki/Serial_port#Speed.

$\text{“9600 baud"}$ means that the serial port is capable of transferring a maximum of $\text{“9600 bits per second".}$

So, transmission rate here $=9600\text{ bps}$.

An eight bit data (which is a char) requires $1$ start bit, $2$ stop bits and $1$ parity bit $=12\text{ bits}$.

So, number of characters transmitted per second $=\dfrac{9600}{12}=800$.

Correct Answer: $B$

edited by
by

4 Comments

Asynchronous mode :

1] baud rate = bit rate

2] Start, stop and parity bits will be considered.

Synchronous mode :

1] baud rate = 2 * bit rate

2] Start, stop and parity bits won't be considered.
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Hi can anyone please tell me why all are considering 800 but not 600, because here 800 means number of frames but not number of 8bit characters.

 

9600/12 =800 frames.

Number of 8 bit characters 800*12(bits) –  800*4(bits)  = 800*6(bits) = 600 Bytes.
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1 vote
1 vote
Baud rate = 2* Bit rate

But this has no link to the question

So unlearn it.

as it is Asynchronous communication,

9600 is the baud rate

so 12 bits you have

No of characters transmitted per second= 9600/12 = 800
–3 votes
–3 votes
baud rate  = 2* bit rate

bit rate = 9600 / 2 = 4800 bps

# of 8-bit character that can be send  =  4800 / (1 + 8 + 2 + 1) = 400 chars

as there is no option matching I will go with 800 chars.

option B

4 Comments

baud rate  = 2* bit rate

How is this?
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Are these ques in syllabus ??
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@Arjun, it seems like he is taking it as differential manchester/manchester encoding. There baud rate = 2 * bit rate. Though as no encoding is given, we got baud rate = bit rate here  !
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waah
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Answer:

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