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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. $1200$
edited | 5k views
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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 $=\dfrac{9600}{12}=800$.

edited
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@ arjun sir here bit rate is depend on what is encoding method ... is it ...?

in case of manchester encoding bit rate = 1/2 *baud rate

or if NRZ encoding then bit rate = baud rate . .

so here nothing is mention about encoding ....so how do you consider baud rate = bit rate ... by default it will be used or what is logic ..?

or is it hit and try ...assume according to answer

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Nopes. This is fixed in "serial communication".
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2 stop bits ????  Why stop bbits , plzzzzzz xplain the concept clearly,,,,
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@Arjun sir, isn't this a synchronous transmission mode? I mean only asynchronous transmission mode requires start bit, stop bit, parity bit.

Synchronous transmission uses external clock to synchronous.

Accordingly, 9600/8= 1200. Here, 8 is the given 8-bits character.

So, option D

P.S. A similar question came in ISRO-2015 Set-A question number-51.
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yeah you are right
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Correct
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
+3
baud rate  = 2* bit rate

How is this?
+2
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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