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Assume that each character code consists of $8$ bits. The number of characters that can be transmitted per second through an asynchronous serial line at $2400$ baud rate, and with two stop bits is

1. $109$
2. $216$
3. $218$
4. $219$
edited | 3.8k views

Total bit per character $\text{= 8 bit data + 2 stop bit +1 start bit (#) = 11 bits}$
no of characters $=\dfrac{2400}{11}= 218.18$

Since it is asked for transmitted characters we take floor and answer is $218$.
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+1
start bit?
–2
But start bit not mentioned in questions..
Without start bit, stop bit is useless. But problem here is how many for start bit ??
+4

Take 1- I don't know if it is a standard but most places I saw 1 start bit and stop bits meaning stop bits can be more than 1 but not start bit.

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Bt sir why that type of constraints regarding start bit only ?? For better (let der is single start bit and that bit lost or corrupted ) reliability cant we put more than one start bit ??
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Thanks for the reply but still I didn't got it why to take 1 start bit this can be infer by ans. But  what about if nothing is given is there any standard for assumption ????
+5

+8
in  asynchronous serial line baud rate is same as bit rate.
+22

it says: There’s always only one start bit, but the number of stop bits is configurable to either one or two (though it’s commonly left at one).

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Then why haven't we considered the #stop bits=1 in the calculation above?
+1
it mentioned in the question that  "with two stop bits". if not then we  can take either 1 or 2!!
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in synchronous serial transfer... is baud rate is same as bit rate like in asynchronous...?
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what about synchronous? in synchronous baud rate = 2*bit rate?

am i right or wrong please confirm.
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in synchronous serial communction baud rate=frequency of clock signal.
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@reena_kandari  do u have a reference ?

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in the @sachin mittal sir comment they mention that

data bit range from 5 to 9

but in the nptel they given from 5 to 8

https://nptel.ac.in/courses/108107029/module12/lecture66.pdf

which one is right ?

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I think for data bit 5 to8 or 5 to 9 doesn't matter these are very much theoretical concepts

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.

"2400 baud" means that the serial port is capable of transferring a maximum of 2400 bits per second."

So, transmission rate here = 2400 bps

An eight bit data (which is a char) requires 1 start bit, 2 stop bits = 11 bits.

So, number of characters transmitted per second = 2400 / 11 = 218.18.

Since we can only count the fully transmitted ones, we take floor = 218.

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I downloaded the question paper from ISRO's website.It is in scanned pdf format sir.

Seems like it has a few mistakes.

one more here https://gateoverflow.in/18478/context-sensitive-grammar-isro-2008-7
+8

Sorry- I missed that part. I gave for asynchronous communication only. For synchronous communication, we don't need start-stop bits. So, this question must be for asynchronous one. For synchronous, answer should be 300.

https://en.wikipedia.org/wiki/Synchronous_serial_communication

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If an asynchronous serial line will be used instead of a synchronous one, what will be the effect on the answer sir?
+1
it must be a typo in question- "an" should be for "asynchronous". They missed "a" and just typed "synchronous".
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Sir, as 218.18 is the count of frames transmitted per second which should be in decimals so can we write 219 ?
+2

No, even if we get 218.99 we should write 218 as the question asks for "frames transmitted". Floor, ceil, round are to be applied as per the situation. Here, we should always do "floor".

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@ Arjun sir what about baud rate in case of synchronous serial transfer? Is it equal to bit rate?
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Depends on type of signalling as Arjuna sir mentioned
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@Mk Utkarsh @HeadShot in case of asynchronous is Baud rate equal to bit rate..

and for synchronous bud rate =2* bitrate ...is above two statements correct if nothing except synchronous or asynchronous is mentioned in the question?

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in the transmission manchster encoding must be performed so bit rate= 1/2*baudrate=1200bps

we have to add start bit so 1200/11=109.09==109

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