GATE CSE 1993 | Question: 6.4, ISRO2008-14

Question

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

• A. $109$
• B. $216$
• C. $218$
• D. $219$

Comments

Is “serial-communication” in syllabus ?

Answer 1

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 $218$.

Comments

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

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from where did you got 8+2+1 bits @Digvijay Pandey

Answer 2

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.

Comments

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