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38 votes
38 votes

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. $219$
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2 Answers

Best answer
41 votes
41 votes
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 $218$.
edited by
36 votes
36 votes

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.

Answer:

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