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a)What is the symbol rate (baud rate) of a binary system if 12 kbits are transmitted in 8 seconds?

1. 1500 baud

2. 1000 baud

3. 2000 baud

4. 2500 baud

b)

What is the percentage of overhead for transimitting a 7-bit code using 2 stop bit and 1 parity bit ?

1. 22%

2. 30%

3. 6%

4. 40%

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Last Post by CimmerianX
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bit rate = baud rate when over 110, so for 1st question, 12000/8 = 1500 baud


For the 2nd question, total bits sent = 7 bit data, 2 stop bits, 1 bit parity = 10 bits. 2 stop bits + 1 parity bit = 3 bits. 3/10 = 30% overhead.

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For question 1 --> i thought isn't the formula baud rate = bit rate /log base 2 M
so --> baud rate 12000/8 = 1500 bits/s / ??? then how you get 1500 baud??? please explain?

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isnt their suppose to have a start bit to calculte the answer?

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For the baud rate, did I incorrectly assume you were talking about modem speed?

The Gross Bitrate can be expressed as R = bitrate*log_base2(M) where R is the rate and M is the number of distinct messages.


For Rs232, each character frame has 1 start bit, the data bits, the parity bit (optional), and the stop bit(s).

The start bit is the transition from negative voltage to positive. All rs232 communication has 1 start bit, it's just assumed to be there. I didn't include the 1 start bit in the math.

7 data bits + 1 Parity bit + 2 stop bit = 10 bit total with 3 being used for control. 30%. If you included the 1 start bit, you'd get 1+7+1+2. 11 bits total with 4 for overhead with is something like 36.36363636% IIRC Assuming the start bit is not included is the only way to get one of the multiple choice answers.

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sorry i have mistakenly type in the wrong question!!
Question 15 of 25

What is the percentage of overhead for transimitting a 7-bit code using 2 stop bit and 1 parity bit ?

22%.
30%.
36%.
40%

Now should i take into consideration of the "start bit" to get the answer?

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If you included the 1 start bit, you'd get 1+7+1+2. 11 bits total with 4 for overhead with is something like 36.36363636%

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