Originally posted by Jesh
Because HUY....
18 is D!!!! NOT A
This is the question I argued about.
Read Heinemann's IPT text.
I answered A and got it right, even though the CSSA answers were D. The majority of my classmates answered A as well, with one or two answering D.
My teacher accepted both A and D, however she knows that A is the 'textbook' answer.
18. Which of the following would be considered as the most common settings for asynchronous transmission?
(A) 8 data bits, no parity, 1 start bit, 1 stop bit
(D) 7 data bits, 1 parity bit (odd or even), 1 stop bit.
Heinemann definition:
Page 87.
"Serial transmission can be either synchronous or asynchronous. Asynchronous transmission is the sending of data by identifying each
byte with special
start and stop bits. It has become the standard for personal computers".
Now, take a look at answer D, which is "correct" (according to CSSA).
Firstly, 7 data bits does not equal 1 byte.
It takes 8 bits to make 1 byte.
Secondly, turn the page to Page 88.
Take a look at Figure 3.10, serial transmission.
Count how many 0's and 1's there are in the figure.
1011110110
With arrows under
stop bit and
start bit.
That makes 10 single bits, correct?
8 bits (or 1 byte) for the character, with
start and stop bits
Finally,
7 data bits, 1 parity bit and 1 stop bit equals
9 data bits.
As mentioned, 7 data bits does not equal 1 byte, and having 1 stop bit does not equate to "special start and stop
bits (plural)". You need 1 start bit, and 1 stop bit.
You cannot have 1 stop bit on it's own (without the start).
B and C are obviously incorrect, because it has "0 stop bits" for (B), uses 7 data bits, and as for (C), you have 8 data bits, and no start bit in use.
(A) is correct as:
There are 8 data bits (to form one byte, by definition),
1 start bit,
1 stop bit. (by definition).