Below is the given question: enter image description here

For my understanding, 1/baud_rate = 104.16 However, when I tried to do reverse division, the result is so different than expected.

  1. My calculation to find baud rate: 1/104.16 =9.6006 * 10^-3

Is my calculation of baud rate wrong?

2.How do I determine the number of stop bits in this case?

  • \$\begingroup\$ The link just provides some dark blue background page, without it I have no idea what you are asking. \$\endgroup\$ – PlasmaHH Dec 8 '16 at 15:51
  • \$\begingroup\$ Any receiver can handle the specified number of stop bits or arbitrarily more. Almost all modern UARTS only need one stop bit on receive. Sending out 2 is occasionally used to pace data but is a RARE need after the days of Teletype terminals, paper tapes punches and primitive protocol converters. \$\endgroup\$ – KalleMP Dec 8 '16 at 19:17

Baud rate is \$\dfrac{1}{104.16\times 10^{-6}}\$ = 9600.6 bits per second.

How do I determine the number of stop bits in this case?

It looks like 2 stop bits on the example you give but it could also be regarded as 1 stop bit and an indeterminate idling period.

  • \$\begingroup\$ Hi, I would like to ask how do you determine is 10^-6 to calculate the baud rate? Is base on experience or have to count the number of bits to determine ? \$\endgroup\$ – beginnerK Dec 8 '16 at 16:01
  • 3
    \$\begingroup\$ @beginnerK doesn't the word "microseconds" clearly written on your picture mean anything to you? 1 micro second is 10^-6 seconds. \$\endgroup\$ – Andy aka Dec 8 '16 at 16:06

You've forgotten the prefix. It's not 104.15 s, it's 104.16 us. So 1/(104.16 us) = 9.60061 kHz. Because 9600 baud is a very common baud rate, I'd round to that.

Your second question is more murky. You appear to have marked the last bit as the stop bit, which I think is correct, because the next bit period is marked as idle. So it would appear that the stop bit length is 1 stop bit. In general, however, without knowing the length of the stop bit, you can't figure it out from a scope trace because you can't differentiate between the stop bit and the line idling.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.