2
\$\begingroup\$

I am analysing some requirements regarding the sampling to be performed on a UART.

There is information provided however I am slightly confused by it and would like some clarification:

"The UART Receiver will sample at least 16 times faster than the fundamental transmission frequency of the UART. E.g for a 2 Mbit/sec bit-rate, the fundamental frequency is 1 MHz, requiring a sampling rate of 16 MHz min. The UART clock will therefore cycle at least eight times within one transmitted bit period."

  1. I understood that 16x Oversampling meant 16 samples per bit (not necessarily all captured depending on the RXR bit validation criteria), not 8.

  2. I understood that a 2 Mbps rate would be a fundamental frequency of 2 MHz, requiring a 32 MHz clock to perform the oversampling (this would therefore require a 1 Mbps rate to O-sample using a 16 MHz clock).

Is the example wrong or am I missing something fundamental here?

\$\endgroup\$
2
\$\begingroup\$

I understood that a 2Mbps rate would be a fundamental frequency of 2MHz

No, that is incorrect - in one period of 2 MHz there are two symbols. Hence, a baud rate of 2 Mbps might have a fundamental frequency of 1 MHz if all the data bits were 10101010101 etc..

I understood that 16x Oversampling meant 16 samples per bit

That's what I understand it to mean.

\$\endgroup\$
  • \$\begingroup\$ Thanks for that, but the penny still hasn't dropped. I understand that if I transmitted @2MHz that would be 1 bit every 500ns, resulting in a bit rate of 2M-bits every 1sec for me to expect at the Rxr... why does the value of the data have an effect on the fundamental transmission frequency? \$\endgroup\$ – user7427901 May 2 '18 at 20:26
  • \$\begingroup\$ There are two data symbols in one cycle of frequency. I can’t be clearer. Draw 101010. Every 10 is one cycle but it contains two data symbols. 2MHz has a period of 500 ns but can accommade two bits of data. Draw it. \$\endgroup\$ – Andy aka May 2 '18 at 22:02
  • \$\begingroup\$ I understand that if I was using a 1MHz clock at the RXR I would get 2 bits per cycle, but, the point I was making is that the wording of the given example suggests that the frequency the sending TXR uses is 1MHz - which would required transmitting using DDR to achieve 2Mbits/per/second rate... thanks anyway.. unless you would like me to draw it for you? ;) \$\endgroup\$ – user7427901 May 3 '18 at 12:20
  • \$\begingroup\$ In an earlier comment you said this: if I transmitted @2MHz that would be 1 bit every 500ns - that is incorrect - it would be 1 bit every 250 ns. That's all I was commenting on. \$\endgroup\$ – Andy aka May 3 '18 at 12:29
  • \$\begingroup\$ Thanks again but if T=1/F and if F= 2x10^6, the period T = 500 nS for one cycle, 250ns for 1/2 a cycle (presuming 50:50 Duty cycle). At the RTL level, I capture on the rising edge only.... I appreciate your time and input. \$\endgroup\$ – user7427901 May 3 '18 at 14:53
2
\$\begingroup\$

Most commercial UARTs do indeed sample 16× per bit, not 8.

The "fundamental frequency" created by an alternating 1-0 sequence is of very little relevance, except to the extent that it relates to the required analog bandwidth of the communication channel.

\$\endgroup\$
  • \$\begingroup\$ Thanks for that. I think it is terminology that is confusing me. \$\endgroup\$ – user7427901 May 2 '18 at 20:18
  • \$\begingroup\$ I clearly don't understand what 'fundamental transmission frequency' is. I took it to be the lowest frequency at which the serial data could be transmitted at which would be 2MHz for a 2Mbits-per-sec bit-rate.... Is the example in the information saying that for a 2Mbps bit-rate, and a Rxr using a 1MHz clock, will require a 16MHz clock to osample. That would make the statement regarding the only 8 samples per bit make sense... or am I still barking up the wrong tree?? \$\endgroup\$ – user7427901 May 2 '18 at 20:36
  • \$\begingroup\$ To be honest, I don't really know what they mean by it either. Based on the context, they seem to be referring to the waveform generated by a 1-0 pattern, but as I said, that isn't really relevant to anything related to the digital design. \$\endgroup\$ – Dave Tweed May 2 '18 at 21:23
  • \$\begingroup\$ Thank you, I discussed this with some old colleagues and it is agreed that the example provided information doesn't quite make it clear that the RXR is operating at 1MHz... thanks again, its good to have further confirmation, I thought I was starting to lose the plot! \$\endgroup\$ – user7427901 May 3 '18 at 12:02

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.