3
\$\begingroup\$

I'm using the Artix-7 FPGA to output a clock signal which is terminated in a resistor bridge. There is a little mismatch which gives the overshoot and undershoot.

My question is, what causes the "overshoot" and "undershoot" marked in red circles in the oscilloscope picture below?

schematic

simulate this circuit – Schematic created using CircuitLab

Clock signal

\$\endgroup\$
5
  • 2
    \$\begingroup\$ how did you measure the signal? Could simply be probe adjustment or the method you've used to probe the signal. \$\endgroup\$
    – Kartman
    Commented Mar 4, 2022 at 9:47
  • \$\begingroup\$ Haven't you answered your own question when stating the 2nd sentence? \$\endgroup\$
    – Andy aka
    Commented Mar 4, 2022 at 10:06
  • \$\begingroup\$ I have never before seen over- and undershoot, on both sides of the signal edge. Looking where the signal goes from high to low, there is something that looks like overshoot, but it cannot be overshoot because the signal is falling. Even more strange, the bump appears just before the signal starts falling. \$\endgroup\$
    – JakobJ
    Commented Mar 4, 2022 at 10:16
  • 2
    \$\begingroup\$ Please show your measurement setup. \$\endgroup\$
    – winny
    Commented Mar 4, 2022 at 10:45
  • \$\begingroup\$ What else is going on the FPGA at this time? Is this on a breadboard or multilayer PCB? It could be ground bounce inside the FPGA. \$\endgroup\$
    – SteveSh
    Commented Mar 4, 2022 at 11:01

1 Answer 1

4
\$\begingroup\$

These undershoots/overshoots are the artifacts of the probing/oscilloscope measurement. They are due to the harmonics. I believe your oscilloscope is a cheap one with a small bandwidth compared to the signal's frequency you're probing.

A square signal has odd harmonics, if you google square wave harmonics, you'll see the exact same waveform with better explanations.

Of course undershoot and overshoot aren't what is actually happening physically, lots of high frequencies harmonics are present which can't be easily seen. That's also why you'll almost never see a perfect square signal.

This article among others explains it: http://blog.teledynelecroy.com/2017/11/probing-techniques-and-tradeoffs-part.html

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Yes, I think this is correct. Fits the sampling bandwidth of my Picoscope. \$\endgroup\$
    – JakobJ
    Commented Mar 4, 2022 at 12:27
  • 1
    \$\begingroup\$ And I have now verified that my Picoscope does not have anti-aliasing filters in hw. \$\endgroup\$
    – JakobJ
    Commented Mar 4, 2022 at 12:40
  • \$\begingroup\$ If anyone else have this problem using PicoScope, I found a solution. Goto: Tools->Preferences->Sampling, deselect Sin(x)/x Interpolation. \$\endgroup\$
    – JakobJ
    Commented Mar 7, 2022 at 13:42
  • \$\begingroup\$ web.archive.org/web/20170829021855/https://m.eet.com/media/… \$\endgroup\$
    – None
    Commented Mar 7, 2022 at 23:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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