-1
\$\begingroup\$

I have read that clock data recovery is essential for decoding signals, for instance decoding output of rotary encoders

But I don't know how it helps to decode signals.

Any hint or reference would be welcome

\$\endgroup\$
  • \$\begingroup\$ Oversampling is meaning that you sample the signal with higher rate than it is expected to change. This way you get the same value several times when it is not expected to change, so every sample which is deviating from this value can be considered as a noise. \$\endgroup\$ – Eugene Sh. Feb 22 '19 at 14:38
  • \$\begingroup\$ But surely the Wiki article can really explain it in much more comprehensive way \$\endgroup\$ – Eugene Sh. Feb 22 '19 at 14:43
  • 2
    \$\begingroup\$ Why have you edited your question to something completely different? Clock recovery is a different concept. Like completely. \$\endgroup\$ – Eugene Sh. Feb 22 '19 at 16:50
  • \$\begingroup\$ @EugeneSh. I was confused, at the decoding circuit I was looking at was written oversampling clock. And I wrote clock oversampling, not just oversampling about which there is too much information on internet \$\endgroup\$ – veronika Feb 22 '19 at 17:51
  • 1
    \$\begingroup\$ So what is the real question you are asking? \$\endgroup\$ – Eugene Sh. Feb 22 '19 at 17:53
1
\$\begingroup\$

I have read that clock data recovery is essential for decoding signals, for instance decoding output of rotary encoders

This example is nonsense. Rotary encoders generate low-speed digital signals which can be decoded using a simple state machine. Clock recovery is neither necessary nor relevant to them.

Clock recovery is a technique that can be used to allow high-speed digital signals -- like USB, SATA, or PCI Express -- to be transmitted without an associated clock signal. Data is encoded using a scheme like 8b/10b which guarantees a minimum density of transitions, and an edge detector and PLL are used to regenerate a data clock in the receiver.

\$\endgroup\$

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.