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Because of redundancy( 4^5>2^8 ) 4D-PAM5 has, I think there is no DC component and synchronization problem but I can't sure.

What is the truth?

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Of course there's an extensive synchronization problem – how would a modulation solve that? Your receiver still has to figure out symbol timing of the transmitter.

And: on average, yes, there should be no DC component, but for short observation times, of course there is a DC component in pulse amplitude modulations.

This is the same answer you got for your NRZ Question. Why should this change?

I think your question is an indication of you missing a solid understanding of the stochastic process' properties and the properties of a time-limited observation. If the term stochastic process doesn't ring a bell: you should read up on that, it will make further studies much, much more easy for you!

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  • $\begingroup$ Thanks. But what I was supposed to say is "Is there DC component can cause 'base wandering' with 4d-PAM5 shceme and problem of 'synchronization of bitfield' ". I think NRZ surely has, so I didn't add aditory comment. $\endgroup$ – wyldecat Oct 19 '16 at 19:15
  • $\begingroup$ and sorry for my bad english. :) $\endgroup$ – wyldecat Oct 19 '16 at 19:15
  • $\begingroup$ your English is very fine! Don't worry. But add that info to your question itself, not as a comment to an answer. You should also clarify what exactly you mean with "base wandering" and "synchronization of bitfield"; I think these terms aren't 100% clear. $\endgroup$ – Marcus Müller Oct 19 '16 at 19:17
  • $\begingroup$ Thanks! Here is the term "base wandering" in my text book. "A long string of 0s or 1s can cause a drift in the baseline (baseline wandering) and make it difficult for the receiver to decode correctly." Here is the term "synchronization" which I want to talk about. "To correctly interpret the signals received from the sender, the receiver’s bit intervals must correspond exactly to the sender’s bit intervals." Thank you for your patience :) $\endgroup$ – wyldecat Oct 19 '16 at 19:22
  • $\begingroup$ Why can't I insert new line in my comment? :( $\endgroup$ – wyldecat Oct 19 '16 at 19:22

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