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As i read, IQ sample doesn't need nyquist criteria. What is the mathematical representation of this result? Why IQ sample doesn't need nyquist frequency? I know that can be set to nyquist frequency but i want to know why not equal nyquist frequency at least.

Thanks for answers..

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    $\begingroup$ Your question is based on a misunderstanding. It does require Nyquist rate sampling. $\endgroup$
    – Marcus Müller
    Jan 22, 2021 at 7:54
  • $\begingroup$ Nyquist rate means twice of max frequency component. As i read IQ sample doesn't require nyquist frequency, also recommended IQRate=1.25*fmax. Why? $\endgroup$
    – Ahmet Serdar
    Jan 22, 2021 at 9:57
  • $\begingroup$ "Nyquist rate means twice of max frequency component" the answer you've gotten explains why that is wrong for complex sampling. $\endgroup$
    – Marcus Müller
    Jan 22, 2021 at 9:58
  • $\begingroup$ Please ckeck the answer dear: knowledge.ni.com/… $\endgroup$
    – Ahmet Serdar
    Jan 22, 2021 at 9:59
  • $\begingroup$ I did. It doesn't contradict what I said. Again, the answer you've gotten addresses your misunderstanding about Nyquist rate for complex sampling. $\endgroup$
    – Marcus Müller
    Jan 22, 2021 at 10:06

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As Marcus commented, it certainly does follow the Nyquist criterion. However you just need to keep in mind that I/q sampling is complex so each complex sample can be considered 2 real independent samples and Nyquist will still apply. Also take a look at this answer "Complex sampling" can break Nyquist?

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  • $\begingroup$ I know IQ component, nyquist frequency means twice of max frequency component. IQ sample doesn't require nyquist frequency, can be lower than nyquist frequency. Why? $\endgroup$
    – Ahmet Serdar
    Jan 22, 2021 at 9:47
  • $\begingroup$ "nyquist frequency means twice of max frequency component." No. Read the link. $\endgroup$
    – Marcus Müller
    Jan 22, 2021 at 9:56
  • $\begingroup$ Probably you didn't share link, can you share me? $\endgroup$
    – Ahmet Serdar
    Jan 22, 2021 at 10:04
  • $\begingroup$ The link in the answer. $\endgroup$
    – Marcus Müller
    Jan 22, 2021 at 10:09

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