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    I Q signals lower sampling rate

    According to this article https://www.dsprelated.com/showarticle/192.php by using IQ signals we have at the end the benifit that Each A/D converter operates at half the sampling rate of standard real-signal sampling. Can somebody explain why? I mean the signal has still the same BW.

    •   Alt6th December 2017, 12:23

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    Re: I Q signals lower sampling rate

    The argument is referring to sampling rate, not signal bandwidth.

    If I understand right, it says that a signal with carrier fc can be either sampled with a single ADC at fs >= 2fc (Nyquist sampling) or quadrature sampled with two ADCs at fs = fc.



    •   Alt6th December 2017, 12:38

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    Re: I Q signals lower sampling rate

    I don't think this is what he means. After all you can still sample in lower sampling rate than the Nyquist rate if you use undersampling techniques https://en.wikipedia.org/wiki/Undersampling.

    It has to do with spliting the signal to two ADCs and in that way the sampling rate is not 2B anymore but B. But still dont understand why.



    •   Alt6th December 2017, 13:47

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    Re: I Q signals lower sampling rate

    With IQ signals, positive and negative frequencies are not the same. This means that the represented range is -fs/2 to +fs/2, so the bandwidth is equal to fs.
    It is the baseband signal that is represented by the IQ samples. The carrier is not visible in the IQ "domain".
    A fixed value of the IQ baseband signal (= zero frequency) corresponds to the carrier frequency in the RF domain.



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    Re: I Q signals lower sampling rate

    Still don't get it. Supposed our signal has a BW of B as in Fig18 what is the sampling rate of each ADC and waht is the benifit over not-complex down convertion?



    •   Alt6th December 2017, 16:26

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    Re: I Q signals lower sampling rate

    The sample rate of each A/D is fs, as usual. With "not-complex" sampling the frequency goes from 0 to fs/2, which means that BW = fs/2.
    With complex IQ signals, the frequency goes from -fs/2 to fs/2, which means that BW = fs.

    Figure 18 assumes complex sampling. With non-complex sampling, both the negative and positive frequencies will map to the range 0 - fs/2, so they can't be separated.


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