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[SOLVED] I Q signals lower sampling rate

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ctzof

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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.
 

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.
 

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.
 

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.
 

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?
 

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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