According to this tutorial, by exploiting quadrature sampling with IQ signals, we then have the benefit that each ADC operates at half the sampling rate of standard real-signal sampling (This conclusion in the link is right below the Figure. 18). Why is that? It seems the bandwidth of the signal is the same.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Does this answer your question? "Complex sampling" can break Nyquist? $\endgroup$– Marcus MüllerOct 29, 2022 at 8:20
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Let’s say you wanted to sample an analog bandpass signal whose bandwidth is 1 Mhz. Using a single A/D converter, to avoid aliasing errors you would have to sample that analog bandpass signal at a sample rate of just a bit greater than 2 MHz (2 megasamples/second). In quadrature sampling using two A/D converters (the top portion of Figure 14), to avoid aliasing errors each converter would operate at a sample rate of just a bit greater than 1 MHz (1 megasamples/second).
-
1$\begingroup$ Thanks! So the benefit of quadrature sampling basically is allowing us to sample with a sample rate just a bit greater than the signal band but not twice of it as the bandpass sampling requires? $\endgroup$– tyrelaOct 30, 2022 at 13:16