Follow up question regarding: “Complex sampling can break Nyquist?”

I'm having some trouble understanding the sample rate limitations when considering a complex baseband signal.

I understand (based on the linked SE questions below), the either (1) physically sampling a signal with frequencies between $$0$$ and $$B$$ using 2 ADCs at sample rate $$B$$, one 90 degrees out of phase, or (2) sampling with one at $$2B$$, can satisfy the Nyquist criteria.

I'm working with a system where we have a real digital signal of bandwidth $$B$$, sampled at $$2B$$. It has then been converted to complex baseband representation by calculating the I/Q components from the existing samples. After that, the center frequency (also carrier in this example) is moved to zero and the signal now occupies $$-B/2$$ to $$B/2$$.

I suspect that even though the complex representation is twice the data, I still cannot down sample without breaking the Nyquist criteria because my I/Q components are dependent. For example, each real sample in the original signal is used to compute an I and Q component for that time.

Is this actually the case? Or can I somehow down sample by 2 and still fully represent the signal in complex form with a sample rate of $$B$$? To clarify, I believe that I could if the I/Q components were physically sampled independently.

I've done a few simple tests of this in python using a chirped signal, I could share some of the code if it will help. So far, if down sampling does violate Nyquist, I'm still able to reconstruct the original signal nearly as well as in the case with no down-sampling.

This question: Complex sampling can break Nyquist?

To represent a complex baseband signal in the range $$[-B/2,B/2]$$ you need either $$2B$$ real samples or $$B$$ complex samples per second.
To take real samples, you write the signal as $$I(t)+jQ(t)$$, with $$I(t)$$ and $$Q(t)$$ real, and sample each one at $$B$$ real samples per second, for a total of $$2B$$ real samples per second. It should be clear that you can't downsample without aliasing.
You can also take complex samples at $$B$$ samples per second (which is just a different way of thinking about the real sampling described above). This allows you to represent the entire band $$[-B/2,B/2]$$. Downsampling in any way will also result in aliasing.
To be more specific regarding your application: You have a bandpass signal with bandwidth $$B$$ and take real samples at rate $$2B$$. When you calculate the complex envelope, the I and Q components can be sampled at a rate of $$B$$ real samples per second each. The complex envelope can also be sampled at a rate of $$B$$ complex samples per second. In any case, you end up with a total of $$2B$$ real samples per second.
• Thank you @MBaz! That mostly makes sense but I'm still a little confused. What if I take my bandpass signal with bandwidth $B$, and take real samples at $2B$. Then using those (let's say it's over a 1 sec duration) $2B$ real samples, I calculate $I[t]$ and $Q[t]$ for each of the $2B$ samples, so that I can write $X[t] = I[t] + jQ[t]$. Can I now down-sample by 2, because my signal is now $2B$ complex samples? It seems like it should not be possible, because the calculation of $I[t]$ and $Q[t]$ where not independent. – Heath Dec 7 '18 at 3:41
• Yes, you can downsample them to $B$ samples per second each, since they are real signals with bandwidth $B/2$. – MBaz Dec 7 '18 at 16:13
• Thanks again @MBaz. Is there any resource you can recommend to help me understand why that works even when the I/Q components are not independent? Here's the 'thought experiement' that's giving my confusion: It seems like we could then take a real signal, with bandwidth $B$, sampled at $B$ and then convert complex (so now we have each component I and Q sampled at $B$). If that complex version is enough to represent the signal - couldn't I now upsample and convert to back to real/bandpass representation - thus avoiding the Nyquist constraint? – Heath Dec 7 '18 at 16:34