# Difference in information between IQ Sampling vs. Real Sampling at $2F_s$

As per my current understanding we use IQ sampling as way to increase the useful bandwidth to Fs rather than $$F_s/2$$ when viewing the FFT because the IQ sampling removes the negative frequencies and the entire $$0-F_s$$ band contains useful frequencies and not images of first half spectrum.

My questions are:

• Is there any difference between this and sampling at twice the original sample rate?
• And if I am sampling at $$2F_s$$ can that data be used to create the same IQ data as sampling I and Q components individually at $$F_s$$ or would there be a difference?
• Yes, in theory, the two are equivalent; they have the same information content. As is often the case, practical considerations can come into play depending on how the system is implemented; for example, SDRs that perform "IQ sampling" are often direct-conversion receivers, which have their own pros and cons. You could instead use a superheterodyne receiver that performs real sampling and postprocess its output in software to generate a complex-valued stream at half the sample rate, with a different set of system tradeoffs. – Jason R Dec 2 '20 at 20:54

Yes. Let’s say we construct both signals such that their frequency domain’s are equivalent from $$0-F_s$$. For the real signal the frequency domain would be limited to $$X(F_s-f) = X^{*}(F_s+f)$$. For the complex signal it would be $$X(f) = X(F_s+f)$$.
Yes. Again, we assume the frequency domains are equal from $$0-F_s$$. First, we filter the real signal to remove the frequency content from $$F_s-2F_s$$, which would result in an interpolated version of the IQ signal. Then we downsample by throwing away every other sample, yielding the IQ signal. There are practical considerations, but that would be one way to go about it.