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More on IQ modulation


From: John Byrne
Subject: More on IQ modulation
Date: Mon, 22 Feb 2021 13:48:45 -0500
User-agent: Cyrus-JMAP/3.5.0-alpha0-141-gf094924a34-fm-20210210.001-gf094924a

 As someone with not much radio knowledge, I'm trying to understand IQ modulation. I have a few questions if anyone can help. I recently read the interesting thread on IQ signals, and I also came across this site which has a few articles that helped (https://www.markimicrowave.com/blog/the-why-and-when-of-iq-mixers-for-beginners/).

But there's still a few things that confuse me. The way I understand it now is in terms of how the sidebands of the modulated I and Q signals add to produce the final RF signal: on the upper side of the carrier, Q is added to I, but on the lower side, Q is subtracted from I (due to the -sin term in the trig identify for cosA*sinB).

So we end up with a signal that has two sidebands, but they contain different info. At the receiver, the process happens again: the signal coming out of the I mixer has 2 positive copies of I (one in each sideband), and one positive copy of Q and one negative copy of Q. Combining all these gives 2I, and the opposite thing happens with the Q mixer, so we end up with 2I and 2Q (or maybe they're not x2 because the original signal's power was divided between 2 sidebands?)

Is that the gist of it? I hope so because it makes sense to me. But what's giving me trouble now is this: what happens if either the Tx or the Rx don't use IQ modulation?

The best answer I can come up with is as follows: if the Tx uses IQ but the Rx doesn't, then the Rx would have to use a local oscillator that's offset from the carrier so that the whole signal appears as a single sideband - otherwise the 2 different sidebands will interfere with each other. Is that right? And conversely, if the Tx just produces 2 identical sidebands, but the Rx uses IQ, then the I and Q signals in the Rx will be identical to each other, except for a phase difference right? And we need to combine them to get the full power?

Thanks
John




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