Re: More on IQ modulation

[Christophe Seguinot]

About the Hilbert transform, it's a mathematical operator which is quite complex don't loose time for this if you don't have the math background. Just remind it has the effect of shifting the phase of the negative frequency components of by +90° (​^π⁄₂ radians) and the phase of the positive frequency components by −90° (my first post was not correct)

When we have 2 sidebands in an AM modulated signal, these two sidebands carry the same information based on I and Q signals. We can recover I(t) and q(t) from only one sideband. This is why transmitting AM can be made with 2 or 1 (SSB) sideband to  lower the spectrum bandwith.  Using only 1 sideband to recover the signal while 2 are present in the spectrum will simply be less efficient then using the 2 sideband from the noise point of view.

And Yes the mixer a quite a complex hardware too. Me have numerous type of mixers ( Mixer Basics Primer ) and upconversion is quite different them downcovnversion. Some uses direct conversion (FIF=0) some use an Intermediate Frequency (IF)...

• the ideal basic mixer is similar to a mathematical product of input signal and Carrier sine wave. Hardware realization use a non linear component such as diode, this is quite far from the mathematical product since is generated o lot of undesired harmonics at nFLO+mFIF (n,m being any positive or negative integer).
  
• in figure 1 the shift betweeen LO and RF is used to get and IF at FIF=|FLO-FRF|. Suppose our "IF" is at 0 Hz FLO=FRF.
  

The SSB mixer is obtained by combining several mixers whose input signal and/or carrier are shifted by 90° or -90°. Suppose we get 2 mixers having each 2 side band "correctly shifted", adding the mixed signal of these 2 mixers can result in an USB SSB, and substracting them to an LSB SSB. Corresponding hardware are Double and Triple balanced mixers (figure 6 and 7 in my first reference).

These is no silly question, we all have to learn things we don't learned before from those who know and like to share.

On 22/02/2021 23:58, John Byrne wrote:

Thanks for putting all this together! Unfortunately it sounds like I have massively underestimated the complexity of the whole thing. I don't know what the Hilbert transform is so I have a lot of reading to do before I can really understand most of what you wrote, or the grc file.

But let me ask this for now: when a receiver uses an LO that's the same as the carrier centre frequency, does't that result in the two sidebands being superimposed on each other? That's what I thought generally happens, because we have the sum and the difference for both sidebands. If the sidebands happen to be identical, then all is well. But if the signal in question was generated IQ modulation, then the sidebands are different, and they would interfere with each other. You'd lose the Q information right? That's why I thought the carrier would have to be offset. But you're saying it doesn't. Have I completely misunderstood how down-mixing works?

Sorry if these are silly questions.

On Mon, 22 Feb 2021, at 5:21 PM, Christophe Seguinot wrote:

Hi

Some rapid answers inside the original email

Christophe

On 22/02/2021 19:48, John Byrne wrote:

 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/).

this article is only related to SSB

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).

You conclusion is quite right, however, this is not due to -sin term. This can be obtained only if I and Q signal are related to each other by a constant phase shift  of -90° at ALL frequencies including positive and negative frequencies .This correspond to Hilbert transform as explained here figure 1  which is not physically existing.

So if Q(t) is the Hilbert transform of I(t), the output of the IQ modulator is and Upper or lower side band SSB.

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?

 Regarding your reference https://www.markimicrowave.com/blog/the-why-and-when-of-iq-mixers-for-beginners/  we found 3 methods of obtaining an SSB signal. You can combine any modulator (1 out of 3) with any demodulator (1 out of 3) to correctly demodulate the incoming signal.

It can be shown that an AM demodulator can be used to demodulate an SSB modulated signal

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?

In any case I mentioned above, the carrier frequency should be the same in modulator and demodulator without any shift.

Thanks

John