[time-nuts] Re: uncertainty/SNR of IQ measurements

[Bob kb8tq]

Hi

Once you get things down to baseband, you (likely) shove the I and Q into some 
sort of arc tangent math. Depending on just how you do that math, there can be 
some bumps in the road. It’s implementation dependent so I don’t know of a 
“generic” answer. I’m not all knowing so there may be one … :)

Bob

> On Aug 26, 2021, at 2:27 PM, Tom Holmes <tholmes at woh.rr.com> wrote:
> 
> HI Jim...
> 
> From my admittedly limited understanding of IQ demodulators, the first thing done is to split the signal power (signal, noise, and all) evenly between two paths, which then ideally feed identical double balanced mixers (I'm thinking of a hardware implementation, obviously) whose only difference is the quadrature phase of the LO. So both paths are seeing the same SNR at that point. So my first guess would be that the relative phase of the LO to the input signal would only affect the phase of the output from each path, but the noise content ( or modulation if there is any) would not be any different between the two paths. I'm not aware that a single DBM used as a downconverting mixer shows any preference to the phase angle of the input to the LO. 
> 
> Tom Holmes, N8ZM
> 
> -----Original Message-----
> From: Lux, Jim <jim at luxfamily.com> 
> Sent: Thursday, August 26, 2021 1:37 PM
> To: Discussion of precise time and frequency measurement <time-nuts at lists.febo.com>
> Subject: [time-nuts] uncertainty/SNR of IQ measurements
> 
> This is sort of tangential to measuring time, really more about 
> measuring phase.
> 
> I'm looking for a simplified treatment of the uncertainty of I/Q 
> measurements.  Say you've got some input signal with a given SNR and you 
> run it into a I/Q demodulator - you get a series of I and Q measurements 
> (which might, later, be turned into mag and phase).
> 
> If the phase of the input happens to be 45 degrees relative to the LO 
> (and at the same frequency), then you get equal I and Q values, with 
> (presumably) equal SNRs.
> 
> But if the phase is 0 degrees, is the SNR of the I term the same as the 
> input (or perhaps, even, better), but what's the SNR of the Q term (or 
> alternately, the sd or variance) - Does the noise power in the input 
> divide evenly between the branches?  Is the contribution of the noise 
> from the LO equally divided? So the I is "input + noise/2" and Q is 
> "zero + noise/2"
> 
> If one looks at it as an ideal multiplier, you're multiplying some "cos 
> (omega t) + input noise" times "cos (omega t) + LO noise" - so the noise 
> in the output is input noise * LO + LO noise *input and a noise * noise 
> term.
> 
> I'm looking for a sort of not super quantitative and analytical 
> treatment that I can point folks to.
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