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 <[email protected]> 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 <[email protected]> > Sent: Thursday, August 26, 2021 1:37 PM > To: Discussion of precise time and frequency measurement > <[email protected]> > 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. > _______________________________________________ > time-nuts mailing list -- [email protected] -- To unsubscribe send an > email to [email protected] > To unsubscribe, go to and follow the instructions there. > > _______________________________________________ > time-nuts mailing list -- [email protected] -- To unsubscribe send an > email to [email protected] > To unsubscribe, go to and follow the instructions there. _______________________________________________ time-nuts mailing list -- [email protected] -- To unsubscribe send an email to [email protected] To unsubscribe, go to and follow the instructions there.
