# Phase noise measurement with an IQ mixer

I have a question regarding a measurement scheme of phase noise that I'm trying to implement. The idea is that I have two identical signal generators (I actually do) that generate a sinusoidal voltage signal $V(t) = V_a(t) \sin\left(2\pi\nu + \phi(t)\right)$ where $V_a(t)$ is the amplitude, $\nu$ the frequency (I'll just assume that this is implemented perfectly for now) and $\phi(t)$ the phase.

Now ideally both the amplitude and phase would be static values, but in practice they are both subjected to noise (think thermal noise, 1/f noise, etc). In general this noise has a vanishing mean, so I could for example write $V_a(t) = V_a + \delta V_a(t)$, $\phi(t) = \phi + \delta \phi(t)$ where $\left<V_a(t)\right> = V_a$ and $\left<\phi(t)\right> = \phi$. With this the signals produced by the generators become $V(t) = \left(V_a+\delta V_a(t)\right) \sin\left(2\pi\nu + \phi + \delta\phi(t)\right)$.

Now, the purpose of my investigation is to characterize $\delta\phi(t)$: I am trying to find its power spectral density. The way I do this is by measuring the voltage signal in a time series and using (discrete) Fourier transforms to find the voltage spectral density and then the power spectral density. However, in general the noise is quite hard to distinguish when it competes with the actual time alternating part due to $\nu$. So to get this out, we use a trick: using an IQ mixer with an LO and RF signal at the same frequency, we use the downconverted signal at their difference ($=0$) frequency to get a DC signal, which is actually slightly time varying due to the phase noise.

Great! So then we would have a signal of the type (setting $\phi = 0$) for convenience) $V(t) = \left(V_a+\delta V_a(t)\right) \sin\left(\delta\phi(t)\right)$.

But here is where my question comes in. We only want to measure the phase noise, not the amplitude noise, and I don't understand how this is achieved. In the scheme we use, one uses an IQ mixer (inputs LO and RF) with outputs I and Q, but we terminate Q and only use the output from I. Somehow this gives a signal in which the amplitude oscillations are not longer relevant. Personally, I don't see this however, and I can't really find any documentation on the whole I-Q picture that's helping me out there.

So my question is, how does using only the I channel from the mixer make it so that our output signal is unaffected by the amplitude noise, and that we only look at phase noise?

• $I^2+Q^2$ gives you the amplitude, of you want it. You can also measure the amplitude of each signal independently and use that to normalize the calculation, which will throw that noise term out. Using just one of the two quadrature signals won't do that for you. – CuriousOne Feb 22 '16 at 10:10
• @CuriousOne Okay, so you're saying that if I simply downconvert to the DC signal and terminate the Q-port (so I only look at the I part, right?) I'll have a signal that contains both the phase noise and the amplitude noise? I'm doing this as a part of an orientation in a new lab, and my supervisor made some argument about how you are insensitive to the amplitude noise if you measure along the I axis in the phase diagram, which I didn't completely follow. But if I understand you correctly, it's also just wrong. How would normalizing throw out the noise term though? Because noise/amp = approx 0? – user129412 Feb 22 '16 at 10:14
• Okay so I just talked to him to ask about this, and what he says is that we set our phase so that the mean value of the signal is along the Q-axis, making us insensitive to the amplitude noise. I don't really see this yet 100% but I will think about it for a moment. – user129412 Feb 22 '16 at 10:35
• If the amplitude noise is very large, it's wrong, but usually it's not large. The other question is... is your second oscillator part of a loop? Are you already phase locked? In that case your supervisor might be right. I would have to think about that. As a general note... it's never a good idea to downconvert to DC, better do this measurement on an IF frequency and use a spectrum around that frequency, otherwise you are folding the DC drift and 1/f spectrum of your DC amplifier in (or use a really, really good DC system...). – CuriousOne Feb 22 '16 at 10:38
• I see... I would still have to think about it, but I would agree with you that this setup is sensitive to amplitude (modulation) noise. I would monitor that by measuring $I^2+Q^2$, if possible. – CuriousOne Feb 22 '16 at 10:57