Phase Noise measurement with Quadrature Detection

[MikeLogix]

I was reading up on Phase noise measurement. In our setup at work we have a N5500 test set. The common technique to measure phase noise is the quadrature detection method where a ref and DUT (device under test) are applied to a phase detector (like a double bal mixer). According to what I read, when the ref and DUT are 90 degree phase difference then the two signals are quadrature and the output of the mixer is zero volts (see image). This confuses me as to why 90 degree, to me it would seem that 180 degree would cause zero beat. Can anyone help me understand why it is 90 degree and not 180?
Thanks

[dmills]

Let An be (1 + the amplitude noise of signal A) and Bn be (1 plus the the amplitude noise of signal B).

Then let A = An* sin wt, and B = Bn * cos wt.

Multiplying these (Which is what a mixer does), you get An * Bn * sin wt * cos wt

applying the double angle identity, sin 2u = (2 sin u cos u) we get 0.5 (An * Bn) sin 2wt so for the quadrature case all of the amplitude nose ends up being mixed up to twice the carrier frequency and rejected by the measurement lowpass filter.

This makes the measured nose purely that due to phase noise and does not include the amplitude noise, providing the phase noise is small compared to the oscillator power (If it is not then the two can no longer be meaningfully said to be in quadrature!). 

Regards, Dan.

[MikeLogix]

Let An be (1 + the amplitude noise of signal A) and Bn be (1 plus the the amplitude noise of signal B).

Then let A = An* sin wt, and B = Bn * cos wt.

Multiplying these (Which is what a mixer does), you get An * Bn * sin wt * cos wt

applying the double angle identity, sin 2u = (2 sin u cos u) we get 0.5 (An * Bn) sin 2wt so for the quadrature case all of the amplitude nose ends up being mixed up to twice the carrier frequency and rejected by the measurement lowpass filter.

This makes the measured nose purely that due to phase noise and does not include the amplitude noise, providing the phase noise is small compared to the oscillator power (If it is not then the two can no longer be meaningfully said to be in quadrature!). 

Regards, Dan.

Hi Dan, thanks for replying. Just to clear some of my confusion. You say let A and B, be 1 + amplitude of noise. Would not that be the main carrier amplitude so A = (1+signal) that we want to cancel out. What I mean, two signals say 10 MHz, would we not want to zero beat the 10 MHz so what remains is the noise to measure?
Also in your formula, you use wt, would that be w= 2piF, and t= time? Just want to make sure I put in the correct numbers.
Thanks

[rfeecs]

Here are some references on using a double balanced mixer as a phase detector.  When the RF is in phase with the LO, you get a maximum (say -1), and when RF is 180 degrees out of phase you get a negative maximum (say +1).  In between, at 90 degrees, the output passes through zero:

https://www.jlab.org/uspas11/Reading/RF/Mixers%20-%20phase%20detectors.pdf

http://www.markimicrowave.com/blog/2015/02/all-about-mixers-as-phase-detectors/

[rfeecs]

Let An be (1 + the amplitude noise of signal A) and Bn be (1 plus the the amplitude noise of signal B).

Then let A = An* sin wt, and B = Bn * cos wt.

Multiplying these (Which is what a mixer does), you get An * Bn * sin wt * cos wt

applying the double angle identity, sin 2u = (2 sin u cos u) we get 0.5 (An * Bn) sin 2wt so for the quadrature case all of the amplitude nose ends up being mixed up to twice the carrier frequency and rejected by the measurement lowpass filter.

This makes the measured nose purely that due to phase noise and does not include the amplitude noise, providing the phase noise is small compared to the oscillator power (If it is not then the two can no longer be meaningfully said to be in quadrature!). 

Regards, Dan.

The upper sideband is at twice the frequency, the lower sideband is at baseband so it is not rejected by the lowpass filter.  I don't see that this rejects AM.

My impression is that the AM is rejected by saturating the mixer with both inputs:

Q. Why must the DBM be operated in a saturated mode when used as a phase
detector?
A. When sufficiently high input level signals are applied to the LO and RF ports, the
DBM operates in a saturated mode (as a limiter) and thus the IF output becomes nearly
independent of input signal level variations. Thus, the IF output of the DBM would have
a DC voltage proportional only to the phase difference between LO and RF inputs.

...

Q. Data sheets of phase detectors indicate that both input signals, RF and LO (or RF1
and RF2) should be equal in amplitude. What happens if they are not?
A. If the signal inputs to the LO and RF ports are of sufficient magnitude, the phase
detector will operate in a saturated mode and provide a DC output proportional to the
phase difference between the two signals. However, if the LO signal is of sufficient
amplitude but the RF signal is low in amplitude, the output of the phase detector will be
proportional to the amplitude as well as the phase of the RF input. 

https://www.minicircuits.com/app/AN41-001.pdf

[Kire Pûdsje]

Another way of looking at it is with phasor diagrams.
If we mix two signals, we get the difference of the frequencies.
A*exp(jwa+j phia)*B*exp(-jwb -j phib) = AB exp(j(wa-wb+phia-phib))
If both are at the same frequency:
AB exp(j(phia-phib))

If both signals are in phase, the phasor will point at +1.
If both signals are 90 deg out of phase, the phasor will point at +j
If both signals are 180 deg out of phase, the phasor will point at -1

Since we can 'only' measure real signals, we measure the projection on the real axis.
Now imagine what a perturbation will do to the projection

in phase, then only the length of the phasor (AM noise) will influence the projection
90 degrees, then only the angel of the phasor (PM noise) will influence the projection
180 degrees, then only the length of the phasor (AM noise) will influence the projection

[dmills]

Thats what I was getting at, in quadrature the AM noise cancels and all ends up around the sum term, only the PN remains, which is why this measurement is made in quadrature.

Modulation is one of those places where a good feel for polar form and phasor diagrams is really useful.

Regards, Dan.

[Kire Pûdsje]

Sorry, I was not implying your view was invalid.
You gave the example based on sine/cosines. I do agree that my example is exactly the same as yours. The 'other' view is more about thinking of it as phasors. Mathematically it is exactly equivalent, but visualizing the concept with phasors might help the OP, since he still did not grasp the idea. By 'projecting' on the real axis, the phasors degrade to the sine/cosines as described in your post.

[rfeecs]

If you had a full IQ mixer you could separate the magnitude and phase.

Here you are just mixing the test signal with a quadrature reference signal.

This gives you an output that is proportional to the phase.  But it is also proportional to the amplitude.  Your mixer output is still AM modulated.  You haven't removed the AM noise.

Of course there is one point where the two signals are exactly 90 degrees out of phase where you have zero AM.  That's just because the output is zero at that point.

As I mentioned, one way to suppress the AM is the saturate the mixer.

[Kire Pûdsje]

Yes, the output is dependant on amplitude. For phase noise measurement, compensating for the amplitude itself is usually not the hardest problem. For the measurements, in general you are not looking to the last dB. (unless you are just over the design limit).
AM and FM/PM noise are two orthogonal properties. by not having exactly the 90 degrees, you are measuring a combination of the two, as you state. Finding the right point where AM cancels is exactly the whole the point of a phase noise measurement setup.
Compression/saturation could also cause al kinds of other effects like for example AM/PM conversion, etc. and is in general not a good idea. I have seen a lot of cases, where the 1/f noise of the mixers due to the rectifying effect of the mixer was the major cause of AM noise. With saturation, you are really pushing the rectifying effect.

[dmills]

Thats a reasonable perspective, you can always use a limiting amplifier to remove AM, no so with PN.

Remember that both AM and PM are small at any reasonable offset, so for the purpose of first order analysis you can consider whichever one you don't care about to be zero providing your detection arrangements do not gratuitously amplify whichever one you are trying to cancel.

Consider, that while the measured amplitude of the PN in a quadrature measurement depends on the level of the measured signal, and so is really the product PN * (1 + AN) where the nominal carrier level is 1, AN the amplitude noise component will generally be very much smaller then the carrier level, so unless you have a spur at a really silly level or something the AN component can be ignored, similarly the in phase measurement of amplitude noise is really AN * (1 + PN) where providing PN is much smaller then 1 it can be ignored.

Regards, Dan.