# Effect of Lock-in Amplifier Phase Noise

Suppose that an oscillator with phase noise is used to drive a modulating process and that the same oscillator is used (with an appropriate phase shift) is used to synchronously demodulate the signal from a detector monitoring the process. I will assume that the oscillator frequency is sufficiently low that the system contributes no relative phase noise between the signal and the reference (i.e. the system is perfectly coherent)

Since the reference modulating waveforms are identical (albeit noisy), will the oscillator phase noise have any influence on the demodulated signal? Does the linewidth and pedestal of the oscillator used affect the SNR of the lock-in detector?

Intuitively I feel that there is no effect, however, most lock-in equipment proclaim exceptionally low phase-noise characteristics.

• Would Electrical Engineering be a better home for this question? Commented Nov 16, 2017 at 19:32

Let your oscillator signal be $r(t)=A(1+m(t))cos(\omega t +\phi(t))$; here $m(t)$ and $\phi(t)$ are the AM (amplitude modulation) and PM (phase modulation) noise processes. If you use this $r(t)$ to detect synchronously another signal, say, $x(t)=B(t)cos(\omega t +\theta(t))$ that is derived from the oscillator then aside from other noise sources you have $$x(t)=B(t)cos(\omega t +\theta(t)) = A(1+m(t-\tau))cos(\omega (t-\tau) +\phi(t-\tau) + \theta_0),$$ where $\tau$ is a time delay and $\theta_0$ is a phase delay, both independent of all the other quantities. Synchronous detection really just what nowadays is more frequently called IQ detection, you multiply the reference and its $\pi/2$ shifted version with the incoming signal and low pass filter the baseband part, then th result is further processed, mostly digitally. The result has two terms $$c(t) \propto cos(\mu(t)+\theta_1) = cos(\phi(t)-\phi(t-\tau)+\theta_1)$$ and $$s(t) \propto sin(\mu(t)+\theta_1))=sin(\phi(t)-\phi(t-\tau)+\theta_1)$$ where $\theta_1=\omega \tau +\theta_0$ is a fixed phase shift and AM noise is ignored.

Notice the phase noise shows up in the expression as $\mu(t)=\phi(t)-\phi(t-\tau)$. For small delays, $\mu(t) \approx \dot\phi(t)\tau$, and here $\dot\phi(t)$ is what conventionally called the FM noise of the oscillator. In short, smaller the delay $\tau$ is the less the effect of the phase noise will be on the measurement. I have ignored the AM noise but usually in a mixer (the synchronous detector - multiplier) there can be AM to PM conversion effects, and these can be quite painful to eliminate if very low noise measurements are needed. Usually, the easy thing to do is to follow the oscillator with a very good amplitude limiter to avoid such noise conversion.

• In your signal x(t) you are not adding any information from the process being probed, just adding a constant phase shift \theta_0 to the time delayed signal?
– Mike
Commented Nov 16, 2017 at 20:00
• I guess I don't understand why you include another phase shift \theta_o in addition to the \omega\tau delay.
– Mike
Commented Nov 16, 2017 at 20:12
• Yes, I left out all modulations (e.g., chopping) to simplify but you can add that in by having $A=A(t)$ and replace $\omega t$ with $\omega t + \psi(t)$ where $A(t)$ and $\psi(t)$ are given functions. The effect of the delay is deterministic and can be explicitly calculated unlike on the stochastic process $\phi(t)$. Commented Nov 16, 2017 at 20:26
• I added the fixed phase shift $\theta_0$ because it is always there. It is not caused by the transfer delay through the circuits, it is a "wideband" essentially frequency independent phase shift that must be compensated for in a synchronous receiver that has only one channel for if that phase wrong you get nothing. In fact, this is the origin of the IQ (2-channel) receiver having outputs in phase quadrature, so you do not have to compensate for that phase shift. Using both channels you get the full picture, so to speak. Commented Nov 16, 2017 at 20:31
• I see, $\theta_0$ is just the knob that nulls out the $\theta_1$. For $\lambda \gg$ system scale the phase noise seems negligible, however, the amplitude noise on the oscillator can be a big problem. $c(t)$ seems to have an $(1+m(t))^2$ term! I have a much better model of this problem now, thank you.
– Mike
Commented Nov 16, 2017 at 21:09