How does phase-sensitive detection eliminate noise?

[ergospherical]

Trying to get my head around how to use phase-sensitive detection to eliminate noise. If the input (say, $\cos{\omega_{\mathrm{in}} t}$) is modulated by a sinusoidal reference signal at $\omega_0$ with phase shift $\phi$, fed through the amplifier (gain $\kappa$) and then de-modulated by a square reference signal also at $\omega_0$ (but in phase with the input, say, for simplicity), then tracking the signal through the circuit should give:\begin{align*}
\cos{(\omega_{\mathrm{in}} t)} &\overset{\mathrm{modulate}}{\longrightarrow} \cos{(\omega_{\mathrm{in}} t)} \cos{(\omega_{0} t+ \phi)} \\ \\

&\overset{\mathrm{add \, noise}}{\longrightarrow} \cos{(\omega_{\mathrm{in}} t)} \cos{(\omega_{0}t+ \phi)} + \cos{(\omega_{\mathrm{noise}} t + \theta_{\mathrm{noise}})} \\ \\

&\overset{\mathrm{amplify}}{\longrightarrow} \kappa \cos{(\omega_{\mathrm{in}} t)} \cos{(\omega_{0} t+ \phi)} + \kappa \cos{(\omega_{\mathrm{noise}} t + \theta_{\mathrm{noise}})} \\ \\

&\overset{\mathrm{de-modulate}}{\longrightarrow} \kappa \cos{(\omega_{\mathrm{in}} t)} \cos{(\omega_{0} t+ \phi)} \mathrm{sgn}(\sin{(\omega_{0} t)}) +\kappa \cos{(\omega_{\mathrm{noise}} t + \theta_{\mathrm{noise}})} \mathrm{sgn}(\sin{(\omega_{0} t)}) \\ \\

\end{align*}Time-averaging over a cycle of the (de-)modulator,\begin{align*}
\langle \mathrm{output} \rangle &\sim \dfrac{\omega_0 \kappa}{2\pi} \int_{0}^{\pi / \omega_0} \left( \cos{(\omega_{\mathrm{in}} t)} \cos{(\omega_{0} t+ \phi)} + \cos{(\omega_{\mathrm{noise}} t + \theta_{\mathrm{noise}})} \right) dt \\ \\
&\quad\quad\quad\quad- \dfrac{\omega_0 \kappa}{2\pi} \int_{\pi/\omega_0}^{2\pi / \omega_0} \left( \cos{(\omega_{\mathrm{in}} t)} \cos{(\omega_{0} t+ \phi)} + \cos{(\omega_{\mathrm{noise}} t + \theta_{\mathrm{noise}})} \right) dt
\end{align*}
This is not a very nice integral, and also it's not clear why the noise should disappear except in the case that $\omega_{\mathrm{noise}} \gg \omega_{0}$. Even then, I'm not sure how the input signal is recovered...

 

[DaveE]

I'll skip the math for now, since I'm in a hurry. Plus you're better at that than me anyway. I have designed a few circuits like this; lock-in amplifiers and tracking stuff. There are, of course, multiple ways of thinking about this heuristically. One of my favorites is to think of this as mixing the desired signal down to DC. Of course one phase is rejected in that process. But the in-phase components make the DC output. So, the big advantage is that it's easier to design a really good LPF, to exclude what you don't want, than an equally good BPF at high frequency. It's hard to make an awesome BPF that accurately tracks the desired signal. It's easy to miss with the center frequency. The phase selectivity is also usually an important benefit if you are working with quadrature signals, 2-D tracking, etc.

Yes, it's a mess. The in-phase, in-frequency, noise isn't eliminated. For noise near the reference frequency it's really a filter design issue; it's mixed down to near DC. Also, I wouldn't say the input signal is recovered as much as measured. You could reconstruct it with the reference frequency, but I'm not sure why you would want that.

 

[tech99]

The demodulating reference signal has to be at carrier frequency, not modulating frequency. As DaveE says, it then allows us to use an LPF to restrict the noise bandwidth. The demodulating reference frequency has to be locked to the carrier frequency. For some purposes it can be unlocked, or slightly off frequency, but this introduces distortion of the demodulated signal.
Noise at the input can be thought of as two quadrature spaced carriers each modulated with noise, When we demodulate using a reference signal, we are demoduolating along one axis only, so the noise power is halved. The signal is not halved. If we use an envelope detector we do not have this 3dB noise advantage.

 

[Twigg]

It's your last equation, the integral expression, that needs a little tweaking.

Integrating from $\pi/\omega_0$ to $2\pi/\omega_0$ is the correct recipe for textbook Fourier stuff, but it's not how you get the best noise suppression. Instead, you want to integrate for as long a time T as possible, $T \gg \frac{2\pi}{\omega_0}$. We call T the "integration time." $$\langle \mathrm{output}(t) \rangle \propto \frac{1}{T} \int_t^{t+T} \left( \cos(\omega_{in} t' ) \cos (\omega_0 t' + \phi) + \cos(\omega_{noise} t' + \theta_{noise}) \cos (\omega_0 t' + \phi) \right) dt'$$ Notice that the noise is suppressed by a factor of $\frac{1}{\omega_{noise} T}$, whereas the signal amplitude stays constant with increasing T. The longer you integrate for, the better your signal-to-noise. (Edit: As @DaveE says, the in-frequency ($\omega_{noise} = \omega_{0}$), in-phase ($\theta_{noise} = \phi$) noise doesn't average down.)
Edit Edit: Fixed some missing stuff in the integral.
Edit edit edit: Fixed some subscripts

Also, this noise-suppressing circuit is famous and has a name: it's called a lock-in amplifier. It's one of very few pieces of equipment that experimentalists in any field will recognize, because it's so darn useful.

 

[ergospherical]

Thanks, this helps. If the noise injected before the lock-in goes something like $\sim \cos{((\omega_0 + \delta \omega)t + \theta_{\mathrm{noise}})}$ then the noise output from the lock-in should be amplified and down-shifted in frequency i.e. $\sim \kappa \cos{((\delta \omega) t + \theta_{\mathrm{noise}})}$

So the questions are a) why does the off-carrier noise get removed and b) what happens to the noise at $\omega_0$? For a), if $\delta \omega \gg 1/RC$ then time-averaging over the noise output (over large integration time, as alluded to by @Twigg) should make it go to zero. It follows also that making $RC$ really big should narrow the frequency band of noise which isn't removed.

I'm not sure about b). Is it possible to get rid of the in-frequency noise?

 

[Twigg]

I'm going to start with question (b) cause it's simpler.

I'm not sure about b). Is it possible to get rid of the in-frequency noise?

No. You need information to discriminate between signal and noise. In lock-in amplification, that information is the frequency of the signal. You'd need some other source of information to distinguish between signal and noise at the same frequency.

a) why does the off-carrier noise get removed

Let me make one change to the formalism. Let me introduce an amplitude for the input signal, so $\cos(\omega_{in} t) \rightarrow A_{in} \cos(\omega_{in} t)$, and let me introduce an amplitude for the noise, so $\cos(\omega_{noise} t + \theta_{noise}) \rightarrow A_{noise} \cos(\omega_{noise} t + \theta_{noise})$. When you evaluate the integral I wrote (and you ignore frequency components at ($2\omega_0$), you get something of the form
$$\langle \mathrm{output}(t) \rangle \propto \frac{1}{2} A_{in} + \frac{A_{noise}}{2T(\omega_0 - \omega_{noise})} \left[ \sin((\omega_0 - \omega_{noise})(t+T) + \phi - \theta_{noise}) - \sin((\omega_0 - \omega_{noise})t + \phi - \theta_{noise})\right]$$
Pulling out just the amplitudes, the final signal-to-noise ratio is $(\omega_0 - \omega_{noise})T\left( \frac{A_{in}}{A_{noise}} \right)$. What's going on is that frequency components that are not synchronized with the modulation frequency $\omega_0$ and phase $\phi$ integrate to a quantity that scales like $\frac{1}{T}$. When the two sinusoids are synchroized, their product has a DC component (i.e., $\cos^2(x) = \frac{1}{2} + \frac{1}{2} \cos(2x)$). It's the same idea as taking a Fourier integral.

 

[ergospherical]

Ahhh, I see! The key idea really is that of orthogonal function decomposition, with the off-carrier frequencies being integrated out to zero. Gonna need a bit of time to get everything straight but I think I'm on the right path now. Thanks!

 

[tech99]

Also notice that if the signal varies in phase, and if the reference frequency is at the same time held constant, we obtain a voltage output which varies with the carrier phase. This is why it is called a phase sensitive detector. However, notice that the output is not linearly proportional to phase.