Removing $1/f$ noise with lock-in-amplifiers

Question

On the Wikipedia page for 1/f noise (at the bottom of the page) it suggests the noise can be reduced if the signal of interest is at DC. DC signals suffer from significant 1/f noise, so one method of removing this is to modulate the signal at some higher frequency and use a lock-in-amplifier to detect the now modulated signal synchronously. However, to my understanding, lock-in-amplifiers then convert the modulated signal back into a DC signal, surely again limited by the 1/f noise as before, so how is this method helpful for DC signals?

Answer 1

Say the original signal is $y$ with noise spectrum $s(f)$. Let $A$ be the gain of an amplifier, and let $z(f)$ be the noise introduced by the amplifier at the output. A d.c. amplifier of gain $A_{dc}$ would produce at its output approximately $$ A_{dc} (y + s(0)) + z(0). $$ There is more than one way to modulate the signal so as to use a lock-in amplifier. One way is to chop the signal, so that it appears and disappears altogether at the chosen frequency and phase. In this case the output is approximately $$ A\left( y + s(f)\right) + z(0). $$ Another method is to modulate some parameter $x$, in which case the output is something like $$ A\left( \frac{dy}{dx} + s(f)\right) + z(0). $$ Broadly speaking, the d.c. amplifier amplifies the d.c. noise whereas the lock-on amplifier amplifies the noise at the chosen frequency $f$. So if the latter is smaller (which it almost always is if you pick $f$ sensibly) then the lock-in method is superior. In practice you would only use the amplifier when $z(0)$ does not dominate $A s(f)$.

(Thanks to Jamie1989 for pointing out an omission in the first version of this answer.)