Lock-In Amplifier: Why multiplication with Cos and Sin?

I have two questions:

1. I was wondering, why a Lock-In Amplifier does multiply the incoming signal s(t) with both a cosine and a sine reference and not just a single one?

2. What problem may eventually arise, if you multiply the signal only with either cosine or sine ?

Trig theory to extract the phase & magnitude.

If you only care about phase...

Say you have a signal $Acos(\omega t + \phi)$ and you want to extract $\phi$. You can use an oscillator of the same frequency to extract this info BUT the issue is the phase.

$V_{sig} = Acos(\omega t + \phi)$

$V_{osc} = cos(\omega t)$

$V_{sig}V_{osc} = Acos(\omega t + \phi) Cos(\omega t)$

By the double angle identity:

$= \frac{1}{2}Acos(\phi) + \frac{1}{2}ACos(2 \omega t + \phi)$

a DC term relating to the phase can be realised as well as a component at twice teh freqency of the carrier. The phase can then be extracted by a moving average filter at the carrier frequency of a simple low pass filter.

$V_{sig}V_{osc}Filt = \frac{1}{2}Acos(\phi)$

If you care about phase and magnitude

To clearly extract the phase and magnitude then two oscillators, in quadrature, are required.

$V_{sig} = Asin(\omega t +\phi)$

$V_{osc0} = Xsin(\omega t)$

$V_{osc90} = Xcos(\omega t)$

$V_0 = Xsin(\omega t)Asin(\omega t +\phi) = \frac{XA}{2}(cos(\phi) - cos(2\omega t + \phi))$ $V_{90} = Xcos(\omega t)Asin(\omega t +\phi) = \frac{XA}{2}(sin(\phi) + sin(2\omega t + \phi))$

Filter these signals to remove the twice carrier component

$V_{0f} = \frac{XA}{2}(cos(\phi) )$

$V_{90f} = \frac{XA}{2}(sin(\phi) )$

via trig:

$\phi = atan( \frac{V_{90f}}{V_{0f}} )$ $A = \frac{2}{X}\sqrt{V_{0f}^2 + V_{90f}^2 }$