Why is no phase information lost when a received signal goes through a mixer?

Question

I'm sorry if the question is too philosophical or makes no sense, but I need to be able to explain why it doesn't make sense. When we take the in-phase and quadrature-phase base-band components of an intermediate frequency signal, it's very easy to show that any one of the two orthogonal components alone would be insufficient to distinguish the variations in a single-carrier phase from variations in its amplitude. This tutorial makes it quite clear. The Q component, although real, carries the imaginary component of the phasor $A(t)\exp(2\pi f_I\cdot t + \phi(t) )$.

However, when we pass the RF signal through a mixer, there is no Q component lost. No phase modulation will be lost in the base-band signal obtained later. Why not? A figure to help create context.

Answer 1

The receiver's mixer introduces a phase-shift to the in-phase and quadrature components. In very general terms, assume we have a signal $$a(t)e^{j(2\pi f_0 t + \phi(t))},$$ which is mixed with a local oscillator of frequency $f_{lo}$ and phase $\delta$: $$e^{j(2\pi f_{lo}t + \delta)}.$$ The output signal is

\begin{align} a(t)e^{j(2\pi f_0 t + \phi(t))}e^{j(2\pi f_{lo}t + \delta)} &= a(t)e^{j(2\pi (f_0+f_{lo})t + \phi(t) + \delta)} \\ &= a(t)e^{j(2\pi (f_0+f_{lo})t + \phi(t))} e^{j\delta}. \end{align}

In other words, even though the frequency has been shifted and a phase-shift equal to $\delta$ has been introduced, $a(t)$ and $\phi(t)$ are still present in the signal and that means that all the information carried by the original signal is still there.

Usually the phase-shift is corrected in the digital back-end in the receiver, or you can use a differential encoding (such as DBSK), in which the phase-shift becomes irrelevant.