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I was wondering what is the physical meaning of IN-PHASE and QUADRATURE components?
In the context of low-pass equivalent of band-pass signals.
Any help is appreciated!

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i will call the bandpass signal:

$$\begin{align} y(t) &= \Re \Big\{ (x(t) + j\hat{x}(t)) \, e^{j \omega_0 t} \Big\} \\ \\ &= x(t) \cos(\omega_0 t) - \hat{x}(t) \sin(\omega_0 t) \\ \end{align}$$

the complex-valued $x(t) + j \hat{x}(t)$ is the low-pass equivalent to the bandpass signal.

the real part $x(t)$ is the in-phase component and the imaginary part $\hat{x}(t)$ is the quadrature component.

the in-phase and quadrature components may have nothing to do with each other and then the spectrum of the low-pass equivalent is arbitrary with no symmetry implied.

if $\hat{x}(t)=0$, the low-pass equivalent is purely real and spectrum of it has hilbert symmetry with the real and imaginary parts.

if the in-phase and quadrature components are related by the Hilbert Transform,

$$\begin{align} \hat{x}(t) &= \mathscr{H} \Big\{ x(t) \Big\} \\ \\ &= \lim_{\epsilon \to 0} \ \int\limits_{-\infty}^{t-\epsilon} x(u) \frac{1}{t-u} du \ + \ \int\limits_{t+\epsilon}^{+\infty} x(u) \frac{1}{t-u} du \\ \end{align}$$

then the spectrum of the low-pass equivalent is zero for all negative frequencies and has only a positive-frequency component.

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