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A band pass signal representation goes by the generalization as $X(t)=XX(t)*e^{j \cdot2\pi \cdot ft}$ where $f$ be the carrier freq. and $XX(t)$ be the complex envelope. On further decomposition it boils down to: $X(t)=XXi(t)\cos(2\pi ft)-XXq(t)\sin(2\pi ft)$
Here is it implicit that $XXi(t)$ and $XXq(t)$ are real functions and their spectra be even symmetric?
$X(t) = Re\{XX(t)e^{j2\pi ft}\}$. So $XX(t)$ can be any generic complex baseband signal with real I and Q components - $XX(t) = XX_i(t)+j XX_q(t)$ and $e^{j2\pi ft} = \cos(2\pi ft) + j\sin(2\pi ft)$
After the complex multiplication, and taking real part, you get $X(t) = XX_i(t)\cos(2\pi ft) - XX_q(t)\sin(2\pi ft)$.
So $XX_i(t)$ and $XX_q(t)$ are indeed real functions and their spectra is even-symmetric.
If $\hat{X}(t)$ is the Hilbert transform of $X(t)$, then $X^+(t) = X(t)+j\hat{X}(t)$. $XX_i(t)=X(t)\cos(2\pi ft)+\hat{X}(t)\sin(2\pi ft); XX_q(t)=\hat{X}(t)\cos(2\pi ft)-X(t)\sin(2\pi ft)$
You have two different definitions of $X(t)$.
$X'(t)=XX(t)*e^{(j2\pi ft)}$ is the tidy theoretical shorthand way to express $X(t)$ -- but note the prime.
The real-valued version is $X(t)=XXi(t)\cos(2\pi ft)-XXq(t)\sin(2\pi ft)$
The difference is that $X'(t)$ does not have any energy in $\omega < 0$, where $X(t)$ has energy centered around $\omega = 2 \pi f$ and $\omega = -2 \pi f$.
The assumption is that $f$ is much greater than the bandwidth of $XX(t)$, so that when you recover a version of $XX(t)$ from $X(t)$ using I/Q demodulation, filtering out the components around $2f$ is trivial. When this assumption holds, you can treat the inphase and quadrature components as the real and imaginary parts of $XX(t)$ for the purposes of processing the signal.
