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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?

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$X(t) = Re\{XX(t)e^{j2\pi ft}\}$. 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 $X^+(t) = X(t)+j\hat{X}(t)$. $XX_i(t)=X(t)\cos(2\pi ft)+\hat{X}(t)\sin(2\pi ft); XX_q(t)=\hat{X}(t)\cos(2\pi ft)-X(t)\sin(2\pi ft)$

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  • $\begingroup$ Hi Jithin,pardon me for not knowing Latex,but i wish some clarification on this: you see XX(t)=XXi(t)+jXXq(t); then X+(t)=X(t)+jX^(t) and X-(t)=X(t)+jX^(t) which are the +ve and -ve pre envelopes....on top that XX(t) is defined by XX(t)=(X+(t))e^(-2pift) as we consider it be a generalized low pass complex signal. So does it mean XXi(t)=X(t)cos(2pift)+X^(t)sin(2pift); XXq(t)=X^(t)cos(2pift)-X(t)sin(2pift) $\endgroup$ – shubhayan de Apr 11 at 15:19
  • $\begingroup$ Yes you are correct $XX_i(t) = X(t)\cos(2\pi ft) + \hat{X}(t)\sin(2\pi ft)$. $XX_q(t) = \hat{X(t)}\cos(2\pi ft) - X(t)\sin(2\pi ft)$ $\endgroup$ – jithin Apr 11 at 15:45
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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.

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  • $\begingroup$ Hi Tim,Thanks for your reply.I would like to know whats the relation among the inphase,quadrature phase with the corresponding +ve and -ve pre envelopes...as i have seen two sources which raise some confusion $\endgroup$ – shubhayan de Apr 11 at 15:09
  • $\begingroup$ I'm not sure where your "+ve and -ve pre envelopes" come from -- I certainly don't see them in the context of this question. Perhaps you should ask a new question. $\endgroup$ – TimWescott Apr 11 at 16:37
  • $\begingroup$ Hi Tim,I have mentioned the details in the comment on the other answer by @jithin ,also I referred this "site.uottawa.ca/~damours/courses/ELG_3175/Lec6.pdf" $\endgroup$ – shubhayan de Apr 11 at 17:12

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