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

  • $\begingroup$ 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 '20 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 '20 at 15:45

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.

  • $\begingroup$ 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 '20 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 '20 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 '20 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.