Are the components of a Bandpass signal namely “inphase” and “quadrature” real functions in time domain?

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

• 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) – shubhayan de Apr 11 '20 at 15:19
• Yes you are correct $XX_i(t) = X(t)\cos(2\pi ft) + \hat{X}(t)\sin(2\pi ft)$. $XX_q(t) = \hat{X(t)}\cos(2\pi ft) - X(t)\sin(2\pi ft)$ – 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.

• 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 – shubhayan de Apr 11 '20 at 15:09
• 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. – TimWescott Apr 11 '20 at 16:37
• 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" – shubhayan de Apr 11 '20 at 17:12