1
$\begingroup$

I am a bit confused if the following signal is baseband or passband:

$f(t) = x(t) \cos(2 \pi f_c t) - j \: y(t) \sin(2 \pi f_c t),$ where $j = \sqrt{-1}$.

Since it is clearly not real, I cannot say it is passband. On the other hand, since it contains the frequency $f_c$, it should not be a baseband. Am I missing something?

Thanks,

Improve this question
$\endgroup$
3
$\begingroup$

Both real and complex signals can be passband.

Assuming $x(t)$ and $y(t)$ are baseband signals with bandwidth much less than $f_c$, then $f(t)$ is a passband signal, even if it is complex.

Improve this answer
$\endgroup$
10
  • $\begingroup$ But passband signals are what we transmit through the channel. How come they could be complex? $\endgroup$ – Noor Oct 5 '15 at 23:21
  • $\begingroup$ Indeed, any physical signal (including those transmitted over a channel) are real. That doesn't preclude a complex signal from being passband; it just precludes it from being physical. In other words, passband transmitted signals are a subset of all passband signals. $\endgroup$ – MBaz Oct 5 '15 at 23:49
  • 2
    $\begingroup$ Complex signals can be physically transmitted: they just have to be over two distinct channels as distinguishable real-valued signals. Many (most?) telecommunication systems transmit conceptually complex signals. $\endgroup$ – Peter K.♦ Oct 6 '15 at 0:22
  • $\begingroup$ @PeterK. Conceptually yes, but the actual signals are real. $\endgroup$ – MBaz Oct 6 '15 at 0:44
  • 1
    $\begingroup$ Understood! Isn't that what I said? ;-) $\endgroup$ – Peter K.♦ Oct 6 '15 at 2:24

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.