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?



1 Answer 1


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.

  • $\begingroup$ But passband signals are what we transmit through the channel. How come they could be complex? $\endgroup$
    – Noor
    Commented Oct 5, 2015 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
    Commented Oct 5, 2015 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.
    Commented Oct 6, 2015 at 0:22
  • $\begingroup$ @PeterK. Conceptually yes, but the actual signals are real. $\endgroup$
    – MBaz
    Commented Oct 6, 2015 at 0:44
  • 1
    $\begingroup$ Understood! Isn't that what I said? ;-) $\endgroup$
    – Peter K.
    Commented Oct 6, 2015 at 2:24

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