Say I want to transmit messages using PSK:

I send the message

$$s_m(t) = \text{Re} \left [g(t) e^{\frac{j 2 \pi (m-1)}{M}}e^{j 2 \pi f_c t} \right]$$

where $f_c$ is the carrier frequency. Now my question is,

  • is this a baseband representation or a passband?
  • So is $f_c$ the center frequency of a certain channel? Or would we need to modulate $s_m(t)$ to the channel frequency?
  • So then if we sent $s_m(t) \text{cos}(2 \pi f_b)$ where $f_b$ is the center frequency of the channel we are communicating on, at the receiver we would need to convert the modulated signal back to its low pass equivalent $s_m(t)$ so that we can apply a detection algorithm to demodulate the signal?

1 Answer 1


Assuming $g(t)$ is a baseband signal, $s_m(t)$ is passband. You can see that it is by multiplying the two complex exponentials, expanding the result into sine and cosine terms, and then taking the real part.

An interesting thing about PSK signals is that they are an example of quadrature modulation, and in consequence, you can represent them either as real passband signals, or complex baseband signals. If you assume $s_m(t)$ is a PSK signal, and given that it is real, it must also be passband.

  • $\begingroup$ We can also have complex passband signals, right? Given any baseband signal $f_b(t)$ multiplied by $e{j f_c t}$ would then be a complex passband signal (as commonly done in digital IF processing of waveforms). $\endgroup$ Nov 7, 2020 at 23:58
  • $\begingroup$ Sure, it's possible. In my experience, complex passband signals are only useful when writing the upconversion/downconversion processes mathematically. $\endgroup$
    – MBaz
    Nov 8, 2020 at 1:24
  • $\begingroup$ An application that is used for purpose of simplifying filtering is to split the analog IF signal with a quadrature splitter (resulting in a complex analog IF) which is then quadrature sampled into a complex digital IF. The anti-alias filters can have twice the transition band as half the aliases are eliminated. You could also describe single sideband (quadrature mixers) as representing complex passband signals. $\endgroup$ Nov 8, 2020 at 2:51
  • $\begingroup$ "complex analog IF" -- I guess it's a matter of preference, or maybe I'm just nitpicky, but I wouldn't call such a signal "complex", since it's actually two real signals. But indeed I see that'd be a great application of the concept. Once you realize signals can be rotated by any angle, interesting techniques become possible. $\endgroup$
    – MBaz
    Nov 8, 2020 at 3:02
  • $\begingroup$ when I say I + jQ, or Ke^j theta, I, Q, K and theta are all real signals! In general, math included, it takes two real signals to describe a complex signal (hence as I describe it “takes two scope probes” to measure a complex signal in the lab—- and my typical counter to anyone that professes that complex signals somehow don’t exist but real signals do. They are both equally mathematical descriptions. $\endgroup$ Nov 8, 2020 at 3:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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