Digital Phase Modulation baseband or passband?

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?

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.
• 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). – Dan Boschen Nov 7 '20 at 23:58