What is the result of taking the real part and imaginary part of a complex signal in the frequency domain?

Question

Suppose that $g(t)$ is a lowpass complex signal with magnitude (solid line) and phase (dashed line)

To modulate $g(t)$ into a bandpass equivalent signal $f(t)$ with center frequency $f_c$, we compute the in-phase and quadrature components of $g(t)$ as \begin{align} g_I(t) &= \text{Re}\{g(t)\} \\ g_Q(t) &= \text{Im}\{g(t)\} \end{align} and then compute $$ f(t) = g_I(t)\cos(2\pi f_c t) - g_Q(t) \sin(2\pi f_c t) \tag{1} \label{eq1} $$ To make sense of \eqref{eq1}, I am curious to know what $g_I(t)$ and $g_Q(t)$ would look like in the frequency domain as a function of $G(f)$, the Fourier transform of $g(t)$ shown in the picture above. In other words, what does taking the real part and imaginary part do to what is shown in the picture above?

Answer 1

Since

$$g_I(t)=\frac12\big[g(t)+g^*(t)\big]$$

and

$$g_Q(t)=\frac{1}{2j}\big[g(t)-g^*(t)\big]$$

the corresponding Fourier transforms are

$$G_I(f)=\frac12\big[G(f)+G^*(-f)\big]$$

which is the even part of $G(f)$, and

$$G_Q(f)=\frac{1}{2j}\big[G(f)-G^*(-f)\big]$$

which is the odd part of $G(f)$ (times $1/j$).

It is easier to visualize what's happening if you write the bandpass signal as

$$f(t)=\textrm{Re}\left\{g(t)e^{j2\pi f_ct}\right\}\tag{1}$$

From $(1)$ you can see that you just shift the spectrum of the complex baseband signal to the center frequency $f_c$, and by taking the real part, you just get a mirror-image copy at $-f_c$, because - as explained above - the real part in the time domain corresponds to the even part in the frequency domain.