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

enter image description here

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?


1 Answer 1






the corresponding Fourier transforms are


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


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.

  • $\begingroup$ Thanks for the great answer! Two questions: (1) in your eq (1), shouldn't $g(t)e^{j2\pi f_ct}$ be multiplied by $2$ before taking the real part? and (2) I am guessing then that the demodulation process would involve expanding your eq (1) and then solving for $g(t)$ in terms of $f(t)$, correct? $\endgroup$
    – mhdadk
    Commented Jan 18, 2022 at 23:27
  • $\begingroup$ And $1/j = -j$. $\endgroup$ Commented Jan 19, 2022 at 3:58
  • 1
    $\begingroup$ @mhdadk: If you expand Eq. (1) then you get exactly your Eq. (1), so there's no factor 2 involved. Demodulation would mean to shift the spectrum down to DC (by multiplying $f(t)$ with a complex carrier). Then you'd apply a lowpass filter to remove the term at twice the carrier frequency. $\endgroup$
    – Matt L.
    Commented Jan 19, 2022 at 8:49

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