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

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?

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.

• 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? Jan 18, 2022 at 23:27
• And $1/j = -j$. Jan 19, 2022 at 3:58
• @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. Jan 19, 2022 at 8:49