# Is it possible to obtain a complex lowpass signal directly from its equivalent bandpass signal?

Given the complex lowpass signal $$x_l(t)$$, its real bandpass equivalent signal with center frequency $$f_c$$ is \begin{align} x_b(t) &= \text{Re}\{x_l(t) \cdot e^{j2\pi f_c t}\} \\ &= \frac{1}{2}\left(x_l(t) \cdot e^{j2\pi f_c t} + x_l^*(t) \cdot e^{-j2\pi f_c t}\right) \label{eq1} \tag{1} \end{align} Similarly, given a real bandpass signal $$x_b(t)$$ with center frequency $$f_c$$, its complex lowpass equivalent signal is \begin{align} x_l(t) &= x_+(t) \cdot e^{-j2\pi f_c t} \label{eq2} \tag{2} \end{align} where $$x_+(t)$$ is the pre-envelope of $$x_b(t)$$ defined as $$x_+(t) = x_b(t) + j\hat{x}_b(t)$$ where $$\hat{x}_b(t)$$ is the Hilbert transform of $$x_b(t)$$ and $$x_+(t)$$ has the Fourier transform $$X_+(f) = \begin{cases} 2X_b(f), &f>0 \\ X_b(f), &f = 0 \\ 0, &f < 0 \end{cases}$$ I am trying to find a way to relate equation \eqref{eq1} directly to equation \eqref{eq2}. Is it possible to solve equation \eqref{eq1} directly for $$x_l(t)$$ to get what is shown in equation \eqref{eq2}? Here is my initial attempt: \begin{align} x_b(t) &= \frac{1}{2}\left(x_l(t) \cdot e^{j2\pi f_c t} + x_l^*(t) \cdot e^{-j2\pi f_c t}\right) \\ 2x_b(t) &= x_l(t) \cdot e^{j2\pi f_c t} + x_l^*(t) \cdot e^{-j2\pi f_c t} \\ \end{align} but this doesn't seem to lead anywhere. I also tried to do it in the frequency domain: \begin{align} X_b(f) &= \frac{1}{2}\left(X_l(f - f_c) + X_l^*(-f - f_c)\right) \\ 2X_b(f) &= X_l(f - f_c) + X_l^*(-f - f_c) \\ 2X_b(f) &= X_l(f - f_c) + X_l^*(-(f + f_c)) \end{align} We can then consider the three cases $$f > f_c, f = f_c,$$ and $$f < f_c$$ for the last line such that $$\begin{cases} 2X_b(f) &= X_l(f - f_c) + X_l^*(-(f + f_c)), &f > f_c \\ 2X_b(f_c) &= 2X_l(0), &f = f_c \\ 2X_b(f) &= X_l(f - f_c) + X_l^*(-(f + f_c)), &f < f_c \end{cases}$$ However, I am not sure how to proceed from here.

• I have an answer that shows moving a baseband (or lowpass) filter to a bandpass here: dsp.stackexchange.com/a/68599/26009 Commented Jan 27, 2022 at 19:20

If you realize that the second term on the right-hand side of Eq. $$(1)$$ is only non-zero for negative frequencies, you note that you can remove that term using a phase splitter, i.e., a complex-valued filter with zero magnitude at negative frequencies and a constant magnitude (of value $$2$$) at positive frequencies. Such a system can be realized using a Hilbert transformer. The complex low pass signal is then obtained by complex demodulation, i.e., by multiplication with $$e^{-j2\pi f_ct}$$.
Another way of doing the same thing is to first demodulate with a complex carrier and then apply a low pass filter to remove the term around $$-2f_c$$.