# A communication system

I am solving the following problem as an exercise. I have a signal $$z(t) = x(t)\cos(wt) + y(t)\cos(ut)$$

and we assume $$x,y$$ to be band-limited continuous time signals. In the frequency domain we have that both signals are $$0$$ if the frequency $$\omega \geq \omega_{max}$$. We try to reconstruct $$x$$ by passing $$r$$ through a bandpass filter $$B(\omega)$$ that is $$1$$ if $$\omega$$ is $$\omega = w$$ or $$\omega = -w$$ (it actually is $$1$$ in an interval centered around $$w$$ and $$-w$$ but I didn't know how to write it since I have no information about the width of the interval. The outcome is then multiplied by $$\cos(wt)$$ and then passed through a lowpass filter $$L(\omega)$$ which is $$2$$ if $$-\omega_{max}\leq \omega\leq \omega_{max}$$. What do we need to assume on $$w,u$$ in order for $$x$$ to be reconstructed?

In the frequency domain, I have that $$Z(\omega)= \frac12\big(X(\omega+w)+X(\omega-w)+Y(\omega-u)+Y(\omega+u)\big)$$ So picotrially I have 2 copies of $$X$$ centred at $$w$$ and $$-w$$ and two copies of $$Y$$ centred at $$-u$$ and $$u$$ with half the amplitude, right?

Now, the first bandpass filter should give me only the two copies of $$X$$, since it is centred around $$w$$, then there is this multiplication with $$\cos(wt)$$ again. What does this do? Doesn't it simply make yet another two copies of $$X$$ centred around $$w$$? Does it give me back the original amplitude getting rid of the $$1/2$$? Then, how can a lowpass filter allow me to reconstruct the signal when I have $$2$$ copies of it?

Let me slightly change notation and denote the two modulation frequencies by $$\omega_1$$ and $$\omega_2$$. You need $$|\omega_1-\omega_2|>2\omega_{max}$$ to be satisfied, otherwise the shifted spectral of $$x(t)$$ and $$y(t)$$ will overlap. If this condition is satisfied, you can retrieve $$x(t)$$ by demodulation (multiplication with $$\cos(\omega_1t)$$) and (scaled) low pass filtering, as you suggested:
\begin{align}2x(t)\cos^2(\omega_1t)=x(t)\big[1+\cos(2\omega_1t)\big]\Big|_{low pass}=x(t)\end{align}\tag{1}