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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?

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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:

$$\begin{align}2x(t)\cos^2(\omega_1t)=x(t)\big[1+\cos(2\omega_1t)\big]\Big|_{low pass}=x(t)\end{align}\tag{1}$$

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