I have the next problem.
$H_{c1}(j\omega )$ is the ideal antialising filter and $H_{c2}(j\omega )$ is a real one. I'm asked to design $H(e^{j\Omega })$ so that $y[n]$ in the second diagram (the one on the right) is exactly the same as $x[n]$ in the first diagram (the one on the left).
I did something but I'm pretty sure that it isn't right. I found the equation for $H_{c2}(j\omega )$ as a function of $\omega$. Then I normalized the $\omega$-axis due to the A/D converter and divided the magnitude by $T$. If I didn't make any mistakes while doing the algebra, we get that
$$H_{c2}(e^{j\Omega })=\frac{9}{10}\cdot \frac{1}{\pi - \omega _{p}T}\cdot \Omega + \frac{9}{10}\cdot \frac{1}{\pi - \omega _{p}T}\cdot \pi + \frac{1}{10}\hspace{0.5cm} for\hspace{0.5cm} -\pi < \Omega < -\omega _{p}T$$
After that, I just thought of finding the inverse function of $H_{c2}(e^{j\Omega })$ for $-\pi < \Omega < -\omega _{p}T$... And that was it. Let $B(\Omega)=\frac{1}{\frac{9}{10}\cdot \frac{1}{\pi - \omega _{p}T}\cdot \Omega + \frac{9}{10}\cdot \frac{1}{\pi - \omega _{p}T}\cdot \pi + \frac{1}{10}}$. Then,
$$H(e^{j\Omega }) = \left\{ \begin{array}{l1} B(\Omega ) & \mbox{if } -\pi <\Omega <-\omega _{p}T \\ 1 & \mbox{if } -\omega _{p}T < \Omega<\omega _{p}T \\ B(-\Omega) & \mbox{if } \omega _{p}T < \Omega<\pi \end{array} \right.$$
The problem with this is that I'm pretty sure that the filter I "designed" has no antitransform. I think that there is a smarter way of approaching this exercise but I just can't see it.