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I have the next problem.

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$H_{c1}(j\omega )$ 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.

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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).

Considering that $H_{c1}$ is per your very own definition $\ne H_{c2}$ means that without restricting your system's input $r_c$, there's nothing you can do. You will have to implement a filter with transfer function $H_{c1}$ if you need a system with transfer function $H_{c1}$; there's no way around that.

Now, $H_{c1}$ being impossible with finitely long filters (as it is infinitely sharp in frequency domain), this is impossible.

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  • $\begingroup$ But couldn't $H_{c1}$ be expressed as a product of $H_{c2}$ and something else? It doesn't seem so impossible, at least to imagine. $\endgroup$ – Tendero Feb 9 '16 at 0:42
  • $\begingroup$ So, you can't allow aliases, because it's impossible to get rid of them. Hence the first filter needs to have reached its stop band at the Nyquist edge of the ADC $\endgroup$ – Marcus Müller Feb 9 '16 at 7:51
  • $\begingroup$ Hence, close to that, your signal will be distorted $\endgroup$ – Marcus Müller Feb 9 '16 at 7:52
  • $\begingroup$ Also, since $H_{c2}$needs to be infinitely long, no, it cannot be considered a product of two finitely long filters. $\endgroup$ – Marcus Müller Feb 9 '16 at 7:53
  • $\begingroup$ And what if $H(e^{j\Omega})$ was allowed to be infinitely long? Would it be possible then? $\endgroup$ – Tendero Feb 9 '16 at 15:07
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I just found this exact problem in Oppenheim-Schafer's Discrete Time Signal Processing (2nd edition). For those interested, it's exercise 4.56.

I've found in Internet the solutions for those problems. The solution for 4.56 that the book gives us is

enter image description here

So the definition of the filter written in the original question was correct.

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