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Hey everyone, I'm stuck on this problem and I've got an exam in a few days. I think the filter H is used here so as to avoid aliasing artifacts, and afterwards went on to the second question.

I've got using my notebook

$g(n) = \int_{-\frac{1}{2}}^{\frac{1}{2}} G(e^{i2\pi v}) \cdot e^{i2\pi vn} \, dv$ $g(n) = \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{i\pi v(2n+1)} \, dv$ $g(n) = -\frac{1}{{i\pi(1+2n)}} \left(e^{i\pi(n+1/2)}-e^{-i\pi(n+1/2)}\right)$

But I have absolutely no idea where to go from here. Please help! Thank you.

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    $\begingroup$ hint: $\sin(x) = \frac{1}{2j}(e^{jx} - e^{-jx})$ $\endgroup$
    – Jdip
    Commented Nov 9, 2023 at 0:11
  • $\begingroup$ Thank you so much! I found that the filter was stable but not causal. Would you have any idea how to do the c) 1) ? $\endgroup$
    – Raidriar
    Commented Nov 9, 2023 at 0:47
  • $\begingroup$ I'll keep editing my answer with hints as you work your way through. $\endgroup$
    – Jdip
    Commented Nov 9, 2023 at 1:19
  • $\begingroup$ I'm finding $U(z) = \frac{1}{2} (X(z) + X(z)z^{-1}G(z^2)$ But I'm not sure how to express $G(z^2)$ Into $G(z)$ $\endgroup$
    – Raidriar
    Commented Nov 9, 2023 at 2:42
  • $\begingroup$ Try leaving it as $G(z^2)$ and move on to the next question $\endgroup$
    – Jdip
    Commented Nov 9, 2023 at 17:40

1 Answer 1

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b 1): hint: $\sin(x) = \frac{1}{2j}(e^{jx} - e^{-jx})$

c 1): hint: do it in the $\mathcal{Z}$ domain: $U(z) = \frac{1}{2} (X(z) + X(z)z^{-1} \cdots)$

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