Z-transform of $\sin{(n\frac{\pi}{2}(-1)^{n})}$

Question

I was trying to do this z transform:

$\sin{(n\frac{\pi}{2}(-1)^{n})}$

but I don't have the solution. I have divided in 4 cases:
for n=0 0
for n=1 -1
for n=2 0
for n=3 1

so 4k, 4k+1, 4k+2 and 4k+3

But now I don't know how to proceed and I don't know if my result is right.
Any help would be appreciated.

Answer 1

You already observed that this signal $x[n]$ is periodic: $x[n+4]=x[n]$ for all integers $n$. Therefore we get $$\begin{align*} \mathcal{Z}\{x[n]\} =\sum_{k=0}^\infty x[k]x^k z^{-k} &= 0 + (-1)z^{-1} + 0 z^{-2} + 1 z^{-3} + \sum_{k=4}^\infty x[k]z^{-k} \\ &= -z^{-1}+z^{-3} + z^{-4} \sum_{k=0} x[k]z^{-k}. \end{align*} $$ The last term is equal to $z^{-4}\mathcal{Z}\{x[n]\}$, so from this equation you can distill the $\mathcal{Z}$-transform.