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I have 2 signals. One is $x(n)=(-0.5)^nu(n)$ and the other one is $y(n)=0.8^{n+2}u(n-1)$. I know that for the first one it is $X(z)= 1/(1+0.5z^{-1})$, but what about the other one? I know $y(n)$ is time shifted but i don't know how to find this z-transform. Any help would be appreciated!

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  • $\begingroup$ Assuming this is homework, here's a hint: put everything in y[n] in terms of n-1 $\endgroup$
    – Juancho
    Jul 17 '20 at 0:02
  • $\begingroup$ Can i separate 0.8^2 Z(0.8^nu(n-1), and treat 0.8^n and u(n-1) separately? $\endgroup$ Jul 17 '20 at 1:54
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HINT:

If you rewrite $y[n]$ as

$$y[n]=(0.8)^3(0.8)^{n-1}u[n-1]\tag{1}$$

does it become easier to find the $\mathcal{Z}$-transform?

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