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!

  • $\begingroup$ 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? $\endgroup$ – Notoriousphd Jul 17 '20 at 1:54


If you rewrite $y[n]$ as


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


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