# What is the Z-transform of $0.8^{n+2}u(n-1)$?

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!

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

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?