What do I with the sampling period when I inverse $\mathcal Z$-transform?

Question

Given the input $$e(kT_s) = {0.4, 0.8,1.2, - 0.9}, \quad k = 0,1,2,3$$

and transfer function $$T\left(z^{-1}\right)=\frac {z^{-1}-0.8z^{-2}}{1-1.1z^{-1}+0.3}$$ Find the output $Y(kT_s),\quad k=0,1,2,3$

(I don't know why $0.3$ and $1$ haven't been added).

Starting with the inverse transform I found $$T(k)=\frac{1}{1.3}\left(\frac{1.1}{1.3}\right)^{k-1}u(k-1)-\frac{0.8}{1.3}\left(\frac{1.1}{1.3}\right)^{k-2}u(k-2)$$

I'm not sure if that is correct as I'm not so familiar with $\mathcal Z$-transform. Anyway, I don't have a sampling period in $T(k)$. Do I just put a $T_s$ next to $k$? Do I ignore $T_s$ and do the following? $$Y(0)=T(0)E(0)$$

$T(0)=0$ so $Y(0)=0$, but why am I given $0.4$ for $E(0)$? This has to be wrong. $$Y(1)=T(1)E(1)=\frac{1}{1.3}0.8=10.4 $$ etc etc.

Answer 1

There should probably be a $z^{-2}$ term with the $0.3$. It looks like a typo.

Once you've transformed to the $z$ domain (or discrete-time), and are not interested in transforming back to the continuous-time domain, there is no real need to carry the $T_s$. Just refer to $e[k], y[k]$, etc.