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I tried to calculate the unilateral Z transform of x[n-2], is it right?

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  • $\begingroup$ Your last X(z) should be unilateral also. $\endgroup$
    – Juancho
    Apr 28 '20 at 22:39
  • $\begingroup$ @Juancho so you'r basically saying Unilateral Z transform of something is equal to ...something...plus(U z transform of itself)? $\endgroup$ Apr 29 '20 at 0:05
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There is nothing wrong with your calculation. However, by definition, $x[n]=0$ for $n<0$, because otherwise we couldn't have used the unilateral $\mathcal{Z}$-transform in the first place. So you get

$$\mathcal{Z}\big\{x[n-k]\big\}=x^{-k}X(z),\quad k> 0\tag{1}$$

Eq. $(1)$ is valid for the unilateral as well as for the bilateral $\mathcal{Z}$-transform.

Things are different for a time advance. The unilateral $\mathcal{Z}$-transform of $x[n+k]$, $k>0$, is

$$\begin{align}\mathcal{Z}\big\{x[n+k]\big\}&=\sum_{n=0}^{\infty}x[n+k]z^{-n}\\&=\sum_{n=k}^{\infty}x[n]z^{-(n-k)}\\&=z^k\left(X(z)-\sum_{n=0}^{k-1}x[n]z^{-n}\right),\quad k>0\tag{2}\end{align}$$

For the bilateral $\mathcal{Z}$-transform we simply have

$$\mathcal{Z}\big\{x[n+k]\big\}=z^kX(z)\tag{3}$$

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