# Unilateral Z transform

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

• Your last X(z) should be unilateral also. Apr 28 '20 at 22:39
• @Juancho so you'r basically saying Unilateral Z transform of something is equal to ...something...plus(U z transform of itself)? Apr 29 '20 at 0:05

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

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