Shift Theorem

The shift theorem says that a delay of $\Delta$ samples in the time domain corresponds to a multiplication by $z^{-\Delta}$ in the frequency domain:

$${\cal Z}_z\{$$SHIFT$$_\Delta\{x\}\} \;=\; z^{-\Delta} X(z), \; \Delta\ge 0,$$

or, using more common notation,

$$\zbox {x(n-\Delta) \;\leftrightarrow\; z^{-\Delta} X(z), \; \Delta\ge 0.}$$

Thus, $x(\cdot - \Delta)$ , which is the waveform $x(\cdot)$ delayed by $\Delta$ samples, has the z transform $z^{-\Delta}X(z)$ .

Proof:

$$\begin{eqnarray*} {\cal Z}_z\{\mbox{{\sc Shift}}_\Delta\{x\}\} &\isdef & \sum_{n=0}^{\infty}x(n-\Delta) z^{-n} \\ &=& \sum_{m=-\Delta}^{\infty} x(m) z^{-(m+\Delta)} \qquad(m\;\isdef \; n-\Delta) \\ &=& \sum_{m=0}^{\infty}x(m) z^{-m} z^{-\Delta} \qquad \hbox{($\Delta\ge 0$, $x(n)=0$\ for $n<0$)}\\ \\ &=& z^{-\Delta } \sum_{m=0}^{\infty}x(m) z^{-m} \\ &\isdef & z^{-\Delta} X(z), % \quad\pfendmath \end{eqnarray*}$$

where we used the causality assumption $x(m)=0$ for $m<0$ .