Shift Theorem for the DTFT

We define the shift operator for sampled signals $x(n)$ by

$$\hbox{\sc Shift}_{l,n}(x) \isdefs x(n-l)$$ (3.18)

where $l$ is any integer ( $l\in\mathbb{Z}$ ). Thus, $\hbox{\sc Shift}_l(x)$ is a right-shift or delay by $l$ samples.

The shift theorem states^3.5

$$\zbox {\hbox{\sc Shift}_l(x) \;\longleftrightarrow\;e^{-j(\cdot)l}X},$$ (3.19)

or, in operator notation,

$$\hbox{\sc DTFT}_\omega[\hbox{\sc Shift}_l(x)] \eqsp \left( e^{-j\omega l} \right) X(\omega)$$ (3.20)

Proof:

$$\begin{eqnarray*} \hbox{\sc DTFT}_\omega[\hbox{\sc Shift}_l(x)] &\isdef & \sum_{n=-\infty}^{\infty}x(n-l) e^{-j \omega n} \\ &=& \sum_{m=-\infty}^{\infty} x(m) e^{-j \omega (m+l)} \qquad(m\isdef n-l) \\ &=& \sum_{m=-\infty}^{\infty}x(m) e^{-j \omega m} e^{-j \omega l} \\ &=& e^{-j \omega l} \sum_{m=-\infty}^{\infty}x(m) e^{-j \omega m} \\ &\isdef & e^{-j \omega l} X(\omega) \end{eqnarray*}$$

Note that $e^{-j\omega l}$ is a linear phase term, so called because it is a linear function of frequency with slope equal to $-l$ :

$$\angle \left(e^{-j \omega l}\right) \eqsp -\omega l$$ (3.21)

The shift theorem gives us that multiplying a spectrum $X(\omega)$ by a linear phase term $e^{-j\omega l}$ corresponds to a delay in the time domain by $l$ samples. If $l<0$ , it is called a time advance by $\vert l\vert$ samples.