Shift Theorem

Theorem: For any $x\in{\bf C}^N$ and any integer $\Delta$,

$$\zbox {\hbox{\sc DFT}_k[\hbox{\sc Shift}_\Delta(x)] = e^{-j\omega_k\Delta} X(k).}$$

Proof:

$$\begin{eqnarray*} \hbox{\sc DFT}_k[\hbox{\sc Shift}_\Delta(x)] &\isdef & \sum_{n... ...}x(m) e^{-j 2\pi mk/N} \\ &\isdef & e^{-j \omega_k \Delta} X(k) \end{eqnarray*}$$

The shift theorem is often expressed in shorthand as

$$\zbox {x(n-\Delta) \leftrightarrow e^{-j\omega_k\Delta}X(\omega_k).}$$

The shift theorem says that a delay in the time domain corresponds to a linear phase term in the frequency domain. More specifically, a delay of $\Delta$ samples in the time waveform corresponds to the linear phase term $e^{-j \omega_k \Delta}$ multiplying the spectrum, where $\omega_k \isdeftext 2\pi k/N$. (To consider $\omega_k$ as radians per second instead of radians per sample, just replace $\Delta$ by $\Delta T$ so that the delay is in seconds instead of samples.) Note that spectral magnitude is unaffected by a linear phase term. That is, $\left\vert e^{-j \omega_k \Delta}X(k)\right\vert = \left\vert X(k)\right\vert$.