Shift Theorem

The shift theorem for Fourier transforms states that delaying a signal $x(t)$ by $\tau$ seconds multiplies its Fourier transform by $e^{-j\omega\tau}$ .

Proof:

$$\begin{eqnarray*} \hbox{\sc FT}_\omega(\hbox{\sc Shift}_\tau(x)) &\isdef & \int_{-\infty}^\infty x(t-\tau) e^{-j\omega t}dt\qquad\mbox{(define $\sigma=t-\tau$)}\\ &=& \int_{-\infty}^\infty x(\sigma) e^{-j\omega (\sigma+\tau)}d\sigma\\ &=& e^{-j\omega \tau}\int_{-\infty}^\infty x(\sigma) e^{-j\omega \sigma}d\sigma\\ &\isdef & e^{-j\omega \tau}X(\omega) \end{eqnarray*}$$

Thus,

$$\zbox {x(t-\tau)\;\longleftrightarrow\;e^{-j\omega \tau}X(\omega).}$$ (B.12)