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:

$$\begin{eqnarray*} \hbox{\sc FT}_\omega(\hbox{\sc Shift}_\tau(x)) &\isdef & \int... ...{-j\omega \sigma}d\sigma\\ &\isdef & e^{-j\omega \tau}X(\omega) \end{eqnarray*}$$

Thus,

$$\zbox {x(t-\tau)\longleftrightarrow e^{-j\omega \tau}X(\omega).}$$ (3.3)

For the inverse Fourier transform, we have

$$\begin{eqnarray*} \hbox{\sc IFT}_\omega(\hbox{\sc Shift}_\nu(X)) &\isdef & \fra... ...nfty X(\sigma) e^{j\sigma t}d\sigma\\ &\isdef & e^{j\nu t} x(t) \end{eqnarray*}$$

or,

$$\zbox {e^{j\nu t} x(t) \longleftrightarrow X(\omega-\nu).}$$ (3.4)