Justin Kietzman - Time Shifting property of F.T and you ECE301Fall2008mboutin - Rhea

Time Shifting Propery

$ F(x(t-t_0)) = e^{-jwt_0}X(w) $

Proof:

$ F(x(t-t_0)) = \int_{-\infty}^{\infty}x(t-t_0)e^{-jwt}dt $

Let $ (t-t_0) = \tau $

Therefore $ d\tau = dt $

$ = \int_{-\infty}^{\infty}x(\tau)e^{-jw(\tau + t_0)}(d\tau) $

$ = \int_{-\infty}^{\infty}x(\tau)e^{-jw\tau}e^{-jwt_0}(d\tau) $

Since $ e^{-jwt_0} $ does not depend on "$ \tau $" it can be taken out of the integral.

$ =e^{-jwt_0} \int_{-\infty}^{\infty}x(\tau)e^{-jw\tau}(d\tau) $

Thus

$ = e^{-jwt_0}X(w) $

Example

Let's take the 2nd question from the Group Quiz.

Prove that $ e^{-jw}X(-w) = F(x(-t +1 )) $

Solution 1

We can solve by the definition of a Fourier Transform

$ F(x(-t + 1)) = \int_{-\infty}^{\infty}x(-t + 1)e^{-jwt}dt $

$ = \int_{\infty}^{-\infty}x(\tau)e^{-jw(-\tau + 1)}(-d\tau) $

Notice the bounds have changed

$ = \int_{-\infty}^{\infty}x(\tau)e^{-jw(-\tau + 1)}(d\tau) $

$ = \int_{-\infty}^{\infty}x(\tau)e^{jw\tau}e^{-jw}(d\tau) $

$ =e^{-jw} \int_{-\infty}^{\infty}x(\tau)e^{-j(-w)\tau}(d\tau) $

$ F(x(-t + 1)) = e^{-jw}X(w) $

Solution 2

$ F(x(-t+1)) = ? $

By the Time Shifting Property proven above

$ = e^{-jw}F(x(-t)) $

$ = e^{-jw}X(-w) $