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) $
