Laplace Transforms
The basic formula is: $ L(x(t)) = \int_{-\infty}^{\infty}x(t)e^{-st}dt $
As you can see, this is very similar to the Fourier Transform. In fact, the Fourier Transform is just the restriction of the Laplace transform to the imaginary axis, in other words, s=jw.
Trying a Laplace Transform
$ L(x(t-t_0)) = \int_{-\infty}^{\infty}x(t-t_0)e^{-st}dt $
- Let $ (t-t_0) = \tau $
- Therefore $ d\tau = dt $
- $ = \int_{-\infty}^{\infty}x(\tau)e^{-s(\tau + t_0)}(d\tau) $
- $ = \int_{-\infty}^{\infty}x(\tau)e^{-s\tau}e^{-st_0}(d\tau) $
- Since $ e^{-st_0} $ does not depend on "$ \tau $" it can be taken out of the integral.
- $ =e^{-st_0} \int_{-\infty}^{\infty}x(\tau)e^{-s\tau}(d\tau) $
Thus
- $ = e^{-st_0}X(s) $
As you can see, this is nearly the same as a time shift in Fourier Transforms.
