Originally Posted by

**M1ke** I've been messing with the equations trying to better understand how the transform works and have done the following:

$\displaystyle f(T-t) \rightarrow e^{jT\omega}F(-\omega)$ and $\displaystyle F(\omega) = \int_{-\infty}^\infty\!\!\!f(t)e^{-j\omega t}dt$

So

$\displaystyle e^{jT\omega}F(-\omega) = e^{jT\omega}\int_{-\infty}^\infty\!\!\!f(T-t)e^{-jt\omega}dt$ (That is correct.)

In that integral, when $\displaystyle t=T \rightarrow f(T-T) \rightarrow f(0)$ because the e values cancel out. With Fourier being about frequencies I'd assume this to indicate a fundamental frequency of a signal. Hopefully that made some sense.