Originally Posted by

**sapsapz** Hello there,

I have a question about the Shift property of the Fourier Transform.

I understood that this is the property:

$\displaystyle $FourierTransform(\ \{\ f(t-t_0)\ \}\ ) = e^{-iwt_0} FourierTransform(\ \{ \ f(t)\ \}\ )$$

However, this doesn't really work for me when I apply it to Sin...

This is the Transform of Sin(t):

$\displaystyle $FourierTransform(\ \{\ Sin(t)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}[ \delta(w-1)-\delta(w+1)]$$

So according to the shift property, I would have expected to see this as the Transform of Sin(t-t0):

$\displaystyle $FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-iwt_0}[\delta(w-1)-\delta(w+1)]$$

However, this is the actual transform of Sin(t-t0):

$\displaystyle $FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-it_0}[e^{2it_0}\delta(w-1)-\delta(w+1)]$$

So the shift property doesn't work? What am I doing wrong?

Thank you (: