# Math Help - The Fourier Transform Shift property

1. ## The Fourier Transform Shift property

Hello there,

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

I understood that this is the property:
$FourierTransform(\ \{\ f(t-t_0)\ \}\ ) = e^{-iwt_0} FourierTransform(\ \{ \ f(t)\ \}\ )$

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

This is the Transform of Sin(t):
$FourierTransform(\ \{\ Sin(t)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}[ \delta(w-1)-\delta(w+1)]$

So according to the shift property, I would have expected to see this as the Transform of Sin(t-t0):
$FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-iwt_0}[\delta(w-1)-\delta(w+1)]$

However, this is the actual transform of Sin(t-t0):
$FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-it_0}[e^{2it_0}\delta(w-1)-\delta(w+1)]$

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

Thank you (:

2. ## Re: The Fourier Transform Shift property

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

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

This is the Transform of Sin(t):
$FourierTransform(\ \{\ Sin(t)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}[ \delta(w-1)-\delta(w+1)]$

So according to the shift property, I would have expected to see this as the Transform of Sin(t-t0):
$FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-iwt_0}[\delta(w-1)-\delta(w+1)]$

However, this is the actual transform of Sin(t-t0):
$FourierTransform(\ \{\ Sin(t-t_0)\ \}\ ) = i\sqrt{\frac{\Pi}{2}}e^{-it_0}[e^{2it_0}\delta(w-1)-\delta(w+1)]$

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

Thank you (:
Where did you obtain the form of the transform you have there? It's incorrect.

3. ## Re: The Fourier Transform Shift property

Here is a short proof I encountered:

Edit:
Unless you're asking about the transforms of the Sin functions, in which case I calculated them in Wolfram Mathematica

4. ## Re: The Fourier Transform Shift property

I was asking about the final form of the FT of Sin[t-t0]

That's not what I get from Mathematica but it doesn't matter.

Multiply your Exp[- i t0] factor through and you obtain the correct answer.