# The Fourier Transform Shift property

• Dec 1st 2013, 08:18 AM
sapsapz
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 (:
• Dec 1st 2013, 12:48 PM
romsek
Re: The Fourier Transform Shift property
Quote:

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.
• Dec 1st 2013, 01:00 PM
sapsapz
Re: The Fourier Transform Shift property
Here is a short proof I encountered:
Attachment 29835

Edit:
Unless you're asking about the transforms of the Sin functions, in which case I calculated them in Wolfram Mathematica
• Dec 1st 2013, 01:27 PM
romsek
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.