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:

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

This is the Transform of Sin(t):

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

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

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

Thank you (:

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:

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

This is the Transform of Sin(t):

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

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

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.

1 Attachment(s)

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

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.