Question regarding Fourier Transform/e^-2pi*i

[pawnwarp]

Hello,

I am a HS Senior and in my AP chem class we discussed Fourier Transform and their applications (didn't actually do any math with it, just how it works and the formula) in spectroscopy. It seemed pretty interesting to me. But I have a simple question about the formula that I just don't get.

I noticed that in the integral, there is a e^-2*pi*i. This will probably sound really stupid but given that I have only taken up to Calc BC and HS Physics, why isn't this simplified to 1 via Euler's formula? And how would you even go about evaluating the integral? My TI-89 was unable to solve the complex integral...

I think it might be related to the fact that frequency doesn't measure cycles per unit time like wiki says...but then what does it measure??

 

[I like Serena]

Welcome to PF, pawnwarp!

The integral would contain something like e^-2*pi*i*t.
I think you dropped off the "t" (or something similar).
Therefore it can not be simplified to 1.

The typical problem with the integral is that it extends to infinity where the exponentiation is not defined properly.
To this end the Dirac delta function is sometimes introduced to represent the result, but I'll not go into that now.

For finite time signals, like a square function:
$$f(t)=\begin{cases}1 & \textrm{if } -1 \le t \le 1 \\ 0 & \textrm{otherwise} \end{cases}$$
you can integrate the formula without special tricks.

Btw, the frequency would measure cycles per unit time.

 

[pawnwarp]

Thank You for the welcome :D

Yes, I saw the t but can't e^(-2*pi*i*t) be changed to (e^(-2*pi*i))^t, which becomes 1^t and then becomes 1? I must surely be missing something here...

 

[mathman]

Thank You for the welcome :D

Yes, I saw the t but can't e^(-2*pi*i*t) be changed to (e^(-2*pi*i))^t, which becomes 1^t and then becomes 1? I must surely be missing something here...

Unless t is an integer, 1^t is not necessarily 1.