# Thread: Request for Explanation of Fourier Transform

1. ## Request for Explanation of Fourier Transform

The introduction to the Laplace transform chapter in my DE textbook says (I'm paraphrasing some of this):

In general, integral transforms address the question: How much is a given function $y(t)$ "like" a particular standard function?

For example, if $y(t)$ represents a radio signal and we want to compare it to $sin \omega t$ with frequency $\frace \omega {2 \pi}$, just integrate:

$\int_{-N}^N y(t)\; sin (\omega t)\; dt$.

If $y(t)$ goes up when $sin(\omega t)$ goes up, and goes down when $sin(\omega t)$ goes down, then the integral will be large, because $y(t)$ is very much "like" $sin(\omega t)$. So far, this makes perfect sense.

Then, it says you can compare $y(t)$ to $e^{-zt}$, where $z=s+i \omega$ is Complex, using:
$\int_{-N}^N y(t)\; e^{-zt}\; dt$

Which can be split into

$\int_{-N}^N y(t)\; e^{-st}\; dt\,\,$ and $\mathbf{\int_{-N}^N y(t)\; e^{-i \omega t}\; dt }$.

The last (bold) of those integrals is called the Fourier Transform (which is what I really need to learn). Then the book says:

The Fourier Transform's value at a particular value of $\omega$ is a measure of the extent to which $y(t)$ oscillates with a frequency of $\frac \omega {2 \pi}$.
Can someone explain how that integral compares $y(t)$ to an oscillation of $\frac \omega {2 \pi}$?

I am aware of Euler's equation:
$e^{it}=cos(t)+i \cdot sin(t)$

If I apply it to just the Fourier integral, I get:

$\int_{-N}^N y(t)\; e^{-i \omega t}\; dt\, = \int_{-N}^N y(t)\; \cos (\omega t)\; dt + i \int_{-N}^N y(t)\; \sin (\omega t)\; dt$.

Unless the $\sin( \omega t)$ is always zero, I don't see how this can make that comparison.

Jeff

2. ## Re: Request for Explanation of Fourier Transform

Off-hand, I would say that $e^{-i\omega t}$ oscillates around the unit circle in the complex plane. So that's the oscillation being described. Does that answer your question?

3. ## Re: Request for Explanation of Fourier Transform

Originally Posted by MSUMathStdnt
....

If I apply it to just the Fourier integral, I get:

$\int_{-N}^N y(t)\; e^{-i \omega t}\; dt\, = \int_{-N}^N y(t)\; \cos (\omega t)\; dt + i \int_{-N}^N y(t)\; \sin (\omega t)\; dt$.

Unless the $\sin( \omega t)$ is always zero, I don't see how this can make that comparison.