Trouble Understanding the Fourier Integral

[ColdFusion85]

[SOLVED] Trouble Understanding the Fourier Integral

I am having a bit of trouble understanding how to use the Fourier Integral Formula.

The integral is $$\int_{0}^{\infty} [A_{\omega}cos(\omega x) + B_{\omega}sin(\omega x)]d\omega$$

Where,

$$A_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)cos(\omega\xi)d\xi$$
and
$$B_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)sin(\omega\xi)d\xi$$

The example in the book doesn't help because f(x) is 1 for -1<=x<=1, and 0 for |x|>1. This doesn't clear up my confusion as to what the integral is supposed to look like with $$f(\xi)$$ inserted. What is $$f(\xi)$$? Is it the function given, f(x)? Like if I had f(x) = sin(x) on [-2, 2], would A_w look like:

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega\xi)d\xi$$,

with sin(x) being treated as a constant that can be brought outside the integral, or am I supposed to make A_w be

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega x)dx$$?

This is where I am confused.

 

[FrogPad]

I am having a bit of trouble understanding how to use the Fourier Integral Formula.

The integral is $$\int_{0}^{\infty} [A_{\omega}cos(\omega x) + B_{\omega}sin(\omega x)]d\omega$$

Where,

$$A_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)cos(\omega\xi)d\xi$$
and
$$B_{\omega} = \frac{1}{\pi}\int_{-\infty}^{\infty} f(\xi)sin(\omega\xi)d\xi$$

The example in the book doesn't help because f(x) is 1 for -1<=x<=1, and 0 for |x|>1. This doesn't clear up my confusion as to what the integral is supposed to look like with $$f(\xi)$$ inserted. What is $$f(\xi)$$? Is it the function given, f(x)? Like if I had f(x) = sin(x) on [-2, 2], would A_w look like:

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega\xi)d\xi$$,

with sin(x) being treated as a constant that can be brought outside the integral, or am I supposed to make A_w be

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega x)dx$$?

This is where I am confused.

Your book gives: f(x) = 1 for values between -1 and 1. Let's just say you wanted to integrate this function:

eg. $$\int_{-\infty}^{\infty}f(x)dx$$
This integral would reduce to: $$\int_{-1}^{1}dx$$ since it is zero outside of x=-1 and x=1 and is unity otherwise.

Note: You could also integrate this function as:
$$\int_{-\infty}^{\infty}f(\xi)d\xi$$
and you would get the same thing.

The $\xi$ is a dummy variable.

For example, imagine I wanted to define a quantity that was the area of an arbitrary rectangle. Let's call this number $$R_{AB}$$ where A and B are the lengths of the rectangle respectively.

Well I could simply define $$R_{AB} = R \times B$$ or I could also write it as:

$$R_{AB} = \int_0^A \int_0^B d\xi_1 d\xi_2$$ Does that make sense? Its kinda like a loop.

 

[FrogPad]

I don't think I fully answered your question:

If you were to integrate:

$$\int_{-\infty}^{\infty} f(\xi)cos(\omega \xi) d\xi$$

Lets let f(x) = 1 for x= 0 to 1 and f(x)=0 otherwise. Thus, your integral would become:

$$A = \int_{a}^{b} (1)cos(\omega \xi) d\xi$$
$$A = \frac{\sin b \omega }{\omega }-\frac{\sin a \omega }{\omega}$$

 

[CompuChip]

What is $$f(\xi)$$? Is it the function given, f(x)? Like if I had f(x) = sin(x) on [-2, 2], would A_w look like:

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega\xi)d\xi$$,

with sin(x) being treated as a constant that can be brought outside the integral, or am I supposed to make A_w be

$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(x)cos(\omega x)dx$$?

This is where I am confused.

Apparently you are confused over basic definitions. The notation $f(\xi)$ refers to the value that the function f takes when you substitute $x = \xi$, just like f(0) means: the value of f in x = 0, and f(12.4872) means: the value of f in x = 12.8472.
So if you put that into the formula you gave, you get
$$A_{\omega} = \frac{1}{\pi}\int_{-2}^{2} sin(\xi)cos(\omega\xi)d\xi$$.

 

[ColdFusion85]

Yeah that's what I was having trouble understanding. Pretty simple actually, but it was late at night and my brain wasn't working. I don't know why I didn't see that in the first place. Thanks.