# Fourier transform

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• Jun 10th 2011, 02:23 PM
Ulysses
Fourier transform
Hi there. I'm starting with the Fourier transforms, and I'm having some trouble with my first exercise on this topic.
The problem says: Given $\displaystyle f(x)=H(x)-H(x-l)$ (H(x) is the Heaviside unit step function).
a) Consider the odd extension for f and find its Fourier integral representation.
b) Using the previous incise calculate the value for $\displaystyle \int_0^{\infty}\frac{1-cos (\omega l)}{\omega}\sin (\omega l) d\omega$

Well, for a) I think I should get the fourier transfor for the sign function, I don't know if this is right, but anyway I've defined the function like this:
$\displaystyle g(x)=\begin{Bmatrix}{ 0}&\mbox{ if }& x>l\\1 & \mbox{if}& 0<x<l\\-1 & \mbox{if}& -l<x<0\\0 & \mbox{if}& x<-l\end{matrix}$

Then the Fourier integral representation:
$\displaystyle g(x)=\displaystyle\frac{1}{2\pi}\displaystyle\int_ {-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}g(x)e^{-i \omega x}dx e^{i \omega x}d\omega$

But g(x) is odd, then:

$\displaystyle g(x)=\displaystyle\frac{1}{2\pi}\displaystyle\int_ {-\infty}^{\infty} 2 \displaystyle\int_{0}^{\infty}g(x)(-i)\sin(\omega x)}dx e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{0}^{\infty}g(x) \sin(\omega x)}dx e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty} \left ( \displaystyle\int_{0}^{\infty} \sin(\omega x)}dx - \displaystyle\int_{-\infty}^{0} \sin(\omega x)}dx\right ) e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty} \left ( \cos(\omega x)} |_0^{\infty} - \cos(\omega x)}|_{-\infty}^{0} \right ) e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty} \left ( \frac{2}{\omega}-\displaystyle\lim_{c \to{+}\infty}{}\cos{\omega c}-\displaystyle\lim_{c \to{-}\infty}{}\cos{\omega c}}\right ) e^{i \omega x}d\omega$

And the thing is that the integral for the cosine diverges as I see it, but I'm probably doing something wrong.

Help please :)
• Jun 10th 2011, 09:34 PM
CaptainBlack
Quote:

Originally Posted by Ulysses
Hi there. I'm starting with the Fourier transforms, and I'm having some trouble with my first exercise on this topic.
The problem says: Given $\displaystyle f(x)=H(x)-H(x-l)$ (H(x) is the Heaviside unit step function).
a) Consider the odd extension for f and find its Fourier integral representation.
b) Using the previous incise calculate the value for $\displaystyle \int_0^{\infty}\frac{1-cos (\omega l)}{\omega}\sin (\omega l) d\omega$

Well, for a) I think I should get the fourier transfor for the sign function, I don't know if this is right, but anyway I've defined the function like this:
$\displaystyle g(x)=\begin{Bmatrix}{ 0}&\mbox{ if }& x>l\\1 & \mbox{if}& 0<x<l\\-1 & \mbox{if}& -l<x<0\\0 & \mbox{if}& x<-l\end{matrix}$

Assuming that $\displaystyle l>0$ you should have:

$\displaystyle g(x)= \left\{ \begin{array}{ll}1,&0\le x<l \\ 0,&\text{otherwise}\end{array} \right.$

What you give is the odd extension (more of less).

Quote:

Then the Fourier integral representation:
$\displaystyle g(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(x)e^{-i \omega x}dx e^{i \omega x}d\omega$

But g(x) is odd, then:

$\displaystyle g(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty} 2 \int_{0}^{\infty}g(x)(-i)\sin(\omega x)}dx e^{i \omega x}d\omega$
OK to this point

Quote:

.................. $\displaystyle =\frac{-i}{\pi}\int_{-\infty}^{\infty}\int_{0}^{\infty}g(x)\sin(\omega x)}dx e^{i \omega x}d\omega$
You have lost the 2, and this last integral becomes (putting the 2 back in) OK found it below now fixed:

.................. $\displaystyle =\frac{-i}{\pi}\int_{-\infty}^{\infty}\int_{0}^{l}\sin(\omega x)}dx e^{i \omega x}d\omega$
• Jun 11th 2011, 09:43 AM
Ulysses
I canceled it out with the 2 in the denominator, are you sure that's wrong? I've found one mistake, it's pretty silly. I was making for -infinity to zero, but I shouldn't use that part. I see now, I shouldn't take from zero to infinity, I must use l. Thanks :)
• Jun 11th 2011, 09:53 AM
CaptainBlack
Quote:

Originally Posted by Ulysses
I canceled it out with the 2 in the denominator, are you sure that's wrong? I've found one mistake, it's pretty silly. I was making for -infinity to zero, but I shouldn't use that part. I see now, I shouldn't take from zero to infinity, I must use l. Thanks :)

You are correct about the 2.

CB