# fourier transform of f(x)=x, for 0<x<2

• Mar 22nd 2010, 07:24 AM
jeneverboy
fourier transform of f(x)=x, for 0<x<2
Hi,

I am trying to find the Fourier transform of this function
$
f(x) = \left\{
\begin{array}{cl}
x & o \leq x \leq 2\\
0 & \text{otherwise}
\end{array}
\right.
$

What I have got so far:
$
\hat{f}(u) = \frac{1}{\sqrt{2\pi}} \int_{0}^{2} x \cdot e^{-iux} dx
$

$
\sqrt{2\pi} \hat{f}(u)= -\int_{0}^{2}\frac{-1}{iu} e^{-iux} dx + \lbrack \frac{-x}{iu} e^{-iux} \rbrack_{0}^{2}
$

$
=\int_{0}^{2}\frac{1}{iu} e^{-iux} dx + \lbrack \frac{-x}{iu} e^{-iux} \rbrack_{0}^{2}
$

$
= \lbrack \frac{1}{u} e^{-iux} \rbrack_{0}^{2}+ \lbrack \frac{-x}{iu} e^{-iux} \rbrack_{0}^{2}
$

$
= \left( \frac{1}{u}e^{-i2u} - \frac{1}{u} \right) + \left( \frac{-2}{iu} e^{-i2u} -0 \right)
$

any thoughts would be very helpfull.

Thanks Alex.
• Mar 22nd 2010, 10:41 PM
CaptainBlack
Quote:

Originally Posted by jeneverboy
Hi,

I am trying to find the Fourier transform of this function
$
f(x) = \left\{
\begin{array}{cl}
x & o \leq x \leq 2\\
0 & \text{otherwise}
\end{array}
\right.
$

What I have got so far:
$
\hat{f}(u) = \frac{1}{\sqrt{2\pi}} \int_{0}^{2} x \cdot e^{-iux} dx
$

$
\sqrt{2\pi} \hat{f}(u)= -\int_{0}^{2}\frac{-1}{iu} e^{-iux} dx + \left[ \frac{-x}{iu} e^{-iux} \right]_{0}^{2}
$

$
=\int_{0}^{2}\frac{1}{iu} e^{-iux} dx + \left[ \frac{-x}{iu} e^{-iux} \right]_{0}^{2}
$

Its OK up to here, but:

$\int_{0}^{2}\frac{1}{iu} e^{-iux} dx=\left[ \frac{1}{u^2}e^{-iux} \right]_0^2$

CB
• Mar 23rd 2010, 02:16 AM
jeneverboy
allright, so
$
\sqrt{2 \pi}\hat{f}(u)= \left[ \frac{1}{u^{2}} e^{-iux} \right]_{0}^{2}+ \left[ \frac{-x}{iu} e^{-iux} \right]_{0}^{2}
$

$
\sqrt{2 \pi}\hat{f}(u)= \left( \frac{1}{u^{2}}e^{-i2u} - \frac{1}{u^{2}} \right) + \left( \frac{-2}{iu} e^{-i2u} -0 \right)
$

now Euler's formula can be used
$
\boxed{
e^{ix}=cos x + i \; sinx
}
$

here
$
e^{-2ui}=cos (-2u) + i \; sin(-2u)
$

plug in equation
$
\sqrt{2 \pi} \hat{f}(u) = -\frac{1}{u^{2}} + \frac{1}{u^{2}} \left( cos(-2u)+i \; sin(-2u) \right) + \frac{-2}{iu} \left( cos(-2u) + i \; sin(-2u) \right)
$

now I use $\frac{-2}{iu}=\frac{2i}{u}$ so equation becomes;
$
\sqrt{2 \pi} \hat{f}(u) = -\frac{1}{u^{2}} + \frac{1}{u^{2}} \left( cos(-2u)+i \; sin(-2u) \right) + \frac{2i}{u} \left( cos(-2u) + i \; sin(-2u) \right)
$

rewrite
$
\sqrt{2 \pi} \hat{f}(u) = -\frac{1}{u^{2}} + \frac{cos(-2u)}{u^{2}} -\frac{2sin(-2u)}{u} + \frac{2i cos(-2u)}{u} + \frac{i sin(-2u)}{u^{2}}
$

so now there is a real part and a imaginary part:
$
\sqrt{2 \pi} \hat{f}(u) = -\frac{1}{u^{2}} + \frac{cos(-2u)}{u^{2}} -\frac{2sin(-2u)}{u} + i \; \left( \frac{2 cos(-2u)}{u} + \frac{ sin(-2u)}{u^{2}} \right)
$

is this correct so far?
does anyone have a suggestion how to simplify this?

thanks, Alex.
• Mar 23rd 2010, 03:57 AM
CaptainBlack
Quote:

Originally Posted by jeneverboy
allright, so
$
\sqrt{2 \pi}\hat{f}(u)= \left[ \frac{1}{u^{2}} e^{-iux} \right]_{0}^{2}+ \left[ \frac{-x}{iu} e^{-iux} \right]_{0}^{2}
$

.

Just put:

$
(\mathfrak{F}f)(u)=\frac{1}{\sqrt{2 \pi}}$
$\left[ \frac{1}{u^2}(e^{-2ui}-1)+\frac{2i}{u}e^{-2ui} \right]
$

and leave it there.

CB
• Mar 23rd 2010, 05:10 AM
jeneverboy
Hi,

I need an expression with sin and cos, because now I must evaluate

$
\int_{-\infty}^{\infty} \frac{sin(u)}{u} du
$

and
$
\int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du
$

I can use the inverse transform here at a certain point $x_{0}$ which is a continuity point of $f$.

inversion formula:
$
\lim_{a\rightarrow \infty} \frac{1}{\sqrt{2 \pi}} \int_{-a}^{a} \hat{f}(u)e^{iux_{0}} du = \frac{1}{2} (f(x_{0}^{+})+f(x_{0}^{-})) =f(x_{0})
$

so this is $f(x_{0})$ because for instance at x=0 f is continuous, but for x=2 f is not continuous so there I can't use it.

But first I need an expression with cos and sin I think.

Alex.
• Mar 23rd 2010, 05:30 AM
CaptainBlack
Quote:

Originally Posted by jeneverboy
Hi,

I need an expression with sin and cos, because now I must evaluate

$
\int_{-\infty}^{\infty} \frac{sin(u)}{u} du
$

and
$
\int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du
$

I can use the inverse transform here at a certain point $x_{0}$ which is a continuity point of $f$.

inversion formula:
$
\lim_{a\rightarrow \infty} \frac{1}{\sqrt{2 \pi}} \int_{-a}^{a} \hat{f}(u)e^{iux_{0}} du = \frac{1}{2} (f(x_{0}^{+})+f(x_{0}^{-})) =f(x_{0})
$

so this is $f(x_{0})$ because for instance at x=0 f is continuous, but for x=2 f is not continuous so there I can't use it.

But first I need an expression with cos and sin I think.

Alex.

Sorry but none of that make the least bit of sense.

Why don't you just post the quaestion you are really trying to address.

CB
• Mar 23rd 2010, 05:38 AM
jeneverboy
ok here is the question I got

1a) Calculate the Fourier transform of the function
$
f(x) = \left\{
\begin{array}{cl}
x & o \leq x \leq 2\\
0 & \text{otherwise}
\end{array}
\right.
$

b)Evaluate
$
\int_{-\infty}^{\infty} \frac{sin(u)}{u} du
$

and
$\int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du$
• Mar 23rd 2010, 05:49 AM
CaptainBlack
Quote:

Originally Posted by jeneverboy
ok here is the question I got

1a) Calculate the Fourier transform of the function
$
f(x) = \left\{
\begin{array}{cl}
x & o \leq x \leq 2\\
0 & \text{otherwise}
\end{array}
\right.
$

b)Evaluate
$
\int_{-\infty}^{\infty} \frac{sin(u)}{u} du
$

and
$\int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du$

For the second consider:

$
\mathfrak{F}[f(x)](u) - \mathfrak{F}[f(-x)](u)
$

For the first consider the use of the translation theorem.

CB