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Math Help - fourier transform of f(x)=x, for 0<x<2

  1. #1
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    fourier transform of f(x)=x, for 0<x<2

    Hi,

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

    What I have got so far:
    <br />
\hat{f}(u) = \frac{1}{\sqrt{2\pi}} \int_{0}^{2} x \cdot e^{-iux} dx<br />
    <br />
\sqrt{2\pi} \hat{f}(u)= -\int_{0}^{2}\frac{-1}{iu} e^{-iux} dx + \lbrack \frac{-x}{iu} e^{-iux} \rbrack_{0}^{2}<br />
    <br />
=\int_{0}^{2}\frac{1}{iu} e^{-iux} dx + \lbrack \frac{-x}{iu} e^{-iux} \rbrack_{0}^{2}<br />
    <br />
= \lbrack \frac{1}{u} e^{-iux} \rbrack_{0}^{2}+ \lbrack \frac{-x}{iu} e^{-iux} \rbrack_{0}^{2}<br />
    <br />
= \left( \frac{1}{u}e^{-i2u} - \frac{1}{u} \right) + \left( \frac{-2}{iu} e^{-i2u} -0 \right)<br />

    any thoughts would be very helpfull.

    Thanks Alex.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jeneverboy View Post
    Hi,

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

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

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

    <br />
=\int_{0}^{2}\frac{1}{iu} e^{-iux} dx + \left[ \frac{-x}{iu} e^{-iux} \right]_{0}^{2}<br />
    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
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  3. #3
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    allright, so
    <br />
\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}<br />
    <br />
\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)<br />
    now Euler's formula can be used
    <br />
\boxed{<br />
e^{ix}=cos x + i \; sinx<br />
}<br />
    here
    <br />
e^{-2ui}=cos (-2u) + i \; sin(-2u)<br />
    plug in equation
    <br />
\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)<br />
    now I use \frac{-2}{iu}=\frac{2i}{u} so equation becomes;
    <br />
\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)<br />
    rewrite
    <br />
\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}}<br />
    so now there is a real part and a imaginary part:
    <br />
\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)<br />

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

    thanks, Alex.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by jeneverboy View Post
    allright, so
    <br />
\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}<br />

    .
    Just put:

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

    and leave it there.

    CB
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  5. #5
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    Hi,

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

    <br />
\int_{-\infty}^{\infty} \frac{sin(u)}{u} du<br />
    and
    <br />
\int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du<br />

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

    inversion formula:
    <br />
\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})<br />
    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.
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  6. #6
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    Quote Originally Posted by jeneverboy View Post
    Hi,

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

    <br />
\int_{-\infty}^{\infty} \frac{sin(u)}{u} du<br />
    and
    <br />
\int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du<br />

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

    inversion formula:
    <br />
\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})<br />
    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
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  7. #7
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    ok here is the question I got

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

    b)Evaluate
    <br />
\int_{-\infty}^{\infty} \frac{sin(u)}{u} du<br />
    and
    \int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by jeneverboy View Post
    ok here is the question I got

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

    b)Evaluate
    <br />
\int_{-\infty}^{\infty} \frac{sin(u)}{u} du<br />
    and
    \int_{-\infty}^{\infty} \frac{u cos(u) -sin(u)}{u^{2}} du
    For the second consider:

     <br />
\mathfrak{F}[f(x)](u) - \mathfrak{F}[f(-x)](u)<br />

    For the first consider the use of the translation theorem.

    CB
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