Fourier transform of church function

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  • #1
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Fourier transform of "church" function

This is an old examn question that I'm trying to solve. There is a solution, but I'm having a hard time getting it since there is only some kind of graphic equation with no explanation. To only test in the solution is "Derivate!"

Homework Statement


Determine the Fourier transform of the following modern church function
http://www.apspektakel.com/bilder/churchf.svg

Homework Equations


[tex]
F(s) = \int ^\infty _{-\infty} f(x) e ^ {-i2\pi xs} dx[/tex]
[tex]
i2\pi sF(s) = \int ^\infty _{-\infty} f'(x) e ^ {-i2\pi xs} dx
[/tex]


The Attempt at a Solution


This is what I think the derivative of the church function would be. I coloured the impulses red, so that their origins are visible. Is the derivative correct?
http://www.apspektakel.com/bilder/churchfd.svg
So, the next step, would it be adding the transform of the first impulse to the transform of the first square, to the second impulse and so on?
I mean would the correct way be something like:

[tex]
i2\pi sF(s) = 2 + \text{sinc}_{something} - 2 \cdots
[/tex]
 

Answers and Replies

  • #2
jbunniii
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Your derivative looks right. Your proposed solution also looks to be on the right track, although the Fourier transform of the second impulse isn't simply -2. There is also a phase component.

Another way to do this without derivatives and impulse functions is to express the function as a sum of various rectangles and triangles. Each rectangle's Fourier transform is a sinc function with some scale factors and a phase factor. How about the triangles?
 
  • #3
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Well, the intergral of a right triangle is just the integral under a straight line function, f(x) = kx+m.

If the left bottom corner of the triangle is a, the bottom right is b and the triangle sits on the positive x-axis: f(x) = kx-ka (since x=a when f(x)=0).
I transform this to
[tex]
F(s) = \frac{k}{i2\pi s} \left ( e^{-i2\pi bs} (1+a-b) -e^{-i2\pi as} \right )
[/tex]
Anyone recognise this to be true? I didn't find this transform tabulated anywhere, so I need to be sure it's correct...
 
  • #4
jbunniii
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I got something more complicated than that, although it's possible that it simplifies to your answer.

It's easier if you define the triangle to sit between x = 0 and x = c. Then if you need to shift or scale it, you can use the standard Fourier transform properties.

In this case, you will have

[tex]\int_{0}^{c} kx e^{-i2\pi s x} dx[/tex]

You can use integration by parts. Wolfram Alpha gives

[tex]\frac{1}{4 \pi^2 s^2} [2\pi i c s k e^{-i2 \pi c s} - k(1 - e^{-i2\pi c s})][/tex]

which can be further simplified using the typical tricks, e.g.,

[tex]1 - e^{-ix} = e^{-ix/2}(e^{ix/2} - e^{-ix/2}) = 2ie^{-ix/2} \sin(x/2)[/tex]
 
  • #5
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What syntax did you use to get wolfram alpha to produce that?
 
  • #6
jbunniii
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Here is what I typed:

integrate (k*x*e^(-i*2*pi*s*x) from x = 0 to x = c
 
  • #7
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Cool! I'll start using wolfram alpha :) Thanks for the help.
 
  • #8
Ray Vickson
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Cool! I'll start using wolfram alpha :) Thanks for the help.
For the function f(x) = (x-a) if a < x < b and f(x) = 0 elsewhere, Maple 14 gets:

f1:=(x-a)*Heaviside(x-a)*Heaviside(b-x);
f1 := (x - a) Heaviside(x - a) Heaviside(b - x)

J2:=fourier(f1,x,w) assuming b>a;

J2:= I*exp(-I b w) (-w a - I + b w)/w^2 - exp(-I w a)/w^2

where I = sqrt(-1).

RGV
 
  • #9
jbunniii
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I might point out that this is easier than you are making it. Write the triangle as a convolution of two rectangles
/\ = rect * rect

Use the convolution theorem to find the transform, and the answer falls right out.
That's true for a triangle shaped like this: /\, but not for one shaped like this /|. The latter type is what appears in this problem.
 
  • #10
marcusl
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My mistake, I didn't see the shape listed anywhere. I've deleted my earlier post.
 

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