# Fourier transform of church function

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

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

## 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:

$$i2\pi sF(s) = 2 + \text{sinc}_{something} - 2 \cdots$$

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jbunniii
Homework Helper
Gold Member

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?

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
$$F(s) = \frac{k}{i2\pi s} \left ( e^{-i2\pi bs} (1+a-b) -e^{-i2\pi as} \right )$$
Anyone recognise this to be true? I didn't find this transform tabulated anywhere, so I need to be sure it's correct...

jbunniii
Homework Helper
Gold Member

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

$$\int_{0}^{c} kx e^{-i2\pi s x} dx$$

You can use integration by parts. Wolfram Alpha gives

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

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

$$1 - e^{-ix} = e^{-ix/2}(e^{ix/2} - e^{-ix/2}) = 2ie^{-ix/2} \sin(x/2)$$

What syntax did you use to get wolfram alpha to produce that?

jbunniii
Homework Helper
Gold Member

Here is what I typed:

integrate (k*x*e^(-i*2*pi*s*x) from x = 0 to x = c

Cool! I'll start using wolfram alpha :) Thanks for the help.

Ray Vickson
Homework Helper
Dearly Missed

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

jbunniii
Homework Helper
Gold Member

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.

marcusl