Integral sinc(a sin theta) knowing sinc, sinc^2 etc.

In summary, the power diagram expresses the power (per area) flowing radially inwards toward the telescope. The aim of the problem is to integrate the power diagram over a surface surrounding the telescope, thereby giving the total energy (per time) reaching the telescope.
  • #1
elizevin
5
0

Homework Statement


Solve main lobe of radio telescope if the power diagram is given as:
P[itex]_{n}[/itex]([itex]\vartheta[/itex], [itex]\varphi[/itex])=sinc[itex]^{2}[/itex](a*sin[itex]\vartheta[/itex])



Homework Equations


Ω[itex]_{m}[/itex]=[itex]\int[/itex][itex]\int[/itex]P[itex]_{n}[/itex] sin[itex]\vartheta[/itex] d[itex]\vartheta[/itex]d[itex]\varphi[/itex]


The Attempt at a Solution


Purely math question - but I have problem solving the integral. I can solve sinc or sinc[itex]^{n}[/itex] etc. Hint is to solve integral in Descartes coordinates

x=asin[itex]\vartheta[/itex]cos[itex]\varphi[/itex]
y=asin[itex]\vartheta[/itex]sin[itex]\varphi[/itex]

Does anybody have an idea how to solve this integral?
 
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  • #2
hey man, welcome to physicsforums! Since they gave that hint, then try to make use of it.
 
  • #3
Why thank you very much, and thank you for your helpful suggestion, but I already tried that, didn't work, so I came here asking for help. I can't seem to transform integral in purely Descartes coordinates, there is always one spheric coordinate left. I need fresh idea.
 
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  • #4
Your integrand doesn't seem to depend on φ. So the specific substitution for x and y doesn't look useful. Is that exactly the hint, or did it just suggest Cartesian co-ordinates without specifying the equations?
 
  • #5
TA just told us to think about those substitutions and similar problem of sinc x *sinc y which are easy to solve (where x and y can be expressed in spheric coordinates).
 
  • #6
Your integrand does not depend on phi, so what happens if you integrate over phi?
 
  • #7
Phi is not a problem, I'll have 2*pi constant before integral over theta, problem is this integral which I don't know how to solve: (sinc(a*sin theta))^2
 
  • #8
mm. That's a very tricky integral. Are you sure you wrote down the problem correctly? hopefully not :)
 
  • #9
Unfortunately I'm sure. To make matters worse, boundaries are the first nulls of sinc, and not infinity, so even if it's perfectly normal sinc integral I would have a problem.
 
  • #10
Maybe stepping back and looking at it a bit differently would be a good idea. Am I right in thinking that the power diagram expresses the power (per area) flowing radially inwards toward the telescope? and 'a' is the radius?

And the aim of the problem is to integrate the power diagram over a surface surrounding the telescope, thereby giving the total energy (per time) reaching the telescope? So your method so far was to try to integrate over a unit sphere, right? So maybe try integrating over another shape which takes advantage of the symmetry around the z axis.
 

1. What is the definition of "sinc" in relation to the integral sinc(a sin theta)?

The "sinc" function, denoted as sinc(x), is defined as the ratio of the sine of x to x, or sinc(x) = sin(x)/x. It is also known as the "sampling function" in signal processing.

2. How is the sinc function related to the sinc^2 function?

The sinc^2 function is simply the squared version of the sinc function, or sinc^2(x) = [sinc(x)]^2. This function is used in Fourier analysis and has a peak value of 1 at x = 0.

3. What is the special significance of the integral sinc(a sin theta)?

The integral sinc(a sin theta) has applications in optics and diffraction theory. It represents the amplitude of a diffraction pattern produced by a circular aperture with a diameter of a units when illuminated by a plane wave at an angle of theta with respect to the aperture's normal axis.

4. Can the integral sinc(a sin theta) be evaluated analytically?

Yes, the integral sinc(a sin theta) can be evaluated analytically using the Bessel function. The result is 2πJ0(a), where J0(a) is the Bessel function of the first kind with order 0.

5. How does the value of a affect the graph of sinc(a sin theta)?

The value of a affects the width of the central peak of the graph of sinc(a sin theta). A larger value of a results in a narrower peak, while a smaller value of a results in a wider peak. This is because the sinc function has a periodic nature and the value of a determines the frequency of this periodicity.

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