Optics Convolution Homework: Finite Sinusoidal Aperture Function

In summary: I'll have to ask my tutor.In summary, the conversation discusses the convolution of a sinusoidal aperture/transmission function with a 'top-hat' aperture function, where the latter is given by a box function with width d. The resulting convolution function is a finite sinusoidal aperture function, except over a certain range where it is zero. However, the formula for the convolution applies over the whole real line and does not vanish outside of the finite interval. The conversation also brings up the possibility that the notes may be incorrect.
  • #1
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0

Homework Statement


According to my notes, if we have a sinusoidal aperture/transmission function of the form a(x)=1+sin(wx) and a 'top-hat' aperture function given by b(x)=1, -0.5d≤x≤0.5d, b(x)=0 otherwise, then their convolution should give a finite sinusoidal aperture function, i.e sinusoidal but only over a certain range - outside of this it is zero.

Homework Equations


Convolution of a(x) and b(x) is ∫-∞a(y)b(x-y)dy

The Attempt at a Solution


We need ∫-∞[1+sin(wy)]b(x-y)dy, and my thinking is that the integrand is zero everywhere apart from between x-0.5d and x+0.5d, where it is simply [1+sin(wy)]. Then the convolution integral reduces to ∫x-0.5dx+0.5d[1+sin(wy)]dy giving a-{cos[w(x+d/2)]+cos[w(x-d/2)]}/w which is not the finite sinusoidal aperture function I've been told I should get. Is this right?
 
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  • #2
The convolution of a signal with a "box" is not the signal restricted to a specific range. If you want that, you have to multiply the two functions.
"a" seems to have two different meanings which is confusing.
 
  • #3
mfb said:
The convolution of a signal with a "box" is not the signal restricted to a specific range. If you want that, you have to multiply the two functions.
"a" seems to have two different meanings which is confusing.

Changed a to d, didn't realize that.

I'm not clear - the function b(x-y) is a box centred on y=x with width d, right? Then the integrand [1+sin(wy)]b(x-y) surely vanishes everywhere but for where the box is non-zero, and so basically we can write b(x-y)=1 as long as we integrate over the box, from x-0.5d to x+0.5d. Then I just do the integral. I'm not sure what you mean...
 
  • #5
physiks said:

Homework Statement


According to my notes, if we have a sinusoidal aperture/transmission function of the form a(x)=1+sin(wx) and a 'top-hat' aperture function given by b(x)=1, -0.5d≤x≤0.5d, b(x)=0 otherwise, then their convolution should give a finite sinusoidal aperture function, i.e sinusoidal but only over a certain range - outside of this it is zero.

Homework Equations


Convolution of a(x) and b(x) is ∫-∞a(y)b(x-y)dy

The Attempt at a Solution


We need ∫-∞[1+sin(wy)]b(x-y)dy, and my thinking is that the integrand is zero everywhere apart from between x-0.5d and x+0.5d, where it is simply [1+sin(wy)]. Then the convolution integral reduces to ∫x-0.5dx+0.5d[1+sin(wy)]dy giving a-{cos[w(x+d/2)]+cos[w(x-d/2)]}/w which is not the finite sinusoidal aperture function I've been told I should get. Is this right?

Yes, except for a sign. If ##C(x)## is the convolution we have
[tex] C(x) = a+\frac{\cos(wx - wa/2) -\cos(wx+wa/2)}{w}[/tex]
However, using standard trigonometric additional formulas, this can be manipulated to give ##C(x) = a + K \sin(w x)##, where ##K## is some constant that I will leave you to compute.

Note, however: the formula for ##C(x)## applies on the whole real line ## -\infty < x < +\infty##; it does not vanish outside of some finite interval. This follows from the fact that
[tex] C(x) = \int_{-\infty}^{\infty} a(y) b(x-y) \, dy = \int_{-\infty}^{\infty} a(x-y) b(y) \, dy [/tex]
(by a simple change of variables). The second form gives
[tex] C(x) = \int_{-a/2}^{a/2} [1 + \sin(w(x-y)) \, dy [/tex]
 
  • #6
Ray Vickson said:
Yes, except for a sign. If ##C(x)## is the convolution we have
[tex] C(x) = a+\frac{\cos(wx - wa/2) -\cos(wx+wa/2)}{w}[/tex]
However, using standard trigonometric additional formulas, this can be manipulated to give ##C(x) = a + K \sin(w x)##, where ##K## is some constant that I will leave you to compute.

Note, however: the formula for ##C(x)## applies on the whole real line ## -\infty < x < +\infty##; it does not vanish outside of some finite interval. This follows from the fact that
[tex] C(x) = \int_{-\infty}^{\infty} a(y) b(x-y) \, dy = \int_{-\infty}^{\infty} a(x-y) b(y) \, dy [/tex]
(by a simple change of variables). The second form gives
[tex] C(x) = \int_{-a/2}^{a/2} [1 + \sin(w(x-y)) \, dy [/tex]

Thanks for your reply! So my notes say that it should give me a sinusoidal function, BUT, only over a finite interval - so we agree that they must be wrong then?
 

1. What is optics convolution?

Optics convolution is a mathematical operation that combines two functions to create a new function. In the context of optics, it is used to calculate the resulting image when light passes through an optical system, such as lenses or mirrors.

2. What is a finite sinusoidal aperture function?

A finite sinusoidal aperture function is a mathematical function that describes the shape of a finite aperture, such as a circular or rectangular opening. It is often used to model the behavior of light passing through a real-world aperture in optics simulations.

3. How is optics convolution used in the study of optics?

Optics convolution is used to predict the behavior of light as it passes through an optical system. By convolving the input light with the finite sinusoidal aperture function of the system, the resulting image can be calculated and analyzed to understand how the system will affect the light.

4. What are the applications of optics convolution?

Optics convolution is used in a variety of applications, including designing and optimizing optical systems, analyzing the performance of imaging systems, and simulating the behavior of light in different scenarios. It is also used in fields such as astronomy, microscopy, and medical imaging.

5. How does optics convolution homework help in understanding optics?

Optics convolution homework helps students to gain a deeper understanding of how light behaves in optical systems. By working through problems and simulations, students can apply the principles of optics convolution to real-world scenarios, improving their understanding of the subject.

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