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Optics convolution

  • Thread starter physiks
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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
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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.
 
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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 realise 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...
 
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LCKurtz
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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
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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?
 

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