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Convolution, Triangle Function

  1. Feb 3, 2014 #1
    1. The problem statement, all variables and given/known data


    Part (a): Find the intensity as function of ##\theta## and sketch it.

    Part (b): Find the intensity as function of ##\theta## and sketch it. Comment on first minima.

    2. Relevant equations

    3. The attempt at a solution


    Convolution Method


    [tex]V_b = \frac{1}{2a}, 0 \leq y \leq a [/tex]

    The convolution ##V_b \otimes V_b ## gives the angular distribution ##V_{b(\beta)}##

    Fourier transform of ##V_b##:

    [tex]\alpha \int_0^{\infty} \frac{1}{2a} e^{-i\beta y} dy[/tex]
    [tex]= \frac{\alpha}{2ai\beta}[e^{-i\beta y}]_a^0 [/tex]
    [tex]= \frac{\alpha}{2ai\beta} [1 - e^{-i\beta a}] [/tex]
    [tex] = \frac{1}{2} \alpha e^{\frac{-i\beta a}{2}} \frac{sin ( \frac{\beta a}{2})}{\frac{\beta a}{2}}[/tex]
    [tex] |A_{\theta}|^2 = \frac {1}{4} \alpha^2 \frac{sin^2(\beta')}{\beta'^2} = I_0 \frac{sin^2(\beta')}{\beta'^2} [/tex]


    Analytical Method

    [tex] A_{\theta} = \alpha \int_{-\infty}^{\infty} T_y e^{-iky sin {\theta}} dy [/tex]

    For 0 < y < a:
    [tex]T_y = \frac{1}{a^2}[/tex]

    For a < y < 2a:
    [tex]T_y = -\frac{1}{a^2}y + \frac{2}{a}[/tex]

    [tex]A_{\theta} = \alpha \int_0^a \frac{1}{a^2} e^{-ikysin\theta} dy + \alpha \int_a^{2a} \left (-\frac{1}{a^2}y + \frac{2}{a}\right )e^{-ikysin\theta} dy[/tex]

    First Integral:

    [tex]\alpha \int_0^a \frac{1}{a^2} e^{-ikysin\theta} dy[/tex]
    [tex]= \alpha \frac{1}{a^2} \frac{1}{iksin\theta} [e^{-ikysin\theta}]_a^0[/tex]
    [tex]= \frac{\alpha}{a^2}\frac{1}{iksin\theta}\left (1 - e^{-ikasin\theta}\right )[/tex]

    Second Integral:

    [tex]\alpha \int_a^{2a} \left (-\frac{1}{a^2}y + \frac{2}{a}\right )e^{-ikysin\theta} dy[/tex]
    [tex]=\frac{-\alpha}{a^2}\int_a^{2a} y e^{-ikysin\theta} dy + \frac{2\alpha}{a}\int_a^{2a}e^{-ikysin\theta} dy [/tex]
    [tex]=\frac{-\alpha}{a^2}\{ \frac{1}{iksin\theta}[y e^{-ikysin\theta}]_{2a}^a + \frac{1}{iksin\theta}\int_a^{2a} e^{-ikysin\theta} dy \} + \frac{2\alpha}{a} \frac{1}{iksin\theta} [e^{-ikysin\theta}]_{2a}^a [/tex]
    [tex]=\frac{-\alpha}{a} \frac{1}{iksin\theta}[e^{-ikasin\theta} - 2e^{-2ikasin\theta}] - \frac{\alpha}{a^2} \frac{1}{k^2sin^2\theta}[e^{-ikysin\theta}]_a^{2a} + \frac{2\alpha}{a}\frac{1}{iksin\theta} [e^{-ikasin\theta} - e^{-2ikasin\theta}][/tex]
    [tex]= -\frac{\alpha}{a}\frac{1}{iksin\theta}[e^{-ikasin\theta} - 2e^{-2ikasin\theta}] - \frac{\alpha}{a^2} \frac{1}{k^2sin^2\theta}[e^{-2ikasin\theta} - e^{-ikasin\theta}] + \frac{2\alpha}{a^2}\frac{1}{iksin\theta}[e^{-ikasin\theta} - e^{-2ikasin\theta}][/tex]
    [tex]= \frac{\alpha}{a}\frac{1}{iksin\theta}(e^{-ikasin\theta}) - \frac{\alpha}{a^2}\frac{1}{k^2sin^2\theta}e^{-\frac{3}{2}ikasin\theta}[e^{\frac{-ikasin\theta}{2}} - e^{\frac{ikasin\theta}{2}}][/tex]
    [tex]\frac{\alpha}{a}\frac{1}{iksin\theta}e^{-ikasin\theta} - \frac{\alpha}{a^2}\frac{1}{k^2sin^2\theta} e^{\frac{=3ikasin\theta}{2}} -2i sin (\frac{ka sin\theta}{2})[/tex]
    [tex]= \frac{-\alpha}{a}\frac{i}{ksin\theta} e^{-ikasin\theta} + \frac{2\alpha}{a^2}\frac{i}{k^2sin^2\theta} e^{\frac{-3ikasin\theta}{2}} sin(\frac{ka sin \theta}{2})[/tex]

    Adding the first and second integral and then multiplying it by its complex conjugate takes me nowhere..



    I'm definitely going with the convolution method with this one.

    Let ##V_c = \frac{1}{2a}## for -2a < y < a and 0 < y < a.

    Fourier transform of ##V_c##=
    [tex]\alpha \int_{-2a}^a \frac{1}{2a} e^{-i\beta y} dy + \alpha \int_0^a \frac{1}{2a} e^{-i\beta y} dy [/tex]

    Second integral is simply ## \frac{1}{2} \alpha e^{\frac{-i\beta a}{2}} \frac{sin ( \frac{\beta a}{2})}{\frac{\beta a}{2}}##

    First Integral:

    [tex]\frac{\alpha}{2a}\int_{-2a}^{a} e^{-i\beta b} dy [/tex]
    [tex]= \frac{\alpha}{2a\beta i}[e^{-i\beta y}]_a^{-2a} [/tex]
    [tex]= \frac{\alpha}{2a\beta i} e^{\frac{i\beta a}{2}} [e^{\frac{3i\beta a}{2} - e^{\frac{-3i\beta a}{2}}}][/tex]
    [tex] = \frac{3}{2} \alpha e^{\frac{i\beta a}{2}} \frac{sin (\frac{3}{2}\beta a)}{\frac{3}{2}\beta a}[/tex]

    Adding both integrals and multiplying them with its complex conjugate:

    [tex]|A_{\theta}|^2 = \left(\frac{9}{4}\alpha^2\right) \frac{sin^2 (\frac{3}{2}\beta a)}{(\frac{3}{2}\beta a)^2} + \left(\frac{\alpha^2}{4}\right) \frac{sin^2 (\frac{\beta a}{2})}{(\frac{\beta a}{2})^2} + \frac{3}{2} cos (\beta a) \alpha^2 \frac{sin (\frac{3}{2}\beta a)}{(\frac{3}{2}\beta a)} \frac{sin(\frac{\beta a}{2})}{(\frac{\beta a}{2})} [/tex]

    First minimum occurs when ##\beta = \pi##, but due to the presence of the cross-term, it is non-zero. Is this explanation right?
    Last edited: Feb 3, 2014
  2. jcsd
  3. Feb 7, 2014 #2
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