1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convolution, Triangle Function

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

    2ppyav6.png

    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

    Part(a)

    Convolution Method

    6s59av.png

    [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]

    1zxo41h.png

    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..

    Part(b)

    wloeq1.png

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Convolution, Triangle Function
  1. Heavyside convolution (Replies: 5)

Loading...