# Convolution, Triangle Function

1. Feb 3, 2014

### unscientific

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

Part(a)

Convolution Method

$$V_b = \frac{1}{2a}, 0 \leq y \leq a$$

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

Fourier transform of $V_b$:

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

Analytical Method

$$A_{\theta} = \alpha \int_{-\infty}^{\infty} T_y e^{-iky sin {\theta}} dy$$

For 0 < y < a:
$$T_y = \frac{1}{a^2}$$

For a < y < 2a:
$$T_y = -\frac{1}{a^2}y + \frac{2}{a}$$

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

First Integral:

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

Second Integral:

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

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

Part(b)

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 $V_c$=
$$\alpha \int_{-2a}^a \frac{1}{2a} e^{-i\beta y} dy + \alpha \int_0^a \frac{1}{2a} e^{-i\beta y} dy$$

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:

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

Adding both integrals and multiplying them with its complex conjugate:

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

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. Feb 7, 2014

bumpp