Convolution, Triangle Function

In summary, the conversation discusses finding the intensity as a function of ##\theta## using the convolution and analytical methods. The Fourier transform of ##V_b## is calculated using the convolution method and yields an angular distribution of ##V_{b(\beta)}##. The analytical method is used to find the Fourier transform of ##V_c##, which is then added to the first integral to find the first minimum at ##\beta = \pi##. However, due to the presence of a cross-term, the first minimum is non-zero.
  • #1
unscientific
1,734
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Homework Statement



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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.

Homework Equations


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?
 
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  • #2
bumpp
 

FAQ: Convolution, Triangle Function

What is convolution in terms of mathematics?

Convolution is a mathematical operation that combines two functions to create a third function. It involves multiplying one function by a reversed and shifted version of the other function, and then integrating the result. The resulting function represents the amount of overlap between the two original functions at each point.

How is convolution used in signal processing?

In signal processing, convolution is used to analyze the effects of a linear system on a signal. It allows us to model how a signal will be altered as it passes through a system by convolving the signal with the system's impulse response function. Convolution also plays a key role in filtering, deconvolution, and other signal processing techniques.

What is the triangle function and how does it relate to convolution?

The triangle function, also known as the triangular pulse, is a mathematical function that has a triangular shape. It is often used as an approximation for more complex functions. In convolution, the triangle function is often used as the impulse response function for a linear system, as it allows for easier calculations and can still provide useful insights into the system's behavior.

How does convolution relate to the concept of time-domain and frequency-domain?

In the time-domain, convolution is used to analyze how a signal changes over time as it passes through a system. In the frequency-domain, convolution plays a key role in understanding how a system affects the frequency components of a signal. The convolution theorem states that convolution in the time-domain is equivalent to multiplication in the frequency-domain, allowing us to easily switch between the two domains for analysis.

Can convolution be applied to non-mathematical systems?

Yes, convolution can be applied to non-mathematical systems as long as they can be represented by a mathematical function. For example, in image processing, convolution is used to apply filters to images by convolving the image with a filter kernel. In this case, the image and the filter kernel are both represented as discrete functions, and the convolution operation is used to combine them and create a new image with the desired effects.

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