I am working on this question, using all sorts of trig identities, but I didn't get very far. I am not sure also how to do the double integral. The problem is:
what is given is:
1. \(\displaystyle L_{s}=\displaystyle{\int} dxdyD_{s}(x,y)Acos(Kxcos(\Theta)+Kysin(\Theta)-\Phi)\)
2. \(\displaystyle D_{s}(x,y)=\frac{1}{2\pi\sigma_{x}\sigma_{y}}e^{(-\frac{x^2}{2\sigma_{x}^2}-\frac{y^2}{2\sigma_{y}^2})}cos(kx-\phi)\)
3. \(\displaystyle \sigma_{x}=\sigma_{y}=\sigma\)
what you have to proove is:
\(\displaystyle L_{s}=\frac{A}{2}e^{(-\frac{\sigma^2(k^2+K^2)}{2})}\displaystyle{(}cos(\phi-\Phi)e^{(\sigma^2kKcos(\Theta))}+cos(\phi+\Phi)e^{(-\sigma^2kKcos(\Theta))}\displaystyle{)}\)
Any help will be greatly appreciated...
Thanks,
eran
what is given is:
1. \(\displaystyle L_{s}=\displaystyle{\int} dxdyD_{s}(x,y)Acos(Kxcos(\Theta)+Kysin(\Theta)-\Phi)\)
2. \(\displaystyle D_{s}(x,y)=\frac{1}{2\pi\sigma_{x}\sigma_{y}}e^{(-\frac{x^2}{2\sigma_{x}^2}-\frac{y^2}{2\sigma_{y}^2})}cos(kx-\phi)\)
3. \(\displaystyle \sigma_{x}=\sigma_{y}=\sigma\)
what you have to proove is:
\(\displaystyle L_{s}=\frac{A}{2}e^{(-\frac{\sigma^2(k^2+K^2)}{2})}\displaystyle{(}cos(\phi-\Phi)e^{(\sigma^2kKcos(\Theta))}+cos(\phi+\Phi)e^{(-\sigma^2kKcos(\Theta))}\displaystyle{)}\)
Any help will be greatly appreciated...
Thanks,
eran