10.2 The convolution theorem and diffraction

To handle more complex cases of diffraction using Fourier transforms we need to know the convolution theorem. Say $g(x)$ is the convolution of two other functions $f$ and $h$ . Then $$g(x)=f\otimes h={\int }_{-\infty }^{\infty }f(x^{\prime })h(x-x^{\prime })d{x}^{\prime }$$ It is probably best to illustrate convolution with some examples. In eachexample, the blue line represents the function $h(x-x^{\prime })$ , the red line the function $f(x)$ and the green line is the convolution. In the animation; follow the vertical green line that is the point where the convolution is being evaluated. Itsvalue is the area under the product of the two curves at that point.

It might be easier to picture what is going on if we capture a couple of frames.

The convolution theorem states that if $$G(k)=Ϝ\{g(x)\}$$ $$F(k)=Ϝ\{f(x)\}$$ and $$H(k)=Ϝ\{h(x)\}$$ and if $$g(x)=f\otimes h$$ then $$G(k)=F(k)H(k)\text{.}$$ We can easily show this $${\displaystyle \begin{array}{c}G(k)={\int }_{-\infty }^{\infty }g(x^{\prime })e^{ik{x}^{\prime }}d{x}^{\prime }\\ ={\int }_{-\infty }^{\infty }{e}^{ik{x}^{\prime }}{\int }_{-\infty }^{\infty }f(x)h(x^{\prime }-x)dxd{x}^{\prime }\\ ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }h(x^{\prime }-x)e^{ik{x}^{\prime }}d{x}^{\prime }f(x)dx\end{array}}$$ now set $w=x^{\prime }-x$ then $x^{\prime }=w+x$ $${\displaystyle \begin{array}{c}G(k)={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }h(w)e^{ikw}dw{e}^{ikx}dx\\ ={\int }_{-\infty }^{\infty }h(w)e^{ikw}dw{\int }_{-\infty }^{\infty }f(x)e^{ikx}dx\\ =H(k)F(k)\end{array}}$$

Now say we want to consider the case of two long slits with width $a$ . This can be described by the convolution of one slit with two delta functions.Unfortunately it is not possible to animate this since the delta function is infinitely narrow. However an extremely narrow Gaussian is a goodapproximation to the Dirac delta function and I have used that for the animationbelow.

To sumarize: Fraunhofer diffraction patterns are the Fourier transform of the aperture function. The Fourier transform of the convolution of functions isthe product of the Fourier transforms of the individual functions. each of our complex diffraction cases could be considered the convolution of simplercases, hence the resulting patterns were the products of those simpler cases.