## Abstract

Optical differentiation and integration are not only interesting from the standpoint of optical computation, but are also important in picture processing. Based on optical correlation, or convolution, and the applications of difference equations, a technique for synthesizing the differentiation and integration filters is presented. This technique compares favorably with other methods which now exist.

© 1971 Optical Society of America

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### Equations (13)

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(1)
$${\mathrm{\Delta}}_{x}f=[f(x+h,y)-f(x,y)]/h,$$
(2)
$$\mathit{\int}\mathit{\int}f(\zeta ,\eta )\delta [\zeta -(x+a),\eta -(y+b)]d\zeta d\eta =f(x+a,y+b).$$
(3)
$${\mathrm{\Delta}}_{x}f=\frac{1}{h}\mathit{\int}\mathit{\int}f(\zeta ,\eta )[\delta (\zeta -x-h,\eta -y)-\delta (\zeta -x,\eta -y)]d\zeta d\eta \equiv (1/h)f(x,y)*{g}_{1}(x,y),$$
(4)
$$\partial f/\partial x=\underset{h\to 0}{lim}{\mathrm{\Delta}}_{x}f.$$
(5)
$$\partial f/\partial y=\underset{h\to 0}{lim}[(1/h)f(x,y)*{g}_{2}(x,y)],$$
(6)
$$\begin{array}{l}{g}_{2}(x,y)=\delta (x,y+h)-\delta (x,y);\hfill \\ \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}=\underset{h\to 0}{lim}\left[\frac{1}{h}f(x,y)*{g}_{3}(x,y)\right],\hfill \end{array}$$
(7)
$$\begin{array}{l}{g}_{3}(x,y)=\delta (x+h,y)+\delta (x,y+h)-2\delta (x,y);\hfill \\ \frac{{\partial}^{2}f}{\partial {x}^{2}}+\frac{{\partial}^{2}f}{\partial {y}^{2}}=\underset{h\to 0}{lim}\left[\frac{1}{{h}^{2}}f(x,y)*{g}_{4}(x,y)\right],\hfill \end{array}$$
(8)
$${g}_{4}(x,y)=\delta (x+h,y)+\delta (x-h,y)+\delta (x,y+h)+\delta (x,y-h)-4\delta (x,y);$$
(9)
$${\partial}^{2}f/\partial x\partial y=\underset{h\to 0}{lim}[(1/{h}^{2})f(x,y)*{g}_{5}(x,y)],$$
(10)
$${g}_{5}(x,y)=\delta (x+h,y+h)+\delta (x,y)-\delta (x,y+h)-\delta (x+h,y).$$
(11)
$$f(x,y)*u(x,0)=\underset{-\infty}{\overset{\infty}{\mathit{\int}\mathit{\int}}}f(\zeta ,\eta )u(x-\zeta ,-\eta )d\zeta d\eta ={\mathit{\int}}_{-\infty}^{x}f(\zeta ,\eta )d\zeta .$$
(12)
$$f(x,y)*u(x,y)=\underset{-\infty}{\overset{\infty}{\mathit{\int}\mathit{\int}}}f(\zeta ,\eta )u(x-\zeta ,y-\eta )d\zeta d\eta =\underset{-\infty}{{\mathit{\int}}^{x}{\mathit{\int}}^{y}}f(\zeta ,\eta )d\zeta d\eta .$$
(13)
$${g}_{1}(x,y)=\delta (x+h,y)-\delta (x,y).$$