# Why is a circular aperture defined by the circle function rather than a heaviside step function?

In the chapter 4 of the text "Introduction To Fourier Optics" by Goodman, the diffraction pattern of light passing through a circular aperture was studied. The diffraction pattern on the image plane $$(x_i, y_i)$$, is given by the fourier transform of the amplitude transmittance $$U(x_o,y_o)$$ on the aperture plane $$(x_o, y_o)$$ with relevant phases; through the Fraunhofer approximation. I.e.,

$$U(x_i,y_i) = \frac{\exp[ikz] \exp[ik(x_i^2 + y_i^2)/2z]}{i\lambda z}\mathcal{F}\left[U(x_o,y_o)\right]\, .$$

For a circular aperture of diameter $$l$$, the author uses the amplitude transmittance, $$U(x_o,y_o)=circ\left(\frac{r}{l/2}\right)$$. Where $$r=\sqrt{x_o^2+y_o^2}$$, and,

$$circ(r) = \begin{cases} 1, & \text{for } r < 1 \\ 1/2, & \text{for } r = 1 \\ 0, & \text{for }r > 1 \, . \end{cases}$$

My question is, what is the reason for the $$1/2$$ when $$r=1$$? Are we not able to use a heaviside step function in the radial coordinate to encapsulate the amplitude transmittance? I.e. $$U(x_o,y_o) = H(r-l/2)$$, where,

$$H(r-r_0) = \begin{cases} 1, & \text{for } r \leq r_0 \\ 0, & \text{for } r > r_0 \,. \end{cases}$$