以 $\lambda$ 为周期的函数 $E(z)$ 的傅里叶级数的复数形式为
$$
E(z) = \sum_{n=-\infty}^{\infty} C_n e^{\mathrm{i} n k z}
$$
其中 $k = \frac{2 \pi}{\lambda}$，而
$$
C_n = \frac{1}{\lambda} \int_{-\lambda / 2}^{\lambda / 2} E(z) e^{-\mathrm{i} n k z} \, \mathrm{d} z
$$

矩形波在一个周期 $\left[-\frac{\lambda}{2}, \frac{\lambda}{2}\right]$ 内的函数式为
$$
E(z) = \begin{cases} 
+1 & \text{当 } 0 < z < \frac{\lambda}{2} \\
-1 & \text{当 } -\frac{\lambda}{2} < z < 0 
\end{cases}
$$

因此
$$
\begin{aligned}
C_n & = \frac{1}{\lambda} \int_{-\lambda / 2}^{\lambda / 2} E(z) e^{-\mathrm{i} n k z} \, \mathrm{d} z \\
& = \frac{1}{\lambda} \int_{-\lambda / 2}^{0} (-1) e^{-\mathrm{i} n k z} \, \mathrm{d} z + \frac{1}{\lambda} \int_{0}^{\lambda / 2} (+1) e^{-\mathrm{i} n k z} \, \mathrm{d} z \\
& = \frac{1}{\mathrm{i} n 2 \pi} \left[ e^{-\mathrm{i} n k z} \right]_{-\lambda / 2}^{0} - \frac{1}{\mathrm{i} n 2 \pi} \left[ e^{-\mathrm{i} n k z} \right]_{0}^{\lambda / 2} \\
& = \frac{1}{\mathrm{i} n 2 \pi} \left[ \left(1 - e^{\mathrm{i} n \pi}\right) + \left(1 - e^{-\mathrm{i} n \pi}\right) \right] \\
& = \frac{1}{\mathrm{i} n \pi} (1 - \cos n \pi) \\
& = \begin{cases} 
0 & \text{当 } n \text{ 为偶数} \\
\frac{2}{\mathrm{i} n \pi} & \text{当 } n \text{ 为奇数}
\end{cases}
\end{aligned}
$$

$$
C_0 = 0
$$

所以矩形波的傅里叶分析表示式为
$$
\begin{aligned}
E(z) & = \sum_{n=-\infty}^{\infty} C_n e^{\mathrm{i} n k z} \\
& = \frac{2}{\mathrm{i} \pi} \left( e^{\mathrm{i} k z} - e^{-\mathrm{i} k z} \right) + \frac{2}{\mathrm{i} 3 \pi} \left( e^{\mathrm{i} 3 k z} - e^{-\mathrm{i} 3 k z} \right) \\
& + \frac{2}{\mathrm{i} 5 \pi} \left( e^{\mathrm{i} 5 k z} - e^{-\mathrm{i} 5 k z} \right) + \cdots \\
& = -\frac{4}{\pi} \left( \sin k z + \frac{1}{3} \sin 3 k z + \frac{1}{5} \sin 5 k z + \cdots \right)
\end{aligned}
$$

半无限大屏的直边与 $\eta$ 轴重合，在 $\xi=0$ 处由挡光跳变为透光，由式 (4.7.6)，仿射式为
$$
\begin{align*}
E(x, y) & = \frac{\exp (\mathrm{j} k z)}{2 \mathrm{j}} \int_{0}^{+\infty} \exp \left[\frac{\mathrm{j} k}{2 z}(x-\xi)^{2}\right] \mathrm{d} \xi \int_{-\infty}^{+\infty} \exp \left[\frac{\mathrm{j} k}{2 z}(y-\eta)^{2}\right] \mathrm{d} \eta \\
& = \frac{\exp (\mathrm{j} k z)}{2 \mathrm{j}} \left[ F\left( \alpha_1 \right) - F\left( \alpha_2 \right) \right] \left[ F\left( \beta_1 \right) - F\left( \beta_2 \right) \right] \\
& = \frac{\exp (\mathrm{j} k z)}{2 \mathrm{j}} \left[ F\left( x \sqrt{2 / (\lambda z)} \right) - F(-\infty) \right] \left[ F(\infty) - F(-\infty) \right]
\end{align*}
$$

式中应用了
$$
\begin{array}{ll}
\alpha_1 = \left. \alpha \right|_{k - \varepsilon_1 - 0} = x \sqrt{\frac{2}{\lambda z}}, \quad \alpha_2 = \left. \alpha \right|_{-\varepsilon_2 - +\infty} = -\infty \\
\beta_1 = \left. \beta \right|_{-n = -\infty} = +\infty, & \beta_2 = \left. \beta \right|_{-\xi_2 = +\infty} = -\infty \\
F(+\infty) = (1 + \mathrm{j}) / 2, & F(-\infty) = -(1 + \mathrm{j}) / 2
\end{array}
$$

所以
$$
E(x, y) = \frac{(1 + \mathrm{j}) \exp (\mathrm{j} k z)}{2 \mathrm{j}} \left[ F\left( x \sqrt{2 / (\lambda z)} \right) - F(-\infty) \right]
$$

或者
$$
\begin{align*}
E(x, y) & = \frac{E_{\infty}}{1 + \mathrm{j}} \left[ F\left( x \sqrt{2 / (\lambda z)} \right) - F(-\infty) \right] \\
E_{\infty} & = \frac{(1 + \mathrm{j}) \exp (\mathrm{j} k z)}{2 \mathrm{j}} \left[ F(+\infty) - F(-\infty) \right]
\end{align*}
$$

是无衍射屏时的仿射场。式 (4.7.14) 说明，半无限大屏的衍射场只随 $x$ 变化，衍射图样是平行于 $y$ 轴的直线条纹。