3.验证 : 粗浅地验证一下, 将两个测量线分别放置在输入输出两个波峰上(红 : 输入, 紫 : 输出),可以看到 : ∵ \because ∵ 输入与输出信号周期 : T = 1 5 s = 200 m s T = \frac{1}{5}s = 200ms T=51​s=200ms 输出滞后输入信号时长 : t = 49.926 m s t = 49.926ms t=49.926ms ∴ \therefore ∴ 迟滞角度 : D 2 = 49.926 m s × 360 ° 200 m s = 89.8668 D_2 = 49.926ms \times\frac{360°}{200ms} =89.8668 D2​=49.926ms×200ms360°​=89.8668