请根据典型二阶系统实验所测数据F:0.160; w:1.005; lgw:0.002; DA:2.000; AD:2.004; M:1.002; DB:0.017; DU:-0.778; Re:1.002; Im:-0.014; F:0.200; w:1.257; lgw:0.099; DA:2.000; AD:2.006; M:1.003; DB:0.026; DU:-1.224; Re:1.003; Im:-0.021; F:0.250; w:1.571; lgw:0.196; DA:2.000; AD:2.011; M:1.006; DB:0.048; DU:-2.610; Re:1.004; Im:-0.046; F:0.320; w:2.011; lgw:0.303; DA:2.000; AD:2.018; M:1.009; DB:0.078; DU:-2.045; Re:1.008; Im:-0.036; F:0.400; w:2.513; lgw:0.400; DA:2.000; AD:2.031; M:1.016; DB:0.134; DU:-2.880; Re:1.014; Im:-0.051; F:0.500; w:3.142; lgw:0.497; DA:2.000; AD:2.050; M:1.025; DB:0.214; DU:-6.120; Re:1.019; Im:-0.109; F:0.630; w:3.958; lgw:0.597; DA:2.000; AD:2.079; M:1.040; DB:0.336; DU:-8.838; Re:1.027; Im:-0.160; F:0.800; w:5.027; lgw:0.701; DA:2.000; AD:2.133; M:1.067; DB:0.559; DU:-8.208; Re:1.056; Im:-0.152; F:1.000; w:6.283; lgw:0.798; DA:2.000; AD:2.219; M:1.110; DB:0.903; DU:-14.040; Re:1.076; Im:-0.269; F:1.260; w:7.917; lgw:0.899; DA:2.000; AD:2.363; M:1.182; DB:1.449; DU:-15.869; Re:1.136; Im:-0.323; F:1.590; w:9.990; lgw:1.000; DA:2.000; AD:2.634; M:1.317; DB:2.392; DU:-24.638; Re:1.197; Im:-0.549; F:2.000; w:12.566; lgw:1.099; DA:2.000; AD:3.056; M:1.528; DB:3.682; DU:-43.200; Re:1.114; Im:-1.046; F:2.520; w:15.834; lgw:1.200; DA:2.000; AD:3.303; M:1.652; DB:4.358; DU:-78.466; Re:0.330; Im:-1.618; F:3.180; w:19.981; lgw:1.301; DA:2.000; AD:2.475; M:1.238; DB:1.851; DU:-116.222; Re:-0.547; Im:-1.110; F:4.000; w:25.133; lgw:1.400; DA:2.000; AD:1.437; M:0.719; DB:-2.871; DU:-140.400; Re:-0.554; Im:-0.458; F:5.000; w:31.416; lgw:1.497; DA:2.000; AD:0.834; M:0.417; DB:-7.597; DU:-156.600; Re:-0.383; Im:-0.166; F:6.340; w:39.835; lgw:1.600; DA:2.000; AD:0.473; M:0.237; DB:-12.523; DU:-161.259; Re:-0.224; Im:-0.076; F:8.000; w:50.265; lgw:1.701; DA:2.000; AD:0.253; M:0.127; DB:-17.958; DU:-166.320; Re:-0.123; Im:-0.030; F:10.000; w:62.832; lgw:1.798; DA:2.000; AD:0.226; M:0.113; DB:-18.938; DU:-165.600; Re:-0.109; Im:-0.028; ，求出相位角φ(w)

根据典型二阶系统的传递函数公式：

H(s) = M / (s^2 + 2ζωn s + ωn^2)

其中 M 为系统增益，ζ为阻尼比，ωn为自然频率。

将传递函数公式转化为极点-零点形式：

H(s) = M * (s - z1) * (s - z2) / [(s - p1) * (s - p2)]

其中 z1、z2为零点，p1、p2为极点。

根据实验数据可得：

M = 1.002，ΔdB = 0.017，Δφ = -0.778

当ω = 1.005时，幅频特性F = 0.160，相位角φ = -0.778

当ω = 1.257时，幅频特性F = 0.200，相位角φ = -1.224

根据相位角的定义，有：

φ = -arctan[(p1 + p2) / (z1 + z2)] - arctan[(ω / ωn) * (z1 + z2) / (p1 + p2)]

将相位角的表达式代入上述实验数据，可得：

-0.778 = -arctan[(p1 + p2) / (z1 + z2)] - arctan[(1.005 / ωn) * (z1 + z2) / (p1 + p2)]

-1.224 = -arctan[(p1 + p2) / (z1 + z2)] - arctan[(1.257 / ωn) * (z1 + z2) / (p1 + p2)]

将两式相减，得：

arctan[(1.005 / ωn) * (z1 + z2) / (p1 + p2)] - arctan[(1.257 / ωn) * (z1 + z2) / (p1 + p2)] = 0.446

根据反正切函数的性质，有：

tan(φ1 - φ2) = [(1.005 / ωn) * (z1 + z2) / (p1 + p2)] / [(1.257 / ωn) * (z1 + z2) / (p1 + p2)] = 0.847

解方程可得：

φ1 - φ2 = arctan(0.847) = 40.380°

因此，相位角φ(w) = φ1 - arctan[(ω / ωn) * (z1 + z2) / (p1 + p2)]，其中φ1 = -0.778 + 40.380° = 39.602°。

根据实验数据可得：

当ω = 1.005时，相位角φ(w) = 39.602° - arctan[(1.005 / ωn) * (z1 + z2) / (p1 + p2)] = -0.778°

同理，可求得其他频率下的相位角φ(w)。