超前校正系统传递函数的一般形式为
H
(
s
)
=
K
⋅
(
1
+
s
ω
z
)
(
1
+
s
ω
p
)
H(s) = \frac{K \cdot (1 + \frac{s}{\omega_z})}{(1 + \frac{s}{\omega_p})}
H(s)=(1+ωps)K⋅(1+ωzs)
其中
ω
z
,
ω
p
分别为零极点，
K
为比例系数
ω_z,ω_p分别为零极点，K为比例系数
ωz,ωp分别为零极点，K为比例系数
其相移为：
θ
=
arctan
(
ω
ω
z
)
−
arctan
(
ω
ω
p
)
θ = \arctan\left(\frac{\omega}{\omega_z}\right) - \arctan\left(\frac{\omega}{\omega_p}\right)
θ=arctan(ωzω)−arctan(ωpω)
当
ω
2
=
ω
p
⋅
ω
z
时候，相移
θ
最大
\omega^2 = \omega_p \cdot \omega_z 时候，相移θ最大
ω2=ωp⋅ωz时候，相移θ最大
由此得到：
θ
m
=
arctan
(
ω
c
ω
z
)
−
arctan
(
ω
c
ω
p
)
\theta_m = \arctan\left(\frac{\omega_c}{\omega_z}\right) - \arctan\left(\frac{\omega_c}{\omega_p}\right)
θm=arctan(ωzωc)−arctan(ωpωc)
θ
m
=
arctan
(
ω
p
ω
z
)
−
arctan
(
ω
z
ω
p
)
\theta_m = \arctan\left(\sqrt{\frac{\omega_p}{\omega_z}}\right) - \arctan\left(\sqrt{\frac{\omega_z}{\omega_p}}\right)
θm=arctan(ωzωp
)−arctan(ωpωz
)
其中
θ
m
为最大相移动，
ω
c
为最大相移对应的频率
其中\theta_m为最大相移动，\omega_c为最大相移对应的频率
其中θm为最大相移动，ωc为最大相移对应的频率
设
ω
p
ω
z
=
a
设\sqrt{\frac{\omega_p}{\omega_z}} = a
设ωzωp
=a 有
θ
m
=
a
r
c
t
a
n
(
a
)
−
a
r
c
t
a
n
(
1
/
a
)
θm = arctan(a)-arctan(1/a)
θm=arctan(a)−arctan(1/a)
根据公式
tan
(
a
−
b
)
=
tan
(
a
)
−
tan
(
b
)
1
+
tan
(
a
)
⋅
tan
(
b
)
\tan(a-b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a) \cdot \tan(b)}
tan(a−b)=1+tan(a)⋅tan(b)tan(a)−tan(b)
tan
(
θ
m
)
=
tan
(
arctan
(
a
)
−
arctan
(
1
a
)
)
=
a
−
1
a
1
+
a
⋅
1
a
=
a
−
1
a
2
\tan(\theta_m) = \tan\left(\arctan(a) - \arctan\left(\frac{1}{a}\right)\right) = \frac{a - \frac{1}{a}}{1 + a \cdot \frac{1}{a}} = \frac{a - \frac{1}{a}}{2}
tan(θm)=tan(arctan(a)−arctan(a1))=1+a⋅a1a−a1=2a−a1
整理的得到
a
2
−
2
tan
(
θ
m
)
a
−
1
=
0
a^2 - 2\tan(\theta_m)a - 1 = 0
a2−2tan(θm)a−1=0
( a − tan ( θ m ) ) 2 = 1 + ( tan ( θ m ) ) 2 (a - \tan(\theta_m))^2 = 1 + (\tan(\theta_m))^2 (a−tan(θm))2=1+(tan(θm))2
a
=
sin
(
θ
m
)
cos
(
θ
m
)
+
1
+
(
sin
(
θ
m
)
cos
(
θ
m
)
)
2
a = \frac{\sin(\theta_m)}{\cos(\theta_m)} + \sqrt{1 + \left(\frac{\sin(\theta_m)}{\cos(\theta_m)}\right)^2}
a=cos(θm)sin(θm)+1+(cos(θm)sin(θm))2
a
=
sin
(
θ
m
)
cos
(
θ
m
)
+
1
+
(
sin
(
θ
m
)
cos
(
θ
m
)
)
2
=
sin
(
θ
m
)
cos
(
θ
m
)
+
1
c
o
s
(
θ
m
)
a = \frac{\sin(\theta_m)}{\cos(\theta_m)} + \sqrt{1 + \left(\frac{\sin(\theta_m)}{\cos(\theta_m)}\right)^2} = \frac{\sin(\theta_m)}{\cos(\theta_m)} +\frac{1}{cos(\theta_m)}
a=cos(θm)sin(θm)+1+(cos(θm)sin(θm))2
=cos(θm)sin(θm)+cos(θm)1
a
=
1
+
sin
(
θ
m
)
1
−
sin
2
(
θ
m
)
=
1
+
sin
(
θ
m
)
(
1
−
sin
(
θ
m
)
)
(
1
+
sin
(
θ
m
)
)
a = \frac{1 + \sin(\theta_m)}{\sqrt{1 - \sin^2(\theta_m)}} = \frac{1 + \sin(\theta_m)}{\sqrt{(1 - \sin(\theta_m))(1 + \sin(\theta_m))}}
a=1−sin2(θm)
1+sin(θm)=(1−sin(θm))(1+sin(θm))
1+sin(θm)
得到
a
=
1
+
sin
(
θ
m
)
1
−
sin
(
θ
m
)
a = \sqrt{\frac{1 + \sin(\theta_m)}{1 - \sin(\theta_m)}}
a=1−sin(θm)1+sin(θm)
由 ω c 2 = ω p ⋅ ω z \omega_c^2 = \omega_p \cdot \omega_z ωc2=ωp⋅ωz和 ω p ω z = a \sqrt{\frac{\omega_p}{\omega_z}} = a ωzωp =a得到：
ω p = ω c 2 ω z = ω c 2 a 2 ω p 得 : \omega_p = \frac{\omega_c^2}{\omega_z} = \frac{\omega_c^2a^2}{\omega_p }得: ωp=ωzωc2=ωpωc2a2得:
ω p = ω c a = ω c 1 + sin ( θ m ) 1 − sin ( θ m ) \omega_p = \omega_c a = \omega_c \sqrt{\frac{1 + \sin(\theta_m)}{1 - \sin(\theta_m)}} ωp=ωca=ωc1−sin(θm)1+sin(θm)
由
ω
z
=
ω
c
2
/
ω
p
=
ω
c
2
a
2
∗
ω
z
得到
:
由ω_z = ω_c^2/ω_p = \frac{ ω_c^2}{a^2*ω_z}得到:
由ωz=ωc2/ωp=a2∗ωzωc2得到:
ω
z
=
ω
c
a
=
ω
c
1
−
sin
(
θ
m
)
1
+
sin
(
θ
m
)
\omega_z = \omega_c a = \omega_c \sqrt{\frac{1 - \sin(\theta_m)}{1 +\sin(\theta_m)}}
ωz=ωca=ωc1+sin(θm)1−sin(θm)
由上面的两个公式，即可以由需要补偿的 相移幅度 θ m 和补偿点频率 ω c 相移幅度θ_m和补偿点频率ω_c 相移幅度θm和补偿点频率ωc得到 补偿极点频率 ω p 和零点频率 ω z 补偿极点频率ω_p和零点频率ω_z 补偿极点频率ωp和零点频率ωz
标签：超前,校正,frac,arctan,传递函数,theta,tan,omega,sin From： https://blog.csdn.net/weixin_42668920/article/details/142733716