派克变换 (也译作帕克变换 ,英语 :Park's Transformation),是目前分析同步电动机 及感應馬達 运行最常用的一种坐标变换,由美国工程师羅伯特·H·帕克 在1929年提出。派克变换将定子的a,b,c三相电流投影到随着转子旋转的直轴(d轴),交轴(q轴)与垂直于dq平面的零轴(0轴)上去,从而实现了对定子电感矩阵的对角化 ,对电动机的运行分析起到了简化作用。
派克正变换:
i
d
q
0
=
P
i
a
b
c
=
2
3
[
cos
θ
cos
(
θ
−
120
∘
)
cos
(
θ
+
120
∘
)
−
sin
θ
−
sin
(
θ
−
120
∘
)
−
sin
(
θ
+
120
∘
)
1
2
1
2
1
2
]
[
i
a
i
b
i
c
]
{\displaystyle {\mathbf {i} }_{dq0}={\mathbf {P} }{\mathbf {i} }_{abc}={\frac {2}{3}}\left[{\begin{array}{*{20}c}{\cos \theta }&{\cos \left({\theta -120^{\circ }}\right)}&{\cos \left({\theta +120^{\circ }}\right)}\\{-\sin \theta }&{-\sin \left({\theta -120^{\circ }}\right)}&{-\sin \left({\theta +120^{\circ }}\right)}\\{\frac {1}{2}}&{\frac {1}{2}}&{\frac {1}{2}}\\\end{array}}\right]\left[{\begin{array}{*{20}c}{i_{a}}\\{i_{b}}\\{i_{c}}\\\end{array}}\right]}
逆变换:
i
a
b
c
=
P
−
1
i
d
q
0
=
[
cos
θ
−
sin
θ
1
cos
(
θ
−
120
∘
)
−
sin
(
θ
−
120
∘
)
1
cos
(
θ
+
120
∘
)
−
sin
(
θ
+
120
∘
)
1
]
[
i
d
i
q
i
0
]
{\displaystyle {\mathbf {i} }_{abc}={\mathbf {P} }^{-1}{\mathbf {i} }_{dq0}=\left[{\begin{array}{*{20}c}{\cos \theta }&{-\sin \theta }&1\\{\cos \left({\theta -120^{\circ }}\right)}&{-\sin \left({\theta -120^{\circ }}\right)}&1\\{\cos \left({\theta +120^{\circ }}\right)}&{-\sin \left({\theta +120^{\circ }}\right)}&1\\\end{array}}\right]\left[{\begin{array}{*{20}c}{i_{d}}\\{i_{q}}\\{i_{0}}\\\end{array}}\right]}
派克变换也作用在定子电压与定子绕组磁链上:
u
d
q
0
=
P
u
a
b
c
{\displaystyle {\mathbf {u} }_{dq0}={\mathbf {P} }{\mathbf {u} }_{abc}}
,
Ψ
d
q
0
=
P
Ψ
a
b
c
{\displaystyle {\mathbf {\Psi } }_{dq0}={\mathbf {P} }{\mathbf {\Psi } }_{abc}}
上图描绘了派克变换的几何意义,定子三相电流互成120度角,
δ
{\displaystyle \delta }
为定子电流落后于它们对应的相电压的角度。直轴与交轴电流分别等于定子三相电流在d轴与q轴上的投影。(图中的比例系数
3
2
{\displaystyle {\sqrt {\frac {3}{2}}}}
是由于图中所采用的是正交形式的派克变换)d-q坐标系在空间中以角速度
ω
{\displaystyle \omega }
逆时针旋转,故
θ
=
ω
t
{\displaystyle \theta =\omega t}
以d轴领先a相轴线的方向为正。当定子电流为三相对称的正弦交流电时,
i
d
{\displaystyle i_{d}}
,
i
q
{\displaystyle i_{q}}
为直流电流,
i
0
=
0
{\displaystyle i_{0}=0}
。
磁链方程:
[
Ψ
a
b
c
Ψ
f
D
Q
]
=
[
L
S
S
L
S
R
L
R
S
L
R
R
]
[
−
i
a
b
c
i
f
D
Q
]
{\displaystyle \left[{\begin{array}{*{20}c}{{\mathbf {\Psi } }_{abc}}\\{{\mathbf {\Psi } }_{fDQ}}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{{\mathbf {L} }_{SS}}&{{\mathbf {L} }_{SR}}\\{{\mathbf {L} }_{RS}}&{{\mathbf {L} }_{RR}}\\\end{array}}\right]\left[{\begin{array}{*{20}c}{-{\mathbf {i} }_{abc}}\\{{\mathbf {i} }_{fDQ}}\\\end{array}}\right]}
上式中的电感系数矩阵
L
S
S
,
L
S
R
,
L
R
S
,
L
R
R
{\displaystyle {{\mathbf {L} }_{SS}},{{\mathbf {L} }_{SR}},{{\mathbf {L} }_{RS}},{{\mathbf {L} }_{RR}}}
事实上都含有随时间变化的角度参数[ 1] ,使得方程求解困难。
现对等式两边同时左乘
[
P
U
]
{\displaystyle \left[{\begin{array}{*{20}c}{\mathbf {P} }&{}\\{}&{\mathbf {U} }\\\end{array}}\right]}
,其中
U
{\displaystyle {\mathbf {U} }}
为三阶单位矩阵 。方程化为:
[
Ψ
d
q
0
Ψ
f
D
Q
]
=
[
P
U
]
[
L
S
S
L
S
R
L
R
S
L
R
R
]
[
P
−
1
U
]
[
−
i
a
b
c
i
f
D
Q
]
{\displaystyle \left[{\begin{array}{*{20}c}{{\mathbf {\Psi } }_{dq0}}\\{{\mathbf {\Psi } }_{fDQ}}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{\mathbf {P} }&{}\\{}&{\mathbf {U} }\\\end{array}}\right]\left[{\begin{array}{*{20}c}{{\mathbf {L} }_{SS}}&{{\mathbf {L} }_{SR}}\\{{\mathbf {L} }_{RS}}&{{\mathbf {L} }_{RR}}\\\end{array}}\right]\left[{\begin{array}{*{20}c}{{\mathbf {P} }^{-1}}&{}\\{}&{\mathbf {U} }\\\end{array}}\right]\left[{\begin{array}{*{20}c}{-{\mathbf {i} }_{abc}}\\{{\mathbf {i} }_{fDQ}}\\\end{array}}\right]}
[
Ψ
d
q
0
Ψ
f
D
Q
]
=
[
P
L
S
S
P
−
1
P
L
S
R
L
R
S
P
−
1
L
R
R
]
[
−
i
d
q
0
i
f
D
Q
]
{\displaystyle \left[{\begin{array}{*{20}c}{{\mathbf {\Psi } }_{dq0}}\\{{\mathbf {\Psi } }_{fDQ}}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{{\mathbf {PL} }_{SS}{\mathbf {P} }^{-1}}&{{\mathbf {PL} }_{SR}}\\{{\mathbf {L} }_{RS}{\mathbf {P} }^{-1}}&{{\mathbf {L} }_{RR}}\\\end{array}}\right]\left[{\begin{array}{*{20}c}{-{\mathbf {i} }_{dq0}}\\{{\mathbf {i} }_{fDQ}}\\\end{array}}\right]}
其中
P
L
S
S
P
−
1
=
[
L
d
L
q
L
0
]
≜
L
d
q
0
{\displaystyle {\mathbf {PL} }_{SS}{\mathbf {P} }^{-1}=\left[{\begin{array}{*{20}c}{L_{d}}&{}&{}\\{}&{L_{q}}&{}\\{}&{}&{L_{0}}\\\end{array}}\right]\triangleq {\mathbf {L} }_{dq0}}
。
① 变换后的电感系数都变为常数,可以假想dd绕组,qq绕组是固定在转子上的,相对转子静止。
② 派克变换阵对定子自感矩阵
L
S
S
{\displaystyle {\mathbf {L} }_{SS}}
起到了对角化的作用,并消去了其中的角度变量。
L
d
,
L
q
,
L
0
{\displaystyle {L_{d}},{L_{q}},{L_{0}}}
为其特征根。
③ 变换后定子和转子间的互感系数不对称,这是由于派克变换的矩阵不是正交矩阵 。
④
L
d
{\displaystyle {L_{d}}}
为直轴同步电感系数,其值相当于当励磁绕组开路,定子合成磁势产生单纯直轴磁场时,任意一相定子绕组的自感系数。
电压方程:
[
U
a
b
c
U
f
D
Q
]
=
[
r
S
r
R
]
[
−
i
a
b
c
i
f
D
Q
]
+
[
Ψ
˙
a
b
c
Ψ
˙
f
D
Q
]
{\displaystyle \left[{\begin{array}{*{20}c}{{\mathbf {U} }_{abc}}\\{{\mathbf {U} }_{fDQ}}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{{\mathbf {r} }_{S}}&{}\\{}&{{\mathbf {r} }_{R}}\\\end{array}}\right]\left[{\begin{array}{*{20}c}{-{\mathbf {i} }_{abc}}\\{{\mathbf {i} }_{fDQ}}\\\end{array}}\right]+\left[{\begin{array}{*{20}c}{{\mathbf {\dot {\Psi }} }_{abc}}\\{{\mathbf {\dot {\Psi }} }_{fDQ}}\\\end{array}}\right]}
现对等式两边同时左乘
[
P
U
]
{\displaystyle \left[{\begin{array}{*{20}c}{\mathbf {P} }&{}\\{}&{\mathbf {U} }\\\end{array}}\right]}
,其中
U
{\displaystyle {\mathbf {U} }}
为三阶单位矩阵 。方程化为:
[
U
d
q
0
U
f
D
Q
]
=
[
r
S
r
R
]
[
−
i
d
q
0
i
f
D
Q
]
+
[
P
Ψ
˙
a
b
c
Ψ
˙
f
D
Q
]
{\displaystyle \left[{\begin{array}{*{20}c}{{\mathbf {U} }_{dq0}}\\{{\mathbf {U} }_{fDQ}}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{{\mathbf {r} }_{S}}&{}\\{}&{{\mathbf {r} }_{R}}\\\end{array}}\right]\left[{\begin{array}{*{20}c}{-{\mathbf {i} }_{dq0}}\\{{\mathbf {i} }_{fDQ}}\\\end{array}}\right]+\left[{\begin{array}{*{20}c}{{\mathbf {P{\dot {\Psi }}} }_{abc}}\\{{\mathbf {\dot {\Psi }} }_{fDQ}}\\\end{array}}\right]}
由
Ψ
d
q
0
=
P
Ψ
a
b
c
{\displaystyle {\mathbf {\Psi } }_{dq0}={\mathbf {P\Psi } }_{abc}}
,
对两边求导,得
Ψ
˙
d
q
0
=
P
˙
Ψ
a
b
c
+
P
Ψ
˙
a
b
c
{\displaystyle {\mathbf {\dot {\Psi }} }_{dq0}={\mathbf {{\dot {P}}\Psi } }_{abc}+{\mathbf {P{\dot {\Psi }}} }_{abc}}
,
所以
P
Ψ
˙
a
b
c
=
Ψ
˙
d
q
0
−
P
˙
Ψ
a
b
c
=
Ψ
˙
d
q
0
−
P
˙
P
−
1
Ψ
d
q
0
{\displaystyle {\mathbf {P{\dot {\Psi }}} }_{abc}={\mathbf {\dot {\Psi }} }_{dq0}-{\mathbf {{\dot {P}}\Psi } }_{abc}={\mathbf {\dot {\Psi }} }_{dq0}-{\mathbf {{\dot {P}}P} }^{-1}{\mathbf {\Psi } }_{dq0}}
其中
P
˙
P
−
1
=
[
ω
−
ω
]
{\displaystyle {\mathbf {{\dot {P}}P} }^{-1}=\left[{\begin{array}{*{20}c}{}&\omega &{}\\{-\omega }&{}&{}\\{}&{}&{}\\\end{array}}\right]}
,令
S
=
P
˙
P
−
1
Ψ
d
q
0
=
[
ω
−
ω
]
[
Φ
d
Φ
q
Φ
0
]
=
[
ω
Ψ
q
−
ω
Ψ
d
]
{\displaystyle {\mathbf {S} }={\mathbf {{\dot {P}}P} }^{-1}{\mathbf {\Psi } }_{dq0}=\left[{\begin{array}{*{20}c}{}&\omega &{}\\{-\omega }&{}&{}\\{}&{}&{}\\\end{array}}\right]\left[{\begin{array}{*{20}c}{\Phi _{d}}\\{\Phi _{q}}\\{\Phi _{0}}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{\omega \Psi _{q}}\\{-\omega \Psi _{d}}\\{}\\\end{array}}\right]}
于是有
[
U
d
q
0
U
f
D
Q
]
=
[
r
S
r
R
]
[
−
i
d
q
0
i
f
D
Q
]
+
[
Ψ
˙
d
q
0
Ψ
˙
f
D
Q
]
−
[
S
]
{\displaystyle \left[{\begin{array}{*{20}c}{{\mathbf {U} }_{dq0}}\\{{\mathbf {U} }_{fDQ}}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{{\mathbf {r} }_{S}}&{}\\{}&{{\mathbf {r} }_{R}}\\\end{array}}\right]\left[{\begin{array}{*{20}c}{-{\mathbf {i} }_{dq0}}\\{{\mathbf {i} }_{fDQ}}\\\end{array}}\right]+\left[{\begin{array}{*{20}c}{{\mathbf {\dot {\Psi }} }_{dq0}}\\{{\mathbf {\dot {\Psi }} }_{fDQ}}\\\end{array}}\right]-\left[{\begin{array}{*{20}c}{\mathbf {S} }\\{}\\\end{array}}\right]}
上式右边第一项为绕组电阻的压降,第二项为变压器电势,第三项为发电机电势或旋转电势。
^ 定子电感矩阵
L
S
S
=
[
L
a
a
M
a
b
M
a
c
M
b
a
L
b
b
M
b
c
M
c
a
M
c
b
L
c
c
]
{\displaystyle {\mathbf {L} }_{SS}=\left[{\begin{array}{*{20}c}{L_{aa}}&{M_{ab}}&{M_{ac}}\\{M_{ba}}&{L_{bb}}&{M_{bc}}\\{M_{ca}}&{M_{cb}}&{L_{cc}}\\\end{array}}\right]}
,
其中
L
a
a
=
l
0
+
l
2
cos
(
2
θ
)
{\displaystyle L_{aa}=l_{0}+l_{2}\cos \left(2\theta \right)}
L
b
b
=
l
0
+
l
2
cos
2
(
θ
−
120
∘
)
{\displaystyle L_{bb}=l_{0}+l_{2}\cos 2\left({\theta -120^{\circ }}\right)}
L
c
c
=
l
0
+
l
2
cos
2
(
θ
+
120
∘
)
{\displaystyle L_{cc}=l_{0}+l_{2}\cos 2\left({\theta +120^{\circ }}\right)}
M
a
b
=
M
b
a
=
−
m
0
−
m
2
cos
2
(
θ
+
30
∘
)
{\displaystyle M_{ab}=M_{ba}=-m_{0}-m_{2}\cos 2\left({\theta +30^{\circ }}\right)}
M
b
c
=
M
c
b
=
−
m
0
−
m
2
cos
2
(
θ
−
90
∘
)
{\displaystyle M_{bc}=M_{cb}=-m_{0}-m_{2}\cos 2\left({\theta -90^{\circ }}\right)}
M
c
a
=
M
a
c
=
−
m
0
−
m
2
cos
2
(
θ
+
150
∘
)
{\displaystyle M_{ca}=M_{ac}=-m_{0}-m_{2}\cos 2\left({\theta +150^{\circ }}\right)}