克拉克变换 (Clarke transformation)也称为
α
β
γ
{\displaystyle \alpha \beta \gamma }
变换 ,是电机工程学 里简化三相电 分析的数学变换 ,常用在三相逆变器 的控制上,可以将平衡的三相系统转换为互相垂直的二相系统,方便信号的处理。
克拉克变换和派克变换 (park transform)都是简化三相电分析用的数学变换,在电机控制时也常一起使用。
历史
伊迪丝·克拉克 在1937年和1938年发表了应用在不平衡三相电力系统的变换及计算,此方法可以简化计算[ 1] 。
定义
伊迪丝·克拉克用在三相电流的克拉克变换如下[ 2] :
i
α
β
γ
(
t
)
=
T
i
a
b
c
(
t
)
=
2
3
[
1
−
1
2
−
1
2
0
3
2
−
3
2
1
2
1
2
1
2
]
[
i
a
(
t
)
i
b
(
t
)
i
c
(
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)
]
{\displaystyle i_{\alpha \beta \gamma }(t)=Ti_{abc}(t)={\frac {2}{3}}{\begin{bmatrix}1&-{\frac {1}{2}}&-{\frac {1}{2}}\\0&{\frac {\sqrt {3}}{2}}&-{\frac {\sqrt {3}}{2}}\\{\frac {1}{2}}&{\frac {1}{2}}&{\frac {1}{2}}\\\end{bmatrix}}{\begin{bmatrix}i_{a}(t)\\i_{b}(t)\\i_{c}(t)\end{bmatrix}}}
其中
i
a
b
c
(
t
)
{\displaystyle i_{abc}(t)}
是一般的三相电流
i
α
β
γ
(
t
)
{\displaystyle i_{\alpha \beta \gamma }(t)}
是经
T
{\displaystyle T}
变换后所得的电流
逆变换为:
i
a
b
c
(
t
)
=
T
−
1
i
α
β
γ
(
t
)
=
[
1
0
1
−
1
2
3
2
1
−
1
2
−
3
2
1
]
[
i
α
(
t
)
i
β
(
t
)
i
γ
(
t
)
]
.
{\displaystyle i_{abc}(t)=T^{-1}i_{\alpha \beta \gamma }(t)={\begin{bmatrix}1&0&1\\-{\frac {1}{2}}&{\frac {\sqrt {3}}{2}}&1\\-{\frac {1}{2}}&-{\frac {\sqrt {3}}{2}}&1\end{bmatrix}}{\begin{bmatrix}i_{\alpha }(t)\\i_{\beta }(t)\\i_{\gamma }(t)\end{bmatrix}}.}
上述的克拉克变换保持了电机变数的量值。考虑以下对称的三相电流信号
i
a
(
t
)
=
2
I
cos
θ
(
t
)
,
i
b
(
t
)
=
2
I
cos
(
θ
(
t
)
−
2
3
π
)
,
i
c
(
t
)
=
2
I
cos
(
θ
(
t
)
+
2
3
π
)
,
{\displaystyle {\begin{aligned}i_{a}(t)=&{\sqrt {2}}I\cos \theta (t),\\i_{b}(t)=&{\sqrt {2}}I\cos \left(\theta (t)-{\frac {2}{3}}\pi \right),\\i_{c}(t)=&{\sqrt {2}}I\cos \left(\theta (t)+{\frac {2}{3}}\pi \right),\end{aligned}}}
其中
I
{\displaystyle I}
是
i
a
(
t
)
{\displaystyle i_{a}(t)}
、
i
b
(
t
)
{\displaystyle i_{b}(t)}
和
i
c
(
t
)
{\displaystyle i_{c}(t)}
的平方平均数
θ
(
t
)
{\displaystyle \theta (t)}
是随时间变化的角度,可以表示为
ω
t
{\displaystyle \omega t}
。
将上述三相电流信号进行变换,可得
i
α
=
2
I
cos
θ
(
t
)
,
i
β
=
2
I
sin
θ
(
t
)
,
i
γ
=
0
,
{\displaystyle {\begin{aligned}i_{\alpha }=&{\sqrt {2}}I\cos \theta (t),\\i_{\beta }=&{\sqrt {2}}I\sin \theta (t),\\i_{\gamma }=&0,\end{aligned}}}
变换后的电流量值和变换前相同。
功率不变变换
电机系统的电压和电流,经过上述的变换后,实功和虚功会和原系统会差一个系数,原因是因为
T
{\displaystyle T}
不是酉矩阵 (unitary matrix)。若要让实功和虚功的值在变换前后相同,需要用以下的变换
i
α
β
γ
(
t
)
=
T
i
a
b
c
(
t
)
=
2
3
[
1
−
1
2
−
1
2
0
3
2
−
3
2
1
2
1
2
1
2
]
[
i
a
(
t
)
i
b
(
t
)
i
c
(
t
)
]
,
{\displaystyle i_{\alpha \beta \gamma }(t)=Ti_{abc}(t)={\sqrt {\frac {2}{3}}}{\begin{bmatrix}1&-{\frac {1}{2}}&-{\frac {1}{2}}\\0&{\frac {\sqrt {3}}{2}}&-{\frac {\sqrt {3}}{2}}\\{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}&{\frac {1}{\sqrt {2}}}\\\end{bmatrix}}{\begin{bmatrix}i_{a}(t)\\i_{b}(t)\\i_{c}(t)\end{bmatrix}},}
变换矩阵是酉矩阵,而且其逆矩阵恰好为其转置矩阵[ 3]
不过在此例中,变换后电流的量值就和变换前不同了,变换后的电流如下
i
α
=
3
I
cos
θ
(
t
)
,
i
β
=
3
I
sin
θ
(
t
)
,
i
γ
=
0.
{\displaystyle {\begin{aligned}i_{\alpha }=&{\sqrt {3}}I\cos \theta (t),\\i_{\beta }=&{\sqrt {3}}I\sin \theta (t),\\i_{\gamma }=&0.\end{aligned}}}
其逆变换为
i
a
b
c
(
t
)
=
2
3
[
1
0
1
2
−
1
2
3
2
1
2
−
1
2
−
3
2
1
2
]
[
i
α
(
t
)
i
β
(
t
)
i
γ
(
t
)
]
.
{\displaystyle i_{abc}(t)={\sqrt {\frac {2}{3}}}{\begin{bmatrix}1&0&{\frac {1}{\sqrt {2}}}\\-{\frac {1}{2}}&{\frac {\sqrt {3}}{2}}&{\frac {1}{\sqrt {2}}}\\-{\frac {1}{2}}&-{\frac {\sqrt {3}}{2}}&{\frac {1}{\sqrt {2}}}\\\end{bmatrix}}{\begin{bmatrix}i_{\alpha }(t)\\i_{\beta }(t)\\i_{\gamma }(t)\end{bmatrix}}.}
简化的变换
在平衡系统中,
i
a
(
t
)
+
i
b
(
t
)
+
i
c
(
t
)
=
0
{\displaystyle i_{a}(t)+i_{b}(t)+i_{c}(t)=0}
,因此,
i
γ
(
t
)
=
0
{\displaystyle i_{\gamma }(t)=0}
。以下是简化版的变换[ 4] [ 5]
i
α
β
(
t
)
=
2
3
[
1
−
1
2
−
1
2
0
3
2
−
3
2
]
[
i
a
(
t
)
i
b
(
t
)
i
c
(
t
)
]
=
[
1
0
1
3
2
3
]
[
i
a
(
t
)
i
b
(
t
)
]
{\displaystyle {\begin{aligned}i_{\alpha \beta }(t)&={\frac {2}{3}}{\begin{bmatrix}1&-{\frac {1}{2}}&-{\frac {1}{2}}\\0&{\frac {\sqrt {3}}{2}}&-{\frac {\sqrt {3}}{2}}\end{bmatrix}}{\begin{bmatrix}i_{a}(t)\\i_{b}(t)\\i_{c}(t)\end{bmatrix}}\\&={\begin{bmatrix}1&0\\{\frac {1}{\sqrt {3}}}&{\frac {2}{\sqrt {3}}}\end{bmatrix}}{\begin{bmatrix}i_{a}(t)\\i_{b}(t)\end{bmatrix}}\end{aligned}}}
是只考虑前二个方程的克拉克变换,其逆变换如下
i
a
b
c
(
t
)
=
3
2
[
2
3
0
−
1
3
3
3
−
1
3
−
3
3
]
[
i
α
(
t
)
i
β
(
t
)
]
{\displaystyle i_{abc}(t)={\frac {3}{2}}{\begin{bmatrix}{\frac {2}{3}}&0\\-{\frac {1}{3}}&{\frac {\sqrt {3}}{3}}\\-{\frac {1}{3}}&-{\frac {\sqrt {3}}{3}}\end{bmatrix}}{\begin{bmatrix}i_{\alpha }(t)\\i_{\beta }(t)\end{bmatrix}}}
几何诠释
相关条目
参考资料
^ O'Rourke, Colm J. A Geometric Interpretation of Reference Frames and Transformations: dq0, Clarke, and Park . IEEE Transactions on Energy Conversion. December 2019, 34, 4 (4): 2070–2083. Bibcode:2019ITEnC..34.2070O . S2CID 203113468 . doi:10.1109/TEC.2019.2941175 . hdl:1721.1/123557 –通过MIT Open Access Articles (英语) .
^ W. C. Duesterhoeft; Max W. Schulz; Edith Clarke. Determination of Instantaneous Currents and Voltages by Means of Alpha, Beta, and Zero Components. Transactions of the American Institute of Electrical Engineers. July 1951, 70 (2): 1248–1255. ISSN 0096-3860 . S2CID 51636360 . doi:10.1109/T-AIEE.1951.5060554 .
^ S. CHATTOPADHYAY; M. MITRA; S. SENGUPTA. Area Based Approach for Three Phase Power Quality Assessment in Clarke Plane . Journal of Electrical Systems. 2008, 04 (1): 62 [2020-11-26 ] .
^ F. Tahri, A.Tahri, Eid A. AlRadadi and A. Draou Senior, "Analysis and Control of Advanced Static VAR compensator Based on the Theory of the Instantaneous Reactive Power," presented at ACEMP, Bodrum, Turkey, 2007.
^ Clarke Transform . www.mathworks.com.
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