克拉克變換 (Clarke transformation)也稱為
α
β
γ
{\displaystyle \alpha \beta \gamma }
變換 ,是電機工程學 裡簡化三相電 分析的數學變換 ,常用在三相逆變器 的控制上,可以將平衡的三相系統轉換為互相垂直的二相系統,方便信號的處理。
克拉克變換和派克变换 (park transform)都是簡化三相電分析用的數學變換,在電機控制時也常一起使用。
歷史
伊迪絲·克拉克 在1937年和1938年發表了應用在不平衡三相電力系統的變換及計算,此方法可以簡化計算[ 1] 。
定義
伊迪絲·克拉克用在三相電流的克拉克變換如下[ 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
(
t
)
]
{\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)}
是一般的三相電流
i
α
β
γ
(
t
)
{\displaystyle i_{\alpha \beta \gamma }(t)}
是經
T
{\displaystyle T}
變換後所得的電流
逆變換為:
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}}.}
上述的克拉克變換保持了電機變數的量值。考慮以下對稱的三相電流信號
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)}
是隨時間變化的角度,可以表示為
ω
t
{\displaystyle \omega t}
。
將上述三相電流信號進行變換,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}}}
變換後的電流量值和變換前相同。
功率不變變換
電機系統的電壓和電流,經過上述的變換後,實功和虛功會和原系統會差一個係數,原因是因為
T
{\displaystyle T}
不是酉矩阵 (unitary matrix)。若要讓實功和虛功的值在變換前後相同,需要用以下的變換
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}},}
變換矩陣是酉矩阵,而且其逆矩陣恰好為其轉置矩陣[ 3]
不過在此例中,變換後電流的量值就和變換前不同了,變換後的電流如下
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}}}
其逆變換為
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}}.}
簡化的變換
在平衡系統中,
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}
。以下是簡化版的變換[ 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}}}
是只考慮前二個方程的克拉克變換,其逆變換如下
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}}}
幾何詮釋
相關條目
參考資料
^ 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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