派克變換（也譯作帕克變換，英語(yǔ)：Park's Transformation），是目前分析同步電動(dòng)機(jī)運(yùn)行最常用的一種坐標(biāo)變換，由美國(guó)工程師派克（R.H.Park）在1929年提出。派克變換將定子的a,b,c三相電流投影到隨著轉(zhuǎn)子旋轉(zhuǎn)的直軸（d軸），交軸（q軸）與垂直于dq平面的零軸（0軸）上去，從而實(shí)現(xiàn)了對(duì)定子電感矩陣的對(duì)角化，對(duì)同步電動(dòng)機(jī)的運(yùn)行分析起到了簡(jiǎn)化作用。

定義

　　派克正變換：

　　{\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_}\\{i_{c}}\\\end{array}}\right]}{\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_}\\{i_{c}}\\\end{array}}\right]}

　　逆變換：

　　{\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_v5fb5dt}\\{i_{q}}\\{i_{0}}\\\end{array}}\right]}{\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_5hb55h5}\\{i_{q}}\\{i_{0}}\\\end{array}}\right]}

　　派克變換也作用在定子電壓與定子繞組磁鏈上：{\displaystyle{\mathbf{u}}_{dq0}={\mathbf{P}}{\mathbf{u}}_{abc}}{\displaystyle{\mathbf{u}}_{dq0}={\mathbf{P}}{\mathbf{u}}_{abc}}，{\displaystyle{\mathbf{\Psi}}_{dq0}={\mathbf{P}}{\mathbf{\Psi}}_{abc}}{\displaystyle{\mathbf{\Psi}}_{dq0}={\mathbf{P}}{\mathbf{\Psi}}_{abc}}

幾何解釋

　　上圖描繪了派克變換的幾何意義，定子三相電流互成120度角，{\displaystyle\delta}\delta為定子電流落后于它們對(duì)應(yīng)的相電壓的角度。直軸與交軸電流分別等于定子三相電流在d軸與q軸上的投影。（圖中的比例系數(shù){\displaystyle{\sqrt{\frac{3}{2}}}}{\displaystyle{\sqrt{\frac{3}{2}}}}是由于圖中所采用的是正交形式的派克變換）d-q坐標(biāo)系在空間中以角速度{\displaystyle\omega}\omega逆時(shí)針旋轉(zhuǎn)，故{\displaystyle\theta=\omega t}{\displaystyle\theta=\omega t}以d軸領(lǐng)先a相軸線的方向?yàn)檎?。?dāng)定子電流為三相對(duì)稱的正弦交流電時(shí)，{\displaystyle i_x5fpl5l}{\displaystyle i_nvtd5h5},{\displaystyle i_{q}}{\displaystyle i_{q}}為直流電流，{\displaystyle i_{0}=0}{\displaystyle i_{0}=0}。

用派克變換化簡(jiǎn)同步發(fā)電機(jī)基本方程

變換后的磁鏈方程

　　磁鏈方程：

　　{\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]}{\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]}

　　上式中的電感系數(shù)矩陣{\displaystyle{{\mathbf{L}}_{SS}},{{\mathbf{L}}_{SR}},{{\mathbf{L}}_{RS}},{{\mathbf{L}}_{RR}}}{\displaystyle{{\mathbf{L}}_{SS}},{{\mathbf{L}}_{SR}},{{\mathbf{L}}_{RS}},{{\mathbf{L}}_{RR}}}事實(shí)上都含有隨時(shí)間變化的角度參數(shù)[1]，使得方程求解困難。

　　現(xiàn)對(duì)等式兩邊同時(shí)左乘{(lán)\displaystyle\left[{\begin{array}{*{20}c}{\mathbf{P}}&{}\\{}&{\mathbf{U}}\\\end{array}}\right]}{\displaystyle\left[{\begin{array}{*{20}c}{\mathbf{P}}&{}\\{}&{\mathbf{U}}\\\end{array}}\right]}，其中{\displaystyle{\mathbf{U}}}{\displaystyle{\mathbf{U}}}為三階單位矩陣。方程化為：

　　{\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]}{\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]}

　　{\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]}{\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{\displaystyle{\mathbf{PL}}_{SS}{\mathbf{P}}^{-1}=\left[{\begin{array}{*{20}c}{L_zxjr5nt}&{}&{}\\{}&{L_{q}}&{}\\{}&{}&{L_{0}}\\\end{array}}\right]\triangleq{\mathbf{L}}_{dq0}}{\displaystyle{\mathbf{PL}}_{SS}{\mathbf{P}}^{-1}=\left[{\begin{array}{*{20}c}{L_d3ffnvp}&{}&{}\\{}&{L_{q}}&{}\\{}&{}&{L_{0}}\\\end{array}}\right]\triangleq{\mathbf{L}}_{dq0}}。

　?、僮儞Q后的電感系數(shù)都變?yōu)槌?shù)，可以假想dd繞組，qq繞組是固定在轉(zhuǎn)子上的，相對(duì)轉(zhuǎn)子靜止。

　　②派克變換陣對(duì)定子自感矩陣{\displaystyle{\mathbf{L}}_{SS}}{\displaystyle{\mathbf{L}}_{SS}}起到了對(duì)角化的作用，并消去了其中的角度變量。{\displaystyle{L_r5jfb55},{L_{q}},{L_{0}}}{\displaystyle{L_rnlhz5z},{L_{q}},{L_{0}}}為其特征根。

　　③變換后定子和轉(zhuǎn)子間的互感系數(shù)不對(duì)稱，這是由于派克變換的矩陣不是正交矩陣。

　?、躿\displaystyle{L_jb5rn55}}{\displaystyle{L_fbvfrxl}}為直軸同步電感系數(shù)，其值相當(dāng)于當(dāng)勵(lì)磁繞組開(kāi)路，定子合成磁勢(shì)產(chǎn)生單純直軸磁場(chǎng)時(shí)，任意一相定子繞組的自感系數(shù)。

變換后的電壓方程

　　電壓方程：

　　{\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]}{\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]}

　　現(xiàn)對(duì)等式兩邊同時(shí)左乘{(lán)\displaystyle\left[{\begin{array}{*{20}c}{\mathbf{P}}&{}\\{}&{\mathbf{U}}\\\end{array}}\right]}{\displaystyle\left[{\begin{array}{*{20}c}{\mathbf{P}}&{}\\{}&{\mathbf{U}}\\\end{array}}\right]}，其中{\displaystyle{\mathbf{U}}}{\displaystyle{\mathbf{U}}}為三階單位矩陣。方程化為：

　　{\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]}{\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{\displaystyle{\mathbf{\Psi}}_{dq0}={\mathbf{P\Psi}}_{abc}}{\displaystyle{\mathbf{\Psi}}_{dq0}={\mathbf{P\Psi}}_{abc}}，

　　對(duì)兩邊求導(dǎo)，得{\displaystyle{\mathbf{\dot{\Psi}}}_{dq0}={\mathbf{{\dot{P}}\Psi}}_{abc}+{\mathbf{P{\dot{\Psi}}}}_{abc}}{\displaystyle{\mathbf{\dot{\Psi}}}_{dq0}={\mathbf{{\dot{P}}\Psi}}_{abc}+{\mathbf{P{\dot{\Psi}}}}_{abc}}，

　　所以{\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}}{\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{\displaystyle{\mathbf{{\dot{P}}P}}^{-1}=\left[{\begin{array}{*{20}c}{}&\omega&{}\\{-\omega}&{}&{}\\{}&{}&{}\\\end{array}}\right]}{\displaystyle{\mathbf{{\dot{P}}P}}^{-1}=\left[{\begin{array}{*{20}c}{}&\omega&{}\\{-\omega}&{}&{}\\{}&{}&{}\\\end{array}}\right]}，令{\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 _5bhpznr}\\{\Phi _{q}}\\{\Phi _{0}}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{\omega\Psi _{q}}\\{-\omega\Psi _hp5vpb5}\\{}\\\end{array}}\right]}{\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 _lthdpph}\\{\Phi _{q}}\\{\Phi _{0}}\\\end{array}}\right]=\left[{\begin{array}{*{20}c}{\omega\Psi _{q}}\\{-\omega\Psi _bxtthbr}\\{}\\\end{array}}\right]}

　　于是有{\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]}{\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]}

　　上式右邊第一項(xiàng)為繞組電阻的壓降，第二項(xiàng)為變壓器電勢(shì)，第三項(xiàng)為發(fā)電機(jī)電勢(shì)或旋轉(zhuǎn)電勢(shì)。

關(guān)于派克變換，小編為大家就分享這些。歡迎聯(lián)系我們合運(yùn)電氣有限公司，以獲取更多相關(guān)知識(shí)。