三角函数基本运算，Clark变换，Park变换，三相系统动态方程

基本展开，倍角公式 { sin ⁡ ( α + β ) = sin ⁡ α ⋅ cos ⁡ β + cos ⁡ α ⋅ sin ⁡ β sin ⁡ ( α − β ) = sin ⁡ α ⋅ cos ⁡ β − cos ⁡ α ⋅ sin ⁡ β cos ⁡ ( α + β ) = cos ⁡ α ⋅ cos ⁡ β − sin ⁡ α ⋅ sin ⁡ β cos ⁡ ( α − β ) = cos ⁡ α ⋅ cos ⁡ β + sin ⁡ α ⋅ sin ⁡ β tan ⁡ ( α + β ) = tan ⁡ α + tan ⁡ β 1 − tan ⁡ α ⋅ tan ⁡ β tan ⁡ ( α − β ) = tan ⁡ α − tan ⁡ β 1 + tan ⁡ α ⋅ tan ⁡ β      { sin ⁡ 2 α = 2 sin ⁡ α cos ⁡ α cos ⁡ 2 α = cos ⁡ 2 α − sin ⁡ 2 α = 2 cos ⁡ 2 α − 1 = 1 − 2 sin ⁡ 2 α tan ⁡ 2 α = 2 tan ⁡ α 1 − tan ⁡ 2 α \left\{ \begin{array}{l} \sin (\alpha + \beta ) = \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \\ \sin (\alpha - \beta ) = \sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta \\ \cos (\alpha + \beta ) = \cos \alpha \cdot \cos \beta - \sin \alpha \cdot \sin \beta \\ \cos (\alpha - \beta ) = \cos \alpha \cdot \cos \beta + \sin \alpha \cdot \sin \beta \\ \tan (\alpha + \beta ) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \cdot \tan \beta }}\\ \tan (\alpha - \beta ) = \frac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \cdot \tan \beta }} \end{array} \right.\ \ \ \ \left\{ \begin{array}{l} \sin 2\alpha = 2\sin \alpha \cos \alpha \\ \cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha = 2{\cos ^2}\alpha - 1 = 1 - 2{\sin ^2}\alpha \\ \tan 2\alpha = \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \end{array} \right. ⎩ ⎨ ⎧​sin(α+β)=sinα⋅cosβ+cosα⋅sinβsin(α−β)=sinα⋅cosβ−cosα⋅sinβcos(α+β)=cosα⋅cosβ−sinα⋅sinβcos(α−β)=cosα⋅cosβ+sinα⋅sinβtan(α+β)=1−tanα⋅tanβtanα+tanβ​tan(α−β)=1+tanα⋅tanβtanα−tanβ​​    ⎩ ⎨ ⎧​sin2α=2sinαcosαcos2α=cos2α−sin2α=2cos2α−1=1−2sin2αtan2α=1−tan2α2tanα​​

积化和差，和差化积 { sin ⁡ α cos ⁡ β = 1 2 [ sin ⁡ ( α + β ) + sin ⁡ ( α − β ) ] cos ⁡ α sin ⁡ β = 1 2 [ sin ⁡ ( α + β ) − sin ⁡ ( α − β ) ] cos ⁡ α cos ⁡ β = 1 2 [ cos ⁡ ( α + β ) + cos ⁡ ( α − β ) ] sin ⁡ α sin ⁡ β = − 1 2 [ cos ⁡ ( α + β ) − cos ⁡ ( α − β ) ]     { sin ⁡ α + sin ⁡ β = 2 sin ⁡ ( α 2 + β 2 ) cos ⁡ ( α 2 − β 2 ) sin ⁡ α − sin ⁡ β = 2 cos ⁡ ( α 2 + β 2 ) sin ⁡ ( α 2 − β 2 ) cos ⁡ α + cos ⁡ β = 2 cos ⁡ ( α 2 + β 2 ) cos ⁡ ( α 2 − β 2 ) cos ⁡ α − cos ⁡ β = − 2 sin ⁡ ( α 2 + β 2 ) sin ⁡ ( α 2 − β 2 ) \left\{ \begin{array}{l} \sin \alpha \cos \beta = \frac{1}{2}\left[ {\sin \left( {\alpha + \beta } \right) + \sin \left( {\alpha - \beta } \right)} \right]\\ \cos \alpha \sin \beta = \frac{1}{2}\left[ {\sin \left( {\alpha + \beta } \right) - \sin \left( {\alpha - \beta } \right)} \right]\\ \cos \alpha \cos \beta = \frac{1}{2}\left[ {\cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right)} \right]\\ \sin \alpha \sin \beta = - \frac{1}{2}\left[ {\cos \left( {\alpha + \beta } \right) - \cos \left( {\alpha - \beta } \right)} \right] \end{array} \right.\ \ \ \left\{ \begin{array}{l} \sin \alpha + \sin \beta = 2\sin \left( {\frac{\alpha }{2} + \frac{\beta }{2}} \right)\cos \left( {\frac{\alpha }{2} - \frac{\beta }{2}} \right)\\ \sin \alpha - \sin \beta = 2\cos \left( {\frac{\alpha }{2} + \frac{\beta }{2}} \right)\sin \left( {\frac{\alpha }{2} - \frac{\beta }{2}} \right)\\ \cos \alpha + \cos \beta = 2\cos \left( {\frac{\alpha }{2} + \frac{\beta }{2}} \right)\cos \left( {\frac{\alpha }{2} - \frac{\beta }{2}} \right)\\ \cos \alpha - \cos \beta = - 2\sin \left( {\frac{\alpha }{2} + \frac{\beta }{2}} \right)\sin \left( {\frac{\alpha }{2} - \frac{\beta }{2}} \right) \end{array} \right. ⎩ ⎨ ⎧​sinαcosβ=21​[sin(α+β)+sin(α−β)]cosαsinβ=21​[sin(α+β)−sin(α−β)]cosαcosβ=21​[cos(α+β)+cos(α−β)]sinαsinβ=−21​[cos(α+β)−cos(α−β)]​   ⎩ ⎨ ⎧​sinα+sinβ=2sin(2α​+2β​)cos(2α​−2β​)sinα−sinβ=2cos(2α​+2β​)sin(2α​−2β​)cosα+cosβ=2cos(2α​+2β​)cos(2α​−2β​)cosα−cosβ=−2sin(2α​+2β​)sin(2α​−2β​)​

三相系统三角函数 { [ cos ⁡ ( α ) cos ⁡ ( α − 2 π 3 ) cos ⁡ ( α − 4 π 3 ) ] [ cos ⁡ ( β ) cos ⁡ ( β − 2 π 3 ) cos ⁡ ( β − 4 π 3 ) ] T = 1.5 cos ⁡ ( α − β ) [ cos ⁡ ( α ) cos ⁡ ( α − 2 π 3 ) cos ⁡ ( α − 4 π 3 ) ] [ cos ⁡ ( β ) cos ⁡ ( β − 4 π 3 ) cos ⁡ ( β − 2 π 3 ) ] T = 1.5 cos ⁡ ( α + β ) [ cos ⁡ ( α ) cos ⁡ ( α − 2 π 3 ) cos ⁡ ( α − 4 π 3 ) ] [ sin ⁡ ( β ) sin ⁡ ( β − 2 π 3 ) sin ⁡ ( β − 4 π 3 ) ] T = − 1.5 sin ⁡ ( α − β ) [ cos ⁡ ( α ) cos ⁡ ( α − 2 π 3 ) cos ⁡ ( α − 4 π 3 ) ] [ sin ⁡ ( β ) sin ⁡ ( β − 4 π 3 ) sin ⁡ ( β − 2 π 3 ) ] T = 1.5 sin ⁡ ( α + β ) [ sin ⁡ ( α ) sin ⁡ ( α − 2 π 3 ) sin ⁡ ( α − 4 π 3 ) ] [ sin ⁡ ( β ) sin ⁡ ( β − 2 π 3 ) sin ⁡ ( β − 4 π 3 ) ] T = 1.5 cos ⁡ ( α − β ) [ sin ⁡ ( α ) sin ⁡ ( α − 2 π 3 ) sin ⁡ ( α − 4 π 3 ) ] [ sin ⁡ ( β ) sin ⁡ ( β − 4 π 3 ) sin ⁡ ( β − 2 π 3 ) ] T = − 1.5 cos ⁡ ( α + β ) [ sin ⁡ ( α ) sin ⁡ ( α − 2 π 3 ) sin ⁡ ( α − 4 π 3 ) ] [ cos ⁡ ( β ) cos ⁡ ( β − 2 π 3 ) cos ⁡ ( β − 4 π 3 ) ] T = 1.5 sin ⁡ ( α − β ) [ sin ⁡ ( α ) sin ⁡ ( α − 2 π 3 ) sin ⁡ ( α − 4 π 3 ) ] [ cos ⁡ ( β ) cos ⁡ ( β − 4 π 3 ) cos ⁡ ( β − 2 π 3 ) ] T = 1.5 sin ⁡ ( α + β ) \left\{ \begin{array}{l} \left[ {\begin{array}{c} {\cos \left( \alpha \right)}&{\cos \left( {\alpha - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\alpha - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]{\left[ {\begin{array}{c} {\cos \left( \beta \right)}&{\cos \left( {\beta - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\beta - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]^T} = 1.5\cos \left( {\alpha - \beta } \right)\\ \left[ {\begin{array}{c} {\cos \left( \alpha \right)}&{\cos \left( {\alpha - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\alpha - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]{\left[ {\begin{array}{c} {\cos \left( \beta \right)}&{\cos \left( {\beta - {\textstyle{{4\pi } \over 3}}} \right)}&{\cos \left( {\beta - {\textstyle{{2\pi } \over 3}}} \right)} \end{array}} \right]^T} = 1.5\cos \left( {\alpha + \beta } \right)\\ \left[ {\begin{array}{c} {\cos \left( \alpha \right)}&{\cos \left( {\alpha - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\alpha - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]{\left[ {\begin{array}{c} {\sin \left( \beta \right)}&{\sin \left( {\beta - {\textstyle{{2\pi } \over 3}}} \right)}&{\sin \left( {\beta - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]^T} = - 1.5\sin \left( {\alpha - \beta } \right)\\ \left[ {\begin{array}{c} {\cos \left( \alpha \right)}&{\cos \left( {\alpha - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\alpha - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]{\left[ {\begin{array}{c} {\sin \left( \beta \right)}&{\sin \left( {\beta - {\textstyle{{4\pi } \over 3}}} \right)}&{\sin \left( {\beta - {\textstyle{{2\pi } \over 3}}} \right)} \end{array}} \right]^T} = 1.5\sin \left( {\alpha + \beta } \right)\\ \left[ {\begin{array}{c} {\sin \left( \alpha \right)}&{\sin \left( {\alpha - {\textstyle{{2\pi } \over 3}}} \right)}&{\sin \left( {\alpha - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]{\left[ {\begin{array}{c} {\sin \left( \beta \right)}&{\sin \left( {\beta - {\textstyle{{2\pi } \over 3}}} \right)}&{\sin \left( {\beta - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]^T} = 1.5\cos \left( {\alpha - \beta } \right)\\ \left[ {\begin{array}{c} {\sin \left( \alpha \right)}&{\sin \left( {\alpha - {\textstyle{{2\pi } \over 3}}} \right)}&{\sin \left( {\alpha - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]{\left[ {\begin{array}{c} {\sin \left( \beta \right)}&{\sin \left( {\beta - {\textstyle{{4\pi } \over 3}}} \right)}&{\sin \left( {\beta - {\textstyle{{2\pi } \over 3}}} \right)} \end{array}} \right]^T} = - 1.5\cos \left( {\alpha + \beta } \right)\\ \left[ {\begin{array}{c} {\sin \left( \alpha \right)}&{\sin \left( {\alpha - {\textstyle{{2\pi } \over 3}}} \right)}&{\sin \left( {\alpha - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]{\left[ {\begin{array}{c} {\cos \left( \beta \right)}&{\cos \left( {\beta - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\beta - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]^T} = 1.5\sin \left( {\alpha - \beta } \right)\\ \left[ {\begin{array}{c} {\sin \left( \alpha \right)}&{\sin \left( {\alpha - {\textstyle{{2\pi } \over 3}}} \right)}&{\sin \left( {\alpha - {\textstyle{{4\pi } \over 3}}} \right)} \end{array}} \right]{\left[ {\begin{array}{c} {\cos \left( \beta \right)}&{\cos \left( {\beta - {\textstyle{{4\pi } \over 3}}} \right)}&{\cos \left( {\beta - {\textstyle{{2\pi } \over 3}}} \right)} \end{array}} \right]^T} = 1.5\sin \left( {\alpha + \beta } \right) \end{array} \right. ⎩ ⎨ ⎧​[cos(α)​cos(α−32π​)​cos(α−34π​)​][cos(β)​cos(β−32π​)​cos(β−34π​)​]T=1.5cos(α−β)[cos(α)​cos(α−32π​)​cos(α−34π​)​][cos(β)​cos(β−34π​)​cos(β−32π​)​]T=1.5cos(α+β)[cos(α)​cos(α−32π​)​cos(α−34π​)​][sin(β)​sin(β−32π​)​sin(β−34π​)​]T=−1.5sin(α−β)[cos(α)​cos(α−32π​)​cos(α−34π​)​][sin(β)​sin(β−34π​)​sin(β−32π​)​]T=1.5sin(α+β)[sin(α)​sin(α−32π​)​sin(α−34π​)​][sin(β)​sin(β−32π​)​sin(β−34π​)​]T=1.5cos(α−β)[sin(α)​sin(α−32π​)​sin(α−34π​)​][sin(β)​sin(β−34π​)​sin(β−32π​)​]T=−1.5cos(α+β)[sin(α)​sin(α−32π​)​sin(α−34π​)​][cos(β)​cos(β−32π​)​cos(β−34π​)​]T=1.5sin(α−β)[sin(α)​sin(α−32π​)​sin(α−34π​)​][cos(β)​cos(β−34π​)​cos(β−32π​)​]T=1.5sin(α+β)​

{ C l a r k 正变换 : [ f α f β f 0 ] = 2 3 [ 1 − 1 2 − 1 2 0 3 2 − 3 2 1 2 1 2 1 2 ] [ f a f b f c ] 紧凑形式 : F α β 0 = P a b c d q 0 ( 0 ) F a b c C l a r k 逆变换 : [ f a f b f c ] = [ 1 0 1 − 1 2 3 2 1 − 1 2 − 3 2 1 ] [ f α f β f 0 ] 紧凑形式 : F a b c = P d q 0 a b c ( 0 ) F α β 0 \left\{ \begin{array}{l} {\rm{Clark正变换:}}\left[ {\begin{array}{c} {{f_\alpha }}\\ {{f_\beta }}\\ {{f_0}} \end{array}} \right] = \frac{2}{3}\left[ {\begin{array}{c} 1&{ - {\textstyle{1 \over 2}}}&{ - {\textstyle{1 \over 2}}}\\ 0&{{\textstyle{{\sqrt 3 } \over 2}}}&{ - {\textstyle{{\sqrt 3 } \over 2}}}\\ {{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}} \end{array}} \right]\left[ {\begin{array}{c} {{f_a}}\\ {{f_b}}\\ {{f_c}} \end{array}} \right]\\ 紧凑形式:{{\bf{F}}_{\alpha \beta 0}} = {\bf{P}}_{abc}^{dq0}\left( 0 \right){{\bf{F}}_{abc}}\\ {\rm{Clark逆变换}}:\left[ {\begin{array}{c} {{f_a}}\\ {{f_b}}\\ {{f_c}} \end{array}} \right] = \left[ {\begin{array}{c} 1&0&1\\ { - {\textstyle{1 \over 2}}}&{{\textstyle{{\sqrt 3 } \over 2}}}&1\\ { - {\textstyle{1 \over 2}}}&{ - {\textstyle{{\sqrt 3 } \over 2}}}&1 \end{array}} \right]\left[ {\begin{array}{c} {{f_\alpha }}\\ {{f_\beta }}\\ {{f_0}} \end{array}} \right]\\ 紧凑形式:{{\bf{F}}_{abc}} = {\bf{P}}_{dq0}^{abc}\left( 0 \right){{\bf{F}}_{\alpha \beta 0}} \end{array} \right. ⎩ ⎨ ⎧​Clark正变换: ​fα​fβ​f0​​ ​=32​ ​1021​​−21​23 ​​21​​−21​−23 ​​21​​ ​ ​fa​fb​fc​​ ​紧凑形式:Fαβ0​=Pabcdq0​(0)Fabc​Clark逆变换: ​fa​fb​fc​​ ​= ​1−21​−21​​023 ​​−23 ​​​111​ ​ ​fα​fβ​f0​​ ​紧凑形式:Fabc​=Pdq0abc​(0)Fαβ0​​

{ P a r k 正变换 : [ f d f q f 0 ] = [ cos ⁡ ( θ ) sin ⁡ ( θ ) 0 − sin ⁡ ( θ ) cos ⁡ ( θ ) 0 0 0 1 ] [ f α f β f 0 ] 紧凑形式 : F d q 0 = P α β 0 d q 0 ( θ ) F α β 0 P a r k 逆变换 : [ f α f β f 0 ] = [ cos ⁡ ( θ ) − sin ⁡ ( θ ) 0 sin ⁡ ( θ ) cos ⁡ ( θ ) 0 0 0 1 ] [ f d f q f 0 ] 紧凑形式 : F α β 0 = P d q 0 α β 0 ( θ ) F d q 0 \left\{ \begin{array}{l} {\rm{Park正变换}}:\left[ {\begin{array}{c} {{f_d}}\\ {{f_q}}\\ {{f_0}} \end{array}} \right] = \left[ {\begin{array}{c} {\cos \left( \theta \right)}&{\sin \left( \theta \right)}&0\\ { - \sin \left( \theta \right)}&{\cos \left( \theta \right)}&0\\ 0&0&1 \end{array}} \right]\left[ {\begin{array}{c} {{f_\alpha }}\\ {{f_\beta }}\\ {{f_0}} \end{array}} \right]\\ 紧凑形式:{{\bf{F}}_{dq0}} = {\bf{P}}_{\alpha \beta 0}^{dq0}\left( \theta \right){{\bf{F}}_{\alpha \beta 0}}\\ {\rm{Park逆变换}}:\left[ {\begin{array}{c} {{f_\alpha }}\\ {{f_\beta }}\\ {{f_0}} \end{array}} \right] = \left[ {\begin{array}{c} {\cos \left( \theta \right)}&{ - \sin \left( \theta \right)}&0\\ {\sin \left( \theta \right)}&{\cos \left( \theta \right)}&0\\ 0&0&1 \end{array}} \right]\left[ {\begin{array}{c} {{f_d}}\\ {{f_q}}\\ {{f_0}} \end{array}} \right]\\ 紧凑形式:{{\bf{F}}_{\alpha \beta 0}} = {\bf{P}}_{dq0}^{\alpha \beta 0}\left( \theta \right){{\bf{F}}_{dq0}} \end{array} \right. ⎩ ⎨ ⎧​Park正变换: ​fd​fq​f0​​ ​= ​cos(θ)−sin(θ)0​sin(θ)cos(θ)0​001​ ​ ​fα​fβ​f0​​ ​紧凑形式:Fdq0​=Pαβ0dq0​(θ)Fαβ0​Park逆变换: ​fα​fβ​f0​​ ​= ​cos(θ)sin(θ)0​−sin(θ)cos(θ)0​001​ ​ ​fd​fq​f0​​ ​紧凑形式:Fαβ0​=Pdq0αβ0​(θ)Fdq0​​

{ P a r k 正变换 : [ f d f q f 0 ] = 2 3 [ cos ⁡ ( θ ) cos ⁡ ( θ − 2 π 3 ) cos ⁡ ( θ + 2 π 3 ) − sin ⁡ ( θ ) − sin ⁡ ( θ − 2 π 3 ) − sin ⁡ ( θ + 2 π 3 ) 1 2 1 2 1 2 ] [ f a f b f c ] 紧凑形式 : F d q 0 = P a b c d q 0 ( θ ) F a b c P a r k 逆变换 : [ f a f b f c ] = [ cos ⁡ ( θ ) − sin ⁡ ( θ ) 1 cos ⁡ ( θ − 2 π 3 ) − sin ⁡ ( θ − 2 π 3 ) 1 cos ⁡ ( θ + 2 π 3 ) − sin ⁡ ( θ + 2 π 3 ) 1 ] [ f d f q f 0 ] 紧凑形式 : F a b c = P d q 0 a b c ( θ ) F d q 0 \left\{ \begin{array}{l} {\rm{Park正变换}}:\left[ {\begin{array}{c} {{f_d}}\\ {{f_q}}\\ {{f_0}} \end{array}} \right] = \frac{2}{3}\left[ {\begin{array}{c} {\cos \left( \theta \right)}&{\cos \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}\\ { - \sin \left( \theta \right)}&{ - \sin \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}\\ {{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}} \end{array}} \right]\left[ {\begin{array}{c} {{f_a}}\\ {{f_b}}\\ {{f_c}} \end{array}} \right]\\ 紧凑形式:{{\bf{F}}_{dq0}} = {\bf{P}}_{abc}^{dq0}\left( \theta \right){{\bf{F}}_{abc}}\\ {\rm{Park逆变换}}:\left[ {\begin{array}{c} {{f_a}}\\ {{f_b}}\\ {{f_c}} \end{array}} \right] = \left[ {\begin{array}{c} {\cos \left( \theta \right)}&{ - \sin \left( \theta \right)}&1\\ {\cos \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&1\\ {\cos \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}&1 \end{array}} \right]\left[ {\begin{array}{c} {{f_d}}\\ {{f_q}}\\ {{f_0}} \end{array}} \right]\\ 紧凑形式:{{\bf{F}}_{abc}} = {\bf{P}}_{dq0}^{abc}\left( \theta \right){{\bf{F}}_{dq0}} \end{array} \right. ⎩ ⎨ ⎧​Park正变换: ​fd​fq​f0​​ ​=32​ ​cos(θ)−sin(θ)21​​cos(θ−32π​)−sin(θ−32π​)21​​cos(θ+32π​)−sin(θ+32π​)21​​ ​ ​fa​fb​fc​​ ​紧凑形式:Fdq0​=Pabcdq0​(θ)Fabc​Park逆变换: ​fa​fb​fc​​ ​= ​cos(θ)cos(θ−32π​)cos(θ+32π​)​−sin(θ)−sin(θ−32π​)−sin(θ+32π​)​111​ ​ ​fd​fq​f0​​ ​紧凑形式:Fabc​=Pdq0abc​(θ)Fdq0​​

{ U d q 0 T I d q 0 = ( P a b c d q 0 ( θ ) U a b c ) T P a b c d q 0 ( θ ) I a b c = U a b c T P a b c d q 0 ( θ ) T P a b c d q 0 ( θ ) I a b c = U a b c T 2 3 [ cos ⁡ ( θ ) − sin ⁡ ( θ ) 1 2 cos ⁡ ( θ − 2 π 3 ) − sin ⁡ ( θ − 2 π 3 ) 1 2 cos ⁡ ( θ + 2 π 3 ) − sin ⁡ ( θ + 2 π 3 ) 1 2 ] 2 3 [ cos ⁡ ( θ ) cos ⁡ ( θ − 2 π 3 ) cos ⁡ ( θ + 2 π 3 ) − sin ⁡ ( θ ) − sin ⁡ ( θ − 2 π 3 ) − sin ⁡ ( θ + 2 π 3 ) 1 2 1 2 1 2 ] I a b c = 4 9 U a b c T [ 1.25 − 0.25 − 0.25 − 0.25 1.25 − 0.25 − 0.25 − 0.25 1.25 ] I a b c ,   A c c o r d i n g   t o   i a + i b + i c = 0 = 4 9 U a b c T [ 1.5 i a 1.5 i b 1.5 i c ] = 2 3 U a b c T I a b c \left\{ \begin{aligned} U_{dq0}^T{I_{dq0}} &= {\left( {{\bf{P}}_{abc}^{dq0}\left( \theta \right){U_{abc}}} \right)^T}{\bf{P}}_{abc}^{dq0}\left( \theta \right){I_{abc}}\\ &= U_{abc}^T{\bf{P}}_{abc}^{dq0}{\left( \theta \right)^T}{\bf{P}}_{abc}^{dq0}\left( \theta \right){I_{abc}}\\ &= U_{abc}^T\frac{2}{3}\left[ {\begin{array}{c} {\cos \left( \theta \right)}&{ - \sin \left( \theta \right)}&{{\textstyle{1 \over 2}}}\\ {\cos \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{{\textstyle{1 \over 2}}}\\ {\cos \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}&{{\textstyle{1 \over 2}}} \end{array}} \right]\frac{2}{3}\left[ {\begin{array}{c} {\cos \left( \theta \right)}&{\cos \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}\\ { - \sin \left( \theta \right)}&{ - \sin \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}\\ {{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}} \end{array}} \right]{I_{abc}}\\ &= \frac{4}{9}U_{abc}^T\left[ {\begin{array}{c} {1.25}&{ - 0.25}&{ - 0.25}\\ { - 0.25}&{1.25}&{ - 0.25}\\ { - 0.25}&{ - 0.25}&{1.25} \end{array}} \right]{I_{abc}},\ According\ to\ {i_a} + {i_b} + {i_c} = 0\\ &= \frac{4}{9}U_{abc}^T\left[ {\begin{array}{c} {1.5i_a}\\ {1.5i_b}\\ {1.5i_c} \end{array}} \right]\\ &= \frac{2}{3}U_{abc}^T{I_{abc}} \end{aligned} \right. ⎩ ⎨ ⎧​Udq0T​Idq0​​=(Pabcdq0​(θ)Uabc​)TPabcdq0​(θ)Iabc​=UabcT​Pabcdq0​(θ)TPabcdq0​(θ)Iabc​=UabcT​32​ ​cos(θ)cos(θ−32π​)cos(θ+32π​)​−sin(θ)−sin(θ−32π​)−sin(θ+32π​)​21​21​21​​ ​32​ ​cos(θ)−sin(θ)21​​cos(θ−32π​)−sin(θ−32π​)21​​cos(θ+32π​)−sin(θ+32π​)21​​ ​Iabc​=94​UabcT​ ​1.25−0.25−0.25​−0.251.25−0.25​−0.25−0.251.25​ ​Iabc​, According to ia​+ib​+ic​=0=94​UabcT​ ​1.5ia​1.5ib​1.5ic​​ ​=32​UabcT​Iabc​​

{ U d q T I d q = ( P a b c d q ( θ ) U a b c ) T P a b c d q ( θ ) I a b c = U a b c T P a b c d q ( θ ) T P a b c d q ( θ ) I a b c = U a b c T 2 3 [ cos ⁡ ( θ ) − sin ⁡ ( θ ) cos ⁡ ( θ − 2 π 3 ) − sin ⁡ ( θ − 2 π 3 ) cos ⁡ ( θ + 2 π 3 ) − sin ⁡ ( θ + 2 π 3 ) ] 2 3 [ cos ⁡ ( θ ) cos ⁡ ( θ − 2 π 3 ) cos ⁡ ( θ + 2 π 3 ) − sin ⁡ ( θ ) − sin ⁡ ( θ − 2 π 3 ) − sin ⁡ ( θ + 2 π 3 ) ] I a b c = 4 9 U a b c T [ 1 − 0.5 − 0.5 − 0.5 1 − 0.5 − 0.5 − 0.5 1 ] I a b c ,   A c c o r d i n g   t o   i a + i b + i c = 0 = 4 9 U a b c T [ 1.5 i a 1.5 i b 1.5 i c ] I a b c = 2 3 U a b c T I a b c \left\{ \begin{aligned} U_{dq}^T{I_{dq}} &= {\left( {{\bf{P}}_{abc}^{dq}\left( \theta \right){U_{abc}}} \right)^T}{\bf{P}}_{abc}^{dq}\left( \theta \right){I_{abc}}\\ &= U_{abc}^T{\bf{P}}_{abc}^{dq}{\left( \theta \right)^T}{\bf{P}}_{abc}^{dq}\left( \theta \right){I_{abc}}\\ &= U_{abc}^T\frac{2}{3}\left[ {\begin{array}{c} {\cos \left( \theta \right)}&{ - \sin \left( \theta \right)}\\ {\cos \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}\\ {\cos \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)} \end{array}} \right]\frac{2}{3}\left[ {\begin{array}{c} {\cos \left( \theta \right)}&{\cos \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)}\\ { - \sin \left( \theta \right)}&{ - \sin \left( {\theta - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\theta + {\textstyle{{2\pi } \over 3}}} \right)} \end{array}} \right]{I_{abc}}\\ &= \frac{4}{9}U_{abc}^T\left[ {\begin{array}{c} 1&{ - 0.5}&{ - 0.5}\\ { - 0.5}&1&{ - 0.5}\\ { - 0.5}&{ - 0.5}&1 \end{array}} \right]{I_{abc}},\ According\ to\ {i_a} + {i_b} + {i_c} = 0\\ &= \frac{4}{9}U_{abc}^T\left[ {\begin{array}{c} {1.5i_a}\\ {1.5i_b}\\ {1.5i_c} \end{array}} \right]{I_{abc}}\\ &= \frac{2}{3}U_{abc}^T{I_{abc}} \end{aligned} \right. ⎩ ⎨ ⎧​UdqT​Idq​​=(Pabcdq​(θ)Uabc​)TPabcdq​(θ)Iabc​=UabcT​Pabcdq​(θ)TPabcdq​(θ)Iabc​=UabcT​32​ ​cos(θ)cos(θ−32π​)cos(θ+32π​)​−sin(θ)−sin(θ−32π​)−sin(θ+32π​)​ ​32​[cos(θ)−sin(θ)​cos(θ−32π​)−sin(θ−32π​)​cos(θ+32π​)−sin(θ+32π​)​]Iabc​=94​UabcT​ ​1−0.5−0.5​−0.51−0.5​−0.5−0.51​ ​Iabc​, According to ia​+ib​+ic​=0=94​UabcT​ ​1.5ia​1.5ib​1.5ic​​ ​Iabc​=32​UabcT​Iabc​​

由于 i a + i b + i c = 0 i_a+i_b+i_c=0 ia​+ib​+ic​=0，零序功率 u 0 i 0 = 0 u_0i_0=0 u0​i0​=0；采用幅值不变变换，变换后的功率是真实功率的 2 3 \frac{2}{3} 32​倍。

{ d d t F d q 0 = d d t [ P a b c d q 0 ( ω t ) F a b c ] = d P a b c d q 0 ( ω t ) d t F a b c + P a b c d q 0 ( ω t ) d F a b c d t = 2 3 { d d t [ cos ⁡ ( ω t ) cos ⁡ ( ω t − 2 π 3 ) cos ⁡ ( ω t + 2 π 3 ) − sin ⁡ ( ω t ) − sin ⁡ ( ω t − 2 π 3 ) − sin ⁡ ( ω t + 2 π 3 ) 1 2 1 2 1 2 ] } [ f a f b f c ] + 2 3 [ cos ⁡ ( ω t ) cos ⁡ ( ω t − 2 π 3 ) cos ⁡ ( ω t + 2 π 3 ) − sin ⁡ ( ω t ) − sin ⁡ ( ω t − 2 π 3 ) − sin ⁡ ( ω t + 2 π 3 ) 1 2 1 2 1 2 ] d d t [ f a f b f c ] = ω 2 3 [ − sin ⁡ ( ω t ) − sin ⁡ ( ω t − 2 π 3 ) − sin ⁡ ( ω t + 2 π 3 ) − cos ⁡ ( ω t ) − cos ⁡ ( ω t − 2 π 3 ) − cos ⁡ ( ω t + 2 π 3 ) 0 0 0 ] [ f a f b f c ] + 2 3 [ cos ⁡ ( ω t ) cos ⁡ ( ω t − 2 π 3 ) cos ⁡ ( ω t + 2 π 3 ) − sin ⁡ ( ω t ) − sin ⁡ ( ω t − 2 π 3 ) − sin ⁡ ( ω t + 2 π 3 ) 1 2 1 2 1 2 ] d d t [ f a f b f c ] = ω [ f q − f d 0 ] + P a b c d q 0 ( ω t ) d d t [ f a f b f c ] \left\{ \begin{aligned} \frac{d}{{dt}}{{\bf{F}}_{dq0}} &= \frac{d}{{dt}}\left[ {{\bf{P}}_{abc}^{dq0}\left( {\omega t} \right){{\bf{F}}_{abc}}} \right]\\ &= \frac{{d{\bf{P}}_{abc}^{dq0}\left( {\omega t} \right)}}{{dt}}{{\bf{F}}_{abc}} + {\bf{P}}_{abc}^{dq0}\left( {\omega t} \right)\frac{{d{{\bf{F}}_{abc}}}}{{dt}}\\ &= \frac{2}{3}\left\{ {\frac{d}{{dt}}\left[ {\begin{array}{c} {\cos \left( {\omega t} \right)}&{\cos \left( {\omega t - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\omega t + {\textstyle{{2\pi } \over 3}}} \right)}\\ { - \sin \left( {\omega t} \right)}&{ - \sin \left( {\omega t - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\omega t + {\textstyle{{2\pi } \over 3}}} \right)}\\ {{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}} \end{array}} \right]} \right\}\left[ {\begin{array}{c} {{f_a}}\\ {{f_b}}\\ {{f_c}} \end{array}} \right] + \frac{2}{3}\left[ {\begin{array}{c} {\cos \left( {\omega t} \right)}&{\cos \left( {\omega t - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\omega t + {\textstyle{{2\pi } \over 3}}} \right)}\\ { - \sin \left( {\omega t} \right)}&{ - \sin \left( {\omega t - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\omega t + {\textstyle{{2\pi } \over 3}}} \right)}\\ {{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}} \end{array}} \right]\frac{d}{{dt}}\left[ {\begin{array}{c} {{f_a}}\\ {{f_b}}\\ {{f_c}} \end{array}} \right]\\ &= \omega \frac{2}{3}\left[ {\begin{array}{c} { - \sin \left( {\omega t} \right)}&{ - \sin \left( {\omega t - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\omega t + {\textstyle{{2\pi } \over 3}}} \right)}\\ { - \cos \left( {\omega t} \right)}&{ - \cos \left( {\omega t - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \cos \left( {\omega t + {\textstyle{{2\pi } \over 3}}} \right)}\\ 0&0&0 \end{array}} \right]\left[ {\begin{array}{c} {{f_a}}\\ {{f_b}}\\ {{f_c}} \end{array}} \right] + \frac{2}{3}\left[ {\begin{array}{c} {\cos \left( {\omega t} \right)}&{\cos \left( {\omega t - {\textstyle{{2\pi } \over 3}}} \right)}&{\cos \left( {\omega t + {\textstyle{{2\pi } \over 3}}} \right)}\\ { - \sin \left( {\omega t} \right)}&{ - \sin \left( {\omega t - {\textstyle{{2\pi } \over 3}}} \right)}&{ - \sin \left( {\omega t + {\textstyle{{2\pi } \over 3}}} \right)}\\ {{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}}&{{\textstyle{1 \over 2}}} \end{array}} \right]\frac{d}{{dt}}\left[ {\begin{array}{c} {{f_a}}\\ {{f_b}}\\ {{f_c}} \end{array}} \right]\\ &= \omega \left[ {\begin{array}{c} {{f_q}}\\ { - {f_d}}\\ 0 \end{array}} \right] + {\bf{P}}_{abc}^{dq0}\left( {\omega t} \right)\frac{d}{{dt}}\left[ {\begin{array}{c} {{f_a}}\\ {{f_b}}\\ {{f_c}} \end{array}} \right] \end{aligned} \right. ⎩ ⎨ ⎧​dtd​Fdq0​​=dtd​[Pabcdq0​(ωt)Fabc​]=dtdPabcdq0​(ωt)​Fabc​+Pabcdq0​(ωt)dtdFabc​​=32​⎩ ⎨ ⎧​dtd​ ​cos(ωt)−sin(ωt)21​​cos(ωt−32π​)−sin(ωt−32π​)21​​cos(ωt+32π​)−sin(ωt+32π​)21​​ ​⎭ ⎬ ⎫​ ​fa​fb​fc​​ ​+32​ ​cos(ωt)−sin(ωt)21​​cos(ωt−32π​)−sin(ωt−32π​)21​​cos(ωt+32π​)−sin(ωt+32π​)21​​ ​dtd​ ​fa​fb​fc​​ ​=ω32​ ​−sin(ωt)−cos(ωt)0​−sin(ωt−32π​)−cos(ωt−32π​)0​−sin(ωt+32π​)−cos(ωt+32π​)0​ ​ ​fa​fb​fc​​ ​+32​ ​cos(ωt)−sin(ωt)21​​cos(ωt−32π​)−sin(ωt−32π​)21​​cos(ωt+32π​)−sin(ωt+32π​)21​​ ​dtd​ ​fa​fb​fc​​ ​=ω ​fq​−fd​0​ ​+Pabcdq0​(ωt)dtd​ ​fa​fb​fc​​ ​​

P a b c d q 0 ( ω t ) d F a b c d t = d F d q 0 d t + ω [ − f q f d 0 ] {\bf{P}}_{abc}^{dq0}\left( {\omega t} \right)\frac{{d{{\bf{F}}_{abc}}}}{{dt}} = \frac{{d{{\bf{F}}_{dq0}}}}{{dt}} + \omega \left[ {\begin{array}{c} { - {f_q}}\\ {{f_d}}\\ 0 \end{array}} \right] Pabcdq0​(ωt)dtdFabc​​=dtdFdq0​​+ω ​−fq​fd​0​ ​

  考虑如下形式的三相系统： { u a − e a = R i a + L d i a d t u b − e b = R i b + L d i b d t u c − e c = R i c + L d i c d t ↔ [ u a u b u c ] − [ e a e b e c ] = R [ i a i b i c ] + L d d t [ i a i b i c ] \left\{ \begin{array}{l} {u_a} - {e_a} = R{i_a} + L\frac{{d{i_a}}}{{dt}}\\ {u_b} - {e_b} = R{i_b} + L\frac{{d{i_b}}}{{dt}}\\ {u_c} - {e_c} = R{i_c} + L\frac{{d{i_c}}}{{dt}} \end{array} \right. \leftrightarrow \left[ \begin{array}{l} {u_a}\\ {u_b}\\ {u_c} \end{array} \right] - \left[ \begin{array}{l} {e_a}\\ {e_b}\\ {e_c} \end{array} \right] = R\left[ \begin{array}{l} {i_a}\\ {i_b}\\ {i_c} \end{array} \right] + L\frac{d}{{dt}}\left[ \begin{array}{l} {i_a}\\ {i_b}\\ {i_c} \end{array} \right] ⎩ ⎨ ⎧​ua​−ea​=Ria​+Ldtdia​​ub​−eb​=Rib​+Ldtdib​​uc​−ec​=Ric​+Ldtdic​​​↔ ​ua​ub​uc​​ ​− ​ea​eb​ec​​ ​=R ​ia​ib​ic​​ ​+Ldtd​ ​ia​ib​ic​​ ​

  左右同乘 P a b c d q 0 {\bf{P}}_{abc}^{dq0} Pabcdq0​， [ u d u q u 0 ] − [ e d e q e 0 ] = R [ i d i q i 0 ] + L d d t [ i d i q i 0 ] + ω L [ − i q i d i 0 ] \left[ \begin{array}{l} {u_d}\\ {u_q}\\ {u_0} \end{array} \right] - \left[ \begin{array}{l} {e_d}\\ {e_q}\\ {e_0} \end{array} \right] = R\left[ \begin{array}{l} {i_d}\\ {i_q}\\ {i_0} \end{array} \right] + L\frac{d}{{dt}}\left[ \begin{array}{l} {i_d}\\ {i_q}\\ {i_0} \end{array} \right] + \omega L\left[ \begin{array}{c} -{i_q}\\ {i_d}\\ {i_0} \end{array} \right] ​ud​uq​u0​​ ​− ​ed​eq​e0​​ ​=R ​id​iq​i0​​ ​+Ldtd​ ​id​iq​i0​​ ​+ωL ​−iq​id​i0​​ ​

  写作标准状态方程形式： d d t [ i d i q i 0 ] = 1 L [ u d u q u 0 ] − 1 L [ e d e q e 0 ] − R L [ i d i q i 0 ] + ω [ i q − i d i 0 ] \frac{d}{{dt}}\left[ \begin{array}{l} {i_d}\\ {i_q}\\ {i_0} \end{array} \right] = \frac{1}{L}\left[ \begin{array}{l} {u_d}\\ {u_q}\\ {u_0} \end{array} \right] - \frac{1}{L}\left[ \begin{array}{l} {e_d}\\ {e_q}\\ {e_0} \end{array} \right] - \frac{R}{L}\left[ \begin{array}{l} {i_d}\\ {i_q}\\ {i_0} \end{array} \right] + \omega \left[ \begin{array}{c} {i_q}\\ -{i_d}\\ {i_0} \end{array} \right] dtd​ ​id​iq​i0​​ ​=L1​ ​ud​uq​u0​​ ​−L1​ ​ed​eq​e0​​ ​−LR​ ​id​iq​i0​​ ​+ω ​iq​−id​i0​​ ​