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+tanβtan(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 α β 0 F a b c = 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 α β 0 a b c F α β 0 = 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}^{\alpha \beta0}{{\bf{F}}_{abc}} = {\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}}_{\alpha \beta0}^{abc}{{\bf{F}}_{\alpha \beta 0}} = {\bf{P}}_{dq0}^{abc}\left( 0 \right){{\bf{F}}_{\alpha \beta 0}} \end{array} \right. ⎩ ⎨ ⎧Clark正变换: fαfβf0 =32 1021−2123 21−21−23 21 fafbfc 紧凑形式:Fαβ0=Pabcαβ0Fabc=Pabcdq0(0)FabcClark逆变换: fafbfc = 1−21−21023 −23 111 fαfβf0 紧凑形式:Fabc=Pαβ0abcFαβ0=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: fdfqf0 = cos(θ)−sin(θ)0sin(θ)cos(θ)0001 fαfβf0 紧凑形式:Fdq0=Pαβ0dq0(θ)Fαβ0Park逆变换: fαfβf0 = cos(θ)sin(θ)0−sin(θ)cos(θ)0001 fdfqf0 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: fdfqf0 =32 cos(θ)−sin(θ)21cos(θ−32π)−sin(θ−32π)21cos(θ+32π)−sin(θ+32π)21 fafbfc 紧凑形式:Fdq0=Pabcdq0(θ)FabcPark逆变换: fafbfc = cos(θ)cos(θ−32π)cos(θ+32π)−sin(θ)−sin(θ−32π)−sin(θ+32π)111 fdfqf0 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. ⎩ ⎨ ⎧Udq0TIdq0=(Pabcdq0(θ)Uabc)TPabcdq0(θ)Iabc=UabcTPabcdq0(θ)TPabcdq0(θ)Iabc=UabcT32 cos(θ)cos(θ−32π)cos(θ+32π)−sin(θ)−sin(θ−32π)−sin(θ+32π)212121 32 cos(θ)−sin(θ)21cos(θ−32π)−sin(θ−32π)21cos(θ+32π)−sin(θ+32π)21 Iabc=94UabcT 1.25−0.25−0.25−0.251.25−0.25−0.25−0.251.25 Iabc, According to ia+ib+ic=0=94UabcT 1.5ia1.5ib1.5ic =32UabcTIabc
{ 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. ⎩ ⎨ ⎧UdqTIdq=(Pabcdq(θ)Uabc)TPabcdq(θ)Iabc=UabcTPabcdq(θ)TPabcdq(θ)Iabc=UabcT32 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=94UabcT 1−0.5−0.5−0.51−0.5−0.5−0.51 Iabc, According to ia+ib+ic=0=94UabcT 1.5ia1.5ib1.5ic Iabc=32UabcTIabc
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 u0i0=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. ⎩ ⎨ ⎧dtdFdq0=dtd[Pabcdq0(ωt)Fabc]=dtdPabcdq0(ωt)Fabc+Pabcdq0(ωt)dtdFabc=32⎩ ⎨ ⎧dtd cos(ωt)−sin(ωt)21cos(ωt−32π)−sin(ωt−32π)21cos(ωt+32π)−sin(ωt+32π)21 ⎭ ⎬ ⎫ fafbfc +32 cos(ωt)−sin(ωt)21cos(ωt−32π)−sin(ωt−32π)21cos(ωt+32π)−sin(ωt+32π)21 dtd fafbfc =ω32 −sin(ωt)−cos(ωt)0−sin(ωt−32π)−cos(ωt−32π)0−sin(ωt+32π)−cos(ωt+32π)0 fafbfc +32 cos(ωt)−sin(ωt)21cos(ωt−32π)−sin(ωt−32π)21cos(ωt+32π)−sin(ωt+32π)21 dtd fafbfc =ω fq−fd0 +Pabcdq0(ωt)dtd fafbfc
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 −fqfd0
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+Ldtdiaub−eb=Rib+Ldtdibuc−ec=Ric+Ldtdic↔
uaubuc
−
eaebec
=R
iaibic
+Ldtd
iaibic
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]
uduqu0
−
edeqe0
=R
idiqi0
+Ldtd
idiqi0
+ωL
−iqidi0
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
idiqi0
=L1
uduqu0
−L1
edeqe0
−LR
idiqi0
+ω
iq−idi0
