Lecture9 - Lecture 9 Transformation from a-b-c frame to 0...

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Lecture 9 Transformation from a-b-c frame to 0- α - β frame . For 3 stator phases Sa , Sb , Sc and 3 rotor phases Ra , Rb , Rc , the flux linkage matrix is + ° ° + ° ° + ° ° ° ° + ° ° + ° ° + = Rc Rb Ra Sc Sb Sa R R R S S S Rc Rb Ra Sc Sb Sa i i i i i i M M M n M n M n M M M M n M n M n M M M M n M n M n M n M n M n M M M M n M n M n M M M M n M n M n M M M M l l l l l l 5 . 0 5 . 0 cos ) 240 cos( ) 120 cos( 5 . 0 5 . 0 ) 120 cos( cos ) 240 cos( 5 . 0 5 . 0 ) 240 cos( ) 120 cos( cos cos ) 120 cos( ) 240 cos( 5 . 0 5 . 0 ) 240 cos( cos ) 120 cos( 5 . 0 5 . 0 ) 120 cos( ) 240 cos( cos 5 . 0 5 . 0 θ λ ( 1 ) Sa Sb Sc Ra Rb Rc
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By including the resistance matrices of the stator and the rotor windings and by applying the transformations = αβ 0 0 0 0 0 0 e e C C e e S abc abc R abc S abc = 0 0 0 0 0 0 i i C C i i S abc abc R abc S abc 1 10 2 21 1 3 32 2 2 11 3 22 2 ⎡⎤ ⎢⎥ =− ⎣⎦ 0 abc C (2) the a-b-c frame equations are transformed to the α - β frame equations: + + + + + + + + = θ R R S S R R m m R R m m m m S S m m S S R R S S i i i i M p R n M p M p M p R n M p n M p n M p n M p M p R n M p n M p M p R e e e e ) ( 0 ) cos ( ) sin ( 0 ) ( ) sin ( ) cos ( ) cos ( ) sin ( ) ( 0 ) sin ( ) cos ( 0 ) ( l l l l ( 3 ) The mechanical equation of motion is:
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] sin ) ( cos ) [( 1 1 m S R S R m S R S R L m m n i i i i n i i i i nM T f p J ω β α = + + (4) m m p θ = Equations (3) has a minor simplification in that the (1,2), (2,1), (3,4) and (4,3) entries are zeroes. This is because the α -axis and the β -axis are at right angles. The mutual inductance coupling between them is zero because cos(90 ° )=0.0.
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This note was uploaded on 02/23/2010 for the course ECSE 462 taught by Professor Bon-tekooi during the Spring '09 term at McGill.

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Lecture9 - Lecture 9 Transformation from a-b-c frame to 0...

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