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Unformatted text preview: Lecture 12 Derivation of Equivalent Circuit from Application of γ δ Frame In from (20) in Lecture 10, the stator current vector in the dq frame is ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ + + = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ ) sin( ) cos( S S S Sq Sd t t I i i ϕ ω ϕ ω (1) In (8) of Lecture 11, the vector is transformed to the γ δ frame as ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ S S S S S I i i ϕ ϕ δ γ sin cos (2) Equations (1) and (2) are related by the transformation matrix as: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ δ γ ω ω ω ω S S Sq Sd i i t t t t i i cos sin sin cos (3) One can define a complex number δ γ S S S j i I + = ~ and express ] ~ Re[ t j S Sd S Sa e I i i i ω α = = = (4) In (4), one is reminded that the aaxis, the αaxis and the daxis of the stator remain in the same position. The γ axis, however, rotates at the speed of the stator frequency ω . Electrical engineers are familiar with ac voltages and currents expressed as complex numbers, such as S I ~ , associated with a synchronously rotating unit phasor t j e ω . This lecture shows how the equivalent circuit of the induction machine is easily derived from (5) of Lecture 11. In order the simplify the notations in (5), let ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 4 3 2 1 e e e e e e e e R R S S δ γ δ γ , ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 4...
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This note was uploaded on 02/23/2010 for the course ECSE 462 taught by Professor Bontekooi during the Spring '09 term at McGill.
 Spring '09
 BonTekOoi

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