Lecture12 - Lecture 12 Derivation of Equivalent Circuit...

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Lecture 12 Derivation of Equivalent Circuit from Application of γ - δ Frame In from (20) in Lecture 10, the stator current vector in the d-q 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 a -axis, the α -axis and the d -axis 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.
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