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The document describes the modeling of a doubly fed induction generator (DFIG) machine. It represents the DFIG as two sets of three-phase windings on the stator and rotor. It then derives the voltage and flux equations for each phase by considering the stator-stator, rotor-rotor, stator-rotor, and rotor-stator inductance terms. The inductances are represented by a 6x6 matrix with submatrices that depend on rotor position and angle due to the changing positional relationship between rotor and stator windings.

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J. McCalley d-q transformation
Machine model 2 Consider the DFIG as two sets of abc windings, one on the stator and one on the rotor. θm ωm
Machine model 3 The voltage equation for each phase will have the form: That is, we can write them all in the following form: dt t d t ri t v ) ( ) ( ) (                                                                                                  cr br ar cs bs as cr br ar cs bs as r r r s s s cr br ar cs bs as dt d i i i i i i r r r r r r v v v v v v       0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 All rotor terms are given on the rotor side in these equations. We can write the flux terms as functions of the currents, via an equation for each flux of the form λ=ΣLkik, where the summation is over all six winding currents. However, we must take note that there are four kinds of terms in each summation.
Machine model 4 • Stator-stator terms: These are terms which relate a stator winding flux to a stator winding current. Because the positional relationship between any pair of stator windings does not change with rotor position, these inductances are not a function of rotor position; they are constants. • Rotor-rotor terms: These are terms which relate a rotor winding flux to a rotor winding current. As in stator-stator-terms, these are constants. • Rotor-stator terms: These are terms which relate a rotor winding flux to a stator winding current. As the rotor turns, the positional relationship between the rotor winding and the stator winding will change, and so the inductance will change. Therefore the inductance will be a function of rotor position, characterized by rotor angle θ. • Stator-rotor terms: These are terms which relate a stator winding flux to a rotor winding current. As described for the rotor-stator terms, the inductance will be a function of rotor position, characterized by rotor angle θ.
Machine model 5 There are two more comments to make about the flux-current relations: • Because the rotor motion is periodic, the functional dependence of each rotor-stator or stator-rotor inductance on θ is cosinusoidal. • Because θ changes with time as the rotor rotates, the inductances are functions of time. We may now write down the flux equations for the stator and the rotor windings.                                                cr br ar cs bs as r rs sr s cr br ar cs bs as i i i i i i L L L L       Each of the submatrices in the inductance matrix is a 3x3, as given on the next slide… Note here that all quantities are now referred to the stator. The effect of referring is straight-forward, given in the book by P. Krause, “Analysis of Electric Machinery,” 1995, IEEE Press, pp. 167-168. I will not go through it here.
Machine model 6                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos Diagonal elements are the self-inductance of each winding and include leakage plus mutual. Off-diagonal elements are mutual inductances between windings and are negative because 120° axis offset between any pair of windings results in flux contributed by one winding to have negative component along the main axis of another winding.             T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos θm ωm
Machine model 7                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1 Summarizing….                                                                                      cr br ar cs bs as cr br ar cs bs as r r r s s s cr br ar cs bs as dt d i i i i i i r r r r r r v v v v v v       0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                                                cr br ar cs bs as r rs sr s cr br ar cs bs as i i i i i i L L L L                   T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos
Machine model 8                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                                                                                         cr br ar cs bs as r rs sr s cr br ar cs bs as r r r s s s cr br ar cs bs as i i i i i i L L L L dt d i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Combining…. It is here that we observe a difficulty – that the stator-rotor and rotor-stator terms, Lsr and Lrs, because they are functions of θr, and thus functions of time, will also need to be differentiated. Therefore differentiation of fluxes results in expressions like The differentiation with respect to L, dL/dt, will result in time-varying coefficients on the currents. This will make our set of state equations difficult to solve. L dt di i dt dL dt d                T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos
Transformation 9 This presents some significant difficulties, in terms of solution, that we would like to avoid. We look for a different approach. The different approach is based on the observation that our trouble comes from the inductances related to the stator-rotor mutual inductances that have time-varying inductances. In order to alleviate the trouble, we project the a-b-c currents onto a pair of axes which we will call the d and q axes or d-q axes. In making these projections, we want to obtain expressions for the components of the stator currents in phase with the and q axes, respectively. Although we may specify the speed of these axes to be any speed that is convenient for us, we will generally specify it to be synchronous speed, ωs. ia a a' id iq d-axis q-axis θ One can visualize the projection by thinking of the a-b-c currents as having sinusoidal variation IN TIME along their respective axes (a space vector!). The picture below illustrates for the a-phase. Decomposing the b-phase currents and the c-phase currents in the same way, and then adding them up, provides us with:   ) 120 cos( ) 120 cos( cos           c b a q q i i i k i   ) 120 sin( ) 120 sin( sin           c b a d d i i i k i Constants kq and kd are chosen so as to simplify the numerical coefficients in the generalized KVL equations we will get.
Transformation 10 We have transformed 3 variables ia, ib, and ic into two variables id and iq, as we did in the α-β transformation. This yields an undetermined system, meaning • We can uniquely transform ia, ib, and ic to id and iq • We cannot uniquely transform id and iq to ia, ib, and ic. We will use as a third current the zero-sequence current: Recall our id and iq equations:   c b a i i i k i    0 0 We can write our transformation more compactly as   ) 120 cos( ) 120 cos( cos           c b a d q i i i k i   ) 120 sin( ) 120 sin( sin           c b a q d i i i k i                                    c b a d d d q q q d q i i i k k k k k k k k k i i i 0 0 0 0 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos      
Transformation 11                                    c b a d d d q q q d q i i i k k k k k k k k k i i i 0 0 0 0 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos       A similar transformation resulted from the work done by Blondel (1923), Doherty and Nickle (1926), and Robert Park (1929, 1933), which is referred to as “Park’s transformation.” In 2000, Park’s 1929 paper was voted the second most important paper of the last 100 years (behind Fortescue’s paper on symmertical components). R, Park, “Two reaction theory of synchronous machines,” Transactions of the AIEE, v. 48, p. 716-730, 1929. G. Heydt, S. Venkata, and N. Balijepalli, “High impact papers in power engineering, 1900-1999, NAPS, 2000. Robert H. Park, 1902-1994 See http://www.nap.edu/openbook.php ?record_id=5427&page=175 for an interesting biography on Park, written by Charles Concordia. Park’s transformation uses a frame of reference on the rotor. In Parks case, he derived this for a synchronous machine and so it is the same as a synchronous frame of reference. For induction motors, it is important to distinguish between a synchronous reference frame and a reference frame on the rotor.
Transformation 12 Here, the angle θ is given by ) 0 ( ) ( 0         t d where ɣ is a dummy variable of integration. The constants k0, kq, and kd are chosen differently by different authors. One popular choice is 1/3, 2/3, and 2/3, respectively, which causes the magnitude of the d-q quantities to be equal to that of the three-phase quantities. However, it also causes a 3/2 multiplier in front of the power expression (Anderson & Fouad use k0=1/√3, kd=kq=√(2/3) to get a power invariant expression). The angular velocity ω associated with the change of variables is unspecified. It characterizes the frame of reference and may rotate at any constant or varying angular velocity or it may remain stationary. You will often hear of the “arbitrary reference frame.” The phrase “arbitrary” stems from the fact that the angular velocity of the transformation is unspecified and can be selected arbitrarily to expedite the solution of the equations or to satisfy the system constraints [Krause].                                    c b a d d d q q q d q i i i k k k k k k k k k i i i 0 0 0 0 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos      
Transformation 13 The constants k0, kq, and kd are chosen differently by different authors. One popular choice is 1/3, 2/3, and 2/3, respectively, which causes the magnitude of the d-q quantities to be equal to that of the three-phase quantities. PROOF (iq equation only):   ) 120 cos( ) 120 cos( cos           c b a d q i i i k i Let ia=Acos(ωt); ib=Acos(ωt-120); ic=Acos(ωt-240) and substitute into iq equation:     ) 120 cos( ) 120 cos( ) 120 cos( ) 120 cos( cos cos ) 120 cos( ) 120 cos( ) 120 cos( ) 120 cos( cos cos                               t t t A k t A t A t A k i d d q Now use trig identity: cos(u)cos(v)=(1/2)[ cos(u-v)+cos(u+v) ]   ) 120 120 cos( ) 120 120 cos( ) 120 120 cos( ) 120 120 cos( ) cos( ) cos( 2                                 t t t t t t A k i d q   ) 240 cos( ) cos( ) 240 cos( ) cos( ) cos( ) cos( 2                           t t t t t t A k i d q Now collect terms in ωt-θ and place brackets around what is left:     ) 240 cos( ) 240 cos( ) cos( ) cos( 3 2                   t t t t A k i d q Observe that what is in the brackets is zero! Therefore:   ) cos( 3 2 3 ) cos( 3 2         t A k t A k i d d q Observe that for 3kdA/2=A, we must have kd=2/3.
Transformation 14 Choosing constants k0, kq, and kd to be 1/3, 2/3, and 2/3, respectively, results in The inverse transformation becomes:                                    0 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos i i i i i i d q c b a                                            c b a d q i i i i i i 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0      
Example 15 Krause gives an insightful example in his book, where he specifies generic quantities fas, fbs, fcs to be a-b-c quantities varying with time on the stator according to: t f t f t f cs bs as sin 2 cos     The objective is to transform them into 0-d-q quantities, which he denotes as fqs, fds, f0s.                                                                  t t t f f f f f f cs bs as s ds qs sin 2 / cos 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0             Note that these are not balanced quantities!
Example 16 This results in Now assume that θ(0)=-π/12 and ω=1 rad/sec. Evaluate the above for t= π/3 seconds. First, we need to obtain the angle θ corresponding to this time. We do that as follows: 4 12 3 ) 12 ( 1 ) 0 ( ) ( 3 / 0 0                      d d t Now we can evaluate the above equations 3A-1, 3A-2, and 3A-3, as follows:
Example 17 This results in
Example 18                                       t t t f f f s ds qs sin 2 / cos 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0       Resolution of fas=cost into directions of fqs and fds for t=π/3 (θ=π/4). Resolution of fbs=t/2 into directions of fqs and fds for t=π/3 (θ=π/4). Resolution of fcs=-sint into directions of fqs and fds for t=π/3 (θ=π/4). Composite of other 3 figures
Inverse transformation 19 The d-q transformation and its inverse transformation is given below.                                      c b a K d q i i i i i i s                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0                                           0 1 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos i i i i i i d q K c b a s                                     2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       s K                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       s K It should be the case that Ks Ks -1=I, where I is the 3x3 identity matrix, i.e.,                                           1 0 0 0 1 0 0 0 1 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2            
Balanced conditions 20 Under balanced conditions, i0 is zero, and therefore it produces no flux at all. Under these conditions, we may write the d-q transformation as                            c b a d q i i i i i ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2                                            c b a d q i i i i i i 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0                                          0 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos i i i i i i d q c b a                                      d q c b a i i i i i ) 120 sin( ) 120 cos( ) 120 sin( ) 120 cos( sin cos      
Rotor circuit transformation 21 We now need to apply our transformation to the rotor a-b-c windings in order to obtain the rotor circuit voltage equation in q-d-0 coordinates. However, we must notice one thing: whereas the stator phase-a winding (and thus it’s a-axis) is fixed, the rotor phase-a winding (and thus it’s a-axis) rotates. If we apply the same transformation to the rotor, we will not account for its rotation, i.e., we will be treating it as if it were fixed.                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       s K Our d-q transformation is as follows: But, what, exactly, is θ? ) 0 ( ) ( 0         t d θ can be observed in the below figure as the angle between the rotating d-q reference frame and the a-axis, where the a-axis is fixed on the stator frame and is defined by the location of the phase-a winding. We expressed this angle analytically using where ω is the rotational speed of the d-q coordinate axes (and in our case, is synchronous speed). This transformation will allow us to operate on the stator circuit voltage equation and transform it to the q-d-0 coordinates.
Rotor circuit transformation To understand how to handle this, consider the below figure where we show our familiar θ, the angle between the stator a-axis and the q-axis of the synchronously rotating reference frame. 22 ia a a' id iq d-axis q-axis θ θm β ω ωm We have also shown • θm, which is the angle between the stator a-axis and the rotor a-axis, and • β, which is the angle between the rotor a-axis and the q-axis of the synchronously rotating reference frame. The stator a-axis is stationary, the q-d axis rotates at ω, and the rotor a-axis rotates at ωm. Consider the iar space vector, in blue, which is coincident with the rotor a-axis. Observe that we may decompose it in the q-d reference frame only by using β instead of θ. Conclusion: Use the exact same transformation, except substitute β for θ, and…. account for the fact that to the rotor windings, the q-d coordinate system appears to be moving at ω-ωm
Rotor circuit transformation We compare our two transformations below.                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       s K ) 0 ( ) ( 0         t d         ) 0 ( 0 ) 0 ( ) 0 ( ) ( ) (           m t m d r                       2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       r K Stator winding transformation, Ks Rotor winding transformation, Kr 23                                      cs bs as s ds qs i i i i i i 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0                                            cr br ar r dr qr i i i i i i 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0       We now augment our notation to distinguish between q-d-0 quantities from the stator and q-d-0 quantities from the rotor:
Transforming voltage equations 24                                                                                   cr br ar cs bs as cr br ar cs bs as r r r s s s cr br ar cs bs as i i i i i i r r r r r r v v v v v v             0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                                                rc rb ra sc sb sa r rs sr s rc rb ra sc sb sa i i i i i i L L L L                                 m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1 Recall our voltage equations: Let’s apply our d-q transformation to it….             T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos
Transforming voltage equations 25 Let’s rewrite it in compact notation                                                                                   cr br ar cs bs as cr br ar cs bs as r r r s s s cr br ar cs bs as i i i i i i r r r r r r v v v v v v             0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                             abcr abcs abcr abcs r s abcr abcs i i r r v v     0 0 Now multiply through by our transformation matrices. Be careful with dimensionality.                              3 2 1 0 0 0 0 0 0 0 0 Term abcr abcs r s Term abcr abcs r s r s Term abcr abcs r s K K i i r r K K v v K K                                                
Transforming voltage equations – Term 1 26 Therefore: the voltage equation becomes                           qdor s qd abcr r abcs s Term abcr abcs r s v v v K v K v v K K 0 1 0 0                              3 2 0 0 0 0 0 0 0 Term abcr abcs r s Term abcr abcs r s r s qdor s qd K K i i r r K K v v                                          
Transforming voltage equations – Term 2 27 What to do with the abc currents? We need q-d-0 currents!                                abcr abcs r r s s Term abcr abcs r s r s i i r K r K i i r r K K 0 0 0 0 0 0 2            Recall:                      r qd s qd r s abcr abcs i i K K i i 0 0 1 1 0 0 and substitute into above.                                        r qd s qd r s r r s s Term abcr abcs r s r s i i K K r K r K i i r r K K 0 0 1 1 2 0 0 0 0 0 0 0 0            Perform the matrix multiplication:                                  r qd s qd r r r s s s Term abcr abcs r s r s i i K r K K r K i i r r K K 0 0 1 1 2 0 0 0 0 0 0            Fact: KRK-1=R if R is diagonal having equal elements on the diagonal. Proof: KRK-1=KrUK-1=rKUK-1=rKK-1=rU=R. Therefore….
Transforming voltage equations – Term 2 28                                               r qd s qd r s r qd s qd r r r s s s Term abcr abcs r s r s i i r r i i K r K K r K i i r r K K 0 0 0 0 1 1 2 0 0 0 0 0 0 0 0            Therefore: the voltage equation becomes            3 0 0 0 0 0 0 0 Term abcr abcs r s r qd s qd r s qdor s qd K K i i r r v v                                    
Transforming voltage equations – Term 3 29            3 0 0 0 0 0 0 0 Term abcr abcs r s r qd s qd r s qdor s qd K K i i r r v v                                     Focusing on just the stator quantities, consider: abcs s s qd K    0 Differentiate both sides abcs s abcs s s qd K K         0 Solve for abcs s K   abcs s s qd abcs s K K         0 Use λabcs =K-1λqd0s: s qd s s s qd abcs s K K K 0 1 0                                 abcr r abcs s Term abcr abcs r s K K K K                  3 0 0 Term 3 is: A similar process for the rotor quantities results in r qd r r r qd abcr r K K K 0 1 0          Substituting these last two expressions into the term 3 expression above results in                                            r qd r r s qd s s r qd s qd abcr r abcs s Term abcr abcs r s K K K K K K K K 0 1 0 1 0 0 3 0 0                          Substitute this back into voltage equation…
Transforming voltage equations – Term 3 30            3 0 0 0 0 0 0 0 Term abcr abcs r s r qd s qd r s qdor s qd K K i i r r v v                                                                                r qd r r s qd s s r qd s qd abcr r abcs s Term abcr abcs r s K K K K K K K K 0 1 0 1 0 0 3 0 0                                                                 r qd r r s qd s s r qd s qd r qd s qd r s qdor s qd K K K K i i r r v v 0 1 0 1 0 0 0 0 0 0 0        
Transforming voltage equations – Term 3 31 Now let’s express the fluxes in terms of currents by recalling that                      abcr abcs r s r qd s qd K K     0 0 0 0                                                                    abcr abcs r rs sr s abcr abcs cr br ar cs bs as r rs sr s cr br ar cs bs as i i L L L L i i i i i i L L L L         and the flux-current relations: Now write the abc currents in terms of the qd0 currents:                      r qd s qd r s abcr abcs i i K K i i 0 0 1 1 0 0 Substitute the third equation into the second:                            r qd s qd r s r rs sr s abcr abcs i i K K L L L L 0 0 1 1 0 0   Substitute the fourth equation into the first:                                    r qd s qd r s r rs sr s r s r qd s qd i i K K L L L L K K 0 0 1 1 0 0 0 0 0 0  
Transforming voltage equations – Term 3 32                                    r qd s qd r s r rs sr s r s r qd s qd i i K K L L L L K K 0 0 1 1 0 0 0 0 0 0   Perform the first matrix multiplication:                              r qd s qd r s r r rs r sr s s s r qd s qd i i K K L K L K L K L K 0 0 1 1 0 0 0 0   and the next matrix multiplication:                          r qd s qd r r r s rs r r sr s s s s r qd s qd i i K L K K L K K L K K L K 0 0 1 1 1 1 0 0  
Transforming voltage equations – Term 3 33                          r qd s qd r r r s rs r r sr s s s s r qd s qd i i K L K K L K K L K K L K 0 0 1 1 1 1 0 0   Now we need to go through each of these four matrix multiplications. I will here omit the details and just give the results (note also in what follows the definition of additional nomenclature for each of the four submatrices): 0 1 0 1 1 0 1 0 0 0 2 3 0 0 0 2 3 0 0 0 0 2 3 0 0 0 2 3 0 0 0 2 3 0 0 0 2 3 rqd r m r m r r r r mqd m m s rs r r sr s sqd s m s m s s s s L L L L L L K L K L L L K L K K L K L L L L L L K L K                                                                                           r qd s qd rqd mqd mqd sqd r qd s qd i i L L L L 0 0 0 0 0 0 0 0   And since our inductance matrix is constant, we can write:                      r qd s qd rqd mqd mqd sqd r qd s qd i i L L L L 0 0 0 0 0 0 0 0       Substitute the above expression for flux derivatives into our voltage equation:
Transforming voltage equations – Term 3                      r qd s qd rqd mqd mqd sqd r qd s qd i i L L L L 0 0 0 0 0 0 0 0                                              r qd r r s qd s s r qd s qd r qd s qd r s qdor s qd K K K K i i r r v v 0 1 0 1 0 0 0 0 0 0 0         Substitute the above expressions for flux & flux derivatives into our voltage equation:                                            r qd r r s qd s s r qd s qd rqd mqd mqd sqd r qd s qd r s qdor s qd K K K K i i L L L L i i r r v v 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0       We still have the last term to obtain. To get this, we need to do two things. 1. Express individual q- and d- terms of λqd0s and λqd0r in terms of currents. 2. Obtain and 1  s s K K  1  r r K K  34
Transforming voltage equations – Term 3 1. Express individual q- and d- terms of λqd0s and λqd0r in terms of currents: 35                      r qd s qd rqd mqd mqd sqd r qd s qd i i L L L L 0 0 0 0 0 0 0 0                                                                          r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3             dr m r ds m dr dr m ds m s ds qr m r qs m qr qr m qs m s qs i L L i L i L i L L i L L i L i L i L L                                     2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3         From the above, we observe:
Transforming voltage equations – Term 3 2. Obtain and 1  s s K K  1  r r K K  36                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       s K                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       r K                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       s K                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       r K To get , we must consider:             ) ( ) 0 ( ) ( ) ( 0 t d t t  m m t m t d r                      ) ( ) 0 ( ) 0 ( ) ( ) ( ) 0 ( 0          s K  Therefore:                   0 0 0 ) 120 cos( ) 120 cos( cos ) 120 sin( ) 120 sin( sin 3 2        s K  Likewise, to get , we must consider: r K  Therefore:                      0 0 0 ) 120 cos( ) 120 cos( cos ) 120 sin( ) 120 sin( sin 3 2         m r K 
Transforming voltage equations – Term 3 2. Obtain 1  s s K K  1  r r K K  37                                                                0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 2 3 0 3 2 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 0 0 0 ) 120 cos( ) 120 cos( cos ) 120 sin( ) 120 sin( sin 3 2 1                 s s K K                                                    0 0 0 0 0 0 ) ( 0 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 0 0 0 ) 120 cos( ) 120 cos( cos ) 120 sin( ) 120 sin( sin 3 2 1 m m m r r K K                    Obtain Substitute into voltage equations…
Transforming voltage equations – Term 3 38              0 0 0 0 0 0 0 1   s s K K                 0 0 0 0 0 0 ) ( 0 1 m m r r K K      Substitute into voltage equations…                                            r qd r r s qd s s r qd s qd rqd mqd mqd sqd r qd s qd r s qdor s qd K K K K i i L L L L i i r r v v 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0                                                                                                                                                                    r dr qr s ds qs m m r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) ( 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                         This results in: Note the “Speed voltages” in the first, second, fourth, and fifth equations. -ωλds ωλqs -(ω- ωm)λdr (ω- ωm) λqs
Transforming voltage equations – Term 3 39 Some comments on speed voltages: -ωλds, ωλqs, -(ω- ωm)λdr, (ω- ωm) λqs: • These speed voltages represent the fact that a rotating flux wave will create voltages in windings that are stationary relative to that flux wave. • Speed voltages are so named to contrast them from what may be called transformer voltages, which are induced as a result of a time varying magnetic field. • You may have run across the concept of “speed voltages” in Physics, where you computed a voltage induced in a coil of wire as it moved through a static magnetic field, in which case, you may have used the equation Blv where B is flux density, l is conductor length, and v is the component of the velocity of the moving conductor (or moving field) that is normal with respect to the field flux direction (or conductor). • The first speed voltage term, -ωλds, appears in the vqs equation. The second speed voltage term, ωλqs, appears in the vds equation. Thus, we see that the d- axis flux causes a speed voltage in the q-axis winding, and the q-axis flux causes a speed voltage in the d-axis winding. A similar thing is true for the rotor winding.
Transforming voltage equations – Term 3 40              0 0 0 0 0 0 0 1   s s K K                 0 0 0 0 0 0 ) ( 0 1 m m r r K K      Substitute the matrices into voltage equation and then expand. This results in:                                            r qd r r s qd s s r qd s qd rqd mqd mqd sqd r qd s qd r s qdor s qd K K K K i i L L L L i i r r v v 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0                                                                                                                                                                    r dr qr s ds qs m m r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) ( 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                         Let’s collapse the last matrix-vector product by performing the multiplication….
Transforming voltage equations – Term 3 41                                                                                                                                                              r dr qr s ds qs m m r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) ( 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                         dr m r ds m dr dr m ds m s ds qr m r qs m qr qr m qs m s qs i L L i L i L i L L i L L i L i L i L L                                     2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3                                                                                                                                                  0 ) ( ) ( 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 qr m dr m qs ds r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v                                                                                                                                                                                                                              0 2 3 2 3 ) ( 2 3 2 3 ) ( 0 2 3 2 3 2 3 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 qr m r qs m m dr m r ds m m qr m qs m s dr m ds m s r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i L L i L i L L i L i L i L L i L i L L i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v                       From slide 35, we have the fluxes expressed as a function of currents  And then substitute these terms in: Results In 
Transforming voltage equations – Term 3 42                                                                                                                                                                                                          0 2 3 2 3 ) ( 2 3 2 3 ) ( 0 2 3 2 3 2 3 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 qr m r qs m m dr m r ds m m qr m qs m s dr m ds m s r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i L L i L i L L i L i L i L L i L i L L i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v                       Observe that the four non-zero elements in the last vector are multiplied by two currents from the current vector which multiplies the resistance matrix. So let’s now expand back out the last vector so that it is a product of a matrix and a current vector.                                                                                                                                                                                                      r dr qr s ds qs m r m m m m r m m m m m s m m s r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 ) ( 0 0 2 ) ( 3 0 2 3 ) ( 0 0 2 ) ( 3 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 2 3 0 0 2 3 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                             Now change the sign on the last matrix.
Transforming voltage equations – Term 3 43                                                                                                                                                                                                      r dr qr s ds qs m r m m m m r m m m m m s m m s r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 ) ( 0 0 2 ) ( 3 0 2 3 ) ( 0 0 2 ) ( 3 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 2 3 0 0 2 3 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                             Notice that the resistance matrix and the last matrix multiply the same vector, therefore, we can combine these two matrices. For example, element (1,2) in the last matrix will go into element (1,2) of the resistance matrix, as shown. This results in the expression on the next slide….
Final Model 44                                                                                                                                                             r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r m r m m m m r m r m m s m s m s m m s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r L L L L L r L r L r L L L L L r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 ) ( 0 0 2 ) ( 3 0 2 3 ) ( 0 2 ) ( 3 0 0 0 0 0 0 0 0 2 3 0 2 3 0 2 3 0 0 2 3                             This is the complete transformed electric machine state-space model in “current form.”
Some comments about the transformation 45 • ids and iqs are currents in a fictitious pair of windings fixed on a synchronously rotating reference frame. • These currents produce the same flux as do the stator a,b,c currents. • For balanced steady-state operating conditions, we can use iqd0s = Ksiabcs to show that the currents in the d and q windings are dc! The implication of this is that: • The a,b,c currents fixed in space (on the stator), varying in time produce the same synchronously rotating magnetic field as • The ds,qs currents, varying in space at synchronous speed, fixed in time! • idr and iqr are currents in a fictitious pair of windings fixed on a synchronously rotating reference frame. • These currents produce the same flux as do the rotor a,b,c currents. • For balanced steady-state operating conditions, we can use iqd0r = Kriabcr to show that the currents in the d and q windings are dc! The implication of this is that: • The a,b,c currents varying in space at slip speed sωs=(ωs- ωm) fixed on the rotor, varying in time produce the same synchronously rotating magnetic field as • The dr,qr currents, varying in space at synchronous speed, fixed in time!
Torque in abc quantities 46 The electromagnetic torque of the DFIG may be evaluated according to m c em W T     m f em W T     The stored energy is the sum of • The self inductances (less leakage) of each winding times one-half the square of its current and • All mutual inductances, each times the currents in the two windings coupled by the mutual inductance Observe that the energy stored in the leakage inductances is not a part of the energy stored in the coupling field. Consider the abc inductance matrices given in slide 6. where Wc is the co-energy of the coupling fields associated with the various windings. We are not considering saturation here, assuming the flux-current relations are linear, in which case the co-energy Wc of the coupling field equals its energy, Wf, so that: We use electric rad/sec by substituting ϴm=θm/p where p is the number of pole pairs. m f em W p T    
Torque in abc quantities 47                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos             T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos The stored energy is given by: abcr r r T abcr abcr sr T abcs abcs s s T abcs f i U L L i i L i i U L L i W ) ( 2 1 ) ( 2 1        Applying the torque-energy relation abcr sr T abcs m m f i L i W        m f em W p T     to the above, and observing that dependence on θm only occurs in the middle term, we get abcr sr T abcs m em i L i p T     So that But only Lsr depend on θm, so abcr m sr T abcs em i L i p T    
Torque in abc quantities 48 We may go through some analytical effort to show that the above evaluates to abcr sr T abcs m em i L i p T                                                           m br ar cs ar cr bs cr br as m ar br cr cs cr ar br bs cr br ar as m em i i i i i i i i i i i i i i i i i i i i i pL T   cos 2 3 sin 2 1 2 1 2 1 2 1 2 1 2 1 To complete our abc model we relate torque to rotor speed according to: m m em T dt d p J T    Inertial torque Mech torque (has negative value for generation) J is inertia of the rotor in kg-m2 or joules-sec2 Negative value for generation
Torque in qd0 quantities However, our real need is to express the torque in qd0 quantities so that we may complete our qd0 model. To this end, recall that we may write the abc quantities in terms of the qd0 quantities using our inverse transformation, according to: r qd r abcr s qd s abcs i K i i K i 0 1 0 1       r qd r sr m T s qd s abcr sr m T abcs em i K L i K p i L i p T 0 1 0 1           Substitute the above into our torque expression: 49
Torque in qd0 quantities 50                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       s K                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       r K                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos                                                                                                    r dr qr m m m m m m m m m m m T s ds qs em i i i L i i i p T 0 0 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos                         r qd r sr m T s qd s em i K L i K p T 0 1 0 1       I will not go through this differentiation but instead provide the result:   qr ds dr qs m em i i i i pL T   4 9
Torque in qd0 quantities 51 Some other useful expressions may be derived from the above, as follows:   qr ds dr qs m em i i i i pL T   4 9   qr dr dr qr em i i p T     2 3   ds qs qs ds em i i p T     2 3 Final comment: We can work with these expressions to show that the electromagnetic torque can be directly controlled by the rotor quadrature current iqr At the same time, we can also show that the stator reactive power Qs can be directly controlled by the rotor direct-axis current idr. This will provide us the necessary means to control the wind turbine.

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dqTransformation.ppt

  • 1.
  • 2.
    Machine model 2 Consider theDFIG as two sets of abc windings, one on the stator and one on the rotor. θm ωm
  • 3.
    Machine model 3 The voltageequation for each phase will have the form: That is, we can write them all in the following form: dt t d t ri t v ) ( ) ( ) (                                                                                                  cr br ar cs bs as cr br ar cs bs as r r r s s s cr br ar cs bs as dt d i i i i i i r r r r r r v v v v v v       0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 All rotor terms are given on the rotor side in these equations. We can write the flux terms as functions of the currents, via an equation for each flux of the form λ=ΣLkik, where the summation is over all six winding currents. However, we must take note that there are four kinds of terms in each summation.
  • 4.
    Machine model 4 • Stator-statorterms: These are terms which relate a stator winding flux to a stator winding current. Because the positional relationship between any pair of stator windings does not change with rotor position, these inductances are not a function of rotor position; they are constants. • Rotor-rotor terms: These are terms which relate a rotor winding flux to a rotor winding current. As in stator-stator-terms, these are constants. • Rotor-stator terms: These are terms which relate a rotor winding flux to a stator winding current. As the rotor turns, the positional relationship between the rotor winding and the stator winding will change, and so the inductance will change. Therefore the inductance will be a function of rotor position, characterized by rotor angle θ. • Stator-rotor terms: These are terms which relate a stator winding flux to a rotor winding current. As described for the rotor-stator terms, the inductance will be a function of rotor position, characterized by rotor angle θ.
  • 5.
    Machine model 5 There aretwo more comments to make about the flux-current relations: • Because the rotor motion is periodic, the functional dependence of each rotor-stator or stator-rotor inductance on θ is cosinusoidal. • Because θ changes with time as the rotor rotates, the inductances are functions of time. We may now write down the flux equations for the stator and the rotor windings.                                                cr br ar cs bs as r rs sr s cr br ar cs bs as i i i i i i L L L L       Each of the submatrices in the inductance matrix is a 3x3, as given on the next slide… Note here that all quantities are now referred to the stator. The effect of referring is straight-forward, given in the book by P. Krause, “Analysis of Electric Machinery,” 1995, IEEE Press, pp. 167-168. I will not go through it here.
  • 6.
    Machine model 6                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                             m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos Diagonal elements are the self-inductance of each winding and include leakage plus mutual. Off-diagonal elements are mutual inductances between windings and are negative because 120° axis offset between any pair of windings results in flux contributed by one winding to have negative component along the main axis of another winding.             T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos θm ωm
  • 7.
    Machine model 7                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1 Summarizing….                                                                                      cr br ar cs bs as cr br ar cs bs as r r r s s s cr br ar cs bs as dt d i i i i i i r r r r r r v v v v v v       0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                                                cr br ar cs bs as r rs sr s cr br ar cs bs as i i i i i i L L L L                  T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos
  • 8.
    Machine model 8                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                                                                                         cr br ar cs bs as r rs sr s cr br ar cs bs as r r r s s s cr br ar cs bs as i i i i i i L L L L dt d i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Combining…. It ishere that we observe a difficulty – that the stator-rotor and rotor-stator terms, Lsr and Lrs, because they are functions of θr, and thus functions of time, will also need to be differentiated. Therefore differentiation of fluxes results in expressions like The differentiation with respect to L, dL/dt, will result in time-varying coefficients on the currents. This will make our set of state equations difficult to solve. L dt di i dt dL dt d                T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos
  • 9.
    Transformation 9 This presents somesignificant difficulties, in terms of solution, that we would like to avoid. We look for a different approach. The different approach is based on the observation that our trouble comes from the inductances related to the stator-rotor mutual inductances that have time-varying inductances. In order to alleviate the trouble, we project the a-b-c currents onto a pair of axes which we will call the d and q axes or d-q axes. In making these projections, we want to obtain expressions for the components of the stator currents in phase with the and q axes, respectively. Although we may specify the speed of these axes to be any speed that is convenient for us, we will generally specify it to be synchronous speed, ωs. ia a a' id iq d-axis q-axis θ One can visualize the projection by thinking of the a-b-c currents as having sinusoidal variation IN TIME along their respective axes (a space vector!). The picture below illustrates for the a-phase. Decomposing the b-phase currents and the c-phase currents in the same way, and then adding them up, provides us with:   ) 120 cos( ) 120 cos( cos           c b a q q i i i k i   ) 120 sin( ) 120 sin( sin           c b a d d i i i k i Constants kq and kd are chosen so as to simplify the numerical coefficients in the generalized KVL equations we will get.
  • 10.
    Transformation 10 We have transformed3 variables ia, ib, and ic into two variables id and iq, as we did in the α-β transformation. This yields an undetermined system, meaning • We can uniquely transform ia, ib, and ic to id and iq • We cannot uniquely transform id and iq to ia, ib, and ic. We will use as a third current the zero-sequence current: Recall our id and iq equations:   c b a i i i k i    0 0 We can write our transformation more compactly as   ) 120 cos( ) 120 cos( cos           c b a d q i i i k i   ) 120 sin( ) 120 sin( sin           c b a q d i i i k i                                    c b a d d d q q q d q i i i k k k k k k k k k i i i 0 0 0 0 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos      
  • 11.
    Transformation 11                                    c b a d d d q q q d q i i i k k k k k k k k k i i i 0 0 0 0 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos       A similar transformationresulted from the work done by Blondel (1923), Doherty and Nickle (1926), and Robert Park (1929, 1933), which is referred to as “Park’s transformation.” In 2000, Park’s 1929 paper was voted the second most important paper of the last 100 years (behind Fortescue’s paper on symmertical components). R, Park, “Two reaction theory of synchronous machines,” Transactions of the AIEE, v. 48, p. 716-730, 1929. G. Heydt, S. Venkata, and N. Balijepalli, “High impact papers in power engineering, 1900-1999, NAPS, 2000. Robert H. Park, 1902-1994 See http://www.nap.edu/openbook.php ?record_id=5427&page=175 for an interesting biography on Park, written by Charles Concordia. Park’s transformation uses a frame of reference on the rotor. In Parks case, he derived this for a synchronous machine and so it is the same as a synchronous frame of reference. For induction motors, it is important to distinguish between a synchronous reference frame and a reference frame on the rotor.
  • 12.
    Transformation 12 Here, the angleθ is given by ) 0 ( ) ( 0         t d where ɣ is a dummy variable of integration. The constants k0, kq, and kd are chosen differently by different authors. One popular choice is 1/3, 2/3, and 2/3, respectively, which causes the magnitude of the d-q quantities to be equal to that of the three-phase quantities. However, it also causes a 3/2 multiplier in front of the power expression (Anderson & Fouad use k0=1/√3, kd=kq=√(2/3) to get a power invariant expression). The angular velocity ω associated with the change of variables is unspecified. It characterizes the frame of reference and may rotate at any constant or varying angular velocity or it may remain stationary. You will often hear of the “arbitrary reference frame.” The phrase “arbitrary” stems from the fact that the angular velocity of the transformation is unspecified and can be selected arbitrarily to expedite the solution of the equations or to satisfy the system constraints [Krause].                                    c b a d d d q q q d q i i i k k k k k k k k k i i i 0 0 0 0 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos      
  • 13.
    Transformation 13 The constants k0,kq, and kd are chosen differently by different authors. One popular choice is 1/3, 2/3, and 2/3, respectively, which causes the magnitude of the d-q quantities to be equal to that of the three-phase quantities. PROOF (iq equation only):   ) 120 cos( ) 120 cos( cos           c b a d q i i i k i Let ia=Acos(ωt); ib=Acos(ωt-120); ic=Acos(ωt-240) and substitute into iq equation:     ) 120 cos( ) 120 cos( ) 120 cos( ) 120 cos( cos cos ) 120 cos( ) 120 cos( ) 120 cos( ) 120 cos( cos cos                               t t t A k t A t A t A k i d d q Now use trig identity: cos(u)cos(v)=(1/2)[ cos(u-v)+cos(u+v) ]   ) 120 120 cos( ) 120 120 cos( ) 120 120 cos( ) 120 120 cos( ) cos( ) cos( 2                                 t t t t t t A k i d q   ) 240 cos( ) cos( ) 240 cos( ) cos( ) cos( ) cos( 2                           t t t t t t A k i d q Now collect terms in ωt-θ and place brackets around what is left:     ) 240 cos( ) 240 cos( ) cos( ) cos( 3 2                   t t t t A k i d q Observe that what is in the brackets is zero! Therefore:   ) cos( 3 2 3 ) cos( 3 2         t A k t A k i d d q Observe that for 3kdA/2=A, we must have kd=2/3.
  • 14.
    Transformation 14 Choosing constants k0,kq, and kd to be 1/3, 2/3, and 2/3, respectively, results in The inverse transformation becomes:                                    0 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos i i i i i i d q c b a                                            c b a d q i i i i i i 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0      
  • 15.
    Example 15 Krause gives aninsightful example in his book, where he specifies generic quantities fas, fbs, fcs to be a-b-c quantities varying with time on the stator according to: t f t f t f cs bs as sin 2 cos     The objective is to transform them into 0-d-q quantities, which he denotes as fqs, fds, f0s.                                                                  t t t f f f f f f cs bs as s ds qs sin 2 / cos 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0             Note that these are not balanced quantities!
  • 16.
    Example 16 This results in Nowassume that θ(0)=-π/12 and ω=1 rad/sec. Evaluate the above for t= π/3 seconds. First, we need to obtain the angle θ corresponding to this time. We do that as follows: 4 12 3 ) 12 ( 1 ) 0 ( ) ( 3 / 0 0                      d d t Now we can evaluate the above equations 3A-1, 3A-2, and 3A-3, as follows:
  • 17.
  • 18.
    Example 18                                       t t t f f f s ds qs sin 2 / cos 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0       Resolution of fas=costinto directions of fqs and fds for t=π/3 (θ=π/4). Resolution of fbs=t/2 into directions of fqs and fds for t=π/3 (θ=π/4). Resolution of fcs=-sint into directions of fqs and fds for t=π/3 (θ=π/4). Composite of other 3 figures
  • 19.
    Inverse transformation 19 The d-qtransformation and its inverse transformation is given below.                                      c b a K d q i i i i i i s                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0                                           0 1 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos i i i i i i d q K c b a s                                     2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       s K                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       s K It should be the case that Ks Ks -1=I, where I is the 3x3 identity matrix, i.e.,                                           1 0 0 0 1 0 0 0 1 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2            
  • 20.
    Balanced conditions 20 Under balancedconditions, i0 is zero, and therefore it produces no flux at all. Under these conditions, we may write the d-q transformation as                            c b a d q i i i i i ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2                                            c b a d q i i i i i i 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0                                          0 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos i i i i i i d q c b a                                      d q c b a i i i i i ) 120 sin( ) 120 cos( ) 120 sin( ) 120 cos( sin cos      
  • 21.
    Rotor circuit transformation 21 Wenow need to apply our transformation to the rotor a-b-c windings in order to obtain the rotor circuit voltage equation in q-d-0 coordinates. However, we must notice one thing: whereas the stator phase-a winding (and thus it’s a-axis) is fixed, the rotor phase-a winding (and thus it’s a-axis) rotates. If we apply the same transformation to the rotor, we will not account for its rotation, i.e., we will be treating it as if it were fixed.                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       s K Our d-q transformation is as follows: But, what, exactly, is θ? ) 0 ( ) ( 0         t d θ can be observed in the below figure as the angle between the rotating d-q reference frame and the a-axis, where the a-axis is fixed on the stator frame and is defined by the location of the phase-a winding. We expressed this angle analytically using where ω is the rotational speed of the d-q coordinate axes (and in our case, is synchronous speed). This transformation will allow us to operate on the stator circuit voltage equation and transform it to the q-d-0 coordinates.
  • 22.
    Rotor circuit transformation Tounderstand how to handle this, consider the below figure where we show our familiar θ, the angle between the stator a-axis and the q-axis of the synchronously rotating reference frame. 22 ia a a' id iq d-axis q-axis θ θm β ω ωm We have also shown • θm, which is the angle between the stator a-axis and the rotor a-axis, and • β, which is the angle between the rotor a-axis and the q-axis of the synchronously rotating reference frame. The stator a-axis is stationary, the q-d axis rotates at ω, and the rotor a-axis rotates at ωm. Consider the iar space vector, in blue, which is coincident with the rotor a-axis. Observe that we may decompose it in the q-d reference frame only by using β instead of θ. Conclusion: Use the exact same transformation, except substitute β for θ, and…. account for the fact that to the rotor windings, the q-d coordinate system appears to be moving at ω-ωm
  • 23.
    Rotor circuit transformation Wecompare our two transformations below.                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       s K ) 0 ( ) ( 0         t d         ) 0 ( 0 ) 0 ( ) 0 ( ) ( ) (           m t m d r                       2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       r K Stator winding transformation, Ks Rotor winding transformation, Kr 23                                      cs bs as s ds qs i i i i i i 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0                                            cr br ar r dr qr i i i i i i 2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2 0       We now augment our notation to distinguish between q-d-0 quantities from the stator and q-d-0 quantities from the rotor:
  • 24.
    Transforming voltage equations 24                                                                                   cr br ar cs bs as cr br ar cs bs as r r r s s s cr br ar cs bs as i i i i i i r r r r r r v v v v v v             0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                                                rc rb ra sc sb sa r rs sr s rc rb ra sc sb sa i i i i i i L L L L                                 m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1 Recallour voltage equations: Let’s apply our d-q transformation to it….             T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos
  • 25.
    Transforming voltage equations 25 Let’srewrite it in compact notation                                                                                   cr br ar cs bs as cr br ar cs bs as r r r s s s cr br ar cs bs as i i i i i i r r r r r r v v v v v v             0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                             abcr abcs abcr abcs r s abcr abcs i i r r v v     0 0 Now multiply through by our transformation matrices. Be careful with dimensionality.                              3 2 1 0 0 0 0 0 0 0 0 Term abcr abcs r s Term abcr abcs r s r s Term abcr abcs r s K K i i r r K K v v K K                                                
  • 26.
    Transforming voltage equations– Term 1 26 Therefore: the voltage equation becomes                           qdor s qd abcr r abcs s Term abcr abcs r s v v v K v K v v K K 0 1 0 0                              3 2 0 0 0 0 0 0 0 Term abcr abcs r s Term abcr abcs r s r s qdor s qd K K i i r r K K v v                                          
  • 27.
    Transforming voltage equations– Term 2 27 What to do with the abc currents? We need q-d-0 currents!                                abcr abcs r r s s Term abcr abcs r s r s i i r K r K i i r r K K 0 0 0 0 0 0 2            Recall:                      r qd s qd r s abcr abcs i i K K i i 0 0 1 1 0 0 and substitute into above.                                        r qd s qd r s r r s s Term abcr abcs r s r s i i K K r K r K i i r r K K 0 0 1 1 2 0 0 0 0 0 0 0 0            Perform the matrix multiplication:                                  r qd s qd r r r s s s Term abcr abcs r s r s i i K r K K r K i i r r K K 0 0 1 1 2 0 0 0 0 0 0            Fact: KRK-1=R if R is diagonal having equal elements on the diagonal. Proof: KRK-1=KrUK-1=rKUK-1=rKK-1=rU=R. Therefore….
  • 28.
    Transforming voltage equations– Term 2 28                                               r qd s qd r s r qd s qd r r r s s s Term abcr abcs r s r s i i r r i i K r K K r K i i r r K K 0 0 0 0 1 1 2 0 0 0 0 0 0 0 0            Therefore: the voltage equation becomes            3 0 0 0 0 0 0 0 Term abcr abcs r s r qd s qd r s qdor s qd K K i i r r v v                                    
  • 29.
    Transforming voltage equations– Term 3 29            3 0 0 0 0 0 0 0 Term abcr abcs r s r qd s qd r s qdor s qd K K i i r r v v                                     Focusing on just the stator quantities, consider: abcs s s qd K    0 Differentiate both sides abcs s abcs s s qd K K         0 Solve for abcs s K   abcs s s qd abcs s K K         0 Use λabcs =K-1λqd0s: s qd s s s qd abcs s K K K 0 1 0                                 abcr r abcs s Term abcr abcs r s K K K K                  3 0 0 Term 3 is: A similar process for the rotor quantities results in r qd r r r qd abcr r K K K 0 1 0          Substituting these last two expressions into the term 3 expression above results in                                            r qd r r s qd s s r qd s qd abcr r abcs s Term abcr abcs r s K K K K K K K K 0 1 0 1 0 0 3 0 0                          Substitute this back into voltage equation…
  • 30.
    Transforming voltage equations– Term 3 30            3 0 0 0 0 0 0 0 Term abcr abcs r s r qd s qd r s qdor s qd K K i i r r v v                                                                                r qd r r s qd s s r qd s qd abcr r abcs s Term abcr abcs r s K K K K K K K K 0 1 0 1 0 0 3 0 0                                                                 r qd r r s qd s s r qd s qd r qd s qd r s qdor s qd K K K K i i r r v v 0 1 0 1 0 0 0 0 0 0 0        
  • 31.
    Transforming voltage equations– Term 3 31 Now let’s express the fluxes in terms of currents by recalling that                      abcr abcs r s r qd s qd K K     0 0 0 0                                                                    abcr abcs r rs sr s abcr abcs cr br ar cs bs as r rs sr s cr br ar cs bs as i i L L L L i i i i i i L L L L         and the flux-current relations: Now write the abc currents in terms of the qd0 currents:                      r qd s qd r s abcr abcs i i K K i i 0 0 1 1 0 0 Substitute the third equation into the second:                            r qd s qd r s r rs sr s abcr abcs i i K K L L L L 0 0 1 1 0 0   Substitute the fourth equation into the first:                                    r qd s qd r s r rs sr s r s r qd s qd i i K K L L L L K K 0 0 1 1 0 0 0 0 0 0  
  • 32.
    Transforming voltage equations– Term 3 32                                    r qd s qd r s r rs sr s r s r qd s qd i i K K L L L L K K 0 0 1 1 0 0 0 0 0 0   Perform the first matrix multiplication:                              r qd s qd r s r r rs r sr s s s r qd s qd i i K K L K L K L K L K 0 0 1 1 0 0 0 0   and the next matrix multiplication:                          r qd s qd r r r s rs r r sr s s s s r qd s qd i i K L K K L K K L K K L K 0 0 1 1 1 1 0 0  
  • 33.
    Transforming voltage equations– Term 3 33                          r qd s qd r r r s rs r r sr s s s s r qd s qd i i K L K K L K K L K K L K 0 0 1 1 1 1 0 0   Now we need to go through each of these four matrix multiplications. I will here omit the details and just give the results (note also in what follows the definition of additional nomenclature for each of the four submatrices): 0 1 0 1 1 0 1 0 0 0 2 3 0 0 0 2 3 0 0 0 0 2 3 0 0 0 2 3 0 0 0 2 3 0 0 0 2 3 rqd r m r m r r r r mqd m m s rs r r sr s sqd s m s m s s s s L L L L L L K L K L L L K L K K L K L L L L L L K L K                                                                                           r qd s qd rqd mqd mqd sqd r qd s qd i i L L L L 0 0 0 0 0 0 0 0   And since our inductance matrix is constant, we can write:                      r qd s qd rqd mqd mqd sqd r qd s qd i i L L L L 0 0 0 0 0 0 0 0       Substitute the above expression for flux derivatives into our voltage equation:
  • 34.
    Transforming voltage equations– Term 3                      r qd s qd rqd mqd mqd sqd r qd s qd i i L L L L 0 0 0 0 0 0 0 0                                              r qd r r s qd s s r qd s qd r qd s qd r s qdor s qd K K K K i i r r v v 0 1 0 1 0 0 0 0 0 0 0         Substitute the above expressions for flux & flux derivatives into our voltage equation:                                            r qd r r s qd s s r qd s qd rqd mqd mqd sqd r qd s qd r s qdor s qd K K K K i i L L L L i i r r v v 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0       We still have the last term to obtain. To get this, we need to do two things. 1. Express individual q- and d- terms of λqd0s and λqd0r in terms of currents. 2. Obtain and 1  s s K K  1  r r K K  34
  • 35.
    Transforming voltage equations– Term 3 1. Express individual q- and d- terms of λqd0s and λqd0r in terms of currents: 35                      r qd s qd rqd mqd mqd sqd r qd s qd i i L L L L 0 0 0 0 0 0 0 0                                                                          r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3             dr m r ds m dr dr m ds m s ds qr m r qs m qr qr m qs m s qs i L L i L i L i L L i L L i L i L i L L                                     2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3         From the above, we observe:
  • 36.
    Transforming voltage equations– Term 3 2. Obtain and 1  s s K K  1  r r K K  36                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       s K                  2 1 2 1 2 1 ) 120 sin( ) 120 sin( sin ) 120 cos( ) 120 cos( cos 3 2       r K                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       s K                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       r K To get , we must consider:             ) ( ) 0 ( ) ( ) ( 0 t d t t  m m t m t d r                      ) ( ) 0 ( ) 0 ( ) ( ) ( ) 0 ( 0          s K  Therefore:                   0 0 0 ) 120 cos( ) 120 cos( cos ) 120 sin( ) 120 sin( sin 3 2        s K  Likewise, to get , we must consider: r K  Therefore:                      0 0 0 ) 120 cos( ) 120 cos( cos ) 120 sin( ) 120 sin( sin 3 2         m r K 
  • 37.
    Transforming voltage equations– Term 3 2. Obtain 1  s s K K  1  r r K K  37                                                                0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 2 3 0 3 2 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 0 0 0 ) 120 cos( ) 120 cos( cos ) 120 sin( ) 120 sin( sin 3 2 1                 s s K K                                                    0 0 0 0 0 0 ) ( 0 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 0 0 0 ) 120 cos( ) 120 cos( cos ) 120 sin( ) 120 sin( sin 3 2 1 m m m r r K K                    Obtain Substitute into voltage equations…
  • 38.
    Transforming voltage equations– Term 3 38              0 0 0 0 0 0 0 1   s s K K                 0 0 0 0 0 0 ) ( 0 1 m m r r K K      Substitute into voltage equations…                                            r qd r r s qd s s r qd s qd rqd mqd mqd sqd r qd s qd r s qdor s qd K K K K i i L L L L i i r r v v 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0                                                                                                                                                                    r dr qr s ds qs m m r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) ( 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                         This results in: Note the “Speed voltages” in the first, second, fourth, and fifth equations. -ωλds ωλqs -(ω- ωm)λdr (ω- ωm) λqs
  • 39.
    Transforming voltage equations– Term 3 39 Some comments on speed voltages: -ωλds, ωλqs, -(ω- ωm)λdr, (ω- ωm) λqs: • These speed voltages represent the fact that a rotating flux wave will create voltages in windings that are stationary relative to that flux wave. • Speed voltages are so named to contrast them from what may be called transformer voltages, which are induced as a result of a time varying magnetic field. • You may have run across the concept of “speed voltages” in Physics, where you computed a voltage induced in a coil of wire as it moved through a static magnetic field, in which case, you may have used the equation Blv where B is flux density, l is conductor length, and v is the component of the velocity of the moving conductor (or moving field) that is normal with respect to the field flux direction (or conductor). • The first speed voltage term, -ωλds, appears in the vqs equation. The second speed voltage term, ωλqs, appears in the vds equation. Thus, we see that the d- axis flux causes a speed voltage in the q-axis winding, and the q-axis flux causes a speed voltage in the d-axis winding. A similar thing is true for the rotor winding.
  • 40.
    Transforming voltage equations– Term 3 40              0 0 0 0 0 0 0 1   s s K K                 0 0 0 0 0 0 ) ( 0 1 m m r r K K      Substitute the matrices into voltage equation and then expand. This results in:                                            r qd r r s qd s s r qd s qd rqd mqd mqd sqd r qd s qd r s qdor s qd K K K K i i L L L L i i r r v v 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0                                                                                                                                                                    r dr qr s ds qs m m r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) ( 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                         Let’s collapse the last matrix-vector product by performing the multiplication….
  • 41.
    Transforming voltage equations– Term 3 41                                                                                                                                                              r dr qr s ds qs m m r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) ( 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                         dr m r ds m dr dr m ds m s ds qr m r qs m qr qr m qs m s qs i L L i L i L i L L i L L i L i L i L L                                     2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3                                                                                                                                                  0 ) ( ) ( 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 qr m dr m qs ds r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v                                                                                                                                                                                                                              0 2 3 2 3 ) ( 2 3 2 3 ) ( 0 2 3 2 3 2 3 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 qr m r qs m m dr m r ds m m qr m qs m s dr m ds m s r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i L L i L i L L i L i L i L L i L i L L i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v                       From slide 35, we have the fluxes expressed as a function of currents  And then substitute these terms in: Results In 
  • 42.
    Transforming voltage equations– Term 3 42                                                                                                                                                                                                          0 2 3 2 3 ) ( 2 3 2 3 ) ( 0 2 3 2 3 2 3 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 qr m r qs m m dr m r ds m m qr m qs m s dr m ds m s r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i L L i L i L L i L i L i L L i L i L L i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v                       Observe that the four non-zero elements in the last vector are multiplied by two currents from the current vector which multiplies the resistance matrix. So let’s now expand back out the last vector so that it is a product of a matrix and a current vector.                                                                                                                                                                                                      r dr qr s ds qs m r m m m m r m m m m m s m m s r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 ) ( 0 0 2 ) ( 3 0 2 3 ) ( 0 0 2 ) ( 3 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 2 3 0 0 2 3 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                             Now change the sign on the last matrix.
  • 43.
    Transforming voltage equations– Term 3 43                                                                                                                                                                                                      r dr qr s ds qs m r m m m m r m m m m m s m m s r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r r s s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L i i i i i i L L L L L L L L L L L L L L i i i i i i r r r r r r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 3 ) ( 0 0 2 ) ( 3 0 2 3 ) ( 0 0 2 ) ( 3 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 2 3 0 0 2 3 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0                             Notice that the resistance matrix and the last matrix multiply the same vector, therefore, we can combine these two matrices. For example, element (1,2) in the last matrix will go into element (1,2) of the resistance matrix, as shown. This results in the expression on the next slide….
  • 44.
    Final Model 44                                                                                                                                                             r dr qr s ds qs r m r m m r m s m m s m m s r dr qr s ds qs r r m r m m m m r m r m m s m s m s m m s s r dr qr s ds qs i i i i i i L L L L L L L L L L L L L L i i i i i i r r L L L L L r L r L r L L L L L r v v v v v v 0 0 0 0 0 0 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 0 0 2 3 0 0 0 2 3 0 0 2 3 0 0 0 0 0 0 2 3 ) ( 0 0 2 ) ( 3 0 2 3 ) ( 0 2 ) ( 3 0 0 0 0 0 0 0 0 2 3 0 2 3 0 2 3 0 0 2 3                             This isthe complete transformed electric machine state-space model in “current form.”
  • 45.
    Some comments aboutthe transformation 45 • ids and iqs are currents in a fictitious pair of windings fixed on a synchronously rotating reference frame. • These currents produce the same flux as do the stator a,b,c currents. • For balanced steady-state operating conditions, we can use iqd0s = Ksiabcs to show that the currents in the d and q windings are dc! The implication of this is that: • The a,b,c currents fixed in space (on the stator), varying in time produce the same synchronously rotating magnetic field as • The ds,qs currents, varying in space at synchronous speed, fixed in time! • idr and iqr are currents in a fictitious pair of windings fixed on a synchronously rotating reference frame. • These currents produce the same flux as do the rotor a,b,c currents. • For balanced steady-state operating conditions, we can use iqd0r = Kriabcr to show that the currents in the d and q windings are dc! The implication of this is that: • The a,b,c currents varying in space at slip speed sωs=(ωs- ωm) fixed on the rotor, varying in time produce the same synchronously rotating magnetic field as • The dr,qr currents, varying in space at synchronous speed, fixed in time!
  • 46.
    Torque in abcquantities 46 The electromagnetic torque of the DFIG may be evaluated according to m c em W T     m f em W T     The stored energy is the sum of • The self inductances (less leakage) of each winding times one-half the square of its current and • All mutual inductances, each times the currents in the two windings coupled by the mutual inductance Observe that the energy stored in the leakage inductances is not a part of the energy stored in the coupling field. Consider the abc inductance matrices given in slide 6. where Wc is the co-energy of the coupling fields associated with the various windings. We are not considering saturation here, assuming the flux-current relations are linear, in which case the co-energy Wc of the coupling field equals its energy, Wf, so that: We use electric rad/sec by substituting ϴm=θm/p where p is the number of pole pairs. m f em W p T    
  • 47.
    Torque in abcquantities 47                           m s m m m m s m m m m s s L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                           m r m m m m r m m m m r r L L L L L L L L L L L L L    2 1 2 1 2 1 2 1 2 1 2 1                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos             T sr m m m m m m m m m m rs L L L                             cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos The stored energy is given by: abcr r r T abcr abcr sr T abcs abcs s s T abcs f i U L L i i L i i U L L i W ) ( 2 1 ) ( 2 1        Applying the torque-energy relation abcr sr T abcs m m f i L i W        m f em W p T     to the above, and observing that dependence on θm only occurs in the middle term, we get abcr sr T abcs m em i L i p T     So that But only Lsr depend on θm, so abcr m sr T abcs em i L i p T    
  • 48.
    Torque in abcquantities 48 We may go through some analytical effort to show that the above evaluates to abcr sr T abcs m em i L i p T                                                           m br ar cs ar cr bs cr br as m ar br cr cs cr ar br bs cr br ar as m em i i i i i i i i i i i i i i i i i i i i i pL T   cos 2 3 sin 2 1 2 1 2 1 2 1 2 1 2 1 To complete our abc model we relate torque to rotor speed according to: m m em T dt d p J T    Inertial torque Mech torque (has negative value for generation) J is inertia of the rotor in kg-m2 or joules-sec2 Negative value for generation
  • 49.
    Torque in qd0quantities However, our real need is to express the torque in qd0 quantities so that we may complete our qd0 model. To this end, recall that we may write the abc quantities in terms of the qd0 quantities using our inverse transformation, according to: r qd r abcr s qd s abcs i K i i K i 0 1 0 1       r qd r sr m T s qd s abcr sr m T abcs em i K L i K p i L i p T 0 1 0 1           Substitute the above into our torque expression: 49
  • 50.
    Torque in qd0quantities 50                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       s K                 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos 1       r K                              m m m m m m m m m m sr L L          cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos                                                                                                    r dr qr m m m m m m m m m m m T s ds qs em i i i L i i i p T 0 0 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos cos 120 cos 120 cos 120 cos cos 120 cos 120 cos 120 cos cos 1 ) 120 sin( ) 120 cos( 1 ) 120 sin( ) 120 cos( 1 sin cos                         r qd r sr m T s qd s em i K L i K p T 0 1 0 1       I will not go through this differentiation but instead provide the result:   qr ds dr qs m em i i i i pL T   4 9
  • 51.
    Torque in qd0quantities 51 Some other useful expressions may be derived from the above, as follows:   qr ds dr qs m em i i i i pL T   4 9   qr dr dr qr em i i p T     2 3   ds qs qs ds em i i p T     2 3 Final comment: We can work with these expressions to show that the electromagnetic torque can be directly controlled by the rotor quadrature current iqr At the same time, we can also show that the stator reactive power Qs can be directly controlled by the rotor direct-axis current idr. This will provide us the necessary means to control the wind turbine.