Chapter 3 reference frame theory

Analysis of Electric Machinery and Drive Systems Editor(s): Paul Krause, Oleg Wasynczuk, Scott Sudhoff, Steven Pekarek

3 REFERENCE-FRAME THEORY 3.1 INTRODUCTION We have found that some of the machine inductances are functions of rotor position, whereupon the coefficients of the differential equations (voltage equations) that describe the behavior of these machines are rotor position dependent A change of variables is often used to reduce the complexity of these differential equations There are several changes of variables that are used, and it was originally thought that each change of variables was unique and therefore they were treated separately [1–4] It was later learned that all changes of variables used to transform actual variables are contained in one [5, 6] This general transformation refers machine variables to a frame of reference that rotates at an arbitrary angular velocity All known real transformations are obtained from this transformation by simply assigning the speed of the rotation of the reference frame In this chapter, this transformation is set forth and, since many of its properties can be studied without the complexities of the machine equations, it is applied to the equations that describe resistive, inductive, and capacitive circuit elements Using this approach, many of the basic concepts and interpretations of this general transformation Analysis of Electric Machinery and Drive Systems, Third Edition Paul Krause, Oleg Wasynczuk, Scott Sudhoff, and Steven Pekarek © 2013 Institute of Electrical and Electronics Engineers, Inc Published 2013 by John Wiley & Sons, Inc 86 BACKGROUND 87 are readily and concisely established Extending the material presented in this chapter to the analysis of ac machines is straightforward, involving a minimum of trigonometric manipulations 3.2 BACKGROUND In the late 1920s, R.H Park [1] introduced a new approach to electric machine analysis He formulated a change of variables that in effect replaced the variables (voltages, currents, and flux linkages) associated with the stator windings of a synchronous machine with variables associated with fictitious windings rotating at the electrical angular velocity of the rotor This change of variables is often described as transforming or referring the stator variables to a frame of reference fixed in the rotor Park’s transformation, which revolutionized electric machine analysis, has the unique property of eliminating all rotor position-dependent inductances from the voltage equations of the synchronous machine that occur due to (1) electric circuits in relative motion and (2) electric circuits with varying magnetic reluctance In the late 1930s, H.C Stanley [2] employed a change of variables in the analysis of induction machines He showed that the varying mutual inductances in the voltage equations of an induction machine due to electric circuits in relative motion could be eliminated by transforming the variables associated with the rotor windings (rotor variables) to variables associated with fictitious stationary windings In this case, the rotor variables are transformed to a frame of reference fixed in the stator G Kron [3] introduced a change of variables that eliminated the position-dependent mutual inductances of a symmetrical induction machine by transforming both the stator variables and the rotor variables to a reference frame rotating in synchronism with the fundamental angular velocity of the stator variables This reference frame is commonly referred to as the synchronously rotating reference frame D.S Brereton et al [4] employed a change of variables that also eliminated the varying mutual inductances of a symmetrical induction machine by transforming the stator variables to a reference frame rotating at the electrical angular velocity of the rotor This is essentially Park’s transformation applied to induction machines Park, Stanley, Kron, and Brereton et al developed changes of variables, each of which appeared to be uniquely suited for a particular application Consequently, each transformation was derived and treated separately in literature until it was noted in 1965 [5] that all known real transformations used in induction machine analysis are contained in one general transformation that eliminates all rotor position-dependent mutual inductances by referring the stator and the rotor variables to a frame of reference that may rotate at any angular velocity or remain stationary All known real transformations may then be obtained by simply assigning the appropriate speed of rotation, which may in fact be zero, to this so-called arbitrary reference frame Later, it was noted that the stator variables of a synchronous machine could also be referred to the arbitrary reference frame [6] However, we will find that the varying inductances of a synchronous machine are eliminated only if the reference frame is rotating at the electrical angular velocity of the rotor (Park’s transformation); consequently, the arbitrary reference frame 88 REFERENCE-FRAME THEORY does not offer the advantages in the analysis of the synchronous machines that it does in the case of induction machines 3.3 EQUATIONS OF TRANSFORMATION: CHANGE OF VARIABLES Although changes of variables are used in the analysis of ac machines to eliminate time-varying inductances, changes of variables are also employed in the analysis of various static, constant-parameter power-system components and control systems associated with electric drives For example, in many of the computer programs used for transient and dynamic stability studies of large power systems, the variables of all power system components, except for the synchronous machines, are represented in a reference frame rotating at synchronous speed, wherein the electric transients are often neglected Hence, the variables associated with the transformers, transmission lines, loads, capacitor banks, and static var units, for example, must be transformed to the synchronous rotating reference frame by a change of variables Similarly, the “average value” of the variables associated with the conversion process in electric drive systems and in high-voltage ac–dc systems are often expressed in the synchronously rotating reference frame Fortunately, all known real transformations for these components and controls are also contained in the transformation to the arbitrary reference frame, the same transformation used for the stator variables of the induction and synchronous machines and for the rotor variables of induction machines Although we could formulate one transformation to the arbitrary reference frame that could be applied to all variables, it is preferable to consider only the variables associated with stationary circuits in this chapter and then modify this analysis for the variables associated with the rotor windings of the induction machine at the time it is analyzed A change of variables that formulates a transformation of the three-phase variables of stationary circuit elements to the arbitrary reference frame may by expressed as fqd s = K s fabcs (3.3-1) where ( fqd s )T = [ fqs fds f0 s ] (3.3-2) ( fabcs ) = [ fas fbs fcs ] (3.3-3) T ⎡ ⎢cos θ ⎢ K s = ⎢ sin θ 3⎢ ⎢ ⎢ ⎢ ⎣ 2π ⎞ ⎛ cos ⎜ θ − ⎟ ⎝ 3⎠ 2π ⎞ ⎛ sin ⎜ θ − ⎟ ⎝ 3⎠ 2π ⎞ ⎤ ⎛ cos ⎜ θ + ⎟ ⎥ ⎝ 3⎠ ⎥ 2π ⎞ ⎛ sin ⎜ θ + ⎟ ⎥ ⎝ ⎠⎥ ⎥ ⎥ ⎥ ⎦ (3.3-4) EQUATIONS OF TRANSFORMATION: CHANGE OF VARIABLES 89 where the angular position and velocity of the arbitrary reference frame are related as dθ =ω dt (3.3-5) It can be shown that the inverse transformation is ⎡ ⎢ cos θ ⎢ 2π ⎞ ⎛ (K s )−1 = ⎢ cos ⎜ θ − ⎟ ⎢ ⎝ 3⎠ ⎢ ⎢cos ⎛ θ + 2π ⎞ ⎟ ⎢ ⎜ 3⎠ ⎣ ⎝ ⎤ 1⎥ sin θ ⎥ 2π ⎞ ⎛ sin ⎜ θ − ⎟ 1⎥ ⎝ 3⎠ ⎥ ⎥ 2π ⎞ ⎛ sin ⎜ θ + ⎟ 1⎥ ⎥ ⎝ 3⎠ ⎦ (3.3-6) In the above equations, f can represent either voltage, current, flux linkage, or electric charge The superscript T denotes the transpose of a matrix The s subscript indicates the variables, parameters, and transformation associated with stationary circuits The angular displacement θ must be continuous; however, the angular velocity associated with the change of variables is unspecified The frame of reference may rotate at any constant or varying angular velocity, or it may remain stationary The connotation of 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 system equations or to satisfy the system constraints The change of variables may be applied to variables of any waveform and time sequence; however, we will find that, for a three-phase electrical system, the transformation given above is particularly appropriate for an abc sequence Although the transformation to the arbitrary reference frame is a change of variables and needs no physical connotation, it is often convenient to visualize the transformation equations as trigonometric relationships between variables as shown in Figure 3.3-1 In particular, the equations of transformation may be thought of as if the fqs and fds variables are “directed” along paths orthogonal to each other and rotating at an angular velocity of ω, whereupon fas, fbs, and fcs may be considered as variables directed along stationary paths each displaced by 120° If fas, fbs, and fcs are resolved into fqs, the first row of (3.3-1) is obtained, and if fas, fbs, and fcs are resolved into fds, the second row is obtained It is important to note that f0s variables are not associated with the arbitrary reference frame Instead, the zero variables are related arithmetically to the abc variables, independent of ω and θ Portraying the transformation as shown in Figure 3.3-1 is particularly convenient when applying it to ac machines where the direction of fas, fbs, and fcs may also be thought of as the direction of the magnetic axes of the stator windings We will find that the direction of fqs and fds can be considered as the direction of the magnetic axes of the “new” windings created by the change of variables It is also important not to confuse fas, fbs, and fcs or fqs and fds with phasors The total instantaneous power of a three-phase system may be expressed in abc variables as 90 REFERENCE-FRAME THEORY w fbs fqs q fas fcs Figure fds 3.3-1 Transformation for stationary circuits portrayed by trigonometric relationships Pabcs = vas ias + vbs ibs + vcs ics (3.3-7) The total power expressed in the qd0 variables must equal the total power expressed in the abc variables, hence using (3.3-1) to replace actual currents and voltages in (3.37) yields Pqd s = Pabcs = (vqs iqs + vds ids + v0 s i0 s ) (3.3-8) The 3/2 factor comes about due to the choice of the constant used in the transformation Although the waveforms of the qs and ds voltages, currents, flux linkages, and electric charges are dependent upon the angular velocity of the frame of reference, the waveform of total power is independent of the frame of reference In other words, the waveform of the total power is the same regardless of the reference frame in which it is evaluated 3.4 STATIONARY CIRCUIT VARIABLES TRANSFORMED TO THE ARBITRARY REFERENCE FRAME It is convenient to treat resistive, inductive, and capacitive circuit elements separately Resistive Elements For a three-phase resistive circuit v abcs = rs i abcs (3.4-1) STATIONARY CIRCUIT VARIABLES TRANSFORMED 91 From (3.3-1) v qd s = K s rs (K s )−1 i qd s (3.4-2) It is necessary to specify the resistance matrix rs before proceeding All stator phase windings of either a synchronous or a symmetrical induction machine are designed to have the same resistance Similarly, transformers, capacitor banks, transmission lines and, in fact, all power-system components are designed so that all phases have equal or near-equal resistances Even power-system loads are distributed between phases so that all phases are loaded nearly equal If the nonzero elements of the diagonal matrix rs are equal, then K s rs (K s )−1 = rs (3.4-3) Thus, the resistance matrix associated with the arbitrary reference variables (fqs, fds, and f0s) is equal to the resistance matrix associated with the actual variables if each phase of the actual circuit has the same resistance If the phase resistances are unequal (unbalanced or unsymmetrical), then the resistance matrix associated with the arbitrary reference-frame variables contains sinusoidal functions of θ except when ω = 0, whereupon Ks is algebraic In other words, if the phase resistances are unbalanced, the transformation yields constant resistances only if the reference frame is fixed where the unbalance physically exists This feature is quite easily illustrated by substituting rs = diag[ras rbs rcs] into Ksrs(Ks)−1 Inductive Elements For a three-phase inductive circuit, we have v abcs = pl abcs (3.4-4) where p is the operator d/dt In the case of the magnetically linear system, it has been customary to express the flux linkages as a product of inductance and current matrices before performing a change of variables However, the transformation is valid for flux linkages and an extensive amount of work can be avoided by transforming the flux linkages directly This is especially true in the analysis of ac machines, where the inductance matrix is a function of rotor position Thus, in terms of the substitute variables, (3.4-4) becomes v qd s = K s p[(K s )−1 l qd s ] (3.4-5) v qd s = K s p[(K s )−1 ]l qd s + K s (K s )−1 pl qd s (3.4-6) which can be written as It is easy to show that 92 REFERENCE-FRAME THEORY ⎡ ⎢ − sin θ ⎢ 2π ⎞ ⎛ p[(K s )−1 ] = ω ⎢ − sin ⎜ θ − ⎟ ⎢ ⎝ 3⎠ ⎢ ⎢ − sin ⎛ θ + 2π ⎞ ⎜ ⎟ ⎢ ⎝ 3⎠ ⎣ cos θ 2π ⎞ ⎛ cos ⎜ θ − ⎟ ⎝ 3⎠ 2π ⎞ ⎛ cos ⎜ θ + ⎟ ⎝ 3⎠ ⎤ 0⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ ⎦ (3.4-7) Therefore, ⎡ 0⎤ K s p[(K s ) ] = ω ⎢ −1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 0⎦ −1 (3.4-8) Trigonometric identities given in Appendix A are helpful in obtaining (3.4-8) Equation (3.4-6) may now be expressed as v qd s = ω l dqs + pl qd s (3.4-9) where ( l dqs )T = [λ ds −λ qs 0] (3.4-10) Equation (3.4-6) is often written in expanded form as vqs = ωλ ds + pλqs (3.4-11) vds = −ωλqs + pλ ds (3.4-12) v0 s = pλ0 s (3.4-13) The first term on the right side of (3.4-11) or (3.4-12) is referred to as a “speed voltage,” with the speed being the angular velocity of the arbitrary reference frame It is clear that the speed voltage terms are zero if ω is zero, which, of course, is when the reference frame is stationary Clearly, the voltage equations for the three-phase inductive circuit become the familiar time rate of change of flux linkages if the reference frame is fixed where the circuit physically exists Also, since (3.4-4) is valid in general, it follows that (3.4-11)–(3.4-13) are valid regardless if the system is magnetically linear or nonlinear and regardless of the form of the inductance matrix if the system is magnetically linear For a linear system, the flux linkages may be expressed l abcs = L s i abcs (3.4-14) Whereupon, the flux linkages in the arbitrary reference frame may be written as l qd s = K s L s (K s )−1 i qd s (3.4-15) 93 STATIONARY CIRCUIT VARIABLES TRANSFORMED As is the case of the resistive circuit, it is necessary to specify the inductance matrix before proceeding with the evaluation of (3.4-15) However, once the inductance matrix is specified, the procedure for expressing any three-phase inductive circuit in the arbitrary reference frame reduces to one of evaluating (3.4-15) and substituting the resulting λqs, λds, and λ0s into the voltage equations (3.4-11)–(3.4-13) This procedure is straightforward, with a minimum of matrix manipulations compared with the work involved if, for a linear system, the flux linkage matrix λabcs is replaced by Lsiabcs before performing the transformation If, for example, Ls is a diagonal matrix with all nonzero terms equal, then K s L s (K s )−1 = L s (3.4-16) A matrix of this form could describe the inductance of a balanced three-phase inductive load, a three-phase set of line reactors used in high-voltage transmission systems or any symmetrical three-phase inductive network without coupling between phases It is clear that the comments regarding unbalanced or unsymmetrical phase resistances also apply in the case of unsymmetrical inductances An inductance matrix that is common is of the form ⎡ Ls Ls = ⎢ M ⎢ ⎢ ⎣M M Ls M M⎤ M⎥ ⎥ Ls ⎦ ⎥ (3.4-17) where Ls is a self inductance and M is a mutual inductance This general form can be used to describe the stator self- and mutual inductance relationships of the stator phases of symmetrical induction machines, and round-rotor synchronous machines with arbitrary winding arrangement, including double-layer and integer and noninteger slot/pole/ phase windings From our work in Chapter 2, we realize that this inductance matrix is of a form that describes the self- and mutual inductances relationships of the stator phases of a symmetrical induction machine and the stator phases of a round-rotor synchronous machine with or without mutual leakage paths between stator windings It can also describe the coupling of a symmetrical transmission line Example diagrams that portray such coupling are shown in Figure 3.4-1 It is left to the reader to show that for Ls given by (3.4-17) ⎡ Ls − M K s L s (K s )−1 = ⎢ ⎢ ⎢ ⎣ Ls − M 0 ⎤ ⎥ ⎥ Ls + M ⎥ ⎦ (3.4-18) Linear three-phase coupled systems are magnetically symmetrical if the diagonal elements are equal and all off-diagonal elements of the inductance matrix are also equal Equation (3.4-17) is of this form We see from (3.4-18) that, for a symmetrical system, KsLs(Ks)−1 yields a diagonal matrix that, in effect, magnetically decouples the substitute variables in all reference frames This is a very important feature of the transformation 94 REFERENCE-FRAME THEORY ias vas + – rs Ls M ibs vbr + – rs M ics Ls M vcs + – rs (a) Ls ics + rs + rs vcs M Ls vbs ibs Ls M M Ls vas rs ias (b) + Figure 3.4-1 Three-phase RL circuit (a) Symmetrical transmission line; (b) wye connection On the other hand, we have seen in Section 1.4 and Chapter that the self- and mutual inductances between the stator phases of the salient-pole synchronous machine form a magnetically unsymmetrical system It will be shown that for this case, there is only one reference frame, the reference frame rotating at the electrical angular velocity of the rotor, wherein the substitute variables are not magnetically coupled Capacitive Elements For a three-phase capacitive circuit, we have i abcs = pq abcs (3.4-19) Incorporating the substitute variables yields i qd s = K s p[(K s )−1 q qd s ] (3.4-20) i qd s = K s p[(K s )−1 ]q qd s + K s (K s )−1 pq qd s (3.4-21) that can be written as 95 STATIONARY CIRCUIT VARIABLES TRANSFORMED Utilizing (3.4-8) yields i qd s = ω q dqs + pq qd s (3.4-22) where (q dqs )T = [qds −qqs 0] (3.4-23) In expanded form, we have iqs = ω qds + pqqs (3.4-24) ids = −ω qqs + pqds (3.4-25) i0 s = pq0 s (3.4-26) Considering the terminology of “speed voltages” as used in the case of inductive circuits, it would seem appropriate to refer to the first term on the right side of either (3.4-24) or (3.4-25) as “speed currents.” Also, as in the case of inductive circuits, the equations revert to the familiar form in the stationary reference frame (ω = 0) Equations (3.4-24)–(3.4-26) are valid regardless of the relationship between charge and voltage For a linear capacitive system, we have q abcs = C s v abcs (3.4-27) Thus, in the arbitrary reference frame we have q qd s = K s C s (K s )−1 v qd s (3.4-28) Once the capacitance matrix is specified, qqs, qds, and q0s can be determined and substituted into (3.4-24)–(3.4-26) The procedure and limitations are analogous to those in the case of the inductive circuits A diagonal capacitance matrix with equal nonzero elements describes, for example, a three-phase capacitor bank used for power factor correction and the series capacitance used for transmission line compensation or any three-phase electrostatic system without coupling between phases A three-phase transmission system is often approximated as a symmetrical system, whereupon the inductance and capacitance matrices may be written in a form similar to (3.4-17) EXAMPLE 3A For the purpose of demonstrating the transformation of variables to the arbitrary reference frame, let us consider a three-phase RL circuit defined by rs = diag[rs rs rs ] (3A-1) 106 REFERENCE-FRAME THEORY ibs = 2Vs Zs 2π ⎞ 2π ⎞ ⎤ ⎛ ⎛ ⎡ −t /τ ⎢ − e cos ⎜ α + ⎟ + cos ⎜ ω e t − α − ⎟ ⎥ ⎝ ⎠ ⎝ ⎠⎦ ⎣ (3.9-5) ics = 2Vs Zs 2π ⎞ 2π ⎞ ⎤ ⎛ ⎛ ⎡ −t /τ ⎢ −e cos ⎜ α − ⎟ + cos ⎜ ω e t − α + ⎟ ⎥ ⎝ ⎠ ⎝ ⎠⎦ ⎣ (3.9-6) where Z s = rs + jω e Ls τ= Ls rs α = tan −1 (3.9-7) (3.9-8) ω e Ls rs (3.9-9) It may at first appear necessary to solve the voltage equations in the arbitrary reference frame in order to obtain the expression for the currents in the arbitrary reference frame This is unnecessary, since once the solution is known in one reference frame, it is known in all reference frames In the example at hand, this may be accomplished by transforming (3.9-4)–(3.9-6) to the arbitrary reference frame For illustrative purposes, let ω be an unspecified constant with θ(0) = 0, then θ = ωt, and in the arbitrary reference frame, we have iqs = 2Vs {−e − t / τ cos(ω t − α ) + cos[(ω e − ω )t − α ]} Zs (3.9-10) ids = 2Vs − t / τ {e sin(ω t − α ) − sin[(ω e − ω )t − α ]} Zs (3.9-11) Clearly, the state of the electric system is independent of the frame of reference from which it is observed Although the variables will appear differently in each reference frame, they will exhibit the same mode of operation (transient or steady state) regardless of the reference frame In general, (3.9-10) and (3.9-11) contain two balanced sets One, which represents the electric transient, decays exponentially at a frequency corresponding to the instantaneous angular velocity of the arbitrary reference frame In this set, the qs variable leads the ds variable by 90° when ω > and lags by 90° when ω < The second balanced set, which represents the steady-state response, has a constant amplitude with a frequency corresponding to the difference in the angular velocity of the voltages applied to the stationary circuits and the angular velocity of the arbitrary reference frame In this set, the qs variable lags the ds by 90° when ω < ωe and leads by 90° when ω > ωe This, of course, leads to the concept of negative frequency when relating phasors that represent qs and ds variables by (3.7-8) There are two frames of reference that not contain both balanced sets In the s stationary reference frame ω = and iqs = ias The exponentially decaying balanced set becomes an exponential decay, and the constant amplitude balanced set varies at ωe In the synchronously rotating reference frame where ω = ωe, the electric transients are 107 VARIABLES OBSERVED FROM SEVERAL FRAMES OF REFERENCE represented by an exponentially decaying balanced set varying at ωe, and the constant amplitude balanced set becomes constants The waveforms of the system variables in various reference frames are shown in Figure 3.9-1, Figure 3.9-2, and Figure 3.9-3 The voltages of the form given by (3.91)–(3.9-3) are applied to the three-phase system with Vs = 10 / V, rs = 0.216 Ω, ωeLs = 1.09 Ω with ωe = 377 rad/s The response, for t > 0, of the electric system in the stationary reference frame is shown in Figure 3.9-1 Since we have selected θ(0) = 0, s s s fas = fqs and the plots of vqs and iqs are νas and ias, respectively The variables for the 10 s v qs, V –10 10 s i qs, A –10 0.01 second 10 s v ds, V –10 10 s i ds, A –10 120 90 P, W 60 30 –30 w, rad/s Figure 3.9-1 Variables of stationary three-phase system in the stationary reference frame 108 REFERENCE-FRAME THEORY 10 e v qs, V 10 e i qs, A 0.01 second e v ds, V 10 e i ds, A 120 90 PW , 60 30 –30 377 w, rad/s Figure 3.9-2 Variables of stationary three-phase system in synchronously rotating reference frame same mode of operation are shown in the synchronously rotating reference frame in Figure 3.9-2 Note from (3.9-1) that we have selected θeν(0) = 0, thus, from (3.6-5) and e e (3.6-6) with θ(0) = 0, vqs = 10 V and vds = In Figure 3.9-3, with θ(0) = the speed of the reference frame is switched from its original value of −377 rad/s to zero and the ramped to 377 rad/s There are several features worthy of note The waveform of the instantaneous electric power is the same in all cases The electric transient is very evident in the waveforms of the instantaneous electric power and the currents in the synchronously e e rotating reference frame (Fig 3.9-2) and since vds is zero iqs is related to the power by e a constant (vqs ) In Figure 3.9-3, we selected θeν(0) = and θ(0) = The voltages were applied, and we observed the solution of the differential equations in the reference frame rotating clockwise at ωe (ω = −ωe) The reference-frame speed was then stepped from −377 rad/s to 0, whereupon the differential equations were solved in the stationary reference frame However, when switching from one reference frame to another the variables must be continuous Therefore, after the switching occurs the solution 109 VARIABLES OBSERVED FROM SEVERAL FRAMES OF REFERENCE 10 v qs, V –10 10 i qs, A –10 0.01 second 10 v ds, V –10 10 i ds, A –10 120 90 PW , 60 30 –30 377 w, rad/s –377 Figure 3.9-3 Variables of stationary three-phase system First with ω = −ωe, then ω = 0, followed by a ramp change in reference-frame speed to ω = ωe continues using the stationary reference-frame differential equations with the initial values determined by the instantaneous values of the variables in the previous reference frame (ω = −ωe) at the time of switching It is important to note the change in frequency of the variables as the reference-frame speed is ramped from zero to ωe Here, the differential equations being solved are continuously changing while the variables remain 110 REFERENCE-FRAME THEORY continuous When the reference-frame speed reaches synchronous speed, the variables have reached steady state, therefore they will be constant corresponding to their values at the instant ω becomes equal to ωe In essence, we have applied a balanced three-phase set of voltages to a symmetrical RL circuit, and in Figure 3.9-3, we observed the actual variables from various reference frames by first “jumping” and then “running” from one reference frame to another 3.10 TRANSFORMATION BETWEEN REFERENCE FRAMES In some derivations and analyses, it is convenient to relate variables in one reference frame to variables in another reference frame directly, without involving the abc variables in the transformation In order to establish this transformation between any two frames of reference, let x denote the reference frame from which the variables are being transformed and y the reference frame to which the variables are being transformed, then y x fqd s = x K y fqd s (3.10-1) x x fqd s = K s fabcs (3.10-2) From (3.3-1), we obtain Substituting (3.10-2) into (3.10-1) yields y x fqd s = x K y K s fabcs (3.10-3) y y fqd s = K s fabcs (3.10-4) x y K yKs = Ks (3.10-5) y x K y = K s (K s ) −1 (3.10-6) However, from (3.3-1), we obtain Thus x from which x The desired transformation is obtained by substituting the appropriate transformations into (3.10-6) Hence x ⎡cos(θ y − θ x ) − sin(θ y − θ x ) ⎤ K y = ⎢ sin(θ y − θ x ) cos(θ y − θ x ) ⎥ ⎢ ⎥ 0 1⎦ ⎢ ⎥ ⎣ (3.10-7) 111 SPECIALTY TRANSFORMATIONS wy y fqs wx x fqs qy qx y fds x fds Figure 3.10-1 Transformation between two reference frames portrayed by trigonometric relationships Several of the trigonometric identities given in Appendix A are useful in obtaining (3.10-7) This transformation, which is sometimes referred to as a “vector rotator” or simply “rotator,” can also be visualized from the trigonometric relationship between x two sets of rotating, orthogonal quantities, as shown in Figure 3.10-1 Resolving fqs y x y x x and fds into fqs yields the first row of (3.10-7), and resolving fqs and fds into fds yields the second row It is left for the reader to show that ( x K y )−1 = ( x K y )T (3.10-8) 3.11 SPECIALTY TRANSFORMATIONS The transformations that have been set forth are valid for any three-phase connection In many cases, the three-phase voltages, currents, and flux linkages must sum to zero This is common, although not the case if a third harmonic is present in the back-emf as in some permanent magnet ac machines There are cases where the neutral is not accessible and only the line-to-line voltages are available, and yet it is desirable to express the νqs and νds If fas + fbs + fcs = (3.11-1) 112 REFERENCE-FRAME THEORY we can express any one of the three variables in terms of the other two; let us start with expressing fcs in (3.3-1) as fcs = − fas − fbs (3.11-2) After some trigonometric manipulation, (3.3-1) may be written as ⎡ fqs ⎤ ⎡ cos θ ⎢ f ⎥ = ⎢ sin θ ⎣ ds ⎦ ⎣ fas ⎡ ⎤ sin θ ⎤ ⎢ ⎥ fas + fbs ⎥ − cos θ ⎥ ⎢ ⎦ ⎣ 3 ⎦ (3.11-3) This is the two-phase transformation to the arbitrary reference frame Since we have reduced the three-wire connection to a two-phase system, we should not be too surprised by this In a problem at the end of this chapter, you are asked to express the voltage vector in the right hand side of (3.11-3) in terms of fas and fcs and in terms of fbs and fcs It is left to the reader to show that (3.11-3) can be written π⎞ ⎡ ⎛ cos ⎜ θ − ⎟ 6⎠ ⎡ fqs ⎤ ⎢ ⎝ ⎢ ⎢f ⎥= π⎞ 3⎢ ⎛ ⎣ ds ⎦ ⎢ sin ⎜ θ − ⎟ ⎝ ⎠ ⎣ ⎤ sin θ ⎥ ⎡ fas ⎤ ⎥⎢ ⎥ ⎣f ⎦ − cos θ ⎥ bs ⎥ ⎦ (3.11-4) It is clear that fas, fbs, and fcs can be voltages, currents, or flux linkages and that the variables can be of any form The variables need not form a balanced set The only constraint is that f0s is zero In a wye-connected system, if the neutral is not available then νas, νbs, and νcs are not available, however, we are still dealing with a three-wire, wye connection If ν0s is zero, we can express νqs and νds in terms of line-to-line voltages This, of course, is important when it is necessary for control or analysis purposes to be able to determine νqs and νds when the phase voltages are not physically available The line-to-line voltages may be expressed in terms of phase voltages as ⎡ vabs ⎤ ⎡ vas − vbs ⎤ ⎢v ⎥ = ⎢v − v ⎥ ⎢ bcs ⎥ ⎢ bs cs ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ vcas ⎦ ⎣ vcs − vas ⎦ (3.11-5) Solving (3.11-5) for νas, νbs, and νcs and substituting into (3.3-1) yields ⎡ cos θ ⎡ vqs ⎤ ⎢ ⎢v ⎥ = ⎢ ⎣ ds ⎦ ⎢ sin θ ⎢ ⎣ 2π ⎞ ⎛ cos ⎜ θ − ⎟ ⎝ 3⎠ 2π ⎞ ⎛ sin ⎜ θ − ⎟ ⎝ 3⎠ 2π ⎞ ⎤ ⎛ cos ⎜ θ + ⎟ ⎥ ⎡ vabs + vbs ⎤ ⎝ 3⎠ ⎢ ⎥ vbcs + vcs ⎥ ⎥ 2π ⎞ ⎥ ⎢ ⎛ sin ⎜ θ + ⎟ ⎥ ⎣ vcas + vas ⎦ ⎢ ⎥ ⎝ ⎠⎦ (3.11-6) 113 SPACE-PHASOR NOTATION Since ν0s is zero, (3.11-6) may be written as ⎡ cos θ ⎡ vqs ⎤ ⎢ = ⎢ ⎢v ⎥ ⎣ ds ⎦ ⎢ sin θ ⎢ ⎣ 2π ⎞ ⎤ ⎛ cos ⎜ θ + ⎟ ⎥ ⎝ ⎠ ⎡ vabs ⎤ ⎥ 2π ⎞ ⎥ ⎢ vcbs ⎥ ⎣ ⎦ ⎛ sin ⎜ θ + ⎟ ⎥ ⎝ ⎠⎦ (3.11-7) 3.12 SPACE-PHASOR NOTATION In the early 1900s, Fortescue [8] developed the method of symmetrical components to analyze the steady-state performance of induction machines during unbalanced operation This theory is set forth later in Chapter 9; however, it is sufficient here to mention that this method of analysis, for a three-phase system, involves a balanced three-phase set of positive abc sequence phasors, a balanced three-phase set of negative acb sequence phasors, and a single-phase zero-sequence phasor It is most convenient for the analysis of steady-state unbalanced operation of symmetrical two- or three-phase machines where in the case of a two-phase system the zero sequence is not present In a critical discussion of Park’s hallmark paper [1], W.V Lyon mentioned that Fortescue’s work could be extended to the transient case Twenty-five years later, Lyon published a book wherein he used instantaneous or time-varying symmetrical components (phasors) to analyze the transient operation of electric machines [9] This approach was considered an alternative to Park’s work; however, it did not gain wide acceptance primarily because it was inconvenient when analyzing systems with unsymmetrical components, such as a synchronous machine Moreover, using Lyon’s approach it is not possible to define circuits that carry complex time-varying currents and a negativesequence instantaneous symmetrical component is necessary in the analysis even when the actual two- or three-phase variables are balanced [9, 10] Nevertheless, Lyon’s work appears to be the first use of instantaneous phasors in machine analysis Later, machine analysts began to combine the qd equations into a single complex expression [11–13] That is, since the q- and d-axes are orthogonal, we can write fqds = fqs − jfds (3.12-1) where the over arrow is used to denote the instantaneous phasor, which is often referred to as the space vector or space phasor Here, the fqs and fds variables are arbitrary reference-frame variables Using this space-phasor formulation, the qd-voltage equations given by (3A-3) and (3A-4) for an rL-circuit become vqds = vqs − jvds = rs iqds + pλqds + jωλqds (3-12-2) Recall from Section 3.7, when we considered the form of the reference-frame variables during balanced steady-state operation, we found that the steady-state q and d variables 114 REFERENCE-FRAME THEORY are sinusoidal in all reference frames except the synchronously rotating Therefore, even for transient conditions during balanced conditions, the components of the complex expression (3.12-2) will be varying amplitude, sinusoidal quantities in all reference frames except the synchronously rotating reference frame where the components will be constant in the steady state but vary in amplitude during a transient Although introducing the space phasor provides a means of expressing the q and d equations more compactly, it does not provide an analytical advantage In fact, the complex expression is separated into its real and imaginary expressions before solving for the q and d variables Therefore, the advantage of the space-phasor expression is primarily compactness; it does not offer any analytical advantage In fact, it would be a disadvantage if an attempt was made to solve the complex space-phasor expression analytically except perhaps in the synchronous reference frame It is important to mention that the phasor expression given by (3.7-14), which e e relates the physical variables to fqs and fds quantities during steady-state balanced conditions, is not a space phasor This is a standard phasor that has been used in steady-state ac circuit theory since developed by Steinmetz [14] There is one last space-phasor expression that warrants mentioning In Section 3.10, we introduced the transformation between reference frames, which is given by (3.10-7) Therein it was mentioned that this is sometimes referred to as a “rotator” or “vector rotator.” If the zero-sequence quantities are assumed to be nonexistent then (3.10-1) may be written y x ⎡ fqs ⎤ ⎡ cos(θ y − θ x ) sin(θ y − θ x ) ⎤ ⎡ fqs ⎤ =⎢ ⎢fy⎥ ⎥ ⎢ x⎥ ⎣ ds ⎦ ⎣ − sin(θ y − θ x ) cos(θ y − θ x ) ⎦ ⎣ fds ⎦ (3.12-3) where x is the reference frame from which variables are being transformed and y the reference frame to which the variables are being transformed If we expand (3.12-3) y x and express fqds and fqds as space phasors, we have y y x x fqs − jfds = ( fqs − jfds ) [cos(θ y − θ x ) + j sin(θ y − θ x )] x x = ( fqs − jfds ) e j (θ y −θ x ) (3.12-4) Equation (3.12-4) may be written as y x fqds = fqds e j (θ y −θ x ) (3.12-5) For a transformation between the stationary and synchronous reference frames, the rotator e j (θ y −θ x ) becomes e jωe t when all time-zero displacements are selected as zero [12] Although the space-phasor notation has advantages in providing a compact means of expressing the machine equations and portraying converter operation, this notation is not used in this text, as it offers no advantage other than compactness Nevertheless, once the student has mastered the qd0-analysis, incorporating the space-phasor notation is straightforward This is presented in Reference 11 115 PROBLEMS REFERENCES [1] R.H Park, “Two-Reaction Theory of Synchronous Machines—Generalized Method of Analysis—Part I,” AIEE Trans., Vol 48, July 1929, pp 716–727 [2] H.C Stanley, “An Analysis of the Induction Motor,” AIEE Trans., Vol 57 (Suppl), 1938, pp 751–755 [3] G Kron, Equivalent Circuits of Electric Machinery, John Wiley & Sons, New York, 1951 [4] D.S Brereton, D.G Lewis, and C.G Young, “Representation of Induction Motor Loads During Power System Stability Studies,” AIEE Trans., Vol 76, August 1957, pp 451–461 [5] P.C Krause and C.H Thomas, “Simulation of Symmetrical Induction Machinery,” IEEE Trans Power Apparatus Syst., Vol 84, November 1965, pp 1038–1053 [6] P.C Krause, F Nozari, T.L Skvarenina, and D.W Olive, “The Theory of Neglecting Stator Transients,” IEEE Trans Power Apparatus Syst., Vol 98, January/February 1979, pp 141– 148 [7] E Clarke, Circuit Analysis of A-C Power Systems, Vol I—Symmetrical and Related Components, John Wiley & Sons., New York, 1943 [8] C.L Fortescue, “Method of Symmetrical Co-ordinates Applied to the Solution of Polyphase Networks,” AIEE Trans., Vol 37, 1918, pp 1027–1115 [9] W.V Lyon, Transient Analysis of Alternating-Current Machinery, The Technology Press of Massachusetts Institute of Technology and John Wiley & Sons, New York, 1954 [10] D.C White and H.H Woodson, Electromechanical Energy Conversion, John Wiley & Sons, New York, 1959, pp 316–317 [11] D.W Novotny and T.A Lipo, Vector Control and Dynamics of AC Drives, Oxford University Press, New York, 1997 [12] R.D Lorenz, T.A Lipo, and D.W Novotny, “Motion Control with Induction Motors,” Proc IEEE, Vol 82, No 8, August 1994, pp 1215–1240 [13] P Vas, Sensorless Vector and Direct Torque Control, Oxford University Press, Oxford; New York; Tokyo, 1998 [14] C.P Steinmetz, Theory and Calculation of Alternating Current Phenomena, W.J Johnston, New York, 1897 PROBLEMS The transformation for a two-phase set to the arbitrary reference frame is fqds = K s fabs where ( fqds )T = [ fqs fds ] ( fabs )T = [ fas fbs ] ⎡cos θ Ks = ⎢ ⎣ sin θ sin θ ⎤ − cos θ ⎥ ⎦ 116 REFERENCE-FRAME THEORY where θ is defined by (3.3-5) (a) Determine (Ks)−1 (b) Depict the transformation similar to that shown in Figure 3.3-1 Using the transformation given in Problem 1, express the voltage and flux linkage equations in the arbitrary reference frame for a system vas vbs λ as λbs = ias + pλ as = rbibs + pλbs = La ias = Lbibs where = rb = rs and La = Lb = Ls Then express the voltage and flux linkage equations for the case in which ≠ rb and La ≠ Lb Using the transformation given in Problem 1, express the current equations in the arbitrary reference frame for a two-phase capacitive circuit that is described by vas = ias rs v Cs pvbs + bs = ibs rs Cs pvas + The phases of a three-phase circuit consist of equal resistances, equal inductances, and equal capacitances connected in series The phases are not coupled Write the voltage equations in the arbitrary reference frame and draw the equivalent circuit Repeat Problem for the circuit elements in each phase connected in parallel In a reference frame x, the q- and d-axis voltages of a wye-connected RL circuit may be expressed as x x x x vqs = 0.1iqs − 100ids + piqs x x x x vds = 0.1ids + 100iqs + pids Determine the resistance and inductance of the circuit, as well as the speed of the reference frame Show that for a two-phase set 2 2 fas + fbs = fqs + fds Clarke’s transformation may be written as fαβ = Cfabcs where 117 PROBLEMS ( fαβ )T = [ fα ⎡ ⎢1 ⎢ 2⎢ C = ⎢0 ⎢ ⎢1 ⎢2 ⎣ fβ f0 ] − 2 − 1⎤ − ⎥ ⎥ 3⎥ ⎥ ⎥ ⎥ ⎥ ⎦ s s Relate fqs, fds, and f0s to fa, fβ, and f0, respectively A transformation that is sometimes used in the case of synchronous machines is one where fds leads fqs in Figure 3.3-1 by 90° with ω = ωr (a) Express the transformation (b) Using this transformation, write the voltage equations for a three-phase inductive circuit 10 The inductance matrix that describes the self- and mutual inductances between the stator windings of a salient-pole synchronous machine with mutual leakage terms neglected is given in the following equation: ⎡ ⎢ Lls + L A − LB cos 2θ r ⎢ π⎞ ⎢ ⎛ L s = ⎢ − L A − LB cos ⎜ θ r − ⎟ ⎝ 3⎠ ⎢ π⎞ ⎢ ⎛ ⎢ − L A − LB cos ⎜ θ r + ⎟ ⎝ ⎠ ⎣ 11 12 13 14 15 π⎞ ⎛ − L A − LB cos ⎜ θ r − ⎟ ⎝ 3⎠ 2π ⎞ ⎛ Lls + L A − LB cos ⎜ θ r − ⎟ ⎝ 3⎠ − L A − LB cos 2(θ r + π ) π⎞ ⎤ ⎛ − L A − LB cos ⎜ θ r + ⎟ ⎥ ⎝ 3⎠ ⎥ ⎥ − L A − LB cos 2(θ r + π ) ⎥ ⎥ 2π ⎞ ⎥ ⎛ Lls + L A − LB cos2 ⎜ θ r + ⎟ ⎝ ⎠⎥ ⎦ Evaluate K r L s (K r )−1 for this case Repeat the evaluation for the case in which s s mutual leakage terms (refer to Chapter 2) are included If A is one reference frame and B another, show that (AKB)−1 = BKA Equations (3.7-1)–(3.7-3) form an abc sequence Express an acb sequence and e transform this set to the arbitrary reference frame using (3.3-1) Express fqd s (ω = ω e ) −e and fqd s (ω = −ω e ) Devise a transformation which yields only constants when ω = ωe for a balanced three-phase set with a phase sequence of acb Relate Fbs and Fcs to Fqs and Fds for a balanced three-phase set with a time sequence of abc For steady-state balanced conditions, the total three-phase power and reactive power may be expressed P = 3Vs I s cos[θ ev (0) − θ ei (0)] Q = 3Vs I s sin[θ ev (0) − θ ei (0)] Show that the following expressions are equal to those given above 118 REFERENCE-FRAME THEORY P= (Vqs I qs + Vds I ds ) Q= (Vqs I ds − Vds I qs ) 16 Write the expressions for the currents in Figure 3.9-1 and Figure 3.9-2 17 Assume the steady-state abc variables are of the form Fas = Fa cos ω e t 2π ⎞ ⎛ Fbs = Fb cos ⎜ ω e t − ⎟ ⎝ 3⎠ 2π ⎞ ⎛ Fcs = Fc cos ⎜ ω e t + ⎟ ⎝ 3⎠ where Fa, Fb, and Fc are unequal constants Show that this unbalanced set of abc variables forms two-phase balanced sets of qs and ds variables in the arbitrary reference frame with the arguments of (ωet−θ) and (ωet + θ) Note the form of the qs and ds variables when ω = ωe and ω = −ωe 18 Repeat Problem 17 with Fas = Fs cos(ω e t + φa ) Fbs = Fs cos(ω e t + φb ) Fcs = Fs cos(ω e t + φc ) where φa, φb, and φc are unequal constants 19 It is often suggested that Ks should be changed so that (Ks)T = (Ks)−1 For example, if Ks = ⎡ ⎢cos θ ⎢ 2⎢ sin θ 3⎢ ⎢ ⎢ ⎢ ⎣ 2π ⎞ ⎛ cos ⎜ θ − ⎟ ⎝ 3⎠ 2π ⎞ ⎛ sin ⎜ θ − ⎟ ⎝ 3⎠ 2π ⎞ ⎤ ⎛ cos ⎜ θ + ⎟ ⎥ ⎝ 3⎠ ⎥ 2π ⎞ ⎥ ⎛ sin ⎜ θ + ⎟ ⎥ ⎝ 3⎠ ⎥ ⎥ ⎥ ⎦ then (Ks)T = (Ks)−1 Show that this is true Also, show that in this case 2 2 fas + fbs + fcs = fqs + fds + f02s 20 A system is described by the following equations: 119 PROBLEMS vbs vcs − + vas = ias 2 v v pvbs − as − cs + vbs = ibs 2 vbs vas + vcs = ics pvcs − − 2 pvas − If the currents are a balanced (abc) sequence with Ias = cos(t), determine e s Vqs , Vas , and Vqs 21 A system is described as = v abcs ⎡ ⎢ ⎢ − i abcs − ⎢ − ⎢ ⎢ ⎢− ⎢ ⎣ − − 1⎤ − ⎥ ⎥ 1⎥ − pi abcs − vCabcs 2⎥ ⎥ ⎥ ⎥ ⎦ pvCabcs = i abcs e Determine I qs , I ds , and, I as if Vas = cos10t Assume balanced voltages 22 A system is described by the following: vas = ias + pλ as vbs = ibs + pλbs vcs = ics + pλcs λ as = (0.1 + 1)ias − 0.5ibs − 0.5ics λbs = −0.5ias + (0.1 + 1)ibs − 0.5ics λcs = −0.5ias − 0.5ibs + (0.1 + 1)ics If λ as = − j, ω e = 10 rad/s, use reference-frame theory to determine Vas 23 A system is described as v abcs = rs i abcs + pl abcs l abcs ⎡ ⎢ 0.1 + − 0.25 cos 2θ r ⎢ π⎞ ⎢ ⎛ = ⎢ − − 0.25 cos ⎜ θ r − ⎟ ⎝ 3⎠ ⎢ π⎞ ⎢ ⎛ ⎢ − − 0.25 cos ⎜ θ r + ⎟ ⎠ ⎝ ⎣ π⎞ ⎛ − − 0.25 cos ⎜ θ r − ⎟ ⎝ 3⎠ 2π ⎞ ⎛ 0.1 + − 0.25 cos ⎜ θ r − ⎟ ⎝ 3⎠ − − 0.25 cos 2(θ r + π ) r r If λqs = λ ds = V ⋅ s, determine I as π⎞ ⎤ ⎛ − − 0.25 cos ⎜ θ r + ⎟ ⎥ ⎝ 3⎠ ⎥ ⎥ − − 0.25 cos 2(θ r + π ) ⎥ i abcs ⎥ 2π ⎞ ⎥ ⎛ 0.1 + − 0.25 cos ⎜ θ r + ⎟ ⎥ ⎝ ⎠⎦ 120 REFERENCE-FRAME THEORY 24 A system is described as: v12 r = rr′i12 r + pl12 r ′ ′ ′ ⎡ cos(θ r ) − sin(θ r ) ⎤ l12 r = Lm ⎢ ′ ⎥ i12 s ⎣ − sin(θ r ) − cos(θ r ) ⎦ Using the definitions s ⎡ fqs ⎤ ⎡1 ⎤ s ⎡ f1s ⎤ s ⎢ f s ⎥ = K s ⎢ f ⎥ , where K s = ⎢0 −1⎥ ⎣ ⎦ ⎣ 2s ⎦ ⎣ ds ⎦ ’ − sin(θ r ) − cos(θ r ) ⎤ dθ r ⎡ f1’r ⎤ ⎡ fqrs ⎤ ⎡ s s ⎢ f ’s ⎥ = K r ⎢ f ’ ⎥ , where K r = ⎢ − cos(θ ) sin(θ ) ⎥ , ω r = dt ⎣ ⎦ r r ⎣ 2r ⎦ ⎣ dr ⎦ transform the voltage and flux linkage equations into the stationary reference frame Assuming the input voltages v12r are balanced and have a frequency of ′ 337 − ωr, determine the frequency of the currents, voltages, and flux linkages in the stationary frame of reference 25 A balanced three-phase system can be described in the stationary reference frame as: s s s piqs + iqs = vqs s s s pids + ids = vds pi0 s + i0 s = v0 s e e If the system is operating with ωe = rad/s, Vqs = V, Vds = − V, use the steady-state equations to determine I as In addition, express the physical abc variables 26 Consider a two-phase machine with the flux linkage equations: ⎡8 − cos(2θ r ) −2 sin(2θ r ) ⎤ l abs = ⎢ ⎥ i abs ⎣ −2 sin(2θ r ) + cos(2θ r ) ⎦ Where θr is the electrical rotor position Determine a set of qd flux linkage equations that are rotor position invariant To this, you will need to establish a suitable reference-frame transformation ... the arbitrary reference frame may by expressed as fqd s = K s fabcs (3. 3-1) where ( fqd s )T = [ fqs fds f0 s ] (3. 3-2) ( fabcs ) = [ fas fbs fcs ] (3. 3 -3) T ⎡ ⎢cos θ ⎢ K s = ⎢ sin θ 3? ?? ⎢ ⎢ ⎢ ⎣... ef (3. 6-1) 2π ⎞ ⎛ fbs = fs cos ⎜ θ ef − ⎟ ⎝ 3? ?? (3. 6-2) 2π ⎞ ⎛ fcs = fs cos ⎜ θ ef + ⎟ ⎝ 3? ?? (3. 6 -3) where fs may be a function of time and dθ ef = ωe dt (3. 6-4) Substituting (3. 6-1)– (3. 6 -3) into... (3. 10-1) x x fqd s = K s fabcs (3. 10-2) From (3. 3-1), we obtain Substituting (3. 10-2) into (3. 10-1) yields y x fqd s = x K y K s fabcs (3. 10 -3) y y fqd s = K s fabcs (3. 10-4) x y K yKs = Ks (3. 10-5)