Average Power Flow through a Synchronous Generator

Average Power Flow through a Synchronous GeneratorSam Chevalier

Guiding Question: When a synchronous generator experiences a forced oscillation, whathappens to the average power flow?

1 Synchronous Generator ModelWe present the dynamic equations for a synchronous generator1 with the following characteristics.

• 2 pole

• salient-pole

• 3-phase

• wye-connected

• symmetric (balanced)

• 1 field winding and 3 damper windings on the rotor

• 3 identical stator windings displaced by exactly 120◦

The physical structure of this model is visualized in Figure 1. We define direct (d) and quadrature (q) axeswhich are in line with and perpendicular to the axis of rotation of the rotor, respectively. We also arbitrarilydefine δr which represents the angle between the fixed a phase axis of the stator and the direction of the qaxis.

𝑎𝑠

𝑎𝑠′

𝑏𝑠 𝑐𝑠

𝑏𝑠′𝑐𝑠′

𝑓𝑑′

𝑘𝑑′

𝑘𝑑

𝑓𝑑

𝑘𝑞1

𝑘𝑞2

𝑘𝑞2′

𝑘𝑞1′

𝑑 axis

𝑞 axis

𝑎𝑠 axis

𝛿𝑟

Figure 1: 3-Phase Salient Pole Synchronous Generator (positive current flows into � and out of �)

1Equations and some figures adapted from Analysis of Electric Machinery by Paul Krause

1

Using Figure 1 as a guide, a simple circuit diagram may be employed to model the interior voltages andcurrents. The system is modeled as though each rotor and stator coil acts as a transformer coil, capable oftransforming flux from the air gap into a voltage potential according to Faraday’s Law.

Damper 2

𝑖𝑘𝑞1𝑖𝑏𝑠 +

−𝑁𝑠

Stator

𝑟𝑠

𝑟𝑠

𝑁𝑠

𝑁𝑠

𝑟𝑠

+ 𝑟𝑠

+ 𝑟𝑠

𝑖𝑐𝑠

𝑖𝑎𝑠

𝑟𝑘𝑞1

+

𝑁𝑘𝑞1

−

𝑣𝑘𝑞1𝑖𝑘𝑞2𝑟𝑘𝑞2

+

𝑁𝑘𝑞2

−𝑣𝑘𝑞2

+

+

−

𝑁𝑘𝑑

𝑁𝑓𝑑

𝑖𝑘𝑑

𝑖𝑓𝑑

−

𝑟𝑓𝑑

𝑟𝑘𝑑

𝑣𝑘𝑑

𝑣𝑓𝑑

Rotor

Damper 3

Field Winding

Damper 1

Figure 2: Circuit Model of the 3-Phase Synchronous Generator. Each of the 7 coils has an associatedresistance rx and coil turn count Nx (with a specified reactance).

TransformationsIn order to simplify the circuit analysis, we make two key transformations. First, we reflect all rotor voltages,currents, and impedances to the stator reference frame. In general, way may transform a voltage from asecondary side of a transformer to the primary by noting that vi = Ni

Njvj . Similar relationships exit for

current and impedance transformations. In a circuit diagram, reflected values all contain a prime (′) symbol.This can be seen in Figure 3. The second transformation we employ is known as Park’s transformation. Insteady state, the rotor spins at synchronous frequency ωs. In order to over come the mathematical challengeof a rotating rotor, we choose to analyze the stator’s circuit in the synchronously rotating reference frame ofthe rotor. This eliminates the time varying inductance values of the system since all relationships are nowstationary. For example, once the q-axis, d-axis, and zero sequence voltages have been calculated, we canconvert them back into physically meaningful phase a, b, and c voltages (or currents, fluxes, etc.). vas

vbsvcs

=

cos (δr) sin (δr) 1cos(δr − 2π

3)

sin(δr − 2π

3)

1cos(δr + 2π

3)

sin(δr + 2π

3)

1

vqsvdsv0s

(1)

Transformed Circuit ModelWithout further derivation, we present the equivalent circuit model for a 3-phase synchronous generatorwhich has undergone both transformations (transformer reflection and Park’s) described in the previoussection. Table 1 interprets the circuit diagram.

2

Subscript Significance Superscript Significancefd field winding 1 in d-axis ′ transformed across transformerkd damper 1 in d-axis r rotor reference framekq1 damper 1 in q-axiskq2 damper 2 in q-axism magnetizingl leakage

Table 1: Figure 3 Interpretation

𝑟𝑠

+

+

−

𝐿𝑙𝑠𝑖0𝑠

𝑟𝑠

Zero Sequence

𝐿𝑙𝑠

𝐿𝑚𝑞

𝐿𝑙𝑘𝑞1′

Q Axis


𝑟𝑘𝑞2′

𝑟𝑘𝑞1′

−

+

𝑣𝑞𝑠𝑟

𝑖𝑞𝑠𝑟

𝑖𝑘𝑞1′𝑟

𝑖𝑘𝑞2′𝑟

+

𝑣𝑘𝑞2′𝑟

𝑣𝑘𝑞1′𝑟

+ −

λ𝑑𝑠𝑟 𝜔𝑟

𝑟𝑠

+

𝐿𝑙𝑠

𝐿𝑚𝑑

𝐿𝑙𝑓𝑑′

𝐿𝑙𝑘𝑑′

𝑟𝑘𝑑′

𝑟𝑓𝑑′

−

+

𝑣𝑑𝑠𝑟

𝑖𝑑𝑠𝑟

𝑖𝑓𝑑′𝑟

𝑖𝑘𝑑′𝑟

+

𝑣𝑘𝑑′𝑟

𝑣𝑓𝑑′𝑟

− +

λ𝑞𝑠𝑟 𝜔𝑟

D Axis

𝑣0𝑠

Figure 3: Transformed Circuit Model of the 3-Phase Synchronous Generator. λ is the flux linkage betweenthe rotor and stator coils. The product λω represents a voltage drop in the d-axis (where the field windingis located) and a voltage increase in the q-axis. If the system is feeding balanced load, then v0s = 0 and thezero sequence circuit can be omitted.

Although a very busy diagram, the circuit simplifies considerably in steady state since the damper windingcurrents will be equal to 0. This reduced circuit is shown in Figure 4.

3

𝑟𝑠 𝐿𝑙𝑠

𝐿𝑚𝑞

Q Axis

−

+

𝑣𝑞𝑠𝑟

𝑖𝑞𝑠𝑟

+ −

λ𝑑𝑠𝑟 𝜔𝑟

𝑟𝑠 𝐿𝑙𝑠

𝐿𝑚𝑑

𝐿𝑙𝑓𝑑′

𝑟𝑓𝑑′

−

+

𝑣𝑑𝑠𝑟

𝑖𝑑𝑠𝑟

𝑖𝑓𝑑′𝑟 +𝑣𝑓𝑑

′𝑟

− +

λ𝑞𝑠𝑟 𝜔𝑟

D Axis

Figure 4: Steady State Circuit Model of the Transformed 3-Phase Synchronous Generator. Damper windingeffects have been removed.

Transformed Circuit Model State EquationsUsing basic KVL, KCL and Newton’s Second Law, state equations can be formulated which describe thedynamics of the full synchronous generator model given in Figure 3. Without proof, we note a simplifiedexpression for the electrical torque τe.

τe = 32 (λdsiqs − λqsids) (2)

This expression for electrical torque is used in constructing the system’s swing equation, where damping isomitted. The variable ωr represents the true angular velocity of the rotor while δr represents the rotorsabsolute angular position (not with respect to a synchronously rotating reference).

τm = Jωr + τe (3)ωr = δr (4)

Using the flux linkage variables (λ), we can write out the seven defined voltages in Figure 3 using basiccircuit theory. First we define the three stator voltages, where p denotes the derivative operator with respectto time.

vqs = −rsiqs + ωrλds + pλqs (5)vds = −rsids − ωrλqs + pλds (6)v0s = −rsi0s + pλ0s (7)

Next we define the four rotor voltages.

v′

kq1 = r′

kq1i′

kq1 + pλ′

kq1 (8)

v′

kq2 = r′

kq2i′

kq2 + pλ′

kq2 (9)

v′

fd = r′

fdi′

fd + pλ′

fd (10)

v′

kd = r′

kdi′

kd + pλ′

kd (11)

The d-axis and q-axis current flows are given below.

4

iqs = − 1Lls

(λqs − λmq) (12)

ids = − 1Lls

(λds − λmd) (13)

The flux linkage terms associated with each winding on the d-axis and q-axis are defined, again, using basiccircuit theory.

λqs = −Llsiqs + Lmq

(−iqs + i

′

kq1 + i′

kq2

)(14)

λds = −Llsids + Lmd

(−ids + i

′

kd + i′

fd

)(15)

λ0s = −Llsi0s (16)

λ′

kq1 = L′

lkq1i′

kq1 + Lmq

(−iqs + i

′

kq1 + i′

kq2

)(17)

λ′

kq2 = L′

lkq2i′

kq2 + Lmq

(−iqs + i

′

kq1 + i′

kq2

)(18)

λ′

fd = L′

lfdi′

fd + Lmd

(−ids + i

′

fd + i′

kd

)(19)

λ′

kd = L′

lkdi′

kd + Lmd

(−ids + i

′

fd + i′

kd

)(20)

It is also useful to define the flux linkages associated with the Lmq and Lmd inductors. We do so by balancingthe currents via KCL, but we write this current balance in terms of inductances and the defined flux linkages.

λmq = Laq

(λqsLls

+λ′

kq1

L′lkq1

+λ′

kq2

L′lkq2

)(21)

λmd = Lad

(λdsLls

+λ′

fd

L′lfd

+ λ′

kd

L′lkd

)(22)

In this case, it is clear that Laq and Lad are effective reactance parameters. These are defined as the parallelcombination of all reactances in each respective (q-axis and d-axis circuit).

Laq =(

1Lmq

+ 1Lls

+ 1L′lkq1

+ 1L′lkq2

)−1

(23)

Lad =(

1Lmq

+ 1Lls

+ 1L′lfd

+ 1L′lkd

)−1

(24)

The flux linkage per second variables ψ can be used at the primary state variables, along with θr and ωr,which drive the dynamics of the system. Therefore, we must calculate their derivatives. This is accomplishedthrough Faraday’s Law and basic algebraic manipulation. The results are shown below.

5

λrqs = vrqs − ωrλrds + rsLls

(λrmq − λrqs

)(25)

λrds = vrds + ωrλrqs + rs

Lls(λrmd − λrds) (26)

λ0s = v0s −rsLls

λ0s (27)

λ′rkq1 = v

′rkq1 +

r′

kq1

L′lkq1

(λrmq − λ

′rkq1

)(28)

λ′rkq2 = v

′rkq2 +

r′

kq2

L′lkq2

(λrmq − λ

′rkq2

)(29)

λ′rfd = v

′rfd +

r′

fd

L′lfd

(λrmd − λ

′rfd

)(30)

λ′rkd = v

′rkd + r

′

kd

L′lkd

(λrmd − λ

′rkd

)(31)

At this point, the synchronous generator may be fully simulated using the state equations which have beendeveloped. There are a variety of methods for simulating this system, but the DC field voltage, the appliedmechanical torque, and the AC stator voltages are typically defined as external model inputs. Stator currentsare the external outputs. The procedure for simulating this system is outlined in Figure 5 (a similar procedureis vaguely recommended in Analysis of Electric Machinery by Paul Krause)

Park Transformation

Input

ҧ𝑣𝑎𝑠

ҧ𝑣𝑏𝑠

ҧ𝑣𝑐𝑠

ҧ𝑥 = 𝑥𝑡𝑖−1𝑥 = 𝑥𝑡𝑖

Variable Key

ҧ𝑣𝑞𝑠

ҧ𝑣𝑑𝑠

ҧ𝑣0𝑠

𝜏𝑚 Swing Equation 𝜔𝑟

𝜏𝑒

Inverse Park Transformation

𝑖𝑎𝑠

𝑖𝑏𝑠

𝑖𝑐𝑠

𝑣𝑓𝑑

Compute Internal Flux Derivatives

ሶ ҧ𝜆𝑞𝑑0Forward Euler: Compute New

Flux States

𝜆 = ሶҧ𝜆 + (∆𝑡) ҧ𝜆

Compute Internal Current States

Compute Electrical Torque

𝜆𝑞𝑑0

𝑖𝑞𝑑0𝑖𝑞𝑑0

Forward Euler: Rotor Angle

𝛿𝑟 = ҧ𝛿𝑟 + (∆𝑡)ഥ𝜔𝑟

ҧ𝛿𝑟

ഥ𝜔𝑟

𝛿𝑟

Output

𝛿𝑟

Figure 5: Simulation Procedure

2 Modeling the Interconnection of 2 Synchronous Generators:Network Dynamics Excluded

We wish to model the interconnection of 2 synchronous generators in order to further trace how oscillationspropagate through the system. We do so by considering the diagram shown in Figure 6.

6

𝑖𝑏𝑠2

𝑖𝑐𝑠2

𝑖𝑎𝑠2

+

−𝑁𝑠

Stator 2

𝑟𝑠

𝑟𝑠

𝑁𝑠

𝑁𝑠

𝑟𝑠

+

+

−𝑁𝑠

Stator 1

𝑁𝑠

𝑁𝑠

𝑟𝑠

𝑟𝑠

𝑟𝑠 𝑖𝑏𝑠1

𝑖𝑐𝑠1

𝑖𝑎𝑠1

3 Lines

𝑖𝑏𝑠2

𝑖𝑐𝑠2

𝑖𝑎𝑠2

+

−𝑁𝑠

𝑟𝑠

𝑟𝑠

𝑁𝑠

𝑟𝑠

+

+

−𝑁𝑠 𝑁𝑠

𝑁𝑠

𝑟𝑠

𝑟𝑠

𝑟𝑠 𝑖𝑏𝑠1

𝑖𝑐𝑠1

𝑖𝑎𝑠1

V1𝑒𝑗𝜃1 V2𝑒

𝑗𝜃2

Stator 2Stator 1 Equivalent Line

𝑁𝑠

(A)

(B)

Figure 6: Double Synchronous Generator Interconnection. Panel (A) shows the full interconnection whilepanel (B) shows the simplified connection.

We model the dynamics of each generator individually, and we use a set of algebraic relationships to couplethese dynamics. We first re-state the simplified flux dynamics of a 2 pole machine. Damper winding voltagesare excluded since they always have a value of 0, and we assume a balanced system, so we eliminate the 0sequence terms entirely.

λqs = −ωrλds + rsLls

(λmq − λqs) + vqs

λds = ωrλqs + rsLls

(λmd − λds) + vds

λ′

kq1 =r′

kq1

L′lkq1

(λmq − λ

′

kq1

)λ′

kq2 =r′

kq2

L′lkq2

(λmq − λ

′

kq2

)λ′

fd =r′

fd

L′lfd

(λrmd − λ

′rfd

)+ v

′rfd

λ′

kd = r′

kd

L′lkd

(λmd − λ

′

kd

)ω = τm − τe

J

δr = ωr − ws= ω

We treat the exciter voltage vfd and the generator’s mechanical torque τm as independent inputs. The flux

7

variables λmq and λmd are defined through the following algebraic relationships.

λmq = Laq

(λqsLls

+λ′

kq1

L′lkq1

+λ′

kq2

L′lkq2

)

λmd = Lad

(λdsLls

+λ′

fd

L′lfd

+ λ′

kd

L′lkd

)

The unknown voltage inputs (vqs and vds) may be solved for in terms of flux state variables and knowninductance terms.

vqs = −rsiqs + ωrλds + λqs

vds = −rsids − ωrλqs + λds

We constrain these differential equations with the following set of algebraic equations.

P = 32 (idsvds + iqsvqs)

Q = 32 (idsvqs − iqsvds)

τe = 32 (λdsiqs − λqsids)

The unknown voltage inputs (iqs and ids) may be solved for in terms of flux state variables and knowninductance terms.

iqs = − 1Lls

(λqs − λmq)

ids = − 1Lls

(λds − λmd)

One last highly important relationship exists between the generator and the network buses. We define the busvoltage phasor V ejθ as the bus voltage at the node interfacing the generator with the system. Additionally,δr is the angle of the rotor with respect to some arbitrary synchronously rotating reference frame.

vds =√

2V sin(δr − θ

)vqs =

√2V cos

(δr − θ

)Finally, we may employ the AC power flow equations to approximate the power transfer between the genera-tors. Since phasor analysis is applied to the transmission line power flow, the line reactance is not consideredto be dependent on the system frequency. Assuming small, low frequency power oscillations, this assumptionsshould have almost no effect on the analysis.

P = V2iGii + ViVj [Gij cos(θij) +Bij sin(θij)]

32 (idsvds + iqsvqs) = V2

iGii + ViVj [Gij cos(θij) +Bij sin(θij)]

Q = −V2iBii + ViVj [Gij sin(θij)−Bij cos(θij)]

32 (idsvqs − iqsvds) = −V2

iBii + ViVj [Gij sin(θij)−Bij cos(θij)]

In simulating this system, we assume both generators are identical. The algebraic and differential equations

8

for each generator are summarized below.

Algebraic Equations =

0 = V2iGii + ViVj [Gij cos(θij) +Bij sin(θij)]− 3

2 (idsvds + iqsvqs)0 = −V2

iBii + ViVj [Gij sin(θij)−Bij cos(θij)]− 32 (idsvqs − iqsvds)

0 = 32 (λdsiqs − λqsids)− τe

0 =√

2V sin(δr − θ

)− vds

0 =√

2V cos(δr − θ

)− vqs

0 = − 1Lls

(λds − λmd)− ids0 = − 1

Lls(λqs − λmq)− iqs

0 = Lad

(λdsLls

+ λ′fd

L′lfd

+ λ′kd

L′lkd

)− λmd

0 = Laq

(λqsLls

+ λ′kq1

L′lkq1

+ λ′kq2

L′lkq2

)− λmq

Differential Equations =


(λmq − λqs) + vqsλds = ωrλqs + rs

Lls(λmd − λds) + vds

λ′

kq1 = r′kq1

L′lkq1

(λmq − λ

′

kq1

)λ′

kq2 = r′kq2

L′lkq2

(λmq − λ

′

kq2

)λ′

fd = r′fd

L′lfd

(λrmd − λ

′rfd

)+ v

′rfd

λ′

kd = r′kd

L′lkd

(λmd − λ

′

kd

)ω = τm−τe

J

δr = ωr − ws = ω

3 Modeling the Interconnection of 2 Synchronous Generators andan Infinite Bus: Network Dynamics Included

In order to model the interconnection of 2 synchronous generators and an infinite bus, we consider thenetwork diagram depicted by figure 7. In this model, network dynamics are NOT excluded, so we no longerare able to employ the steady state power flow equations to model the transfer of power.

𝑣∞𝑎

𝑣∞𝑏

𝑣∞𝑐

𝑖1𝑎 𝑖2𝑎

𝑖2𝑏

𝑖2𝑐

+

−𝑁𝑠

𝑟𝑠

𝑟𝑠

𝑁𝑠

𝑁𝑠

𝑟𝑠

+

+

−𝑁𝑠

Stator 1

𝑁𝑠

𝑁𝑠

𝑟𝑠

𝑟𝑠 𝑖1𝑏

𝑖1𝑐

𝑟𝑠

Stator 2

𝑖∞𝑎

𝑖∞𝑏

𝑖∞𝑐

𝑟∞ 𝐿∞

𝑟𝑙 𝐿𝑙

Figure 7: Double Synchronous Generator & Infinite Bus System

9

Analysis of a Multi-Generator System with qd0 Network Transfor-mationIn order to overcome the challenges of analyzing a network in the abc sequence while analyzing each machinein qd0 sequence (with each sequence in its own reference frame too), we choose to transform the networksuch that we mathematically eliminate all abc terminal voltages and currents. This approach is given byFigure 8.

𝑖𝑞𝑑0𝑟

Generator𝑣𝑞𝑑0𝑟

(𝐾𝑟)−1𝑖𝑞𝑑0𝑒

𝑣𝑞𝑑0𝑒

Network

Analyze Generator in rotor reference frame

Transform from rotor to synchronous reference frame

Analyze Network in qd0 sequence of synchronous reference

(𝐾𝑠)−1

Inverse Park Transform to for post analysis

𝑖𝑞𝑑0𝑒

𝑣𝑞𝑑0𝑒

𝑖𝑎𝑏𝑐

𝑣𝑎𝑏𝑐

Figure 8: Reference Frame Transformations

To perform this analysis, we first consider the balanced 3 phase network depicted in Figure 9 where eachvoltage and current variable represents an instantaneous time domain value.

𝑟𝑙 𝐿𝑙

𝑖𝑎

𝑣1𝑎 𝑣2𝑎

𝑟𝑙 𝐿𝑙

𝑖𝑏

𝑣1𝑏 𝑣2𝑏

𝑟𝑙 𝐿𝑙

𝑖𝑐

𝑣1𝑐 𝑣2𝑐

Figure 9: Reference Frame Transformations

We write out the network voltage equations.

v2a = v1a − iarl − Lldiadt

v2b = v1b − ibrl − Lldibdt

v2c = v1c − icrl − Lldicdt

We then apply a Park’s Transformation Ks(θe) to the system equations, where θe is chosen to be theinstantaneous angle of the synchronously rotating reference frame.

vqd0 = Ksvabc

Ks =(

23

) cos (θe) cos(θe − 2π

3)

cos(θe + 2π

3)

sin (θe) sin(θe − 2π

3)

sin(θe + 2π

3)

12

12

12

We choose to apply this transformation to the entire network, where rl and Ll are diagonal resistance and

10

inductance matrices of the network, respectively.

Ksv2abc = Ksv1abc −Ksrliabc −KsLldiabc

dt

v2qd0 = v1qd0 −Ksrl(K−1s Ks

)iabc −KsLl

ddt(K−1s Ksiabc

)v2qd0 = v1qd0 −

(KsrlK

−1s

)iqd0 −KsLl

ddt(K−1s iqd0

)If the network is balanced, than we have that rl = rlI3, where I3 is the 3 by 3 identity matrix. The same istrue for the inductance matrix.

KsrlK−1s = rlKsI3K

−1s

= rl

KsLlK−1s = LlKsI3K

−1s

= Ll

We apply these identities, along with the product differentiation rule, to the network voltage expression.

v2qd0 = v1qd0 − rliqd0 −KsLl

((ddtK

−1s

)iqd0 +K−1

s

(ddtiqd0

))= v1qd0 − rliqd0 − Ll

((Ks

ddtK

−1s

)iqd0 +KsK

−1s

(ddtiqd0

))= v1qd0 − rliqd0 − Ll

((Ks

ddtK

−1s

)iqd0 + d

dtiqd0

)Without showing the algebra (see Krause), we make the following statement.

KsddtK

−1s = ωe

0 1 0−1 0 00 0 0

Therefore, we arrive at the following simplified expression, where idq0 =

[id −iq 0

]>.

v2qd0 = v1qd0 − rliqd0 − Ll(ωeidq0 + d

dtiqd0

)We can write this out explicitly for the voltage of each sequence. Since we assume a balanced system, weneglect the 0 sequence voltages and currents.

v2q = v1q − rliq − ωeLlid − Lldiqdt (32)

v2d = v1d − rlid + ωeLliq − Lldiddt (33)

Using (32) and (33) as a guide, we may write the network equations for the system shown in 7. We firstconsider the connection between the infinite bus and generator 1.

v1q = v∞q − r∞i∞q − ωeL∞i∞d− L∞

di∞q

dt

v1d = v∞d− r∞i∞d

+ ωeL∞i∞q− L∞

di∞d

dtSince infinite bus currents are not known explicitly, we write them in terms of i1 and i2 since we know thati1 + i2 + i∞ = 0. Therefore, i∞ = −i1 − i2.

v1q = v∞q + r∞(i1q + i2q

)+ ωeL∞ (i1d + i2d) + L∞

(di1qdt +

di2qdt

)(34)

v1d = v∞d+ r∞ (i1d + i2d)− ωeL∞

(i1q + i2q

)+ L∞

(di1ddt + di2d

dt

)(35)

11

We next consider the connection between bus 1 and bus 2.

v2q = v1q + rli2q + ωeLli2d + Lldi2qdt (36)

v2d = v1d + rli2d − ωeLli2q + Lldi2ddt (37)

Therefore, equations (34), (35), (36), and (37) define the network equations which are necessary to simulatethe system. We note that in order to determine v∞q

and v∞d, we are required to apply a Park’s Transfor-

mation on the infinite bus voltages. Since, by definition, the infinite bus voltage frequency is constant forall time, θe = ωet. v∞q

v∞d

v∞0

= Ks

v∞a

v∞b

v∞c

=(

23

) cos (θe) cos(θe − 2π

3)

cos(θe + 2π

3)

sin (θe) sin(θe − 2π

3)

sin(θe + 2π

3)

12

12

12

v∞a

v∞b

v∞c

To relate the network variables to the generator variables, we must define the transformation matrix rKe

which transforms voltages and currents in the rotor reference frame to voltages and currents in the syn-chronous reference frame (or vice versa, depending on the need). We define δr as the rotor angle for a givengenerator.

veqd0 =r Kevrqd0

rKe =

cos(θe − δr) − sin(θe − δr) 0sin(θe − δr) cos(θe − δr) 0

0 0 1

Moving forward, we denote veqd0 = vqd0 with no superscript. These are the voltages (or currents) in thesynchronous reference frame. Alternatively, we denote vrqd0 as the voltages (or currents) in the rotor referenceframe. Finally, we may specify the differential equations and the accompanying algebraic constraints whichare necessary to simulate each generator along with the network interconnections.

12

Generator Algebraic Equations =

0 = [rKe] irqd0 − iqd00 = [rKe]vrqd0 − vqd00 = − 1

Lls(λds − λmd)− irds

0 = − 1Lls

(λqs − λmq)− irqs0 = 3

2(λdsi

rqs − λqsirds

)− τe

0 = Lad

(λdsLls

+ λ′fd

L′lfd

+ λ′kd

L′lkd

)− λmd

0 = Laq

(λqsLls

+ λ′kq1

L′lkq1

+ λ′kq2

L′lkq2

)− λmq

(38)

Generator Differential Equations =


(λmq − λqs) + vrqsλds = ωrλqs + rs

Lls(λmd − λds) + vrds

λ′

kq1 = r′kq1

L′lkq1

(λmq − λ

′

kq1

)λ′

kq2 = r′kq2

L′lkq2

(λmq − λ

′

kq2

)λ′

fd = r′fd

L′lfd

(λmd − λ

′

fd

)+ v

′

fd

λ′

kd = r′kd

L′lkd

(λmd − λ

′

kd

)ωr = τm−τe

J

δr = ωr

(39)

Network Equations =

v1q = v∞q

+ r∞(i1q + i2q

)+ ωeL∞ (i1d + i2d) + L∞

(di1qdt + di2q

dt

)v1d = v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q

)+ L∞

(di1d

dt + di2ddt

)v2q = v1q + rli2q + ωeLli2d + Ll

di2qdt

v2d = v1d + rli2d − ωeLli2q + Lldi2d

dt(40)

Specifically, for each generator, we have 10 state variables: λqs, λds, λ′

kq1, λ′

kq2, λ′

fd, λ′

kd, vrqs, vrds, δr, andωr. Therefore, each generator adds ten additional state variables. The network equations, in fact, do notintroduce any new state variables: they simply represent a set of constraints on the aforementioned statevariables.

Steady State Power Flow SolutionBefore attempting to simulate this entire system, we first choose to write out the solution for the system insteady state. To do so, we treat the infinite bus as if it was a swing bus (where RMS voltage magnitude andphase angle are both specified), and we treat both generators like PV buses, where RMS voltage magnitudeand real power are both specified (we could alternatively specify reactive power instead of voltage magnitude).In the time domain, we note that the active power injections have the following definitions.

P1 = p1(t)= i1av1a + i1bv1b + i1cv1c

P2 = p2(t)= i2av2a + i2bv2b + i2cv2c

P∞ = p3(t)= i∞a

v∞a+ i∞b

v∞b+ i∞c

v∞c

Since this system is operating in steady state, we may use phasor analysis on each phase of the generator

13

where V = Vejδ.

P1 + jQ1 = V1a

[V1a − V2arl + jωsll

+ V1a − V∞a

r∞ + jωsl∞

]∗+ V1b

[V1b − V2brl + jωsll

+ V1b − V∞b

r∞ + jωsl∞

]∗+ V1c

[V1c − V2crl + jωsll

+ V1c − V∞c

r∞ + jωsl∞

]∗P2 + jQ2 = V2a

[V2a − V1arl + jωsll

]∗+ V2b

[V2b − V1brl + jωsll

]∗+ V2c

[V2c − V1crl + jωsll

]∗P∞ + jQ∞ = V∞a

[V∞a

− V1ar∞ + jωsl∞

]∗+ V∞b

[V∞b

− V1br∞ + jωsl∞

]∗+ V∞c

[V∞c

− V1cr∞ + jωsl∞

]∗If we separate these equations into their real and imaginary parts, we obtain 6 equations with 6 unknowns:δs1 , δs2 , Q1, Q2, P∞, and Q∞ (we arbitrarily specify δsi = 0). Therefore, an iterative method may be usedto solve for these variables. This is the first step used to determine the initial conditions for this system.Once the unknown values have been solved for, the terminal abc voltages and currents can be solved for. Forexample, the phase a voltages on generator 1 is given below.

v1a(t) =√

2 |V1a | cos(2πωst+ δs1)

To determine the phase a current on generator 1, we solve the current phasor expression and then extrapolateto the time domain.

I1a = V1a − V2arl + jωsll

+ V1a − V∞a

r∞ + jωsl∞

i1a(t) =√

2 |I1a | cos(2πωst+ ∠I1a)

Initializing Other State VariablesOnce power flow is solved, we initialize the other state variables by setting all derivatives equal to 0 (exceptfor the rotor angles, since their derivatives are equal to synchronous frequency) and solving the associatedsystem of equations. We are particularly interested in solving for τm, the input mechanical torque and v′fd,the field voltage (transformed across the machine windings). Both of these parameters represent explicitcontrol settings which are adjusted by generator operators.

Generator Power and Energy AnalysisTo consider how a forced oscillation affects the power and energy oscillations in the system, we first balancethe powers in the stationary circuits given by figure (10).

𝑟𝑠

+

𝐿𝑙𝑠

𝐿𝑚𝑞


Q Axis


𝑟𝑘𝑞2′

𝑟𝑘𝑞1′

−

+

𝑣𝑞𝑠𝑟

𝑖𝑞𝑠𝑟

𝑖𝑘𝑞1′

𝑖𝑘𝑞2′

+

𝑣𝑘𝑞2′

𝑣𝑘𝑞1′

+ −

λ𝑑𝑠𝜔𝑟 𝑟𝑠

+

𝐿𝑙𝑠

𝐿𝑚𝑑

𝐿𝑙𝑓𝑑′

𝐿𝑙𝑘𝑑′

𝑟𝑘𝑑′

𝑟𝑓𝑑′

−

+

𝑣𝑑𝑠𝑟

𝑖𝑑𝑠𝑟

𝑖𝑓𝑑′

𝑖𝑘𝑑′

+

𝑣𝑘𝑑′

𝑣𝑓𝑑′

− +

λ𝑞𝑠𝜔𝑟

D Axis

Figure 10: Transformed Circuit Model of the 3-Phase Synchronous Generator. The zero sequence circuit isomitted. All variables are given in the rotor’s local reference frame, but the r superscript is only given tothe terminal voltages and currents.

14

According to Tellegen’s theorem, the instantaneous powers in each circuit must sum to 0. We now balancethe powers in each circuit, beginning with the Q-axis. Employing the passive sign convention (PSC), powerconsumption is positive while power generation is negative.

Resistive Dissipation ⇒(i′

kq1

)2r′

kq1 +(i′

kq2

)2r′

kq2 +(irqs)2rs

Inductor Power Injection ⇒(i′

kq1

)L′

lkq1

(ddt i

′

kq1

)+(i′

kq2

)L′

lkq2

(ddt i

′

kq2

)+

(irqs)Lls

(ddt i

rqs

)+ (imq)Lmq

(ddt imq

)Power Tranfer ⇒ − λdsωr

Terminal Power Output ⇒ irqsvrqs

Terminal Power Input ⇒ 0

The D-axis circuit power may be balanced similarly, but this circuit contains a power input term (from thefield excitation).

Resistive Dissipation ⇒(i′

kd

)2r′

kd +(i′

fd

)2r′

fd + (irds)2rs

Inductor Power Injection ⇒(i′

kd

)L′

lkd

(ddt i

′

kd

)+(i′

fd

)L′

lfd

(ddt i

′

fd

)+

(irds)Lls(

ddt i

rds

)+ (imd)Lmd

(ddt imd

)Power Tranfer ⇒ λqsωr

Terminal Power Output ⇒ irdsvrds

Terminal Power Input ⇒ − i′

fdv′

fd

Full Machine Power BalanceWe next consider the power balance of the entire machine. To do so, we first consider the conservative forcespresent in a single generator.

Rotational Kinetic Energy of the Turbine ⇒ 12 δ

2rJ

Magentic Potential Energy of the Coils ⇒ 12 q>L(δr)q

where L(δr) is given by (68). We can therefore state the Lagrangian.

L (xj , xj , qkq) = T ∗ − V +W ∗m −We

= 12 δ

2rJ − 0 + 1

2 q>L(δr)q − 0

For the generalized force, we determine Lagrange’s equation:

τm = ddt

(∂L∂δr

)−(∂L∂δr

)= δrJ −

12 q>L′(δr)q (41)

15

And for the generalized voltages, we determine Lagrange’s equation. The voltages in (42) represent thevoltage drop across the inductors in Figure 2.

L′(δr)δrq + L(δr)q =

va + iaravb + ibrbvc + icrc

vfd − ifdrfd−ikdrkd−ikq1rkq1−ikq2rkq2

(42)

We may calculate the energy supplied by the shaft by noting that, in general, energy is equal to theproduct of torque and angular displacement: E = τ (4θ).

Eshaft = τm (δr − δr0)

where δr is the rotor angle as a function of time and δr0 is the initial angular position of the rotor. Towrite this expression more carefully, we must consider the situation where τm is a function of time instead ofconstant. In this case, we integrate the product of the shaft power and the derivative of the angular velocity.

Eshaft =∫ tf

t0

τmδrdt

We are further interested in the flow of energy, or the power, produced by the shaft, where Pshaft = ddtEshaft.

Pshaft = τmδr

=[δrJ −

12 q>L′(δr)q

]δr

The power provided by the turbine shaft represents one power input. The second power input is providedby the product of the excitation field current and the excitation field voltage.

Pexc = 32v′

fdi′

fd

= vfdifd

Therefore, the system has the following total input power.

Pin = Pexc + Pshaft

The input power must be equal to the sum of four factors: 1) output power, 2) resistive dissipation, 3)turbine inertia power, and 4) inductive power. The output power is defined as the product of the terminalvoltage and current (the following identity is shown in the Appendix).

Pout = v1ai1a + v1bi1b + v1ci1c

= 32(vrq1irq1

+ vrd1ird1

)Resistive dissipation consumes energy on both the rotor and stator side of the generator.

Pdiss = q>Rq

As the turbine accelerates and decelerates, it injects positive and negative power into the system. Therotational kinetic energy of the turbine has the following value.

Eturb = 12Jω

2r

16

The derivative of energy yields and expression for turbine power. Clearly, a positive power value correspondsto a power consumption.

Pturb = ddtEturb

= Jωrωr

Finally, we must consider the instantaneous power consumed by the inductors. We note that the energystored in the inductors has the following expression.

EL = 12 q>L(δr)q

The derivative of this expression will compute the power injection of the inductors.

PL = ddtEL

= 12[q>(L(δr)q + L(δr)q

)+ q>L(δr)q

]= 1

2[q>L(δr)q + q>L(δr)q + q>L(δr)q

]If we use Park’s Transform to eliminate the abc variables, the inductor matrix becomes a static matrix, asis shown in equation (69) in the appendix. We use Faraday’s law to determine the power drop across theseinductors.

PL =(

32

)(ddt λ

)>˙q

=(

32

)(ddt L(δr) ˙q

)>˙q

=(

32

)(L(δr)¨q + ˙L(δr) ˙q

)> ˙q

Since L(δr) is a static inductance matrix, its time derivative is 0. Therefore, the expression for inductorpower simplifies.

PL =(

32

)¨q>L(δr)> ˙q

We may now state the power balance for an entire generator system, where PL = PL.

Pin = Pout + Pdiss + Pturb + PL

This may be stated mathematically.

vfdifd + τmωr = 32 (vq1iq1 + vd1id1) + q>Rq + Jωrωr + 1

2[q>L(δr)q + q>L(δr)q + q>L(δr)q

]We note, though, that ωrJ − 1

2 q>L′(δr)q = τm. We make this substitution.

vfdifd +[ωrJ −

12 q>L′(δr)q

]ωr = 3

2 (vq1iq1 + vd1id1) + q>Rq + Jωrωr + 12[q>L(δr)q + q>L(δr)q + q>L(δr)q

]vfdifd −

12 q>L′(δr)qωr = 3

2 (vq1iq1 + vd1id1) + q>Rq + 12[q>L(δr)q + q>L(δr)q + q>L(δr)q

]

17

Setting up the ODEs for LinearizationIn order to perform linearization and eigenvalue analysis, we present a slightly alternative simulation approachwhere we entirely eliminate the terminal qd axis voltages. This in turn eliminates the algebraic variablespresent in the DAE system given by equations (38-40). The reason for doing so it as follows. In order to buildthe Jacobian of the linearized the system, we wish to deal with a set of ODEs. Currently, our expressionsfor q and d axis flux derivatives contain voltage terms. For example:


(λmq − λqs) + vrqs

In our system, since infinite bus voltage is fixed, we can express all unknown voltages in terms of currentsand the known infinite bus voltage. Deriving an expression for voltage in the rotor reference frame, such asvrqs, becomes a slightly arduous task since we must incorporate reference frame transformations in order toeliminate all of the algebraic constraints. In particular, we consider the first two equations of (38), but weentirely ignore the 0 sequence portion of the transformation.[

iqid

]=[

cos(θe − δr) − sin(θe − δr)sin(θe − δr) cos(θe − δr)

] [irqird

][vqvd

]=[

cos(θe − δr) − sin(θe − δr)sin(θe − δr) cos(θe − δr)

] [vrqvrd

]We treat vrq and vrd as independent state variables, so we may write vq and vd explicitly as functions of vrqand vrd. We write out these functions for generators 1 and 2.

v1q = cos(θe − δr1)vr1q − sin(θe − δr1)vr1d (43)v1d = sin(θe − δr1)vr1q + cos(θe − δr1)vr1d (44)v2q = cos(θe − δr2)vr2q − sin(θe − δr2)vr2d (45)v2d = sin(θe − δr2)vr2q + cos(θe − δr2)vr2d (46)

We write out a similar set of equations for the terminal generator currents.

i1q = cos(θe − δr1)ir1q − sin(θe − δr1)ir1d (47)i1d = sin(θe − δr1)ir1q + cos(θe − δr1)ir1d (48)i2q = cos(θe − δr2)ir2q − sin(θe − δr2)ir2d (49)i2d = sin(θe − δr2)ir2q + cos(θe − δr2)ir2d (50)

Unfortunately, irq and ird are functions of flux state variables, so iq and id must be written in term of the fluxstate variables (moving forward, we drop the subscript s which was used on terminal voltages and currentsin the previous section, such that vrqs = vrq , for example). For a generic generator, we have the followingdefinitions for the terminal currents in the rotor reference frame.

ird = − 1Lls

(λds − Lad

(λdsLls

+λ′

fd

L′lfd

+ λ′

kd

L′lkd

))

irq = − 1Lls

(λqs − Laq

(λqsLls

+λ′

kq1

L′lkq1

+λ′

kq2

L′lkq2

))We now turn our attention to the network equations which are originally given by (40). We first substitutein the updated expressions for voltages (43-46).

Network Equations =

cos(θe − δr1)vr1q − sin(θe − δr1)vr1d = v∞q + r∞

(i1q + i2q

)+ ωeL∞ (i1d + i2d) + L∞

(di1qdt + di2q

dt

)sin(θe − δr1)vr1q + cos(θe − δr1)vr1d = v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q

)+ L∞

(di1d

dt + di2ddt

)cos(θe − δr2)vr2q − sin(θe − δr2)vr2d = cos(θe − δr1)vr1q − sin(θe − δr1)vr1d + rli2q + ωeLli2d + Ll

di2qdt

sin(θe − δr2)vr2q + cos(θe − δr2)vr2d = sin(θe − δr1)vr1q + cos(θe − δr1)vr1d + rli2d − ωeLli2q + Lldi2d

dt

18

Next, we choose to write each of these expression in terms of a terminal voltage variable.

vr1q =v∞q+r∞(i1q+i2q )+ωeL∞(i1d+i2d)+L∞

(di1q

dt +di2q

dt

)+sin(θe−δr1 )vr1d

cos(θe−δr1 )

vr1d =v∞d+r∞(i1d+i2d)−ωeL∞(i1q+i2q )+L∞

(di1d

dt +di2d

dt

)−sin(θe−δr1 )vr1q

cos(θe−δr1 )

vr2q =cos(θe−δr1 )vr1q−sin(θe−δr1 )vr1d+rli2q+ωeLli2d+Ll

di2qdt +sin(θe−δr2 )vr2d

cos(θe−δr2 )

vr2d =sin(θe−δr1 )vr1q+cos(θe−δr1 )vr1d+rli2d−ωeLli2q+Ll

di2ddt −sin(θe−δr2 )vr2q

cos(θe−δr2 )

(51)

We are able to eliminate all of the voltage terms on the right hand side of each line of (51) via substitution.

vr1q =

v∞q + r∞(i1q + i2q

)+ ωeL∞

(i1d + i2d

)+ L∞

(di1q

dt +di2q

dt

)+ tan(θe − δr1 )

[v∞d + r∞

(i1d + i2d

)− ωeL∞

(i1q + i2q

)+ L∞

(di1d

dt +di2d

dt

)]cos(θe − δr1 )

(1 + tan2(θe − δr1 )

)(52)

vr1d

=

v∞d + r∞(i1d + i2d

)− ωeL∞

(i1q + i2q

)+ L∞

(di1d

dt +di2d

dt

)− tan(θe − δr1 )

[v∞q + r∞

(i1q + i2q

)+ ωeL∞

(i1d + i2d

)+ L∞

(di1q

dt +di2q

dt

)]cos(θe − δr1 )

(1 + tan2(θe − δr1 )

)(53)

vr2q =

cos(θe − δr1 )vr1q− sin(θe − δr1 )vr1d

+ rli2q + ωeLli2d + Ll

di2qdt + tan(θe − δr2 )

[sin(θe − δr1 )vr1q

+ cos(θe − δr1 )vr1d+ rli2d − ωeLli2q + Ll

di2ddt

]cos(θe − δr2 )

(1 + tan2(θe − δr2 )

)(54)

vr2d

=

sin(θe − δr1 )vr1q+ cos(θe − δr1 )vr1d

+ rli2d − ωeLli2q + Ll

di2ddt − tan(θe − δr2 )

[cos(θe − δr1 )vr1q

− sin(θe − δr1 )vr1d+ rli2q + ωeLli2d + Ll

di2qdt

]cos(θe − δr2 )

(1 + tan2(θe − δr2 )

)(55)

At this point, these voltage expressions may be substituted into the equations for λqs and λds in (??) forboth generators 1 and 2. In order to build the Jacobian for the system though, we must determine values forthe current derivatives in terms of flux values. This is because these current derivative themselves containand depend on terminal voltage variables. We consider these individually, beginning with i2q and i2d sincethese are the only two derivatives which show up in (54) and (55).

ddt i2q = d

dt cos(θe − δr2)ir2q −ddt sin(θe − δr2)ir2d

= cos(θe − δr2)ir2q − ir2q sin(θe − δr2)

[θe − δr2

]− sin(θe − δr2)ir2d − cos(θe − δr2)ir2d

[θe − δr2

]= cos(θe − δr2)ir2q −

[1Lls

(λ2mq − λ2qs)]

sin(θe − δr2)[θe − δr2

](56)

− sin(θe − δr2)ir2d − cos(θe − δr2)[

1Lls

(λ2md − λ2ds)] [θe − δr2

](57)

ddt i2d = d

dt sin(θe − δr2)ir2q + ddt cos(θe − δr2)ir2d

= sin(θe − δr2)ir2q + cos(θe − δr2)[θe − δr2

]ir2q + cos(θe − δr2)ir2d − sin(θe − δr2)

[θe − δr2

]ir2d


] [ 1Lls

(λ2mq − λ2qs)]

(58)

+ cos(θe − δr2)ir2d − sin(θe − δr2)[θe − δr2

] [ 1Lls

(λ2md − λ2ds)]

(59)

Next, we define terminal current derivatives in terms of flux variables.

ir2d =[

1Lls

(λ2md − λ2ds)]

ir2q =[

1Lls

(λ2mq − λ2qs)]

19

We take the derivative and expand.

ir2d = LadLls

(λ′

2fd

L′lfd

+ λ′

2kdL′lkd

)+ λ2ds

(LadL2ls

− 1Lls

)

ir2q = LaqLls

(λ′

2kq1

L′lkq1

+λ′

2kq2

L′lkq2

)+ λ2qs

(LaqL2ls

− 1Lls

)We now substitute in values for the flux derivatives.

ir2d = LadLls

[r′fd

L′lfd

(λ2md − λ

′

2fd

)+ v

′

2fd

]L′lfd

+

[r′kd

L′lkd

(λ2md − λ

′

2kd

)]L′lkd

+[ω2rλ2qs + rs

Lls(λ2md − λ2ds) + vr2d

](LadL2ls

− 1Lls

)

ir2q = LaqLls

[r′kq1

L′lkq1

(λ2mq − λ

′

2kq1

)]L′lkq1

+

[r′kq2

L′lkq2

(λ2mq − λ

′

2kq2

)]L′lkq2

+[−ω2rλ2ds + rs

Lls(λ2mq − λ2qs) + vr2q

](LaqL2ls

− 1Lls

)Finally, we pull out the terminal voltage terms and we introduce two new variables, Q∗2d and Q∗2q , in orderto simplify the notation.

ir2d = LadLls

[r′fd

L′lfd

(λ2md − λ

′

2fd

)+ v

′

2fd

]L′lfd

+

[r′kd

L′lkd

(λ2md − λ

′

2kd

)]L′lkd

+[ω2rλ2qs + rs

Lls(λ2md − λ2ds)

](LadL2ls

− 1Lls

)+ vr2d

(LadL2ls

− 1Lls

)= Q∗2d + vr2d

(LadL2ls

− 1Lls

)

ir2q = LaqLls

[r′kq1

L′lkq1

(λ2mq − λ

′

2kq1

)]L′lkq1

+

[r′kq2

L′lkq2

(λ2mq − λ

′

2kq2

)]L′lkq2

+[−ω2rλ2ds + rs

Lls(λ2mq − λ2qs)

](LaqL2ls

− 1Lls

)+ vr2q

(LaqL2ls

− 1Lls

)= Q∗2q + vr2q

(LaqL2ls

− 1Lls

)These expressions can be placed back into (56) and (58). Performing all of these substitutions will beextremely laborious, so we simply choose to pull out the voltage terms and leave the substitution work to

20

MATLAB.

ddt i2q = cos(θe − δr2)

[Q∗2q + vr2q

(LaqL2ls

− 1Lls

)]−[

1Lls

(λ2mq − λ2qs)]


]− sin(θe − δr2)

[Q∗2d + vr2d

(LadL2ls

− 1Lls

)]− cos(θe − δr2)

[1Lls


]ddt i2d = sin(θe − δr2)

[Q∗2q + vr2q

(LaqL2ls

− 1Lls

)]+ cos(θe − δr2)

[θe − δr2

] [ 1Lls

(λ2mq − λ2qs)]

+ cos(θe − δr2)[Q∗2d + vr2d

(LadL2ls

− 1Lls

)]− sin(θe − δr2)

[θe − δr2

] [ 1Lls

(λ2md − λ2ds)]

We again pull out the voltage terms and we introduce two more variables, Q∗∗2q and Q∗∗2d in order to simplifynotation.


[Q∗2q

]−[

1Lls

(λ2mq − λ2qs)]



[Q∗2d

]− cos(θe − δr2)

[1Lls


]+ vr2q cos(θe − δr2)

(LaqL2ls

− 1Lls

)− vr2d sin(θe − δr2)

(LadL2ls

− 1Lls

)=Q∗∗2q + vr2q cos(θe − δr2)

(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls

)ddt i2d = sin(θe − δr2)

[Q∗2q

]+ cos(θe − δr2)

[θe − δr2

] [ 1Lls

(λ2mq − λ2qs)]

+ cos(θe − δr2)[Q∗2d


[θe − δr2

] [ 1Lls

(λ2md − λ2ds)]

+ vr2q sin(θe − δr2)(LaqL2ls

− 1Lls

)+ vr2d cos(θe − δr2)

(LadL2ls

− 1Lls

)=Q∗∗2d + vr2q sin(θe − δr2)

(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls

)We now return to our expressions for terminal voltage in (54) and (55) and we pull out the current derivativeterms.

vr2q =

cos(θe − δr1 )vr1q − sin(θe − δr1 )vr

1d + rli2q + ωeLli2d + tan(θe − δr2 )[sin(θe − δr1 )vr

1q + cos(θe − δr1 )vr1d + rli2d − ωeLli2q

]cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

+Ll

cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

[di2qdt

]+

Ll tan(θe − δr2 )cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

[di2ddt

]vr

2d =sin(θe − δr1 )vr

1q + cos(θe − δr1 )vr1d + rli2d − ωeLli2q − tan(θe − δr2 )

[cos(θe − δr1 )vr

1q − sin(θe − δr1 )vr1d + rli2q + ωeLli2d

]cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

+Ll

cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

[di2ddt

]−


[di2qdt

]

21

We then substitute in our newly derived current derivative expressions.

vr2q =




]cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

+Ll

cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

[Q∗∗

2q + vr2q cos(θe − δr2 )

(Laq

L2ls

−1Lls

)− vr

2d sin(θe − δr2 )(Lad

L2ls

−1Lls

)]+


[Q∗∗

2d + vr2q sin(θe − δr2 )

(Laq

L2ls

−1Lls

)+ vr

2d cos(θe − δr2 )(Lad

L2ls

−1Lls

)]vr





]cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

+Ll

cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

[Q∗∗


(Laq

L2ls

−1Lls

)+ vr

2d cos(θe − δr2 )(Lad

L2ls

−1Lls

)]−


[Q∗∗


(Laq

L2ls

−1Lls

)− vr

2d sin(θe − δr2 )(Lad

L2ls

−1Lls

)]Once again, we pull out voltage terms.

vr2q

1 −Ll cos(θe − δr2 )

(LaqL2ls

− 1Lls

)cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

−Ll tan(θe − δr2 ) sin(θe − δr2 )

(LaqL2ls

− 1Lls

)cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

=




]cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

+Ll

cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

[Q∗∗

2q − vr2d sin(θe − δr2 )

(Lad

L2ls

−1Lls

)]+


[Q∗∗

2d + vr2d cos(θe − δr2 )

(Lad

L2ls

−1Lls

)]

vr2d

1 −Ll cos(θe − δr2 )

(LadL2ls

− 1Lls

)cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

−Ll tan(θe − δr2 ) sin(θe − δr2 )

(LadL2ls

− 1Lls

)cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

=

sin(θe − δr1 )vr1q + cos(θe − δr1 )vr

1d + rli2d − ωeLli2q − tan(θe − δr2 )[cos(θe − δr1 )vr


]cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

+Ll

cos(θe − δr2 ) (1 + tan2(θe − δr2 ))

[Q∗∗


(Laq

L2ls

−1Lls

)]−


[Q∗∗


(Laq

L2ls

−1Lls

)]We are able to perform on last set of substitutions and rearrangements in order to isolate the terminal

22

voltages.

vr2q =




]cos(θe − δr2 ) (1 + tan2(θe − δr2 )) − Ll cos(θe − δr2 )

(LaqL2ls

− 1Lls

)− Ll tan(θe − δr2 ) sin(θe − δr2 )

(LaqL2ls

− 1Lls

)+

Ll

[Q∗∗

2q + tan(θe − δr2 )Q∗∗2d


(LaqL2ls

− 1Lls


(LaqL2ls

− 1Lls

)+

Ll

[− sin(θe − δr2 )

(LadL2ls

− 1Lls

)+ tan(θe − δr2 ) cos(θe − δr2 )

(LadL2ls

− 1Lls

)]cos(θe − δr2 ) (1 + tan2(θe − δr2 )) − Ll cos(θe − δr2 )

(LaqL2ls

− 1Lls


(LaqL2ls

− 1Lls

) [vr2d

]vr






(LadL2ls

− 1Lls


(LadL2ls

− 1Lls

)+

Ll

[Q∗∗

2d − tan(θe − δr2 )Q∗∗2q


(LadL2ls

− 1Lls


(LadL2ls

− 1Lls

)+

Ll

[sin(θe − δr2 )

(LaqL2ls

− 1Lls

)− tan(θe − δr2 ) cos(θe − δr2 )

(LaqL2ls

− 1Lls

)]cos(θe − δr2 ) (1 + tan2(θe − δr2 )) − Ll cos(θe − δr2 )

(LadL2ls

− 1Lls


(LadL2ls

− 1Lls

) [vr2q

]In order to simplify the expressions, we develop a new set of denominator coefficient variables.

D2q = cos(θe − δr2)(1 + tan2(θe − δr2)

)− Ll cos(θe − δr2)

(LaqL2ls

− 1Lls

)− Ll tan(θe − δr2) sin(θe − δr2)

(LaqL2ls

− 1Lls

)D2d = cos(θe − δr2)

(1 + tan2(θe − δr2)

)− Ll cos(θe − δr2)

(LadL2ls

− 1Lls

)− Ll tan(θe − δr2) sin(θe − δr2)

(LadL2ls

− 1Lls

)We develop numerator coefficients as well, and we pull out all terminal voltage terms.

T12q =

rli2q + ωeLli2d + tan(θe − δr2)[rli2d − ωeLli2q

]D2q

Tvq2qvr1q =

(cos(θe − δr1) + tan(θe − δr2) sin(θe − δr1)

D2q

)vr1q

Tvd2qvr1d =

(tan(θe − δr2) cos(θe − δr1)− sin(θe − δr1)

D2q

)vr1d

T22q =

Ll

[Q∗∗2q + tan(θe − δr2)Q∗∗2d

]D2q

T32q =

Ll

[− sin(θe − δr2)

(LadL2ls

− 1Lls

)+ tan(θe − δr2) cos(θe − δr2)

(LadL2ls

− 1Lls

)]D2q

T12d =

rli2d − ωeLli2q − tan(θe − δr2)[rli2q + ωeLli2d

]D2d

Tvq2dvr1q =

(sin(θe − δr1)− tan(θe − δr2) cos(θe − δr1)

D2q

)vr1q

Tvd2dvr1d =

(cos(θe − δr1) + tan(θe − δr2) sin(θe − δr1)

D2q

)vr1d

T22d =

Ll

[Q∗∗2d − tan(θe − δr2)Q∗∗2q

]D2d

T32d =

Ll

[sin(θe − δr2)

(LaqL2ls

− 1Lls

)− tan(θe − δr2) cos(θe − δr2)

(LaqL2ls

− 1Lls

)]D2d

23

We substitute in these coefficients.vr

2q =T12q + Tvq

2q

[vr

1q

]+ Tvd

2q

[vr

1d

]+ T2

2q + T32q

[vr

2d

]vr

2d =T12d + Tvq

2d

[vr

1q

]+ Tvd

2d

[vr

1d

]+ T2

2d + T32d

[vr

2q

]Next, we solve for generator 2 terminal voltages in terms of generator 1 terminal voltages.

vr2q =T1

2q + Tvq2q

[vr

1q

]+ Tvd

2q

[vr

1d

]+ T2

2q + T32q

[T1

2d + Tvq2d

[vr

1q

]+ Tvd

2d

[vr

1d

]+ T2

2d + T32d

[vr

2q

]]vr

2d =T12d + Tvq

2d

[vr

1q

]+ Tvd

2d

[vr

1d

]+ T2

2d + T32d

[T1

2q + Tvq2q

[vr

1q

]+ Tvd

2q

[vr

1d

]+ T2

2q + T32q

[vr

2d

]]⇓

vr2q =

T12q + Tvq

2q

[vr

1q

]+ Tvd

2q

[vr

1d

]+ T2

2q + T32q

[T1

2d + Tvq2d

[vr

1q

]+ Tvd

2d

[vr

1d

]+ T2

2d

](1 − T3

2qT32d

)=

T12q + T2

2q + T32q

[T1

2d + T22d

](1 − T3

2qT32d

) +Tvq

2q + T32qTvq

2d(1 − T3

2qT32d

) [vr1q

]+

Tvd2q + T3

2qTvd2d(

1 − T32qT3

2d

) [vr1d

]vr

2d =T1

2d + Tvq2d

[vr

1q

]+ Tvd

2d

[vr

1d

]+ T2

2d + T32d

[T1

2q + Tvq2q

[vr

1q

]+ Tvd

2q

[vr

1d

]+ T2

2q

](1 − T3

2dT3

2q

)=

T12d + T2

2d + T32d

[T1

2q + T22q

](1 − T3

2dT3

2q

) +Tvq

2d+ T3

2dTvq2q(

1 − T32d

T32q

) [vr1q

]+

Tvd2d + T3

2dTvd2q(

1 − T32d

T32q

) [vr1d

]We now move on to consider the terminal voltages at generator 1. We state with the current derivatives inthe synchronous reference frame.

ddt i1q = d

dt cos(θe − δr1)ir1q −ddt sin(θe − δr1)ir1d

= cos(θe − δr1)ir1q − ir1q sin(θe − δr1)

[θe − δr1

]− sin(θe − δr1)ir1d − cos(θe − δr1)ir1d

[θe − δr1

]= cos(θe − δr1)ir1q −

[1Lls

(λ1mq − λ1qs)]


](60)

− sin(θe − δr1)ir1d − cos(θe − δr1)[

1Lls


](61)

ddt i1d = d

dt sin(θe − δr1)ir1q + ddt cos(θe − δr1)ir1d


]ir1q + cos(θe − δr1)ir1d − sin(θe − δr1)

[θe − δr1

]ir1d


] [ 1Lls

(λ1mq − λ1qs)]

(62)

+ cos(θe − δr1)ir1d − sin(θe − δr1)[θe − δr1

] [ 1Lls

(λ1md − λ1ds)]

(63)

Next, we define terminal current derivatives in terms of flux variables.

ir1d =[

1Lls

(λ1md − λ1ds)]

ir1q =[

1Lls

(λ1mq − λ1qs)]

We take the derivative of each expression and expand.

ir1d = LadLls

(λ′

1fd

L′lfd

+ λ′

1kdL′lkd

)+ λ1ds

(LadL2ls

− 1Lls

)

ir1q = LaqLls

(λ′

1kq1

L′lkq1

+λ′

1kq2

L′lkq2

)+ λ1qs

(LaqL2ls

− 1Lls

)

24

We now substitute in values for the flux derivatives.

ir1d = LadLls

[r′fd

L′lfd

(λ1md − λ

′

1fd

)+ v

′

1fd

]L′lfd

+

[r′kd

L′lkd

(λ1md − λ

′

1kd

)]L′lkd

+[ω1rλ1qs + rs

Lls(λ1md − λ1ds) + vr1d

](LadL2ls

− 1Lls

)

ir1q = LaqLls

[r′kq1

L′lkq1

(λ1mq − λ

′

1kq1

)]L′lkq1

+

[r′kq2

L′lkq2

(λ1mq − λ

′

1kq2

)]L′lkq2

+[−ω1rλ1ds + rs

Lls(λ1mq − λ1qs) + vr1q

](LaqL2ls

− 1Lls

)

Finally, we pull out the terminal voltage terms.

ir1d = LadLls

[r′fd

L′lfd

(λ1md − λ

′

1fd

)+ v

′

1fd

]L′lfd

+

[r′kd

L′lkd

(λ1md − λ

′

1kd

)]L′lkd

+[ω1rλ1qs + rs

Lls(λ1md − λ1ds)

](LadL2ls

− 1Lls

)+ vr1d

(LadL2ls

− 1Lls

)= Q∗1d + vr1d

(LadL2ls

− 1Lls

)

ir1q = LaqLls

[r′kq1

L′lkq1

(λ1mq − λ

′

1kq1

)]L′lkq1

+

[r′kq2

L′lkq2

(λ1mq − λ

′

1kq2

)]L′lkq2

+[−ω1rλ1ds + rs

Lls(λ1mq − λ1qs)

](LaqL2ls

− 1Lls

)+ vr1q

(LaqL2ls

− 1Lls

)= Q∗1q + vr1q

(LaqL2ls

− 1Lls

)Finally, these expressions can be placed back into (60) and (62). Performing all of these substitutions willbe extremely laborious, so we simply choose to pull out the voltage terms and leave the substitution workto MATLAB.


[Q∗1q + vr1q

(LaqL2ls

− 1Lls

)]−[

1Lls

(λ1mq − λ1qs)]



[Q∗1d + vr1d

(LadL2ls

− 1Lls

)]− cos(θe − δr1)

[1Lls


]ddt i1d = sin(θe − δr1)

[Q∗1q + vr1q

(LaqL2ls

− 1Lls

)]+ cos(θe − δr1)

[θe − δr1

] [ 1Lls

(λ1mq − λ1qs)]

+ cos(θe − δr1)[Q∗1d + vr1d

(LadL2ls

− 1Lls

)]− sin(θe − δr1)

[θe − δr1

] [ 1Lls

(λ1md − λ1ds)]

25

We again pull out the voltage terms.


[Q∗1q

]−[

1Lls

(λ1mq − λ1qs)]



[Q∗1d

]− cos(θe − δr1)

[1Lls


]+ vr1q cos(θe − δr1)

(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls

)=Q∗∗1q + vr1q cos(θe − δr1)

(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls

)ddt i1d = sin(θe − δr1)

[Q∗1q

]+ cos(θe − δr1)

[θe − δr1

] [ 1Lls

(λ1mq − λ1qs)]

+ cos(θe − δr1)[Q∗1d


[θe − δr1

] [ 1Lls

(λ1md − λ1ds)]

+ vr1q sin(θe − δr1)(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls

)=Q∗∗1d + vr1q sin(θe − δr1)

(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls

)We may substitute these current derivative values (and the previously derived ones) into the network voltageequations.

vr1q =v∞q

+ r∞(i1q + i2q

)+ ωeL∞ (i1d + i2d) + tan(θe − δr1)

[v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q

)]cos(θe − δr1)

(1 + tan2(θe − δr1)

) +

L∞

cos(θe − δr1)(1 + tan2(θe − δr1)

) [di1qdt

]+ L∞


) [di2qdt

]+ tan(θe − δr1)L∞


) [di1ddt

]+ tan(θe − δr1)L∞


) [di2ddt

]vr1d =

v∞d+ r∞ (i1d + i2d)− ωeL∞

(i1q + i2q

)− tan(θe − δr1)

[v∞q

+ r∞(i1q + i2q

)+ ωeL∞ (i1d + i2d)

]cos(θe − δr1)

(1 + tan2(θe − δr1)

) +

L∞


) [di1ddt

]+ L∞


) [di2ddt

]− tan(θe − δr1)L∞


) [di1qdt

]− tan(θe − δr1)L∞


) [di2qdt

]⇓

26

vr1q =v∞q

+ r∞(i1q + i2q


[v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q

)]cos(θe − δr1)

(1 + tan2(θe − δr1)

)+ L∞


) [Q∗∗1q + vr1q cos(θe − δr1)(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls

)]+ L∞



− 1Lls


(LadL2ls

− 1Lls

)]+ tan(θe − δr1)L∞


) [Q∗∗1d + vr1q sin(θe − δr1)(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls




− 1Lls


(LadL2ls

− 1Lls

)]vr1d =

v∞d+ r∞ (i1d + i2d)− ωeL∞

(i1q + i2q


[v∞q + r∞

(i1q + i2q

)+ ωeL∞ (i1d + i2d)

]cos(θe − δr1)

(1 + tan2(θe − δr1)

)+ L∞



− 1Lls


(LadL2ls

− 1Lls

)]+ L∞



− 1Lls


(LadL2ls

− 1Lls

)]− tan(θe − δr1)L∞



− 1Lls


(LadL2ls

− 1Lls




− 1Lls


(LadL2ls

− 1Lls

)]Now we pull out the generator 1 terminal voltage terms.

27

vr1q

1−L∞ cos(θe − δr1)

(LaqL2ls

− 1Lls

)cos(θe − δr1)

(1 + tan2(θe − δr1)

) − tan(θe − δr1)L∞ sin(θe − δr1)(LaqL2ls

− 1Lls

)cos(θe − δr1)

(1 + tan2(θe − δr1)

)

=v∞q

+ r∞(i1q + i2q


[v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q

)]cos(θe − δr1)

(1 + tan2(θe − δr1)

)+ L∞


) [Q∗∗1q − vr1d sin(θe − δr1)(LadL2ls

− 1Lls

)]+ L∞



− 1Lls


(LadL2ls

− 1Lls



) [Q∗∗1d + vr1d cos(θe − δr1)(LadL2ls

− 1Lls




− 1Lls


(LadL2ls

− 1Lls

)]

vr1d

1−L∞ cos(θe − δr1)

(LadL2ls

− 1Lls

)cos(θe − δr1)

(1 + tan2(θe − δr1)

) − sin(θe − δr1) tan(θe − δr1)L∞(LadL2ls

− 1Lls

)cos(θe − δr1)

(1 + tan2(θe − δr1)

)

=v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q


[v∞q

+ r∞(i1q + i2q

)+ ωeL∞ (i1d + i2d)

]cos(θe − δr1)

(1 + tan2(θe − δr1)

)+ L∞



− 1Lls

)]+ L∞



− 1Lls


(LadL2ls

− 1Lls




− 1Lls




− 1Lls


(LadL2ls

− 1Lls

)]⇓

28

vr1q

(cos(θe − δr1)

(1 + tan2(θe − δr1)

)− L∞ cos(θe − δr1)

(LaqL2ls

− 1Lls

)− tan(θe − δr1)L∞ sin(θe − δr1)

(LaqL2ls

− 1Lls

))=v∞q

+ r∞(i1q + i2q


[v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q

)]+ L∞

[Q∗∗1q − v

r1d sin(θe − δr1)

(LadL2ls

− 1Lls

)]+ L∞

[Q∗∗2q + vr2q cos(θe − δr2)

(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls


[Q∗∗1d + vr1d cos(θe − δr1)

(LadL2ls

− 1Lls


[Q∗∗2d + vr2q sin(θe − δr2)

(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls

)]vr1d

(cos(θe − δr1)

(1 + tan2(θe − δr1)


(LadL2ls

− 1Lls

)− sin(θe − δr1) tan(θe − δr1)L∞

(LadL2ls

− 1Lls

))=v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q


[v∞q

+ r∞(i1q + i2q

)+ ωeL∞ (i1d + i2d)

]+ L∞


(LaqL2ls

− 1Lls

)]+ L∞


(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls



(LaqL2ls

− 1Lls



(LaqL2ls

− 1Lls


(LadL2ls

− 1Lls

)]We now choose to define two denominator values.

D1q = cos(θe − δr1)(1 + tan2(θe − δr1)


(LaqL2ls

− 1Lls

)− tan(θe − δr1)L∞ sin(θe − δr1)

(LaqL2ls

− 1Lls

)D1d = cos(θe − δr1)

(1 + tan2(θe − δr1)


(LadL2ls

− 1Lls

)− sin(θe − δr1)L∞ tan(θe − δr1)

(LadL2ls

− 1Lls

)

29

We use these denominator definitions to simplify the expressions.

vr1q =[v∞q

+ r∞(i1q + i2q


[v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q

)]D1q

]

+[L∞Q

∗∗1q + L∞Q

∗∗2q + tan(θe − δr1)L∞Q∗∗1d + tan(θe − δr1)L∞Q∗∗2d

D1q

]

+

tan(θe − δr1)L∞ cos(θe − δr1)(LadL2ls

− 1Lls

)− L∞ sin(θe − δr1)

(LadL2ls

− 1Lls

)D1q

[vr1d]

+

L∞ cos(θe − δr2)(LaqL2ls

− 1Lls

)+ tan(θe − δr1)L∞ sin(θe − δr2)

(LaqL2ls

− 1Lls

)D1q

[vr2q]

+

tan(θe − δr1)L∞ cos(θe − δr2)(LadL2ls

− 1Lls

)− L∞ sin(θe − δr2)

(LadL2ls

− 1Lls

)D1q

[vr2d]vr1d =

[v∞d

+ r∞ (i1d + i2d)− ωeL∞(i1q + i2q


[v∞q

+ r∞(i1q + i2q

)+ ωeL∞ (i1d + i2d)

]D1d

]

+[

+L∞Q∗∗1d + L∞Q∗∗2d −Q

∗∗1q tan(θe − δr1)L∞ − tan(θe − δr1)L∞Q∗∗2q

D1d

]

+

L∞ sin(θe − δr1)(LaqL2ls

− 1Lls

)− tan(θe − δr1)L∞ cos(θe − δr1)

(LaqL2ls

− 1Lls

)D1d

[vr1q]

+

L∞ sin(θe − δr2)(LaqL2ls

− 1Lls

)− tan(θe − δr1)L∞ cos(θe − δr2)

(LaqL2ls

− 1Lls

)D1d

[vr2q]

+

tan(θe − δr1)L∞ sin(θe − δr2)(LadL2ls

− 1Lls

)+ L∞ cos(θe − δr2)

(LadL2ls

− 1Lls

)D1d

[vr2d]We assign variable names to the coefficients attached to the voltage variables (we employ alternative alpha-betic variables for simplified notation).

vr1q = Tq1 + Tq2[vr1d]

+ Tq3[vr2q

]+ Tq4

[vr2d]

= a+ b[vr1d]

+ c[vr2q

]+ d

[vr2d]

vr1d = Td1 + Td2[vr1q

]+ Td3

[vr2q

]+ Td4

[vr2d]

= e+ f[vr1q

]+ g

[vr2q

]+ h

[vr2d]

At this point, we import the expressions for vr2q and vr2d (we again employ alternative alphabetic variablesfor simplified notation).

vr2q =

T12q + T2

2q + T32q

[T1

2d + T22d

](1 − T3

2qT32d

) +Tvq

2q + T32qTvq

2d(1 − T3

2qT32d

) [vr1q

]+

Tvd2q + T3

2qTvd2d(

1 − T32qT3

2d

) [vr1d

]=i+ j

[vr

1q

]+ k[vr

1d

]vr

2d =T1

2d + T22d + T3

2d

[T1

2q + T22q

](1 − T3

2dT3

2q

) +Tvq

2d+ T3

2dTvq2q(

1 − T32d

T32q

) [vr1q

]+

Tvd2d + T3

2dTvd2q(

1 − T32d

T32q

) [vr1d

]=l +m

[vr

1q

]+ n[vr

1d

]30

With these four equations, we may solve for the terminal voltage values purely in terms of currents and otherstate variables.

vr1q = a+ b[vr1d]

+ c[i+ j

[vr1q

]+ k

[vr1d]]

+ d[l +m

[vr1q

]+ n

[vr1d]]

vr1d = e+ f[vr1q

]+ g

[i+ j

[vr1q

]+ k

[vr1d]]

+ h[l +m

[vr1q

]+ n

[vr1d]]

We group like terms to solve the expressions.

vr1q = a+ ci+ dl + (cj + dm)[vr1q

]+ (b+ ck + dn)

[vr1d]

vr1d = e+ gi+ hl + (f + gj + hm)[vr1q

]+ (gk + hn)

[vr1d]

⇓

vr1q =a+ ci+ dl + (b+ ck + dn)

[vr1d]

(1− (cj + dm))Now we are able to perform substitution.

vr1d = e+ gi+ hl + (f + gj + hm)[a+ ci+ dl + (b+ ck + dn)

[vr1d]

(1− (cj + dm))

]+ (gk + hn)

[vr1d]

= e+ gi+ hl + (f + gj + hm)(a+ ci+ dl

1− cj − dm

)+[(f + gj + hm)

(b+ ck + dn

1− cj − dm

)+ (gk + hn)

] [vr1d]

vr1d

(1−

[(f + gj + hm)

(b+ ck + dn

1− cj − dm

)+ (gk + hn)

])= e+ gi+ hl + (f + gj + hm)

(a+ ci+ dl

1− cj − dm

)

vr1d =e+ gi+ hl + (f + gj + hm)

(a+ci+dl

1−cj−dm

)(

1−[(f + gj + hm)

(b+ck+dn1−cj−dm

)+ (gk + hn)

])We may now solve for vr1q .

vr1q =a+ ci+ dl + (b+ ck + dn)

[e+gi+hl+(f+gj+hm)( a+ci+dl

1−cj−dm )(1−[(f+gj+hm)( b+ck+dn

1−cj−dm )+(gk+hn)])

](1− cj − dm)

Generator 2 terminal voltages follow similarly.

vr2q =i+ j

[(a+ ci+ dl + (b+ ck + dn)

(1 − (cj + dm))

)e+ gi+ hl(

1 −[(f + gj + hm)

[a+ci+dl+(b+ck+dn)

(1−(cj+dm))

]+ (gk + hn)

])]

+ k

[e+ gi+ hl(

1 −[(f + gj + hm)

[a+ci+dl+(b+ck+dn)

(1−(cj+dm))

]+ (gk + hn)

])]

vr2d =l +m

[(a+ ci+ dl + (b+ ck + dn)

(1 − (cj + dm))

)e+ gi+ hl(

1 −[(f + gj + hm)

[a+ci+dl+(b+ck+dn)

(1−(cj+dm))

]+ (gk + hn)

])]

+ n

[e+ gi+ hl(

1 −[(f + gj + hm)

[a+ci+dl+(b+ck+dn)

(1−(cj+dm))

]+ (gk + hn)

])]

31

Linearization and Eigenvalue AnalysisAssume we have the following set of ODEs.

x = f(x, y)y = g(x, y)

We perform Taylor Series expansion about an equilibrium point (or a steady state power flow solution inour context) x0, y0.

x ≈ f(x0, y0) +[

ddxf(x, y)

]x0,y0

∆x+[

ddy f(x, y)

]x0y0

∆y

y ≈ g(x0, y0) +[

ddxg(x, y)

]x0,y0

∆x+[

ddy g(x, y)

]x0,y0

∆y

We subtract the operating point from both sides in order to determine how the derivatives change, and webuild the Jacobian matrix. [

∆x∆y

]≈

[d

dxf(x, y) ddyf(x, y)

ddxg(x, y) d

dy g(x, y)

]x0,y0

[∆x∆y

]We perform a similar process with our multi-generator system. For our given set of system parameters, webuild the system Jacobian and we find the 16 eigenvalues (each machine requires 8 differential equations,and the network differential equations have been written in terms of the flux variables of the generator).Only two of these eigenvalues have oscillatory (imaginary) components.

λ9,10 = −0.0025± j0.0379λ12,13 = −0.0003± j0.0162

These oscillatory modes correspond to 0.6032 Hz and 0.2578 Hz, respectively. Of course, λ12,13 is the moredominant mode of the two since is has a larger real part (thus a smaller damping component). We nextperform the following experiment. We put a oscillation on the shaft of generator 1 such that the mechanicalpower input oscillates at frequency fosc at magnitude ατm0 where α = 0.1 (10% of the nominal torque).

τm = τm0 + α · τm0 · sin (2πfosct)

In the first plot (Figure 11), we trace the standard deviation of four variables: δ1, δ2, ω1, and ω2 and wecompare their differences. This effectively represents the magnitude of the phase oscillations.

32

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Forced Oscillation Frequncy (Hz)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Sta

ndar

d D

evia

tion

</1</2<!1

<!2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1


-0.3

-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

Sta

ndar

d D

evia

tion

</1 ! </2<!1

! <!2

zero

Figure 11: The left panel shows the standard deviation (a measure of magnitude) of variablesδ1, δ2, ω1, andω2, and the right panel shows the difference in standard deviations (the difference in magnitudes) of thephase angles and the frequencies of the generators. There are noticeable peaks at both oscillatory modes of0.2578 Hz and 0.6032 Hz.

In the next plot, we trace the phase of the phase oscillations. When both angles are oscillating in phase, thephase shift is close to 0, and when both angles are out of phase, the phase shift is close to π.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1


0

0.5

1

1.5

2

2.5

3

3.5

Pha

se S

hift

betw

een

Vol

tage

Pha

ses

?/1 ! ?/2:

Figure 12: The phase shift between the voltage phase angles is shown. Between modes, the phase shift flopsfrom 0 to π.

Linearization of a Single MachineWhen oscillatory mechanical torque (or power) is applied to a synchronous machine, the integral of themachine’s power output may tend positive or negative. We wish to determine the conditions for when thispower integral will tend positive or negative. To do so, we fix the three phase terminal voltage of the machineand study its output. In doing so, we define the three phase voltages accordingly, where Vm is the peakvoltage magnitude of a given phase voltage.

33

va(t) = Vm cos (ωst)

vb(t) = Vm cos(ωst−

2π3

)vc(t) = Vm cos

(ωst+ 2π

3

)To determine the generator’s qd axis voltages, we apply park’s transformation.

vqs = 23

[cos (δr) va(t) + cos

(δr −

2π3

)vb(t) + cos

(δr + 2π

3

)vc(t)

]= 2

3

[cos (δr) Vm cos (ωst) + cos

(δr −

2π3

)Vm cos

(ωst−

2π3

)+ cos

(δr + 2π

3

)Vm cos

(ωst+ 2π

3

)]= 2

3Vm

[cos (δr) cos (ωst) + cos

(δr −

2π3

)cos(ωst−

2π3

)+ cos

(δr + 2π

3

)cos(ωst+ 2π

3

)]vds = 2

3

[sin (δr) va(t) + sin

(δr −

2π3

)vb(t) + sin

(δr + 2π

3

)vc(t)

]= 2

3

[sin (δr) Vm cos (ωst) + sin

(δr −

2π3

)Vm cos

(ωst−

2π3

)+ sin

(δr + 2π

3

)Vm cos

(ωst+ 2π

3

)]= 2

3Vm

[sin (δr) cos (ωst) + sin

(δr −

2π3

)cos(ωst−

2π3

)+ sin

(δr + 2π

3

)cos(ωst+ 2π

3

)]In order to further simplify these identities, we may incorporate the following two identities.

cos(x) cos(y) = 12 [cos (x− y) + cos (x+ y)]

sin(x) cos(y) = 12 [sin (x+ y) + sin (x− y)]

We apply the first identity to the expression for vqs.

vqs = 23Vm

[cos (δr) cos (ωst) + cos

(δr −

2π3

)cos(ωst−

2π3

)+ cos

(δr + 2π

3

)cos(ωst+ 2π

3

)]= 2

6Vm

[cos (δr − ωst) + cos (δr + ωst) + cos (δr − ωst) + cos

(ωst+ δr −

4π3

)+ cos (δr − ωst) + cos

(ωst+ δr + 4π

3

)]= Vm cos (δr − ωst)= Vm cos

(δr)

We are able to make this last substitution since δr ≡ δr − ωst. We now perform similar analysis on theexpression for vds.

vds = 23Vm

[sin (δr) cos (ωst) + sin

(δr −

2π3

)cos(ωst−

2π3

)+ sin

(δr + 2π

3

)cos(ωst+ 2π

3

)]= 2

6Vm

[sin (δr + ωst) + sin (δr − ωst) + sin

(δr + ωst−

4π3

)+ sin (δr − ωst) + sin

(δr + ωst+ 4π

3

)+ sin (δr − ωst)

]= Vm sin (δr − ωst)= Vm sin

(δr)

34

Using these simplified expressions for vqs and vds, we may state the differential equations which model thesystem.


(λmq − λqs) + Vm cos(δr)

λds = ωrλqs + rsLls

(λmd − λds) + Vm sin(δr)

λ′

kq1 = r′kq1

L′lkq1

(λmq − λ

′

kq1

)λ′

kq2 = r′kq2

L′lkq2

(λmq − λ

′

kq2

)λ′

fd = r′fd

L′lfd

(λrmd − λ

′rfd

)+ v

′rfd

λ′

kd = r′kd

L′lkd

(λmd − λ

′

kd

)ω = τm−τe

J

δr = ωr − ωs

(64)

There are three nonlinear terms in this set of equations: f1 = ωrλds, f2 = ωrλqs, and f3 = τe. To linearizethis system, we apply the first order Taylor Series. We start with f1 and f2. The 0 superscript indicates avariable evaluated at its steady state value.

f1 ≈ f1

∣∣∣∣ω0r ,λ

0ds

+ ∂f1

∂ωr

∣∣∣∣ω0r ,λ

0ds

4ωr + ∂f1

∂λds

∣∣∣∣ω0r ,λ

0ds

4λds

≈ ω0rλ

0ds + λ0

ds

(ωr − ω0

r

)+ ω0

r

(λds − λ0

ds

)≈ λ0

dsωr + ω0rλds − ω0

rλ0ds

f2 ≈ f2

∣∣∣∣ω0r ,λ

0qs

+ ∂f2

∂ωr

∣∣∣∣ω0r ,λ

0qs

4ωr + ∂f2

∂λds

∣∣∣∣ω0r ,λ

0qs

4λqs

≈ ω0rλ

0qs + λ0

qs

(ωr − ω0

r

)+ ω0

r

(λqs − λ0

qs

)≈ λ0

qsωr + ω0rλqs − ω0

rλ0qs

The electrical torque expression can be expanded and then linearized

f3 = τe

= 32 (λdsiqs − λqsids)

= 32

(λqsλds + λdsλmq − λqsλmd − λdsλqs

Lls

)

= 32

λqsλds + Laqλds

(λqsLls

+ λ′kq1

L′lkq1

+ λ′kq2

L′lkq2

)− Ladλqs

(λdsLls

+ λ′fd

L′lfd

+ λ′kd

L′lkd

)− λdsλqs

Lls

In this expression, there are 6 flux variables, so the Taylor Series Linearization must include derivatives.Each of these derivatives are evaluated at a steady state equilibrium point.

f3 ≈ f03 + ∂f3

∂λqs4λqs + ∂f3

∂λds4λ

′

kq1 + ∂f3

∂ωr4λ

′

kq2 + ∂f3

∂λds4λds + ∂f3

∂ωr4λ

′

fd + ∂f3

∂λds4λ

′

kd

We may now insert the approximations for f1, f2, and f3 into the set of differential equations in (64). Oncedone, we simulate the full model and the linearized model in order to determine its effectiveness.

35

Linearization of a Single Machine: Damper Windings RemovedWe wish to remove the 3 damper windings present in the generator and then perform a linearization. Indoing so, we end up with the following sets of magnetizing flux variables.

λmq = Laq

(λqsLls

)(65)

λmd = Lad

(λdsLls

+λ′

fd

L′lfd

)(66)

Our initial set of differential equations has the following form, where θv is the angle of the generator busvoltage. When tied to an infinite bus, θv = 0 for all time.

λqs = − (ωr + ωs)λds + rsLls

(λmq − λqs) + Vm cos(δr − θv

)λds = (ωr + ωs)λqs + rs

Lls(λmd − λds) + Vm sin

(δr − θv

)λ′

fd = r′fd

L′lfd

(λmd − λ

′

fd

)+ v

′rfd

ω = τm−τeJ

δr = ωr − ωs

(67)

First, we assume that δr − θv � 1, so cos(δr − θv

)≈ 1 and sin

(δr − θv

)≈ δr − θv. Next, we apply a Taylor

Series Expansion to each element. We begin by writing the electrical torque in terms of flux state variables.

τe = 32 (λdsiqs − λqsids)

= 32

(λdsλmq − λqsλmd

Lls

)= 3

2

(λqsλds

(Laq − Lad

L2ls

)− λqsλ

′

fd

(Lad

LlsL′lfd

))

Now we apply Taylor Series.

4λqs ≈ −[ω0r + ωs

]4λds −

[λ0ds

]4ωr +

[rsLaqL2ls

− rsLls

]4λqs

4λds ≈[ω0r + ωs

]4λqs +

[λ0qs

]4ωr +

[rsLadL2ls

− rsLls

]4λds +

[rsLadLlsL

′lfd

]4λ

′

fd + [Vm]4δr

4λ′

fd ≈

[r′

fd

L′lfd

LadLls

]4λds +

[r′

fd

L′lfd

LadL′lfd

−r′

fd

L′lfd

]4λ

′

fd

4ωr ≈ −3J2

[[λ0ds

(Laq − Lad

L2ls

)− λ

′0fd

(Lad

LlsL′lfd

)]4λqs +

[λ0qs

(Laq − Lad

L2ls

)]4λds −

[λ0qs

(Lad

LlsL′lfd

)]4λ

′

fd

]4δr ≈ 4ωr

We can transform these expressions into the form x = Ax+Bu, where α is a scaling factor for the oscillatinginput torque.

36

4λqs4λds4λ′fd4ωr4δr

=

[rsLaqL2ls

− rsLls

] [−ω0

r − ωs]

0[−λ0

ds

]0[

ω0r + ωs

] [rsLadL2ls

− rsLls

] [rsLadLlsL

′lfd

] [λ0qs

][Vm]

0[r′fd

L′lfd

LadLls

] [r′fd

L′lfd

LadL′lfd

− r′fd

L′lfd

]0 0[

3J2λ

0ds

(Lad−Laq

L2ls

)+ 3

J2λ′0fd

(Lad

LlsL′lfd

)] [3J2λ

0qs

(Lad−Laq

L2ls

)] [3J2λ

0qs

(Lad

LlsL′lfd

)]0 0

0 0 0 1 0

4λqs4λds4λ′fd4ωr4δr

+

0001J0

τ0mα sin (ωdt)

We may determine the linearized electrical power output.

Pe = 32 (vqsiqs + vdsids)

= 32

(Vm cos

(δr)(Laq

L2ls

λqs −λqsLls

)+ Vm sin

(δr)(Lad

L2ls

λds + LadLlsL

′lfd

λ′

fd −λdsLls

))

= 32

(Vm

(LaqL2ls

λqs −λqsLls

)+ Vmδr

(LadL2ls

λds + LadLlsL

′lfd

λ′

fd −λdsLls

))

Appendix I - Terminal Active PowerThe instantaneous power at the terminals of a generator may be written in terms of the qd sequence variablesin the synchronous reference frame through an algebraic relationship.p1 =v1a i1a + v1b i1b + v1c i1c

=[cos (ωst) vs

q1 + sin (ωst) vsd1

] [cos (ωst) isq1 + sin (ωst) isd1

]+[

cos(ωst−

2π3

)vs

q1 + sin(ωst−

2π3

)vs

d1

] [cos(ωst−

2π3

)isq1 + sin

(ωst−

2π3

)isd1

]+[

cos(ωst+

2π3

)vs

q1 + sin(ωst+

2π3

)vs

d1

] [cos(ωst+

2π3

)isq1 + sin

(ωst+

2π3

)isd1

]=[cos2 (ωst) vs

q1 isq1 + sin2 (ωst) vs

d1isd1

+ sin (ωst) vsd1

cos (ωst) isq1 + cos (ωst) vsq1 sin (ωst) isd1

]+[

cos2(ωst−

2π3

)vs

q1 isq1 + sin2

(ωst−

2π3

)vs

d1isd1

+ sin(ωst−

2π3

)vs

d1cos(ωst−

2π3

)isq1 + cos

(ωst−

2π3

)vs

q1 sin(ωst−

2π3

)isd1

]+[

cos2(ωst+

2π3

)vs

q1 isq1 + sin2

(ωst+

2π3

)vs

d1isd1

+ sin(ωst+

2π3

)vs

d1cos(ωst+

2π3

)isq1 + cos

(ωst+

2π3

)vs

q1 sin(ωst+

2π3

)isd1

]We employ the following trig identity: cos2 (x) + cos2 (x+ 2π

3)

+ cos2 (x− 2π3)

= 1.5. This holds for the sineand the cosine terms. We simplify this expression by leveraging this identity and gathering like terms.

p1 =1.5[vsq1isq1

]+ 1.5

[vsd1

isd1

]+

isq1vsd1

[sin (ωst) cos (ωst) + sin

(ωst−

2π3

)cos(ωst−

2π3

)+ sin

(ωst+ 2π

3

)cos(ωst+ 2π

3

)]+

isd1vsq1

[cos (ωst) sin (ωst) + cos

(ωst−

2π3

)sin(ωst−

2π3

)+ cos

(ωst+ 2π

3

)sin(ωst+ 2π

3

)]=3

2(vsq1isq1

+ vsd1isd1

)=3

2(vrq1irq1

+ vrd1ird1

)

37

Appendix II - Time Varying Inductance MatrixThe induction matrix for the salient pole generator is given by equation (68). Its component sub-matricesfollow.

L(δr) =(

Ls LsrL>sr Lr

)(68)

Ls =

Lls + LA − LB cos 2δr − 12LA − LB cos 2

(δr − π

3)

− 12LA − LB cos 2

(δr + π

3)


(δr − π

3)

Lls + LA − LB cos 2(δr − 2π

3)

− 12LA − LB cos 2 (δr + π)


(δr + π

3)

− 12LA − LB cos 2 (δr + π) Lls + LA − LB cos 2

(δr + 2π

3)

Lr =

Llfd + Lmfd Lfdkd 0 0

Lfdkd Llkd + Lmkd 0 00 0 Llkq1 + Lmkq1 Lkq1kq20 0 Lkq1kq2 Llkq2 + Lmkq2

Lsr =

Lsfd sin (δr + 0) Lskd sin (δr + 0) Lskq1 cos (δr + 0) Lskq2 cos (δr + 0)Lsfd sin

(δr − 2π

3)

Lskd sin(δr − 2π

3)

Lskq1 cos(δr − 2π

3)

Lskq2 cos(δr − 2π

3)

Lsfd sin(δr + 2π

3)

Lskd sin(δr + 2π

3)

Lskq1 cos(δr + 2π

3)

Lskq2 cos(δr + 2π

3)

We use this matrix to determine the flux linkages associated with each of the generator’s seven coils.

λ = L(δr)q

The vector λis the magnetic flux linkage associated with each of the seven coils, and q is the correspondingcurrent flow through each coil. All values are instantaneous values.

q = i =[−ia −ib −ic ikq1 ikq2 ifd ikd

]>λ =

[λa λb λc λkq1 λkq2 λfd λkd

]>Appendix III - Static Inductance MatrixIf we apply a Park’s Transform, it can be shown (see Krause) that a constant, time invariant matrix can beused to link the flux linkage and the current variables. All variables are given in the reference frame of therotor δr.

λqsλdsλ′

kq1λ′

kq2λ′

fd

λ′

kd

=

Lls + Lmq 0 Lmq Lmq 0 0

0 Lls + Lmd 0 0 Lmd LmdLmq 0 Llfd + Lmfd Lfdkd 0 0Lmq 0 Lfdkd Llkd + Lmkd 0 0

0 Lmd 0 0 Llkq1 + Lmkq1 Lkq1kq20 Lmd 0 0 Lkq1kq2 Llkq2 + Lmkq2

−iqs−idsi′

kq1i′

kq2i′

fd

i′

kd

(69)

λ = L(δr) ˙q

We define λ and ˙q.

˙q = i =[−iqs −ids i

′

kq1 i′

kq2 i′

fd i′

kd

]>λ =

[λqs λds λ

′

kq1 λ′

kq2 λ′

fd λ′

kd

]>And we denote the constant inductance matrix as L(δr):

38

L(δr) =

Lls + Lmq 0 Lmq Lmq 0 0

0 Lls + Lmd 0 0 Lmd LmdLmq 0 Llfd + Lmfd Lfdkd 0 0Lmq 0 Lfdkd Llkd + Lmkd 0 0

0 Lmd 0 0 Llkq1 + Lmkq1 Lkq1kq20 Lmd 0 0 Lkq1kq2 Llkq2 + Lmkq2

(70)

Appendix IV - Resistive MatrixThe diagonal matrix of resistances for the rotor and stator coils are given below.

R =

ra 0 0 0 0 0 00 rb 0 0 0 0 00 0 rc 0 0 0 00 0 0 rfd 0 0 00 0 0 0 rkd 0 00 0 0 0 0 rkq1 00 0 0 0 0 0 rkq2

(71)

39

Average Power Flow through a Synchronous Generator

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