# A Three-phase Power Flow Model and Balanced Network Analysis

Steven H. Low
CMS, EE, Caltech

###### Abstract

First we present an approach to formulate unbalanced three-phase power flow
problems for general networks that explicitly separates device models and
network models. A device model consists of (i) an internal model and (ii) a
conversion rule. The conversion rule relates the internal variables (voltage,
current, and power) of a device to its terminal variables through a conversion
matrix $\Gamma$ and these terminal variables are related by network equations.
Second we apply this approach to balanced three-phase networks to formalize
per-phase analysis and prove its validity for general networks using the
spectral property of the conversion matrix $\Gamma$.

###### Index Terms:

Unbalanced three-phase power flow models, balanced networks, per-phase
analysis.

##  I Introduction

Motivation. Unbalanced three-phase load flow problems are becoming
increasingly important as we decarbonize the grid. Such problems are more
difficult for several reasons; see, e.g., [1, Chapter 11] for transmission
systems and [2] for distribution systems. First, a network model is more
complicated because three-phase lines couple currents and voltages in
different phases when lines are not transposed or loads are unbalanced, e.g.,
as in most distribution systems. If the network is symmetric, then a
similarity transformation due to Fortescue [3] from the phase coordinate to a
sequence coordinate produces network models that are decoupled in the sequence
coordinate and can therefore be analyzed in a way similar to a single-phase
network; see, e.g., [4]. Without symmetry, this transformation however offers
no simplification. Second, the voltages and currents across the single-phase
devices internal to $\Delta$ configuration are observed externally only
through a linear map $\Gamma$ that is not invertible. While we are typically
interested in solving for or optimizing the _internal_ currents or power flows
across the single-phase devices, e.g., controlling the charging currents of
electric vehicle chargers in $\Delta$ configuration, a network model, such as
$I=YV$ or $s_{j}=\sum_{k}y_{jk}^{s}\left(|V_{j}|^{2}-V_{j}V_{k}^{\sf
H}\right)$, relates only the _terminal_ voltages and currents observable
externally of three-phase devices. The interplay between internal and external
variables of a three-phase device sometimes seems confusing. Third, load flow
formulations sometimes implicitly assume that the neutrals of all
$Y$-configured devices are at zero potential and the zero-sequence components
of the terminal voltages of all $\Delta$-configured devices are zero. This
limits their applicability as, e.g., they exclude the case where some
$Y$-configured loads are ungrounded or grounded with nonzero earthing
imepdances. Solutions to the last two difficulties both lie in a careful
accounting of the conversion between internal and external variables of
$\Delta$-configured devices, using the conversion matrix $\Gamma$.

Summary. In this paper we present such a modeling approach that separates
three-phase device models and network networks. In this approach, a device
model consists of two components: (i) an internal model and (ii) a conversion
rule. The internal model describes how each of the single-phase device behaves
regardless of their configuration. The conversion rule, on the other hand,
depends only on their configuration regardless of the type of devices. It maps
internal voltages, currents, and powers across these single-phase devices to
terminal voltages, currents, and powers observable externally. Since the
network model relates only the terminal variables regardless of the type of
devices or their configurations, the explicit separation of device and network
models allows mix and match of equivalent models, enhancing modeling
flexibility. We present our model in Section II and use it to formulate an
unbalanced three-phase analysis problem for general networks in Section III.

In Section IV we illustrate our model by showing formally that the analysis
problem can be solved using per-phase analysis when the network is balanced.
It is well known that a balanced three-phase device in $\Delta$ configuration
has an $Y$ equivalent that has the same external behavior. The standard way to
justify per-phase analysis is by analyzing specific three-phase circuits,
often simple circuits; e.g., [5, 6, 7], by first converting all
$\Delta$-configured devices into their $Y$ equivalents, and then showing that
all neutrals in the equivalent circuit are at the same potential and that all
phases are decoupled. This implies that the original three-phase circuit can
be solved by analyzing a simpler per-phase circuit. This process has two
limitations.

First, it is not clear how to extend circuit analysis methods, e.g., loop
analysis or mesh analysis [5, Chapter 12], from specific (and simple) circuits
to an arbitrary balanced network and prove that all neutrals are at the same
potential. This is simple to show, however, in our model by expressing the
network equation $I=YV$ in terms of the Kronecker product and using the
spectral property of $\Gamma$ (see Theorems 2 and 3). The intuition is as
follows. In a balanced three-phase network, positive-sequence voltages and
currents are in span$(\alpha_{+})$ where $\alpha_{+}$ is an eigenvector of
$\Gamma$ and $\Gamma^{\sf T}$. This means that the transformation of balanced
voltages and currents under $\left(\Gamma,\Gamma^{\sf T}\right)$ reduces to a
scaling of these variables by their eigenvalues $1-\alpha$ and $1-\alpha^{2}$
respectively. The voltage and current at every point in a network can be
written as linear combinations of transformed source voltages and source
currents, transformed by $\left(\Gamma,\Gamma^{\sf T}\right)$ and line
admittance matrices. Therefore if the source voltages and source currents are
in span$(\alpha_{+})$ and if lines are identical and phase-decoupled, then the
transformed voltages and currents remain in span$(\alpha_{+})$ and hence are
balanced positive-sequence sets. This is explained in Sections IV-B, IV-C and
IV-E.

Second, all neutrals are at the same potential only if the neutral voltages of
all $Y$-configured devices (voltage sources, current sources and impedances)
are assumed zero and the zero-sequence voltages of all $\Delta$-configured
voltage sources are assumed zero. Roughly, this requires that all neutrals are
grounded directly (i.e., with zero earthing impedances). The standard analysis
of specific circuits often makes this assumption sometimes implicitly. Without
this assumption, the neutral voltages on the circuit are generally different.
Yet, per-phase analysis can be extended to the general case without this
assumption as long as the network is balanced, except that per-phase analysis
is needed not only on a per-phase positive-sequence network, but also on a
per-phase zero-sequence network. This is explained in Section IV-D.

We close our summary with two remarks on per-phase analysis. First, if the
network is unbalanced but symmetric, i.e., impedances are balanced and lines
are (phase coupled but) symmetric, then Fortescue’s similarity transformation
[3] from the phase coordinate to the sequence coordinate leads to decoupled
device models and network models. The network equation is therefore decoupled
in the sequence coordinate and can be interpreted as defining three separate
sequence networks, to which the per-phase analysis of Section IV can be
applied. Second, with today’s abundant computing power the smaller problem
size may not be an important advantage of per-phase analysis. Rather, per-
phase analysis illustrates the application of our modeling approach to three-
phase power flow. It also clarifies the simple structure underlying a balanced
network and enhances our conceptual understanding of three-phase networks in
general, balanced or unbalanced.

Literature. Single-phase models are a good approximation for many transmission
network applications where lines are symmetric and loads are balanced so that
the zero and negative-sequence components are negligible compared with the
positive-sequence component. They may not be negligible when lines are not
transposed or equally spaced, e.g., as in distribution systems, and when loads
are unbalanced or nonlinear, e.g., AC furnaces, high-speed trains, power
electronics, or single or two-phase laterals in distribution networks. This
can cause power quality issues such as voltage imbalances and harmonics.
Furthermore single-phase analysis can produce incorrect power flow solutions.

There is a large literature on three-phase power flow analysis and we only
make a few brief remarks. Three-phase load flow solvers have been developed
since at least the 1960s, e.g., see [8] for solution in the sequence
coordinate and [9, 10] in the phase coordinate. A three-phase network is
equivalent to a single-phase circuit where each node in the equivalent circuit
is indexed by a (bus, phase) pair [10]. Single-phase power flow algorithms
such as Newton Raphson [11] or Fast Decoupled methods [12] can be directly
applied to the equivalent circuit. The main difference with a single-phase
network is the circuit models of three-phase devices in the equivalent
circuit, such as models for three-phase lines [13, 2], transformers and co-
generators [10, 14], constant-power devices [1, Chapter 11], as well as
voltage regulators, and loads [2], etc. A state-of-the-art algorithm in [1,
Chapter 11] expresses currents in terms of voltages for both $PQ$ and $PV$
buses, applies the Newton-Raphson algorithm to the resulting nonlinear current
balance equation $I=YV$ in the sequence domain. It allows both grounded and
ungrounded loads in $Y$ and $\Delta$ configurations. For transmission
networks, computing in the sequence domain has the advantage that, when most
lines in the network are symmetric and thus have decoupled representation in
the sequence coordinate, the Jacobian matrix is sparse. Sometimes an
approximate solution is computed by ignoring the coupling across zero,
positive, and negative-sequence variables and solving the three sequence
networks separately as single-phase networks, e.g., [15]. Distribution
networks usually does not enjoy such simplification and hence computation is
usually done in the phase coordinate.

While the papers above study general networks that may contain cycles, another
set of power flow methods are tailored for three-phase radial networks [9, 16,
2, 17, 18, 19]. In particular, the tree topology leads to a spatially
recursive structure that enables iterative algorithms called backward forward
sweep (BFS), apparently first developed in [9]. Different BFS algorithms are
developed in [16][2, Chapter 10.1.3] [17] for three-phase networks ([17]
generalizing the BFS algorithm of [20] from single-phase to three-phase
networks). For single-phase radial networks, a solution method based on the
DistFlow model is developed in [21] that uses one-time forward sweep (to
express all variables in terms of the voltages at the feeder head and all
branch points) followed by a Newton-Raphson algorithm to solve for these
voltages. By exploiting the approximate sparsity of the Jacobian matrix in
[21], approximate fast decoupled methods are developed and their convergence
properties analyzed in [22]. These methods are extended to three-phase radial
networks in [18]. The existence and uniqueness of power flow solutions of
three-phase DistFlow model is analyzed in [19]. The advantage of BFS is that
it does not need to compute Jacobian nor solve a linear system to compute
iteration updates. Newton-Raphson, on the other hand, tends to converge in a
smaller number of iterations.

Notation. Let $\mathbb{C}$ denote the set of complex numbers. For
$a\in\mathbb{C}$, Re$\,\,a$ and Im$\,\,a$ denote its real and imaginary parts
respectively, and $\bar{a}$ or $a^{\sf H}$ denotes its complex conjugate. We
use $\mathbf{i}$ to denote $\sqrt{-1}$. A vector $x\in\mathbb{C}^{n}$ is a
column vector and is denoted in one of two ways:

$\displaystyle x$ $\displaystyle\ \ =\ \ \begin{bmatrix}x_{1}\\\ \vdots\\\
x_{n}\end{bmatrix},$ $\displaystyle x$ $\displaystyle\ \ =\ \
(x_{1},\dots,x_{n})$

Its componentwise complex conjugate is denoted by $\overline{x}$. For any
matrix $A$, $A^{\sf T}$, $A^{\sf H}$, $A^{\dagger}$ denote its transpose,
Hermitian transpose, and pseudo-inverse respectively. If $x$ is a matrix then
$\mathrm{diag}\left(x\right)$ is the vector whose components are the diagonal
entries of $x$, whereas if $x$ is a vector then $\mathrm{diag}(x)$ is a
diagonal matrix with $x_{i}$ as its diagonal entries. Finally
$\textbf{1}\in\mathbb{C}^{3}$ is the column vector of size 3 whose entries are
all 1s and $\mathbb{I}$ is the identity matrix of size 3.

##  II Three-phase network model

A three-phase network connects generators to loads, each of which is modeled
by a single-terminal device. Each terminal has three wires (or ports or
conductors) indexed by its phases $a,b,c$, and possibly a neutral wire indexed
by $n$.111For notational simplicity, we assume all devices and lines have
three phases. Generalization to the case where some devices or lines have only
one or two phases is straightforward.  Internally, the device can be in $Y$ or
$\Delta$ configuration, and the $Y$ configuration may have a neutral wire that
may be grounded. A three-phase line has two terminals, each terminal with
three or four wires, and it connects two single-terminal devices, one at each
end of the line. Its neutral wire may be grounded at regular spacing along the
line. The overall network model consists of three components (see Figure 1):

  1. 1.

_Device model._ The internal behavior of a single-terminal device is defined
by the relationship between the voltages $V^{Y}/V^{\Delta}$, currents
$I^{Y}/I^{\Delta}$, and powers $s^{Y}/s^{\Delta}$ across each of the single-
phase devices that make up the three-phase device. This relationship is
independent of whether the device is in $Y$ or $\Delta$ configuration. The
configuration defines a conversion rule that maps internal variables
$\left(V^{Y/\Delta},I^{Y/\Delta},s^{Y/\Delta}\right)\in\mathbb{C}^{9}$ to
terminal voltages, currents, and powers $(V,I,s)\in\mathbb{C}^{9}$, regardless
of the type of the devices. The internal behavior and the conversion rule
jointly determine the external behavior of the three-phase device, i.e., the
relationship between the terminal variables $(V,I,s)$ that can be observed
externally.

  2. 2.

_Line model._ It relates the terminal voltages, line currents, and line power
flows $(V_{j},I_{jk},S_{jk})\in\mathbb{C}^{9}$ and
$(V_{k},I_{kj},S_{kj})\in\mathbb{C}^{9}$, at each end of the line $(j,k)$.

  3. 3.

_Network model._ It relates the terminal variables of all devices on the
network. This is defined by current or power balance at every bus in the
network.

Figure 1: Overall three-phase network model.

In this section we present circuit models of four types of single-terminal
devices, a voltage source, a current source, a power source, and an impedance.
Then we present a circuit model of a three-phase line. Finally we compose an
overall model of a network of such devices.

###  II-A Conversion matrices $\Gamma,\Gamma^{\sf T}$

We start by defining _conversion matrices_ $\Gamma,\Gamma^{\sf T}$ that maps
between internal and external variables in a $\Delta$ configuration:

$\displaystyle\Gamma$ $\displaystyle:=$ $\displaystyle\begin{bmatrix}1&-1&0\\\
0&1&-1\\\ -1&0&1\end{bmatrix}$ (1a) $\displaystyle\Gamma^{\sf T}$
$\displaystyle:=$ $\displaystyle\begin{bmatrix}1&0&-1\\\ -1&1&0\\\
0&-1&1\end{bmatrix}$ (1b) As we will see,

the spectral properties of $\Gamma,\Gamma^{\sf T}$ underlie much of the
behavior of three-phase systems, balanced or unbalanced. Here we recall some
basic facts on $\Gamma,\Gamma^{\sf T}$ that are useful in the rest of the
paper.

It can be shown that $\Gamma$ and $\Gamma^{\sf T}$ are normal matrices and
their spectral decompositions are

$\displaystyle\Gamma$ $\displaystyle\ \ =\ \
F\Lambda\overline{F},\qquad\qquad\Gamma^{\sf T}\ \ =\ \ \overline{F}\Lambda F$

where $\Lambda$ is a diagonal matrix and $F$ is a unitary matrix defined as:

$\displaystyle\Lambda$ $\displaystyle\ \ :=\ \ \begin{bmatrix}0&&\\\
&1-\alpha&\\\ &&1-\alpha^{2}\end{bmatrix},$ $\displaystyle F$ $\displaystyle\
\ :=\ \
\frac{1}{\sqrt{3}}\begin{bmatrix}\textbf{1}&\alpha_{+}&\alpha_{-}\end{bmatrix}$
(2)

with $\alpha:=e^{-\mathbf{i}2\pi/3}$. The positive-sequence and negative-
sequence vectors $\left(\alpha_{+},\alpha_{-}\right)$ are

$\displaystyle\alpha_{+}$ $\displaystyle:=$ $\displaystyle\begin{bmatrix}1\\\
\alpha\\\ \alpha^{2}\end{bmatrix},\qquad\qquad\alpha_{-}\ \ :=\ \
\begin{bmatrix}1\\\ \alpha^{2}\\\ \alpha\end{bmatrix}$

The eigenvectors of $\Gamma$ are $\textbf{1},\alpha_{+},\alpha_{-}$ and they
are orthogonal. Here $\overline{F}$ is the complex conjugate of $F$
componentwise. Since $F$ is symmetric, the pseudo inverses of
$\Gamma,\Gamma^{\sf T}$ are

$\displaystyle\Gamma^{\dagger}$ $\displaystyle\ \ =\ \
F\Lambda^{\dagger}\overline{F},\qquad\qquad\Gamma^{\sf T{\dagger}}\ \ =\ \
\overline{F}\Lambda^{\dagger}F$

where
$\Lambda^{\dagger}:=\mathrm{diag}\left(0,(1-\alpha)^{-1},(1-\alpha^{2})^{-1}\right)$.
This yields the following properties.

######  Lemma 1 (Pseudo inverses of $\Gamma,\Gamma^{\sf T}$).

  1. 1.

The null spaces of $\Gamma$ and $\Gamma^{\sf T}$ are both span$(\textbf{1})$.

  2. 2.

Their pseudo-inverses are

$\displaystyle\Gamma^{\dagger}\ \ =\ \ \frac{1}{3}\,\Gamma^{\sf
T},\qquad\quad\Gamma^{\sf T{\dagger}}\ \ =\ \ \frac{1}{3}\,\Gamma$

  3. 3.

Consider $\Gamma x=b$ where $b,x\in\mathbb{C}^{3}$. Solutions $x$ exist if and
only if $\textbf{1}^{\sf T}b=0$, in which case the solutions $x$ are given by

$\displaystyle x$ $\displaystyle=$ $\displaystyle\frac{1}{3}\,\Gamma^{\sf T}b\
+\ \gamma\textbf{1},\qquad\qquad\gamma\in\mathbb{C}$

  4. 4.

Consider $\Gamma^{\sf T}x=b$ where $b,x\in\mathbb{C}^{3}$. Solutions $x$ exist
if and only if $\textbf{1}^{\sf T}b=0$, in which case the solutions $x$ are
given by

$\displaystyle x$ $\displaystyle=$ $\displaystyle\frac{1}{3}\,\Gamma b\ +\
\gamma\textbf{1},\qquad\qquad\gamma\in\mathbb{C}$

  5. 5.

$\Gamma\Gamma^{\dagger}=\Gamma^{\dagger}\Gamma=\frac{1}{3}\,\Gamma\Gamma^{\sf
T}\ =\ \frac{1}{3}\,\Gamma^{\sf T}\Gamma\ \ =\ \ \mathbb{I}\ -\
\frac{1}{3}\,\textbf{1}\textbf{1}^{\sf T}$ where $\mathbb{I}$ is the identity
matrix of size 3.

In this paper, we will use Lemma 1 repeatedly, sometimes without explicit
reference.

###  II-B Devices: internal models and conversion rules

(a) $Y$ configuration (b) $\Delta$ configuration

Figure 2: Internal and external variables associated with a single-terminal
device in $Y$ and $\Delta$ configurations. (Even though Figure 2(a) shows a
grounded device, our model allows each $Y$-configured device to be grounded or
ungrounded and, if grounded, $z^{n}$ to be zero or nonzero.)

The internal behavior of a single-terminal device shown in Figure 2 is
described in terms of its internal variables:

  * •

$V^{Y}:=\left(V^{an},V^{bn},V^{cn}\right)\in\mathbb{C}^{3}$,
$I^{Y}:=\left(I^{an},I^{bn},I^{cn}\right)\in\mathbb{C}^{3}$,
$s^{Y}:=\left(s^{an},s^{bn},s^{cn}\right)\in\mathbb{C}^{3}$ : line-to-neutral
voltages, currents, and power across the single-phase devices in $Y$
configuration. By definition $s^{an}:=V^{an}\left(I^{an}\right)^{\sf H}$ is
the power across the phase-$a$ device, etc. The neutral voltage (with respect
to a common reference point) is denoted by $V^{n}$ and is generally nonzero. A
$Y$-configured device may or may not have a neutral line which may or may not
be grounded (Figure 2 shows the case where the device is grounded through an
impedance $z^{n}$). When present, the current on the neutral line is denoted
by $I^{n}$ in the direction away from the neutral.

  * •

$V^{\Delta}:=\left(V^{ab},V^{bc},V^{ca}\right)\in\mathbb{C}^{3}$,
$I^{\Delta}:=\left(I^{ab},I^{bc},I^{ca}\right)\in\mathbb{C}^{3}$,
$s^{\Delta}:=\left(s^{ab},s^{bc},s^{ca}\right)\in\mathbb{C}^{3}$ : line-to-
line voltages, currents, and power across the single-phase devices in $\Delta$
configuration. By definition $s^{ab}:=V^{ab}\left(I^{ab}\right)^{\sf H}$ is
the power across the phase-$a$ device, etc.

Note that the direction of power $s^{an}$ or $s^{ab}$ across a single-phase
device is defined in the _direction of the current through_ the device.

#### Internal models.

The internal behavior of a three-phase device is described by the relation
between the internal variables $\left(V^{Y},I^{Y},s^{Y}\right)$ or between
$\left(V^{\Delta},I^{\Delta},s^{\Delta}\right)$. It depends on the property of
the single-phase device, but not on their configuration nor the presence of a
neutral line:

  1. 1.

_Voltage source:_ An ideal voltage source fixes the internal voltage
$V^{Y/\Delta}$ to be a given constant $E^{Y/\Delta}$.

  2. 2.

_Current source:_ An ideal current source fixes the internal current
$I^{Y/\Delta}$ to be a given constant $J^{Y/\Delta}$.

  3. 3.

_Power Source:_ An ideal power source fixes the internal power $s^{Y/\Delta}$
to be a given constant $\sigma^{Y/\Delta}$.

  4. 4.

_Impedance:_ An impedance $z^{Y/\Delta}$ is a $3\times 3$ complex matrix. It
fixes the relationship between its internal variables to be:

$\displaystyle V^{Y/\Delta}$ $\displaystyle\ \ =\ \
z^{Y/\Delta}\,I^{Y/\Delta},$ $\displaystyle s^{Y/\Delta}$ $\displaystyle\ \
:=\ \ \mathrm{diag}\left(V^{Y/\Delta}\left(I^{Y/\Delta}\right)^{\sf H}\right)$

The voltage and current $\left(V^{Y/\Delta},I^{Y/\Delta}\right)$ of a power
source $\sigma^{Y/\Delta}$ are related quadratically by

$\displaystyle\sigma^{Y/\Delta}$ $\displaystyle\ \ =\ \
\mathrm{diag}\left(V^{Y/\Delta}\left(I^{Y/\Delta}\right)^{\sf H}\right)$

The specification as well as internal and external variables of a three-phase
device are summarized in Table I.

| Voltage | Current | Power | Impedances
---|---|---|---|---
Device specification | $E^{Y/\Delta}$ | $J^{Y/\Delta}$ | $\sigma^{Y/\Delta}$ |  $z^{Y/\Delta}$, $y^{Y/\Delta}$, $z^{n}$
Internal variables | $V^{Y/\Delta}$ | $I^{Y/\Delta}$ | $s^{Y/\Delta}$ |
External variables | $V$ | $I$ | $s$ |
TABLE I: Notation for three-phase single-terminal devices
$(y^{Y/\Delta}:=\left(z^{Y/\Delta}\right)^{-1})$.

As noted above, the internal model does not depend on the configuration nor
the presence of a neutral line.

#### Conversion rules.

The external behavior of the single-terminal device shown in Figure 2 is
described in terms of its terminal variables:

  * •

$V:=\left(V^{a},V^{b},V^{c}\right)\in\mathbb{C}^{3}$,
$I:=\left(I^{a},I^{b},I^{c}\right)\in\mathbb{C}^{3}$,
$s:=\left(s^{a},s^{b},s^{c}\right)\in\mathbb{C}^{3}$ : terminal voltages,
currents, and power. The terminal voltage $V$ is defined with respect to an
arbitrary but common reference point, e.g., the ground. The terminal current
$I$ is defined in the direction coming out of the device, i.e., $I$ is defined
to be the current injection from the device to the rest of the network when it
is connected to a bus bar, regardless of whether it generates or consumes
power. By definition $s^{a}:=V^{a}\,I^{a\sf H}$ is the power across terminal
$a$ and the common reference point, etc.

The external model of a device is the relationship between its terminal
variables $(V,I,s)$. We now derive the conversion rules (3) and (II-B) that
maps internal variables $\left(V^{Y/\Delta},I^{Y/\Delta},s^{Y/\Delta}\right)$
to external variables $(V,I,s)$ for devices in $Y$ and $\Delta$ configurations
respectively. These conversion rules depend only on the configuration and not
on the type of devices. In Section II-C, we apply these conversion rules to
the internal model of each device to derive its external model.

_Conversion in $Y$ configuration._ The terminal voltage, current, and power
$(V,I,s)$ of a $Y$-configured device are related to the internal variables
$(V^{Y},I^{Y},s^{Y})$ by:

$\displaystyle V$ $\displaystyle\ =\ V^{Y}\ +\ V^{n}\textbf{1},$
$\displaystyle I$ $\displaystyle\ =\ -I^{Y},$ $\displaystyle s$
$\displaystyle\ =\ -\left(s^{Y}\ +\ V^{n}\overline{I}^{Y}\right)$ (3)

where $\overline{I}$ denotes the componentwise complex conjugate of the vector
$I^{Y}\in\mathbb{C}^{3}$. The negative sign on the current and power
conversions is due to the definition of $\left(I^{Y},s^{Y}\right)$ as internal
current and power delivered to the single-phase devices whereas $(I,s)$ is
defined as the terminal current and power injections out of the three-phase
device.

In general the neutral voltage $V^{n}$ with respect to a common reference
point is nonzero whether or not there is a neutral line and whether or not the
neutral is grounded. If the neutral is grounded with zero neutral impedance
and voltages are defined with respect to the ground, then $V^{n}=0$ and
$V=V^{Y}$ and $s=-s^{Y}$. It is important to explicitly include $V^{n}$ in a
network model because not every device in a network may be grounded or
grounded with zero neutral impedance.

######  Remark 1 (Total power).

The total terminal power is

$\displaystyle\textbf{1}^{\sf T}s$ $\displaystyle\ \ =\ \ -\textbf{1}^{\sf
T}s^{Y}\ -\ V^{n}\left(\textbf{1}^{T}\overline{I}^{Y}\right)$

The first term $\textbf{1}^{\sf T}s^{Y}$ is the total power delivered across
the single-phase devices. The second term $\textbf{1}^{T}I^{Y}$ is the sum of
internal line-to-neutral current. If the neutral is grounded through an
impedance then $V^{n}\left(\textbf{1}^{T}\overline{I}^{Y}\right)$ is the power
delivered to the neutral impedance. If the neutral is ungrounded then
$\textbf{1}^{T}I^{Y}=0$ by KCL and the second term
$V^{n}\left(\textbf{1}^{T}\overline{I}^{Y}\right)=0$. ∎

_Conversion in $\Delta$ configuration._ The relationship between terminal
voltage and current $(V,I)$ and internal voltage and current
$\left(V^{\Delta},I^{\Delta}\right)$ is:

$\displaystyle V^{\Delta}$ $\displaystyle\ \ =\ \ \Gamma\,V,\qquad\quad I\ \
=\ \ -\Gamma^{\sf T}\,I^{\Delta}$ (4a) Given appropriate vectors $V^{\Delta}$
and $I$, solutions $V$ and $I^{\Delta}$ to (4a) is provided by Lemma 1.

  1. 1.

Given $V^{\Delta}$, there is a solution $V$ to (4a) if and only if
$V^{\Delta}$ is orthogonal to 1, i.e.,

$\displaystyle V^{ab}+V^{bc}+V^{ca}$ $\displaystyle=$ $\displaystyle 0$

which expresses Kirchhoff’s voltage law. In that case, there is a subspace of
solutions $V$ given by

$\displaystyle V$ $\displaystyle=$ $\displaystyle\frac{1}{3}\,\Gamma^{\sf
T}V^{\Delta}\ +\ \gamma\,\textbf{1},\quad\gamma\in\mathbb{\mathbb{missing}}C$
(4b)

This amounts to an arbitrary reference voltage for $V$. The quantity
$\gamma:=\frac{1}{3}\textbf{1}^{\sf T}V$ is the (scaled) zero-sequence voltage
of $V$. In most applications we are given a reference voltage (e.g.,
$V_{0}:=\alpha_{+}$ at the reference bus 0) which will fix the constant
$\gamma$.

  2. 2.

Given $I$, there is a solution $I^{\Delta}$ to (4a) if and only if $I$ is
orthogonal to 1, i.e.,

$\displaystyle I^{a}+I^{b}+I^{c}$ $\displaystyle=$ $\displaystyle 0$

which expresses Kirchhoff’s current law. In that case, there is a subspace of
$I^{\Delta}$ that satisfy (4a), given by

$\displaystyle I^{\Delta}$ $\displaystyle=$ $\displaystyle-\frac{1}{3}\,\Gamma
I\ +\ \beta\,\textbf{1},\quad\beta\in\mathbb{C}$ (4c)

where $\beta$ specifies the amount of loop flow in $I^{\Delta}$ and does not
affect the terminal current $I$ since $\Gamma^{\sf T}\beta\textbf{1}=0$. The
quantity $\beta:=\frac{1}{3}\textbf{1}^{\sf T}I^{\Delta}$ is the (scaled)
zero-sequence current of $I^{\Delta}$.

The terminal power injection from the device is $s:=\mathrm{diag}\left(VI^{\sf
H}\right)$ and the internal power delivered across the single-phase devices in
the direction $ab$, $bc$, $ca$ is
$s^{\Delta}:=\mathrm{diag}\left(V^{\Delta}I^{\Delta\sf H}\right)$. Given
internal voltage and current $\left(V^{\Delta},I^{\Delta}\right)$ with
$\textbf{1}^{\sf T}V^{\Delta}=0$, the terminal power $s$ is (from (4a) (4b)):

$\displaystyle s$ $\displaystyle\ =\
-\mathrm{diag}\left(\Gamma^{\dagger}\left(V^{\Delta}I^{\Delta\sf
H}\right)\Gamma\right)+\,\gamma\overline{I},$ $\displaystyle\textbf{1}^{\sf
T}V^{\Delta}$ $\displaystyle\ =\ 0$ (4d)

where $\overline{I}$ is the complex conjugate of the terminal current
$I=-\Gamma^{\sf T}I^{\Delta}$ and $\gamma\in\mathbb{C}$ is determined by a
reference voltage. Conversely, given terminal voltage and current $(V,I)$ with
$\textbf{1}^{\sf T}I=0$, the internal power $s^{\Delta}$ is (from (4a) (4c)):

$\displaystyle s^{\Delta}$ $\displaystyle\
=\\-\mathrm{diag}\left(\Gamma\left(VI^{\sf
H}\right)\Gamma^{\dagger}\right)+\,\beta V^{\Delta},$
$\displaystyle\textbf{1}^{\sf T}I$ $\displaystyle\ =\ 0$ (4e)

where $V^{\Delta}=\Gamma V$ and $\beta\in\mathbb{C}$ is determined by the
zero-sequence current of $I^{\Delta}$.

######  Remark 2 (Total power).

  1. 1.

Given an internal voltage and current $\left(V^{\Delta},I^{\Delta}\right)$,
the terminal power vector $s$ does not depend on the zero-sequence current
$\beta:=\frac{1}{3}\textbf{1}^{\sf T}I^{\Delta}$ but does depend on the zero-
sequence voltage $\gamma:=\frac{1}{3}\textbf{1}^{\sf T}V$. Since
$I=-\Gamma^{\sf T}I^{\Delta}$ and hence $\textbf{1}^{\sf T}I=0$, the total
terminal power however is independent of $\gamma$:

$\displaystyle\textbf{1}^{\sf T}\,s$ $\displaystyle\ \ =\ \ -\textbf{1}^{\sf
T}\mathrm{diag}\left(\Gamma^{\dagger}\left(V^{\Delta}I^{\Delta\sf
H}\right)\Gamma\right)$

  2. 2.

Given a terminal voltage and current $(V,I)$, from (4e), the internal power
vector $s^{\Delta}$ depends on zero-sequence current $\beta$. Since
$V^{\Delta}=\Gamma\,V$ and hence $\textbf{1}^{\sf T}V^{\Delta}=0$, the total
internal power however is independent of the loop flow:

$\displaystyle\textbf{1}^{\sf T}s^{\Delta}$ $\displaystyle=$
$\displaystyle-\textbf{1}^{\sf T}\mathrm{diag}\left(\Gamma\left(VI^{\sf
H}\right)\Gamma^{\dagger}\right)$

∎

In summary a complete model of a three-phase device is given by its internal
model specifying the relationship among its internal variables
$\left(V^{Y/\Delta},I^{Y/\Delta},s^{Y/\Delta}\right)$ and the conversion rules
(3) and (II-B) between its internal variables and external variables
$(V,I,s)$, for $Y$ and $\Delta$ configuration respectively. This model is
required to fully specify a network model (see below) when the application
under study needs to determine or optimize some of the internal variables such
as the current $I_{j}^{Y/\Delta}$ or power $s_{j}^{Y/\Delta}$ of each of the
single-phase devices connected at a bus $j$.

When the application does not require internal variables, we can apply the
conversion rules (3) (II-B) to the internal models to eliminate the internal
variables and obtain a relationship between the external variables $(V,I,s)$
in terms of device parameters, such as $E^{Y/\Delta}$ for an ideal voltage
source or $z^{Y/\Delta}$ of an impedance, as we explain next.

###  II-C Devices: external models

Since we do not need power sources in this paper, to save space, we omit the
derivation of their external model.

_$Y$ configuration._ Application of the conversion rule (3) to the internal
models of a voltage source, a current source, and an impedance yields the
external models that relate their terminal variables. The result is summarized
in Table II.

Device |  $Y$ configuration
---|---
Voltage source | $V=E^{Y}+\gamma\textbf{1}$ | $s=\mathrm{diag}\left(E^{Y}I^{\sf H}\right)+\gamma\overline{I}$
Current source | $\ I=-J^{Y}$ | $s=-\mathrm{diag}\left(VJ^{Y\sf H}\right)$
Power source | $\mathrm{diag}\left(I^{\sf H}\right)(V-\gamma\textbf{1})=-\sigma$ | $s=-\sigma^{Y}+\gamma\overline{I}$
Impedance | $V=-z^{Y}I+\gamma\textbf{1}$ | $s=-\mathrm{diag}\left(V\left(V-\gamma\textbf{1}\right)^{\sf H}y^{Y\sf H}\right)$
TABLE II: External models of ideal single-terminal devices in $Y$
configuration $(\gamma=V^{n})$.

In the table $\gamma=V^{n}$ is the neutral voltage. These models for ideal
sources do not rely on a common and often implicit assumption that all
neutrals are grounded through an impedance $z^{n}\in\mathbb{C}$, which may or
may not be zero, and voltages are defined with respect to the ground. If this
assumption holds then $\gamma=V^{n}=-z^{n}\left(\textbf{1}^{\sf T}I\right)$ by
KCL. If, in addition, all neutrals are directly grounded, i.e., $z^{n}=0$,
then $\gamma=V^{n}=0$ for all $Y$-configured devices. For a typical three-
phase analysis problem, $\gamma$ for all $Y$-configured device needs to be
specified (see Remark 5).

_$\Delta$ configuration._ The external models of $\Delta$-configured devices
can be derived by applying the conversion rule (II-B) to their internal
models.

  1. 1.

_Voltage source $\left(E^{\Delta},\gamma\right)$:_ Applying the conversion
rule $V^{\Delta}=\Gamma\,V$ in (4a) to the internal model
$V^{\Delta}=E^{\Delta}$ of an ideal voltage source, we obtain the following
external model that relate the terminal voltage, current and power $(V,I,s)$:

$\displaystyle V$ $\displaystyle\ =\ \frac{1}{3}\Gamma^{\sf T}E^{\Delta}\ +\
\gamma\textbf{1},\quad\textbf{1}^{\sf T}I\ =\ 0$ (5a) $\displaystyle s$
$\displaystyle\ =\ \frac{1}{3}\mathrm{diag}\left(\Gamma^{\sf
T}E^{\Delta}I^{\sf H}\right)\ +\ \gamma\overline{I}$ (5b) provided
$\textbf{1}^{\sf T}E^{\Delta}=0$,

where $\gamma\in\mathbb{C}$ is fixed by a given reference voltage. To specify
the external model of an ideal voltage source is to fix the two parameters
$\left(E^{\Delta},\gamma\right)$. Its terminal current and power $(I,s)$ will
be determined by the interaction of its external model (5) with those of other
devices on the network through current or power balance equations.

  2. 2.

_Current source $J^{\Delta}$:_ Multiplying $-\Gamma^{\sf T}$ to both sides of
the internal model $I^{\Delta}=J^{\Delta}$ of an ideal current source and
applying the conversion rule $I=-\Gamma^{\sf T}I^{\Delta}$ in (4a), we obtain
the external model:

$\displaystyle I$ $\displaystyle\ =\ -\Gamma^{\sf T}\\!J^{\Delta},$
$\displaystyle s$ $\displaystyle\ =\ -\mathrm{diag}\left(VJ^{\Delta\sf
H}\Gamma\right)$ (6)

To specify the external model of an ideal current source is to fix the
internal current $J^{\Delta}$ (which also fixes its zero-sequence current
$\beta:=\frac{1}{3}\textbf{1}^{\sf T}J^{\Delta}$). Its terminal voltage and
power $(V,s)$ will be determined by the interaction of its external model (6)
with those of other devices on the network through current or power balance
equations.

  3. 3.

_Impedance $z^{\Delta}$:_ Define the admittance matrix
$y^{\Delta}:=\left(z^{\Delta}\right)^{-1}$. Substituting into the internal
model $y^{\Delta}V^{\Delta}=I^{\Delta}$ of an impedance, multiplying both
sides by $-\Gamma^{\sf T}$ and applying the conversion rule $I=-\Gamma^{\sf
T}I^{\Delta}$, we get

$\displaystyle I$ $\displaystyle\ \ =\ \ -{Y^{\Delta}}V$ (7a) where
$Y^{\Delta}$ is a complex symmetric Laplacian matrix given by $\displaystyle
Y^{\Delta}\ :=\ \Gamma^{\sf T}\,y^{\Delta}\,\Gamma\ =\
\begin{bmatrix}y^{ca}+y^{ab}&-y^{ab}&-y^{ca}\\\
-y^{ab}&y^{ab}+y^{bc}&-y^{bc}\\\ -y^{ca}&-y^{bc}&y^{bc}+y^{ca}\end{bmatrix}$
Note that the terminal current $I$ given by (7a) satisfies $\textbf{1}^{\sf
T}I=0$. The terminal power injection $s$ can be expressed in terms of $V$:
$\displaystyle s$ $\displaystyle\ \ =\ \ \mathrm{diag}\left(VI^{\sf H}\right)\
\ =\ \ -\mathrm{diag}\left(VV^{\sf H}Y^{\Delta\sf H}\right)$ (7b)

The external models (5) (6) (7) of ideal $\Delta$-configured devices are
summarized in Table III.

Device |  $\Delta$ configuration
---|---
Voltage source |  $V=\frac{1}{3}\Gamma^{\sf T}E^{\Delta}+\gamma\textbf{1}$, $\textbf{1}^{\sf T}I=0$ | $s=\frac{1}{3}\mathrm{diag}\left(\Gamma^{\sf T}E^{\Delta}I^{\sf H}\right)+\gamma\overline{I}$
Current source | $I=-\Gamma^{\sf T}J^{\Delta}$ | $s=-\mathrm{diag}\left(VJ^{\Delta\sf H}\Gamma\right)$
Power source | $\sigma^{\Delta}=\mathrm{diag}\left(\Gamma VI^{\Delta\sf H}\right)$ |
Impedance | $I=-Y^{\Delta}V$ | $s=-\mathrm{diag}\left(VV^{\sf H}Y^{\Delta\sf H}\right)$
TABLE III: External models of ideal single-terminal devices in $\Delta$
configuration $\left(\gamma:=\frac{1}{3}\textbf{1}^{\sf
T}V,\beta:=\frac{1}{3}\textbf{1}^{\sf T}I^{\Delta}\right)$.

######  Remark 3 (Non-ideal devices).

For simplicity of exposition, we have presented in this paper the external
models of only ideal devices where the internal series impedances of voltage
sources and shunt admittances of current sources are assumed zero. These
models can be extended to non-ideal devices (see [23]). ∎

######  Remark 4 ($\Delta$-$Y$ transformation).

From the external model (5) of an ideal $\Delta$-configured voltage source and
that of an $Y$-configured voltage source in Table II, the $Y$ equivalent of
$\left(E^{\Delta},\gamma\right)$, _not_ necessarily balanced, is given by

$\displaystyle E^{Y}$ $\displaystyle\ \ :=\ \ \frac{1}{3}\Gamma^{\sf
T}E^{\Delta},$ $\displaystyle V^{n}$ $\displaystyle\ \ :=\ \ \gamma$

If $E^{\Delta}$ is balanced then $\Gamma^{\sf
T}\,E^{\Delta}=(1-\alpha^{2})E^{\Delta}={\sqrt{3}}\,e^{-\mathbf{i}\pi/6}\,E^{\Delta}$
and the $Y$ equivalent $E^{Y}$ reduces to the familiar expression:

$\displaystyle E^{Y}$ $\displaystyle\ \ =\ \
\frac{1}{\sqrt{3}\,e^{\mathbf{i}{\pi}/6}}\,E^{\Delta}$

Similarly an ideal $\Delta$-configured current source $J^{\Delta}$ has an $Y$
equivalent $J^{Y}$ given by

$\displaystyle J^{Y}$ $\displaystyle\ \ =\ \ -\Gamma^{\sf T}J^{\Delta}$

If $J^{\Delta}$ is balanced then

$\displaystyle J^{Y}$ $\displaystyle\ \ =\ \ -(1-\alpha^{2})J^{\Delta}\ \ =\ \
-\frac{\sqrt{3}}{e^{\mathbf{i}{\pi}/6}}\,J^{\Delta}$

∎

###  II-D Three-phase line model

A three-phase line has three wires one for each phase $a,b,c$. It may also
have a neutral wire which may be grounded at one or both ends if the device
connected to that end of the line is in $Y$ configuration. The electromagnetic
interactions among the electric charges in wires of different phases couple
the voltages on and currents in these wires. The relation between the voltages
and currents in these phases can be modeled by a linear mapping that depends
on the line characteristics. For simplicity we will restrict ourselves to a
three-wire line model that takes into account the effect of neutral or earth
return on the impedance of a transmission line. All analysis extends to four-
wire models (including a neutral line) or five-wire models (including a
neutral line and the ground return) almost without change with proper
definitions that include neutral and ground variables.

A three-phase line $(j,k)$ is characterized by three $3\times 3$ matrices
$\left(y_{jk}^{s},y_{jk}^{m},y_{kj}^{m}\right)$ where $y_{jk}^{s}$ is the
_series admittance matrix_ and $\left(y^{m}_{jk},y^{m}_{kj}\right)$ are the
_shunt admittance matrices_ , not necessarily equal. The terminal voltages
$\left(V_{j},V_{k}\right)$ and the sending-end currents
$\left(I_{jk},I_{kj}\right)$ respectively are related according to

$\displaystyle I_{jk}$ $\displaystyle=$ $\displaystyle
y^{s}_{jk}\left(V_{j}-V_{k}\right)\ +\ y^{m}_{jk}\,V_{j}$ (8a) $\displaystyle
I_{kj}$ $\displaystyle=$ $\displaystyle y^{s}_{jk}\left(V_{k}-V_{j}\right)\ +\
y^{m}_{kj}\,V_{k}$ (8b) Note that the voltages $(V_{j},V_{k})$ and currents
$(I_{jk},I_{kj})$ are terminal voltages and currents regardless of whether the
three-phase devices connected to terminals $j$ and $k$ are in $Y$ or $\Delta$
configuration.

To describe the relationship between the sending-end line power and the
voltages $\left(V_{j},V_{k}\right)$, define the matrices
$S_{jk},S_{kj}\in\mathbb{C}^{3\times 3}$ by

$\displaystyle S_{jk}$ $\displaystyle\ \ :=\ \ V_{j}\left(I_{jk}\right)^{\sf
H},$ $\displaystyle S_{kj}$ $\displaystyle\ \ :=\ \
V_{k}\left(I_{kj}\right)^{\sf H}$ (8c)

The three-phase sending-end line power from terminals $j$ to $k$ along the
line is the vector $\mathrm{diag}\left(S_{jk}\right)$ of diagonal entries and
that in the opposite direction is the vector
$\mathrm{diag}\left(S_{kj}\right)$. The off-diagonal entries of these matrices
represent electromagnetic coupling between phases.

###  II-E Network model

Let
$(V,I,s):=\left(V_{j},I_{j},s_{j},j\in\overline{N}\right)\in\mathbb{C}^{3(N+1)}$
be terminal (nodal) variables over the entire network. A network equation is a
relationship between the terminal voltage and current $(V,I)$ or a
relationship between the terminal voltage and power $(V,s)$, independent of
the internal $Y$ or $\Delta$ configurations of the three-phase devices that
are connected by the lines. In both cases the extension of the line model
(II-D) to a network is simply the nodal current or power balance equations:

$\displaystyle I_{j}$ $\displaystyle\ \ =\ \ \sum_{k:j\sim k}I_{jk},$
$\displaystyle s_{j}$ $\displaystyle\ \ =\ \ \sum_{k:j\sim
k}\mathrm{diag}\left(S_{jk}\right),$ $\displaystyle j$ $\displaystyle\ \ \in\
\ \overline{N}$

where $S_{jk}$ are matrices defined in (8c). In this paper we focuses on the
current balance equation which, using (8a), is:

$\displaystyle I_{j}$ $\displaystyle=$ $\displaystyle\sum_{k:j\sim
k}\left(y^{s}_{jk}+{y_{jk}^{m}}\right)V_{j}-\sum_{k:j\sim
k}y^{s}_{jk}V_{k},\quad j\in\overline{N}$ (9a) Note that $I_{j}$ is the net
current injection.222If there is a nodal shunt admittance load
$y_{j}^{\text{sh}}$, e.g., a capacitor bank, in addition to a device whose
terminal injection is $\tilde{I}_{j}$, then the net injection from bus $j$ to
the rest of the network is $I_{j}=\tilde{I}_{j}-y_{j}^{\text{sh}}V_{j}$. This
assumes that $y_{j}^{\text{sh}}$ connects bus $j$ to the ground and the
terminal voltage $V_{j}$ is defined with respect to the ground. In vector
form, this relates the bus current vector $I:=(I_{0},\dots,I_{N})$ to the bus
voltage vector $V:=(V_{0},\dots,V_{N})$: $\displaystyle I$ $\displaystyle=$
$\displaystyle YV$ (9b) in terms of a $3(N+1)\times 3(N+1)$ admittance matrix
$Y$ where its $3\times 3$ submatrices are given by $\displaystyle Y_{jk}$
$\displaystyle=$ $\displaystyle\left\\{\begin{array}[]{lcl}-y^{s}_{jk},&&j\sim
k\ \ (j\neq k)\\\ \sum_{l:j\sim l}\,y^{s}_{jl}\ +\ y_{jj}^{m},&&j=k\\\
0&&\text{otherwise}\end{array}\right.$ (9f)

An overall network model consists of (see Figure 1):

  1. 1.

A network model (9) that relates terminal voltage and current $(V,I)$.

  2. 2.

A device model for each three-phase device $j$. This can either be:

     * •

An internal model together with the conversion rules (3)(II-B) in Section
II-B; or

     * •

An external model summarized in Tables II and III in Section II-C when only
terminal quantities are needed.

##  III Three-phase analysis

We now formulate a general three-phase analysis problem using the overall
model of Section II. Consider a three-phase network $G:=(\overline{N},E)$
where each line $(j,k)\in E$ is characterized by $3\times 3$ series and shunt
admittance matrices $\left(y_{jk}^{s},y_{jk}^{m},y_{kj}^{m}\right)$. At each
bus $j\in\overline{N}$ we assume, without loss of generality, there is a
single three-wire device in either $Y$ or $\Delta$ configuration.

_Three-phase devices._ Partition $\overline{N}$ into 6 disjoint subsets:

  * •

$N_{v}^{Y/\Delta}$: buses with ideal voltage sources in $Y$ or $\Delta$
configurations.

  * •

$N_{c}^{Y/\Delta}$: buses with ideal current sources in $Y$ or $\Delta$
configurations.

  * •

$N_{i}^{Y/\Delta}$: buses with impedances in $Y$ or $\Delta$ configurations.

Buses $j$ | Specification | External model | Unknowns
---|---|---|---
$N_{v}^{Y}$ |  ${\color[rgb]{0,0,1}{V_{j}^{Y}:=E_{j}^{Y}}}$, ${\color[rgb]{0,0,1}{\gamma_{j}}}$ | $V_{j}\ =\ E_{j}^{Y}+\gamma_{j}\textbf{1}$ | $\left(I_{j},I_{j}^{Y}\right)$
$N_{v}^{\Delta}$ |  ${\color[rgb]{0,0,1}{V_{j}^{\Delta}:=E_{j}^{\Delta}}}$, $\gamma_{j},\beta_{j}$, | $V_{j}\ =\ \Gamma^{\dagger}E_{j}^{\Delta}+\gamma_{j}\textbf{1}$ | $\left(I_{j},I_{j}^{\Delta}\right)$
$N_{c}^{Y}$ | ${\color[rgb]{0,0,1}{I_{j}^{Y}:=J_{j}^{Y},\gamma_{j}}}$ | $I_{j}\ =\ -J_{j}^{Y}$ | $\left(V_{j},V_{j}^{Y}\right)$
$N_{c}^{\Delta}$ | ${\color[rgb]{0,0,1}{I_{j}^{\Delta}:=J_{j}^{\Delta}}}$ | $I_{j}\ =\ -\Gamma^{\sf T}J_{j}^{\Delta}$ | $\left(V_{j},V_{j}^{\Delta}\right),\left(\gamma_{j},\beta_{j}\right)$
$N_{i}^{Y}$ | ${\color[rgb]{0,0,1}{z_{j}^{Y},\ \gamma_{j}}}$ | $I_{j}\ =\ -{\color[rgb]{0,0,1}{y_{j}^{Y}}}\left(V_{j}-{\color[rgb]{0,0,1}{\gamma_{j}}}\textbf{1}\right)$ | $\left(V_{j},I_{j}\right),\left(V_{j}^{Y},I_{j}^{Y}\right)$
$N_{i}^{\Delta}$ | $z_{j}^{\Delta},\ \beta_{j}$ | $I_{j}\ =\ -{\color[rgb]{0,0,1}{Y_{j}^{\Delta}}}V_{j}$ | $\left(V_{j},I_{j}\right),\left(V_{j}^{\Delta},I_{j}^{\Delta}\right),\gamma_{j}$
TABLE IV: Internal and external models of three-phase sources and impedances
from Tables II and III. The three-phase analysis problem is: given the
specification in blue, compute the remaining unknowns in black.

###  III-A Device specification

Associated with each device $j$ are the internal variables
$\left(V_{j}^{Y/\Delta},I_{j}^{Y/\Delta},s_{j}^{Y/\Delta}\right)\in\mathbb{C}^{9}$,
the terminal variables $\left(V_{j},I_{j},s_{j}\right)\in\mathbb{C}^{9}$, and
the variables $\left(\gamma_{j},\beta_{j}\right)\in\mathbb{C}^{2}$. Some of
these variables are specified in a three-phase analysis problem and the others
are computed from network equations and conversion rules. We now describe
which of these variables are specified for each device in a typical three-
analysis problem (without constant-power devices). The result is summarized in
Table IV. These requirements may need to be modified depending on the details
of a problem.

  1. 1.

_Voltage source $j$_: It is specified by its internal voltage and a parameter
$\left(E_{j}^{Y/\Delta},\gamma_{j}\right)$ where $\gamma_{j}:=V_{j}^{n}$ is
the neutral voltage if $j$ is in $Y$ configuration and
$\gamma_{j}:=\frac{1}{3}\textbf{1}^{\sf T}V_{j}$ is the zero-sequence
component of the terminal voltage if $j$ is in $\Delta$ configuration. For a
$\Delta$-configured voltage source, the zero-sequence current $\beta_{j}$ also
needs to be specified in order to determine the internal current
$I_{j}^{\Delta}$ from the terminal current $I_{j}$.

  2. 2.

_Current source $j$_: It is specified by its internal current
$J_{j}^{Y/\Delta}$. For a $Y$-configured current source, its neutral voltage
$\gamma_{j}$ is also specified.

  3. 3.

_Impedance $j$_: A $Y$-configured impedance $j$ is specified by its internal
impedance $z_{j}^{Y}$ and the neutral voltage $\gamma_{j}:=V_{j}^{n}$. A
$\Delta$-configured impedance $j$ is specified by $z_{j}^{\Delta}$ and its
zero-sequence current $\beta_{j}$.

The specification of each device also comes with an external model that
relates its terminal variables in terms of the specified parameters, as shown
in Table IV.

###  III-B Analysis problem

A three-phase analysis problem is: given devices specified as in Table IV
connected by three-phase lines with given admittance matrices
$\left(y_{jk}^{s},y_{jk}^{m},y_{jk}^{m}\right)$, compute the remaining
unknowns for each bus $j$ listed in the last column of the table. The general
solution strategy is to use the external models in Table IV and the network
equation $I=YV$ in (9) to compute terminal voltages and currents
$\left(V_{j},I_{j}\right)$. Internal variables
$\left(V_{j}^{Y/\Delta},I_{j}^{Y/\Delta}\right)$ as well as
$\left(\gamma_{j},\beta_{j}\right)$ can then be determined by the conversion
rules.

Specifically let $N_{v}:=N_{v}^{Y}\cup V_{v}^{\Delta}$, $N_{c}:=N_{c}^{Y}\cup
V_{c}^{\Delta}$, and $N_{i}:=N_{i}^{Y}\cup V_{i}^{\Delta}$ be the set of buses
with, respectively, voltage sources, current sources, and impedances. With a
slight abuse of notation define the following (column) vectors of terminal
voltages and currents:

$\displaystyle\left(V_{v},I_{v}\right)$ $\displaystyle\ \ :=\ \
\left(V_{j},I_{j},\,j\in N_{v}\right)$ $\displaystyle\left(V_{c},I_{c}\right)$
$\displaystyle\ \ :=\ \ \left(V_{j},I_{j},\,j\in N_{c}\right)$
$\displaystyle\left(V_{i},I_{i}\right)$ $\displaystyle\ \ :=\ \
\left(V_{j},I_{j},\,j\in N_{i}\right)$

Then $I=YV$ becomes

$\displaystyle\begin{bmatrix}I_{v}\\\ I_{c}\\\ I_{i}\end{bmatrix}$
$\displaystyle\ \ =\ \ \underbrace{\begin{bmatrix}Y_{vv}&Y_{vc}&Y_{vi}\\\
Y_{cv}&Y_{cc}&Y_{ci}\\\
Y_{iv}&Y_{ic}&Y_{ii}\end{bmatrix}}_{Y}\begin{bmatrix}V_{v}\\\ V_{c}\\\
V_{i}\end{bmatrix}$ (10)

where the admittance matrix $Y$ is defined in (9). The three-phase network
analysis problem is then:

  1. 1.

Given the specification of voltage sources, current sources and impedances in
Table IV, solve (10) for the terminal voltage
$V_{-v}:=\left(V_{c},V_{i}\right)$ and current
$I_{-c}:=\left(I_{v},I_{i}\right)$.

  2. 2.

From the terminal voltage $V:=\left(V_{v},V_{c},V_{i}\right)$ and current
$I:=\left(I_{v},I_{c},I_{i}\right)$, the internal voltages, currents and
powers of $Y$-configured devices are obtained from the conversion rule (3)
$\left(\gamma_{j}=V_{j}^{n}\right)$:

$\displaystyle V_{j}^{Y}$ $\displaystyle\ \ =\ \ V_{j}\ -\
\gamma_{j}\textbf{1},$ $\displaystyle I_{j}^{Y}$ $\displaystyle\ =\ -I_{j}$
$\displaystyle s_{j}^{Y}$ $\displaystyle\ \ =\ \
\mathrm{diag}\left(V_{j}I_{j}^{\sf H}\right),$ $\displaystyle j$
$\displaystyle\ \in\ N_{v}^{Y}\cup N_{c}^{Y}\cup N_{i}^{Y}$

  3. 3.

Those of $\Delta$-configured devices can be computed by applying the
conversion rule (II-B) to $(V,I)$:

$\displaystyle V_{j}^{\Delta}$ $\displaystyle\ \ =\ \ \Gamma V_{j},$
$\displaystyle I_{j}^{\Delta}$ $\displaystyle\ =\ -\Gamma^{\sf
T{\dagger}}I_{j}\ +\ \beta_{j}\textbf{1}$ $\displaystyle s_{j}^{\Delta}$
$\displaystyle\ \ =\ \ \mathrm{diag}\left(V_{j}^{\Delta}I_{j}^{\Delta\sf
H}\right),$ $\displaystyle j$ $\displaystyle\ \in\ N_{v}^{\Delta}\cup
N_{c}^{\Delta}\cup N_{i}^{\Delta}$

The main task is to solve the network equation (10) in Step 1 for terminal
voltage and current $(V,I)$ (see [23] for more details). This generally can
only be solved numerically.

The result of the analysis determines both internal and terminal variables
$\left(V_{j}^{Y/\Delta},I_{j}^{Y/\Delta},s_{j}^{Y/\Delta}\right)$,
$\left(V_{j},I_{j},s_{j}\right)$ and $\left(\gamma_{j},\beta_{j}\right)$ at
every bus $j$. We make a few remarks.

######  Remark 5 (Voltage $\gamma_{j}$).

  1. 1.

_Parameter $\gamma_{j}$ for $Y$-configured devices._ The voltage parameter
$\gamma_{j}$ needs to be specified for every $Y$-configured device. By that,
we mean information additional to the models in Table IV is available to
determine the value of $\gamma_{j}$ for that device. It may be specified
directly, or more likely, indirectly. For instance if the neutral of a
$Y$-configured device is grounded and all voltages are defined with respect to
the ground, then $\gamma_{j}=V_{j}^{n}=-z_{j}^{n}\left(\textbf{1}^{\sf
T}I_{j}\right)$, which allows the elimination of $\gamma_{j}$ from the model.
If the neutral is grounded directly (i.e., $z_{j}^{n}=0$), then
$\gamma_{j}=0$. If the neutral is not grounded but the internal voltage
$V_{j}^{Y}$ is known to be balanced, i.e., $\textbf{1}^{\sf T}V_{j}^{Y}=0$,
then $\gamma_{j}:=\frac{1}{3}\textbf{1}^{\sf T}V_{j}$. For a $Y$-configured
current source, $\gamma_{j}$ is usually not needed to determine its terminal
voltage $V_{j}$, but needed to compute its internal voltage
$V_{j}^{Y}=V_{j}-\gamma_{j}\textbf{1}$ from the terminal voltage $V_{j}$.

  2. 2.

_Variable $\gamma_{j}$ for $\Delta$-configured devices._ For a
$\Delta$-configured voltage source, the zero-sequence voltage
$\gamma_{j}:=\frac{1}{3}\textbf{1}^{\sf T}V_{j}$ needs to be specified, e.g.,
by specifying one of its terminal voltages, say, $V_{j}^{a}$. For a
$\Delta$-configured current source or impedance, $\gamma_{j}$ can be
determined once its terminal voltage $V_{j}$ is determined from the network
equation $I=YV$.

  3. 3.

_Neutral voltage $\gamma_{j}$ and zero-sequence voltage._ For any
$Y$-configured device, we have

$\displaystyle V_{j}$ $\displaystyle\ \ =\ \ V_{j}^{Y}\ +\
V_{j}^{n}\textbf{1}$

The parameter $\gamma_{j}:=V_{j}^{n}$ may or may not equal the zero-sequence
voltage $\frac{1}{3}\textbf{1}^{\sf T}V_{j}$. They are equal if and only if
the internal voltages have no zero-sequence component since
$\frac{1}{3}\textbf{1}^{\sf T}V_{j}=\frac{1}{3}\textbf{1}^{\sf
T}V_{j}^{Y}+V_{j}^{n}$.

∎

##  IV Balanced network

In this section we show that, if the voltage sources, current sources, and
impedances are balanced and the lines are decoupled, then the three-phase
network is equivalent to a per-phase network and the analysis problem in
Section III can be solved by analyzing the simpler per-phase network.

The intuition is as follows. In a balanced three-phase network, positive-
sequence voltages and currents are in span$(\alpha_{+})$ and $\alpha_{+}$ is
an eigenvector of $\Gamma$ and $\Gamma^{\sf T}$. This means that the
transformation of balanced voltages and currents under $\Gamma,\Gamma^{\sf T}$
reduces to a scaling of these variables by their eigenvalues $1-\alpha$ and
$1-\alpha^{2}$ respectively. The voltage and current at every point in a
network can be written as linear combinations of transformed source voltages
and source currents, transformed by $\left(\Gamma,\Gamma^{\sf T}\right)$ and
line admittance matrices. Therefore if the source voltages and source currents
are balanced positive-sequence sets and lines are identical and phase-
decoupled, then the transformed voltages and currents remain in
span$(\alpha_{+})$ and hence are balanced positive-sequence sets.

In Section IV-A we describe how the balanced nature of voltage and current
sources simplifies the three-phase analysis problem formulated in Section III.
In Section IV-B we describe the positive-sequence per-phase network. In
Section IV-C we describe per-phase analysis under the assumption that the
neutral voltages $V_{j}^{n}$ of all $Y$-configured devices are zero and the
zero-sequence voltages $\gamma_{j}$ of all $\Delta$-configured voltage sources
are zero, and justify the procedure in Theorem 2. In Section IV-D we extend
the per-phase analysis and Theorem 2 to the case without this assumption. In
Section IV-E we prove Theorem 2.

###  IV-A Problem formulation

Balanced devices. Three-phase devices are balanced positive-sequence sets if
the voltage and current sources are in span$(\alpha_{+})$ and impedances are
balanced (identical) across phases. Then their internal models in Table IV
reduce to those specified in Table V with parameters
$\lambda_{j},\mu_{j},\epsilon_{j}\in\mathbb{C}$. The external models in Table
V are obtained by substituting these specifications into the external models
in Table IV and applying (Theorem 1)

$\displaystyle\Gamma\alpha_{+}$ $\displaystyle\ \ =\ \ (1-\alpha)\alpha_{+},$
$\displaystyle\Gamma^{\sf T}\alpha_{+}$ $\displaystyle\ \ =\ \
\left(1-\alpha^{2}\right)\alpha_{+}$ $\displaystyle\Gamma^{\dagger}$
$\displaystyle\ \ =\ \ \frac{1}{3}\Gamma^{\sf T},$ $\displaystyle\Gamma^{\sf
T{\dagger}}$ $\displaystyle\ \ =\ \ \frac{1}{3}\Gamma$

For example the external model of a $\Delta$-configured impedance is
$I_{j}=-Y^{\Delta}V_{j}$ where the effective matrix
$Y_{j}^{\Delta}:=\Gamma^{\sf T}y_{j}^{\Delta}\,\Gamma$. Since a balanced
impedance is $z^{\Delta}_{j}=\epsilon_{j}^{-1}\mathbb{I}$, we have

$\displaystyle Y_{j}^{\Delta}$ $\displaystyle\ \ =\ \ \Gamma^{\sf
T}y_{j}^{\Delta}\,\Gamma\ \ =\ \
\epsilon_{j}\left(3\mathbb{I}-\textbf{1}\textbf{1}^{\sf T}\right)$

so that, since $\textbf{1}^{\sf T}V_{j}=:3\gamma_{j}$, the external models of
an impedance in $\Delta$ configuration reduces to:

$\displaystyle I_{j}$ $\displaystyle\ \ =\ \ -Y^{\Delta}V_{j}\ \ =\ \
-3\epsilon_{j}\left(V_{j}\ -\ \gamma_{j}\textbf{1}\right)$

Buses $j$ |  Specification | External model | Vars | Internal vars
---|---|---|---|---
$N_{v}^{Y}$ |  $V_{j}^{Y}=\lambda_{j}\alpha_{+}$, $\ \gamma_{j}$ | ${\color[rgb]{0,0,1}{V_{j}\ =\ \lambda_{j}\alpha_{+}+\gamma_{j}\textbf{1}}}$ | $I_{j}$ | $I_{j}^{Y}=-I_{j}$
$N_{v}^{\Delta}$ | $V_{j}^{\Delta}=\lambda_{j}\alpha_{+}$, $\gamma_{j}$, $\beta_{j}$ | ${\color[rgb]{0,0,1}{V_{j}\ =\ \frac{1}{3}(1-\alpha^{2})\lambda_{j}\alpha_{+}+\gamma_{j}\textbf{1}}}$ | $I_{j}$ | $I_{j}^{\Delta}\ =\ -\Gamma^{\sf T{\dagger}}I_{j}+\beta_{j}\textbf{1}$
$N_{c}^{Y}$ | $I_{j}^{Y}\ =\ \mu_{j}\alpha_{+},\gamma_{j}$ | $I_{j}\ =\ -\mu_{j}\alpha_{+}$ | $V_{j}$ | $V_{j}^{Y}\ =\ V_{j}-\gamma_{j}\textbf{1}$
$N_{c}^{\Delta}$ | $I_{j}^{\Delta}\ =\ \mu_{j}\alpha_{+}$ | $I_{j}\ =\ -(1-\alpha^{2})\mu_{j}\alpha_{+}$ | $V_{j}$ | $V_{j}^{\Delta}\ =\ \Gamma V_{j},\ \gamma_{j}:=\frac{1}{3}\textbf{1}^{\sf T}V_{j}$
|  |  |  | $\beta_{j}:=\frac{1}{3}\textbf{1}^{\sf T}I_{j}^{\Delta}$
$N_{i}^{Y}$ | $z_{j}^{Y}\ =\ \epsilon_{j}^{-1}\mathbb{I}$, $\ \gamma_{j}$ | $I_{j}\ =\ -{\color[rgb]{0,0,1}{\epsilon_{j}}}\left(V_{j}-{\color[rgb]{0,0,1}{\gamma_{j}}}\textbf{1}\right)$ | $\left(V_{j},I_{j}\right)$ |  $V_{j}^{Y}\ =\ V_{j}-\gamma_{j}\textbf{1}$, $I_{j}^{Y}=-I_{j}$
$N_{i}^{\Delta}$ | $z_{j}^{\Delta}\ =\ \epsilon_{j}^{-1}\mathbb{I}$, $\ \beta_{j}$ | $I_{j}=-{\color[rgb]{0,0,1}{3\epsilon_{j}}}\left(V_{j}\ -\ \gamma_{j}\textbf{1}\right)$ | $\left(V_{j},I_{j}\right)$ |  $V_{j}^{\Delta}\ =\ \Gamma V_{j}$, $\gamma_{j}:=\frac{1}{3}\textbf{1}^{\sf T}V_{j}$
|  |  |  | $I_{j}^{\Delta}\ =\ -\Gamma^{\sf T{\dagger}}I_{j}+\beta_{j}\textbf{1}$
TABLE V: Internal and external models of balanced positive-sequence sources
and impedances.

Balanced admittance matrix $Y$. We assume all lines are balanced, i.e.,

$\displaystyle y_{jk}^{s}$ $\displaystyle\ \ =\ \ \eta_{jk}^{s}\mathbb{I},$
$\displaystyle y_{jk}^{m}$ $\displaystyle\ \ =\ \ \eta_{jk}^{m}\mathbb{I},$
$\displaystyle y_{kj}^{m}$ $\displaystyle\ \ =\ \ \eta_{kj}^{m}\mathbb{I}$
(11a) for some constants
$\eta_{jk}^{s},\eta_{jk}^{m},\eta_{kj}^{m}\in\mathbb{C}$. The terminal
voltages and currents $V:=(V_{0},\dots,V_{N})$ and $I:=(I_{0},\dots,I_{N})$
are described by (9) which, with balanced lines, reduces to $\displaystyle
I_{j}$ $\displaystyle=$ $\displaystyle\sum_{k:j\sim k}\eta_{jk}V_{j}\ -\
\sum_{k:j\sim k}\eta^{s}_{jk}V_{k},\quad j\in\overline{N}\quad\quad$ (11b)
where

$\eta_{jk}:=\eta_{jk}^{s}+\eta_{jk}^{m}$ and $V_{j},I_{j}\in\mathbb{C}^{3}$.
This in vector form is $I=YV$. Define the
$(\overline{N}+1)\times(\overline{N}+1)$ _per-phase admittance matrix_
$Y^{1\phi}$ by

$\displaystyle Y^{1\phi}_{jk}$ $\displaystyle\ \ :=\ \
\left\\{\begin{array}[]{lcl}-\eta^{s}_{jk},&&(j,k)\in E,\ \ (j\neq k)\\\
\sum_{k:j\sim k}\left(\eta^{s}_{jk}\ +\ \eta_{jk}^{m}\right),&&j=k\\\
0&&\text{otherwise}\end{array}\right.$ (12d) Substituting (11a) into the
admittance matrix $Y$ in (9) for the three-phase network, we can write $Y$ in
terms of the per-phase admittance matrix $Y^{1\phi}$ using the Kronecker
product: $\displaystyle Y$ $\displaystyle\ \ =\ \ Y^{1\phi}\otimes\mathbb{I}$
(12e) The relationship $I=YV$ for the three-phase network becomes
$\displaystyle I$ $\displaystyle\ \ =\ \
\left(Y^{1\phi}\otimes\mathbb{I}\right)V$ (12f)

Three-phase analysis problem. The analysis problem in Section III reduces to
the following problem. To simplify notation define

$\displaystyle\hat{\alpha}_{j}$ $\displaystyle\ \ :=\ \
\left\\{\begin{array}[]{lcl}1&\text{ if }&j\in N_{v}^{Y}\cup N_{c}^{Y}\cup
N_{i}^{Y}\\\ (1-\alpha^{2})/3&\text{ if }&j\in
N_{v}^{\Delta}\qquad(\text{voltage sources})\\\ (1-\alpha^{2})&\text{ if
}&j\in N_{c}^{\Delta}\qquad(\text{current sources})\\\ 3&\text{ if }&j\in
N_{i}^{\Delta}\qquad(\text{admittance})\\\ \end{array}\right.$

Given the following balanced voltage and current sources
$\left(V_{v},I_{c}\right)$ and impedance model (from Table V):

$\displaystyle V_{v}$ $\displaystyle\ \ =\ \
\left(\hat{\alpha}_{j}\lambda_{j}\alpha_{+}+\gamma_{j}\textbf{1},j\in
N_{v}\right)$ (13a) $\displaystyle I_{c}$ $\displaystyle\ \ =\ \
\left(-\hat{\alpha}_{j}\mu_{j}\alpha_{+},j\in N_{c}\right)$ (13b)
$\displaystyle I_{j}$ $\displaystyle\ \ =\ \
-\hat{\alpha}_{j}\epsilon_{j}\left(V_{j}-\gamma_{j}\textbf{1}\right),\ \ j\in
N_{i}$ (13c) our objective

is to solve (12) for the terminal voltage $V_{-v}:=\left(V_{c},V_{i}\right)$
and current $I_{-c}:=\left(I_{v},I_{i}\right)$ and then calculate internal
voltages and currents as well as $\left(\gamma_{j},\beta_{j}\right)$.

The problem can be solved by substituting (13) into (12) and computing
numerically $\left(V_{-v},I_{-c}\right)$. This is Step 1 of the solution
procedure in Section III-B. Steps 2 and 3 will compute the internal variables
given the terminal variables $(V,I)$.

######  Remark 6 ($\Delta$-$Y$ transformation).

The specification (13) corresponds to the step of converting all $\Delta$
configured devices to their $Y$ equivalents. It generalizes the standard
practice of assuming $\gamma_{j}=0$ to the case where $\gamma_{j}$ may be
nonzero, because some $Y$-configured devices on the network are not grounded,
some are grounded through nonzero earthing impedances, and some
$\Delta$-configured devices have nonzero zero-sequence voltages (cf. Remark
4). ∎

###  IV-B Per-phase network

We now formalize the alternative solution that solves (12)(13) using per-phase
analysis. We describe a per-phase positive-sequence network in this subsection
and a per-phase analysis procedure in the next subsection. We make the
following simplifying assumptions:

  * CIV-B: The neutral voltages $\gamma_{j}:=V_{j}^{n}=0$ for all $Y$-configured devices $j\in N_{v}^{Y}\cup N_{c}^{Y}\cup N_{i}^{Y}$.

  * CIV-B: The zero-sequence voltages $\gamma_{j}:=\frac{1}{3}\textbf{1}^{\sf T}V_{j}=0$ for all $\Delta$-configured voltage sources $j\in N_{v}^{\Delta}$.

While $\gamma_{j}$ in CIV-B and CIV-B are part of the device specification,
the zero-sequence voltages $\gamma_{j}$ of $\Delta$-configured current sources
and impedances $j\in N_{c}^{\Delta}\cup N_{i}^{\Delta}$ are not specified but
need to be determined through the network equation (12). We will prove in
Lemma 4 below that CIV-B and CIV-B indeed imply $\gamma_{j}=0$ for $j\in
N_{c}^{\Delta}\cup N_{i}^{\Delta}$. We will explain in Section IV-D how the
results extend to the general case where these assumptions do not hold.

Network equation. Consider a network whose graph is $G=(\overline{N},E)$ as
before but each line $(j,k)\in E$ is characterized by the complex _scalar_
admittances
$\left(\eta_{jk}^{s},\eta_{jk}^{m},\eta_{kj}^{m}\right)\in\mathbb{C}^{3}$ in
(11a), instead of $3\times 3$ admittance matrices in the three-phase network.
Associated with each bus $j\in\overline{N}$ is a scalar voltage and a scalar
current injection $\left(v_{j},i_{j}\right)\in\mathbb{C}^{2}$. The current
vector $i:=(i_{j},j\in\overline{N})$ and the voltage vector
$v:=(v_{j},j\in\overline{N})$ are related by the $(N+1)\times(N+1)$ per-phase
admittance matrix $Y^{1\phi}$ defined in (12d) according to
$i\,=\,Y^{1\phi}v$. This relationship defines a _per-phase positive-sequence
network_. On this per-phase network, the single-phase devices on buses $j\in
N_{v}^{\Delta}\cup N_{c}^{\Delta}\cup N_{i}^{\Delta}$ are specified as (from
Table V):

Voltage source: $\displaystyle v_{j}$ $\displaystyle\ =\
\hat{\alpha}_{j}\lambda_{j},$ $\displaystyle j$ $\displaystyle\ \in\ N_{v}$
Current source: $\displaystyle i_{j}$ $\displaystyle\ =\
-\hat{\alpha}_{j}\mu_{j},$ $\displaystyle j$ $\displaystyle\ \in\ N_{c}$
Impedance: $\displaystyle i_{j}$ $\displaystyle\ =\
-\hat{\alpha}_{j}\epsilon\,v_{j},$ $\displaystyle j$ $\displaystyle\ \in\
N_{i}$

Note that the per-phase impedance model does not involve $\gamma_{j}$. To
express this specification in vector form, define the following (column)
vectors:

$\displaystyle\left(v_{v},i_{v}\right)$ $\displaystyle\ \ :=\ \
\left(v_{j},i_{j},\,j\in N_{v}\right)$ $\displaystyle\left(v_{c},i_{c}\right)$
$\displaystyle\ \ :=\ \ \left(v_{j},i_{j},\,j\in N_{c}\right)$
$\displaystyle\left(v_{i},i_{i}\right)$ $\displaystyle\ \ :=\ \
\left(v_{j},i_{j},\,j\in N_{i}\right)$

The voltage sources and impedance modeled are then specified as:

$\displaystyle v_{v}$ $\displaystyle\ =\
\left(\hat{\alpha}_{j}\lambda_{j},\,j\in N_{v}\right),\qquad i_{c}\ =\
\left(-\hat{\alpha}_{j}\mu_{j},\,j\in N_{c}\right)$ (14a) $\displaystyle
i_{i}$ $\displaystyle\ =\ -Y_{i}^{1\phi}v_{i}\ \ \text{ with }\ \
Y_{i}^{1\phi}:=\mathrm{diag}\left(\hat{\alpha}_{j}\epsilon_{j},\,j\in
N_{i}\right)$ (14b) This

is the per-phase version of the specification (13) of the corresponding three-
phase devices.

A key step in per-phase analysis is to solve a per-phase version of the three-
phase problem (12)(13): given the specification in (14), compute the remaining
variables

$\displaystyle v_{-v}$ $\displaystyle\ \ :=\ \ \left(v_{j},\,j\not\in
N_{v}\right),$ $\displaystyle i_{-c}$ $\displaystyle\ \ :=\ \
\left(i_{j},\,j\not\in N_{c}\right)$

from $i=Y^{1\phi}v$, or equivalently, from

$\displaystyle\begin{bmatrix}i_{v}\\\ i_{c}\\\ i_{i}\end{bmatrix}$
$\displaystyle\ \ =\ \
\underbrace{\begin{bmatrix}Y^{1\phi}_{vv}&Y^{1\phi}_{vc}&Y^{1\phi}_{vi}\\\
Y^{1\phi}_{cv}&Y^{1\phi}_{cc}&Y^{1\phi}_{ci}\\\
Y^{1\phi}_{iv}&Y^{1\phi}_{ic}&Y^{1\phi}_{ii}\end{bmatrix}}_{Y^{1\phi}}\begin{bmatrix}v_{v}\\\
v_{c}\\\ v_{i}\end{bmatrix}$ (15)

where the admittance matrix $Y^{1\phi}$ is defined in (12d). We will first
compute $v_{-v}$ from (15) and then substitute $v_{-v}$ back into (15) to
compute $i_{-c}$.

Computation of $\left(v_{-v},i_{-c}\right)$. To compute $v_{-v}$ define the
following matrices from the per-phase admittance matrix $Y^{1\phi}$:

$\displaystyle\underbrace{\left[\begin{array}[]{c | c c}Y^{1\phi}_{vv}&Y^{1\phi}_{vc}&Y^{1\phi}_{vi}\\\ \hline\cr Y^{1\phi}_{cv}&Y^{1\phi}_{cc}&Y^{1\phi}_{ci}\\\ Y^{1\phi}_{iv}&Y^{1\phi}_{ic}&Y^{1\phi}_{ii}\end{array}\right]}_{Y^{1\phi}}$ $\displaystyle\ \ =:\ \ \left[\begin{array}[]{c | c}A_{11}&A_{21}^{\sf T}\\\ \hline\cr A_{21}&A_{22}\end{array}\right]$ (16f) $\displaystyle\begin{bmatrix}Y^{1\phi}_{cc}&Y^{1\phi}_{ci}\\\ Y^{1\phi}_{ic}&Y^{1\phi}_{ii}+Y_{i}^{1\phi}\end{bmatrix}$ $\displaystyle\ \ =:\ \ A_{22}^{\prime}$ (16g) where the matrix $Y_{i}^{1\phi}$ is defined in (14b). Note that both $A_{22}$ are $A_{22}^{\prime}$ are complex symmetric and therefore legitimate admittance matrices (they will be interpreted below as admittance matrices of a reduced network in (17a)). For the computation of $i_{-c}$ in (17b) below, define the following submatrices of $Y^{1\phi}$: $\displaystyle Y_{-c}^{1\phi}$ $\displaystyle\ \ :=\ \ \left[\begin{array}[]{c | c c}Y^{1\phi}_{vv}&Y^{1\phi}_{vc}&Y^{1\phi}_{vi}\\\ Y^{1\phi}_{iv}&Y^{1\phi}_{ic}&Y^{1\phi}_{ii}\end{array}\right]\ \ =:\ \ \left[\begin{array}[]{c | c}Y^{1\phi}_{-c,v}&Y^{1\phi}_{-c,-v}\end{array}\right]$ (16k)

Substituting the specification (14) into (15) yields a system of
$(|N_{c}|+|N_{i}|)$ linear equations in $(|N_{c}|+|N_{i}|)$ unknown voltages
$v_{-v}$:

$\displaystyle\underbrace{\begin{bmatrix}Y^{1\phi}_{cc}&Y^{1\phi}_{ci}\\\
Y^{1\phi}_{ic}&Y^{1\phi}_{ii}+Y_{i}^{1\phi}\end{bmatrix}}_{A_{22}^{\prime}}\begin{bmatrix}v_{c}\\\
v_{i}\end{bmatrix}$ $\displaystyle\ \ =\ \ \underbrace{\begin{bmatrix}i_{c}\\\
0\end{bmatrix}}_{i_{-v}}\ -\ \underbrace{\begin{bmatrix}Y^{1\phi}_{cv}\\\
Y^{1\phi}_{iv}\end{bmatrix}}_{A_{21}}v_{v}$

or in abbreviation:

$\displaystyle A_{22}^{\prime}v_{-v}$ $\displaystyle\ \ =\ \ i_{-v}\ -\
A_{21}v_{v}\ \ =:\ \ b$ (17a) where $i_{-v}:=(i_{c},0)$ and the matrices
$A_{21},A_{22}^{\prime}$ are defined in (16g). If $A_{22}^{\prime}$ is
invertible, this yields a unique $v_{-v}$ and hence a unique $i_{-v}$. Since
$A_{22}^{\prime}$ can be treated as an admittance matrix, the equation (17a)
can be interpreted as a $V$-$I$ relationship on a reduced network consisting
only of buses in $N_{c}\cup N_{i}$ with current sources and impedances. The
current sources inject currents $i_{c}$. The effect of voltage sources $j\in
N_{v}$ on this reduced network is summarized as additional current injections
$-A_{21}v_{v}$ at these buses, so that the net current injections are
$b:=i_{-v}-A_{21}v_{v}$.

Substituting $v_{-v}$ from (17a) into (15) and using (16k) we have

$\displaystyle i_{-c}$ $\displaystyle\ \ =\ \ Y_{-c}^{1\phi}v\ \ =\ \
Y_{-c,v}^{1\phi}\,v_{v}\ +\ Y_{-c,-v}^{1\phi}\left(A_{22}^{\prime-1}b\right)$
(17b)

assuming $A_{22}^{\prime}$ is invertible.

###  IV-C Per-phase analysis

Suppose the matrices $A_{22}$ and $A_{22}^{\prime}$ are invertible. The
following per-phase analysis procedure offers a simpler alternative to solving
the three-phase problem (12)(13) directly:

  1. 1.

Solve the positive-sequence network problem (IV-B) to obtain
$v_{-v}:=\left(v_{j},j\not\in N_{v}\right)$ and $i_{-c}:=\left(i_{j},j\not\in
N_{c}\right)$. The terminal variables of the three-phase problem are then

$\displaystyle V$ $\displaystyle\ \ =\ \ v\otimes\alpha_{+},$ $\displaystyle
I$ $\displaystyle\ \ =\ \ i\otimes\alpha_{+}$

  2. 2.

From the terminal voltage and current $(V,I)$, the internal voltages, currents
and powers of $Y$-configured devices are obtained from the conversion rule
(3):

$\displaystyle V_{j}^{Y}$ $\displaystyle\ \ =\ \ v_{j}\,\alpha_{+},$
$\displaystyle I_{j}^{Y}$ $\displaystyle\ =\ -i_{j}\,\alpha_{+}$
$\displaystyle s_{j}^{Y}$ $\displaystyle\ \ =\ \
-v_{j}\overline{i}_{j}\textbf{1},$ $\displaystyle j$ $\displaystyle\ \in\
N_{v}^{Y}\cup N_{c}^{Y}\cup N_{i}^{Y}$

  3. 3.

Those of $\Delta$-configured devices can be computed by applying the
conversion rule (II-B) to $(V,I)$:

$\displaystyle V_{j}^{\Delta}$ $\displaystyle\ \ =\ \
(1-\alpha)v_{j}\,\alpha_{+},$ $\displaystyle I_{j}^{\Delta}$ $\displaystyle\
=\ -\frac{1}{3}\left(1-\alpha\right)i_{j}\,\alpha_{+}$ $\displaystyle
s_{j}^{\Delta}$ $\displaystyle\ \ =\ \ -v_{j}\overline{i}_{j}\textbf{1},$
$\displaystyle j$ $\displaystyle\ \in\ N_{v}^{\Delta}\cup N_{c}^{\Delta}\cup
N_{i}^{\Delta}$

where we have used $\mathrm{diag}\left(\alpha_{+}\alpha_{+}^{\sf
H}\right)=\textbf{1}$ and
$\left(1-\alpha\right)\left(1-\overline{\alpha}\right)=3$ in computing the
internal variables.

To prove this per-phase analysis procedure solves the three-phase problem
(12)(13) when the network is balanced, it suffices to justify Step 1 of the
procedure. This is stated in the following theorem.

######  Theorem 2 (Per-phase analysis).

Suppose $A_{22}$, $A_{22}^{\prime}$ are invertible and assumptions CIV-B and
CIV-B hold. Let $\left(v,i\right)$ be the unique solution of the positive-
sequence network (IV-B). Then $\left(V,I\right)$ is the unique three-phase
solution to (12)(13) if and only if $V=v\otimes\alpha_{+}$ and
$I=i\otimes\alpha_{+}$.

###  IV-D Extension

Theorem 2 says that when the voltage and current sources
$\left(V_{v},I_{c}\right)$ are balanced positive-sequence sets, all terminal
voltages and currents $(V,I)$ are balanced positive-sequence sets, as long as
assumptions CIV-B and CIV-B hold. Without assumptions CIV-B and CIV-B, Theorem
2 needs to be modified to the following.

######  Theorem 3 (Per-phase analysis).

Suppose $A_{22}$, $A_{22}^{\prime}$ are invertible. Let $\left(v,i\right)$ be
the unique solution of the positive-sequence network (IV-B). Then
$\left(V,I\right)$ is the unique three-phase solution to (12)(13) if and only
if

$\displaystyle V$ $\displaystyle\ \ =\ \
v\otimes\alpha_{+}+\tilde{\gamma}\otimes\textbf{1}$ $\displaystyle I$
$\displaystyle\ \ =\ \ i\otimes\alpha_{+}\ +\ \tilde{\beta}\otimes\textbf{1}$

for some $\left(\tilde{\gamma},\tilde{\beta}\right)\in\mathbb{C}^{2(N+1)}$.

The theorem says that, without assumptions CIV-B and CIV-B, $(V,I)$ also has a
zero-sequence component $\left(\tilde{\gamma},\tilde{\beta}\right)$ in
addition to the positive-sequence component $(v,i)$. While $(v,i)$ is still
computed from the per-phase positive-sequence network as described in Section
IV-B, $\left(\tilde{\gamma},\tilde{\beta}\right)$ needs to be computed from a
separate per-phase zero-sequence network that is driven by the zero-sequence
voltages $\gamma_{v}$ of both $Y$ and $\Delta$-configured voltage sources. The
computation of $\left(\tilde{\gamma},\tilde{\beta}\right)$ is more
complicated; see [23].

Given the terminal voltage and current $(V,I)$, Steps 2 and 3 of the per-phase
analysis procedure in Section IV-C are modified to:

  1. 2\.

From the terminal voltage and current $(V,I)$, the internal voltages, currents
and powers of $Y$-configured devices are obtained from the conversion rule (3)
$\left(\gamma_{j}=V_{j}^{n}\right)$: for $j\in N_{v}^{Y}\cup N_{c}^{Y}\cup
N_{i}^{Y}$

$\displaystyle V_{j}^{Y}\ \ =\ \ $ $\displaystyle V_{j}\ -\
\gamma_{j}\textbf{1}\ \ =\ \ v_{j}\,\alpha_{+}\ -\
\left(\gamma_{j}-\tilde{\gamma}_{j}\right)\textbf{1}$ $\displaystyle
I_{j}^{Y}\ \ =\ \ $ $\displaystyle-I_{j}\ \ =\ \ -i_{j}\,\alpha_{+}\ +\
\tilde{\beta}_{j}\textbf{1}$ $\displaystyle s_{j}^{Y}\ \ =\ \ $
$\displaystyle-\left(v_{j}i_{j}^{\sf
H}+\left(\gamma_{j}-\tilde{\gamma}_{j}\right)\tilde{\beta}_{j}^{\sf
H}\right)\textbf{1}$ $\displaystyle+v_{j}\tilde{\beta}_{j}^{\sf
H}\,\alpha_{+}\ +\ \left(\gamma_{j}-\tilde{\gamma}_{j}\right)i_{j}^{\sf
H}\,\alpha_{-}$

  2. 3\.

Those of $\Delta$-configured devices can be computed by applying the
conversion rule (II-B) to $(V,I)$: for $j\in N_{v}^{\Delta}\cup
N_{c}^{\Delta}\cup N_{i}^{\Delta}$

$\displaystyle V_{j}^{\Delta}$ $\displaystyle\ \ =\ \ \Gamma V_{j}\ \ =\ \
(1-\alpha)v_{j}\,\alpha_{+}$ $\displaystyle I_{j}^{\Delta}$ $\displaystyle\ \
=\ \ -\Gamma^{\sf T{\dagger}}I_{j}\ +\ \beta_{j}\textbf{1}\ \ =\ \
-\frac{1}{3}\left(1-\alpha\right)i_{j}\,\alpha_{+}\,+\ \beta_{j}\textbf{1}$
$\displaystyle s_{j}^{\Delta}$ $\displaystyle\ \ =\ \ -v_{j}i_{j}^{\sf
H}\textbf{1}\ +\ \left(1-\alpha\right)v_{j}\beta_{j}^{\sf H}\,\alpha_{+}$

In particular, whereas the internal powers $s^{Y}=v_{j}i_{j}^{\sf
H}\textbf{1}$ are identical across the three single-phase devices if
assumptions CIV-B and CIV-B hold, they are generally different otherwise due
to the presence of both positive and negative-sequence components.

###  IV-E Proof of Theorem 2

Consider $V_{-v}:=\left(V_{j},j\not\in N_{v}\right)$. Recall that
$\gamma_{j}:=V_{j}^{n}$ denote the neutral voltages for $Y$-configured devices
$j\in N_{c}^{Y}\cup N_{i}^{Y}$ and $\gamma_{j}:=\frac{1}{3}\textbf{1}^{\sf
T}V_{j}$ denote the zero-sequence voltages of $\Delta$-configured devices
$j\in N_{c}^{\Delta}\cup N_{i}^{\Delta}$. Under assumption CIV-B,
$\gamma_{j}=0$ for $j\in N_{c}^{Y}\cup N_{i}^{Y}$. The main ingredient of the
proof of Theorem 2 is the following lemma that implies that $\gamma_{j}=0$ for
$j\in N_{c}^{\Delta}\cup N_{i}^{\Delta}$ as well and that $V_{-v}$ is
balanced.

######  Lemma 4 (Balanced voltage).

Suppose $(V,I)$ satisfies (12)(13). Under assumptions CIV-B and CIV-B:

  1. 1.

If $A_{22}$ is invertible, then $\gamma_{j}=0$ for $j\in N_{c}^{\Delta}\cup
N_{i}^{\Delta}$ and hence $\gamma_{j}=0$ for all $j\in\overline{N}$.

  2. 2.

The terminal voltage $V=(V_{v},V_{-v})$ satisfies

$\displaystyle\left(A_{22}^{\prime}\otimes\mathbb{I}\right)V_{-v}$
$\displaystyle\ \ =\ \ \left(i_{-v}-A_{21}v_{v}\right)\otimes\alpha_{+}\ \ =:\
\ b\otimes\alpha_{+}$

where $A_{22}^{\prime}$ is defined in (16g). If $A_{22}^{\prime}$ is
invertible, then $V_{-v}\,=\,\left(A_{22}^{\prime-1}b\right)\otimes\alpha_{+}$
is balanced.

###### Proof of Lemma 4.

For part 1 of the lemma, observe from Table V that $\textbf{1}^{\sf T}I_{j}=0$
for $j\not\in N_{v}^{Y}$. Define $\hat{\gamma}_{j}\ :=\
\frac{1}{3}\textbf{1}^{\sf T}V_{j}$. Note that $\hat{\gamma}_{j}=\gamma_{j}$
for $\Delta$-configured devices but not necessarily for $Y$-configured current
sources or impedances. Multiplying both sides of (11b) by $\textbf{1}^{\sf T}$
gives

$\displaystyle\sum_{k:j\sim k}\left(y^{s}_{jk}\ +\
{y_{jk}^{m}}\right)\hat{\gamma}_{j}\ -\sum_{k:j\sim
k}y^{s}_{jk}\hat{\gamma}_{k}$ $\displaystyle\ \ =\ \
\left\\{\begin{array}[]{lcl}\frac{1}{3}\textbf{1}^{\sf T}I_{j}&\text{ if
}&j\in N_{v}^{Y}\\\ 0&\text{ if }&j\not\in N_{v}^{Y}\end{array}\right.$

We can write this for $j\not\in N_{v}$ in vector form using (16g) as:

$\displaystyle\left[\begin{array}[]{c | c}A_{21}&A_{22}\end{array}\right]\begin{bmatrix}\hat{\gamma}_{v}\\\ \hat{\gamma}_{-v}\end{bmatrix}\ \ =\ \ 0$

Note that $\hat{\gamma}_{v}=\gamma_{v}$ because for $j\in N_{v}^{Y}$,
$\hat{\gamma}_{j}:=\frac{1}{3}\textbf{1}^{\sf
T}\left(V_{j}^{Y}+V_{j}^{n}\textbf{1}\right)=V_{j}^{n}=\gamma_{j}$ and for
$j\in N_{v}^{\Delta}$, $\hat{\gamma}_{j}=\gamma_{j}$ by definition. This
implies that $A_{21}\gamma_{v}+A_{22}\hat{\gamma}_{-v}\ =\ 0$. When $A_{22}$
is invertible,

$\displaystyle\hat{\gamma}_{-v}$ $\displaystyle\ \ =\ \
-A_{22}^{-1}A_{21}\gamma_{v}$

Under assumptions CIV-B and CIV-B, $\gamma_{v}=0$ and hence
$\hat{\gamma}_{-v}=0$. This implies $\gamma_{j}=\hat{\gamma}_{j}=0$ for $j\in
N_{c}^{\Delta}\cup N_{i}^{\Delta}$ and hence $\gamma_{j}=0$ for all
$j\in\overline{N}$. This establishes part 1 of the lemma.

We will prove part 2 in three steps. First we will express the three-phase
specification (13) in terms of per-phase variables. Then we will write the
three-phase equation (12) using per-phase parameters. Finally we will
establish the equation for $V_{-v}$ in part 2 of the lemma.

First, using (14), the three-phase quantities in (13) can be written as (under
CIV-B and CIV-B):

$\displaystyle V_{v}$ $\displaystyle\ \ =\ \
\left(\hat{\alpha}_{j}\lambda_{j}\alpha_{+}+\gamma_{j}\textbf{1},\ j\in
N_{v}\right)\ \ =\ \ v_{v}\otimes\alpha_{+}$ (18a) $\displaystyle I_{c}$
$\displaystyle\ \ =\ \ \left(-\hat{\alpha}_{j}\mu_{j}\alpha_{+},\ j\in
N_{c}\right)\ \ =\ \ i_{c}\otimes\alpha_{+}$ (18b) $\displaystyle I_{i}$
$\displaystyle\ \ =\ \
\left(-\hat{\alpha}_{j}\epsilon_{j}\left(V_{j}-\gamma_{j}\textbf{1}\right),\
j\in N_{i}\right)\ \ =\ \ -\left(Y_{i}^{1\phi}\otimes\mathbb{I}\right)V_{i}$
(18c) where

the matrix $Y_{i}^{1\phi}$ is defined in (14b).

Second, write (12) in terms of submatrices of $Y^{1\phi}$ in (15) (since
$Y=Y^{1\phi}\otimes\mathbb{I}$):

$\displaystyle\begin{bmatrix}I_{v}\\\ I_{c}\\\ I_{i}\end{bmatrix}$
$\displaystyle\ \ =\ \ \underbrace{\begin{bmatrix}Y_{vv}&Y_{vc}&Y_{vi}\\\
Y_{cv}&Y_{cc}&Y_{ci}\\\
Y_{iv}&Y_{ic}&Y_{ii}\end{bmatrix}}_{Y}\begin{bmatrix}V_{v}\\\ V_{c}\\\
V_{i}\end{bmatrix}$ $\displaystyle\ \ =\ \
\left(\underbrace{\begin{bmatrix}Y^{1\phi}_{vv}&Y^{1\phi}_{vc}&Y^{1\phi}_{vi}\\\
Y^{1\phi}_{cv}&Y^{1\phi}_{cc}&Y^{1\phi}_{ci}\\\
Y^{1\phi}_{iv}&Y^{1\phi}_{ic}&Y^{1\phi}_{ii}\end{bmatrix}}_{Y^{1\phi}}\otimes\
\mathbb{I}\right)\begin{bmatrix}V_{v}\\\ V_{c}\\\ V_{i}\end{bmatrix}$

Therefore the voltages $(V_{c},V_{i})$ satisfies

$\displaystyle I_{c}$ $\displaystyle\ \ =\ \
\left(Y^{1\phi}_{cv}\otimes\mathbb{I}\right)V_{v}\ +\
\left(Y^{1\phi}_{cc}\otimes\mathbb{I}\right)V_{c}\ +\
\left(Y^{1\phi}_{ci}\otimes\mathbb{I}\right)V_{i}$ $\displaystyle I_{i}$
$\displaystyle\ \ =\ \ \left(Y^{1\phi}_{iv}\otimes\mathbb{I}\right)V_{v}\ +\
\left(Y^{1\phi}_{ic}\otimes\mathbb{I}\right)V_{c}\ +\
\left(Y^{1\phi}_{ii}\otimes\mathbb{I}\right)V_{i}$

Substitute (18) to get

$\displaystyle\left(Y^{1\phi}_{cc}\otimes\mathbb{I}\right)V_{c}+\left(Y^{1\phi}_{ci}\otimes\mathbb{I}\right)V_{i}\
=\ $ $\displaystyle
i_{c}\otimes\alpha_{+}-\left(Y^{1\phi}_{cv}\otimes\mathbb{I}\right)\left(v_{v}\otimes\alpha_{+}\right)$
$\displaystyle\left(Y^{1\phi}_{ic}\otimes\mathbb{I}\right)V_{c}+\left(Y^{1\phi}_{ii}\otimes\mathbb{I}\right)V_{i}\
=\ $ $\displaystyle-\left(Y_{i}^{1\phi}\otimes\mathbb{I}\right)V_{i}$
$\displaystyle-\left(Y^{1\phi}_{iv}\otimes\mathbb{I}\right)\left(v_{v}\otimes\alpha_{+}\right)$

Since
$\left(Y^{1\phi}_{cv}\otimes\mathbb{I}\right)\left(v_{v}\otimes\alpha_{+}\right)=\left(Y^{1\phi}_{cv}v_{v}\right)\otimes\alpha_{+}$
and
$\left(Y^{1\phi}_{iv}\otimes\mathbb{I}\right)\left(v_{v}\otimes\alpha_{+}\right)=\left(Y^{1\phi}_{iv}v_{v}\right)\otimes\alpha_{+}$,
the voltage $(V_{c},V_{i})$ satisfies

$\displaystyle\left(\underbrace{\begin{bmatrix}Y^{1\phi}_{cc}&Y^{1\phi}_{ci}\\\
Y^{1\phi}_{ic}&Y^{1\phi}_{ii}+Y_{i}^{1\phi}\end{bmatrix}}_{A_{22}^{\prime}}\otimes\
\mathbb{I}\right)\begin{bmatrix}V_{c}\\\ V_{i}\end{bmatrix}$ $\displaystyle\
=\ \ \underbrace{\left(\begin{bmatrix}i_{c}\\\
0\end{bmatrix}-A_{21}v_{v}\right)}_{b}\otimes\ \alpha_{+}$

This is the equation in part 2 of the lemma, abbreviated as

$\displaystyle\left(A_{22}^{\prime}\otimes\mathbb{I}\right)V_{-v}$
$\displaystyle\ \ =\ \ b\otimes\alpha_{+}$ (19)

If $A_{22}^{\prime}$ is invertible, so is $A_{22}^{\prime}\otimes\mathbb{I}$.
Then $V_{-v}$ is uniquely determined as

$\displaystyle V_{-v}$ $\displaystyle\ \ =\ \
\left(A_{22}^{\prime}\otimes\mathbb{I}\right)^{-1}\left(b\otimes\alpha_{+}\right)\
\ =\ \
\left(A_{22}^{\prime-1}\otimes\mathbb{I}\right)\left(b\otimes\alpha_{+}\right)$
$\displaystyle\ \ =\ \ \left(A_{22}^{\prime-1}b\right)\otimes\alpha_{+}$

This completes part 2 of the lemma. ∎

###### Proof of Theorem 2.

Under the assumption that $A_{22}$ and $A_{22}^{\prime}$ are invertible, the
solution $v_{-v}$ to the per-phase network (17a) is unique. Lemma 4 implies
that $V_{-v}$ is the unique solution to (19). We now show that the expression
$V_{-v}=v_{-v}\otimes\alpha_{+}$ in the theorem satisfies this equation and
therefore must be its unique solution. We have

$\displaystyle\left(A_{22}^{\prime}\otimes\mathbb{I}\right)\left(v_{-v}\otimes\alpha_{+}\right)$
$\displaystyle\ \ =\ \ \left(A_{22}^{\prime}v_{-v}\right)\otimes\alpha_{+}$

But $A_{22}^{\prime}v_{-v}=i_{-v}-A_{21}v_{v}=b$ from (17a). This proves the
expression for $V_{-v}$ in the theorem.

To prove the expression for $I_{-c}$ in the theorem, let
$V_{-c}:=\left(V_{v},V_{i}\right)$. Substituting into (12) we have (since
$Y=Y^{1\phi}\otimes\mathbb{I}$)

$\displaystyle I_{-c}$ $\displaystyle\ \ =\ \
\left(\underbrace{\begin{bmatrix}Y^{1\phi}_{vv}&Y^{1\phi}_{vc}&Y^{1\phi}_{vi}\\\
Y^{1\phi}_{iv}&Y^{1\phi}_{ic}&Y^{1\phi}_{ii}\end{bmatrix}}_{Y_{-c}^{1\phi}}\otimes\
\mathbb{I}\right)\begin{bmatrix}V_{v}\\\ V_{c}\\\ V_{i}\end{bmatrix}\ \ =:\ \
Y_{-c}^{1\phi}\,V$

where $Y_{-c}^{1\phi}$ is the matrix defined in (16k). Substitute the
expression $V=v\otimes\alpha_{+}$ from Theorem 2 we have

$\displaystyle I_{-c}$ $\displaystyle\ \ =\ \
\left(Y_{-c,v}^{1\phi}v_{v}+Y_{-c,-v}^{1\phi}v_{-v}\right)\otimes\alpha_{+}\ \
=\ \ i_{-c}\otimes\alpha_{+}$

where the last equality follows from $v_{-v}=A_{22}^{\prime-1}b$ and (17b).
This completes the proof of Theorem 2. ∎

##  V Conclusion

In this paper we have introduced a three-phase power flow model. Its key
feature is to model a three-phase device by (i) an internal model that
describes the behavior of the constituent single-phase devices, and (ii) a
conversion rule that maps internal variables to terminal variables observable
externally of the three-phase device based on its configuration. The
flexibility allows a unified approach to compose general network models where
devices can be ungrounded, grounded with zero or nonzero earthing impedances,
or have nonzero zero-sequence voltages. We have illustrated our model by
studying a general analysis problem when the network is balanced, formalizing
per-phase analysis, and proving its validity using the spectral properties of
the linear conversion matrix $\Gamma$.

## Acknowledgment

We would like to thank Janusz Bialek, Frederik Geth, Pierluigi Mancarella, and
Danny Tsang for helpful discussions.

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