Contents
\[\newcommand{\i}{\mathbf{i}} % complex number i \renewcommand{\Re}{\mathsf{Re}} % real component \renewcommand{\Im}{\mathsf{Im}} % imaginary component \DeclareMathOperator{\diag}{diag} % diagonal \newcommand{\Imax}{I_\max} % current amplitude \newcommand{\Vmax}{V_\max} % voltage amplitude\]This post assumes familiarity with:
- complex numbers and trigonometry
- basic electromagnetism physics (e.g., Fundamentals of Physics)
- basic linear algebra (e.g., Linear Algebra Done Right, 3e)
This post is based on excellent lecture notes on power systems by Prof. Steven Low, found here: http://netlab.caltech.edu/book/book.html.
Review of complex numbers
Here is a quick refresher on complex numbers. Let \(\i = \sqrt{-1}\), and let \(x,y \in \C\), with \(x = a + \i b\).
- The real component is denoted \(\Re(x) := a\).
- The imaginary component is denoted \(\Im(x) := b\).
- The magnitude is \(\abs{x} := \sqrt{a^2 + b^2}\).
- The complex conjugate is \(x^* := a - \i b\).
- The angle of \(x\) is \(\angle x := \tan^{-1}(b/a)\)
- Euler’s formula: \(e^{\i \theta} = \cos\theta + \i \sin\theta\)
- If \(\theta = \angle x\), then \(x = \abs{x} e^{\i \theta}\).
- \(\abs{e^{\i \theta}} = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1\) for all \(\theta\)
- Useful identities:
- binomial expansion: \(\abs{x+y}^2 = \abs{x}^2 + 2 \Re(xy^*) + \abs{y}^2\)
- complex conjugate is distributive: \((xy)^* = x^* y^*\)
Lemma (equivalent complex numbers): Let \(c_1 = a_1 + \i b_1\) and \(c_2 = a_2 + \i b_2\) be two complex numbers, and let \(\omega \in \R \setminus \{0\}\) be any nonzero scalar. \(c_1 = c_2\) if and only if \(\forall t \in \R:\ \Re(c_1 e^{\i \omega t}) = \Re(c_2 e^{\i \omega t})\).
Proof
Clearly, if \(c_1 = c_2\), then \(\forall t \in \R:\ \Re(c_1 e^{\i \omega t}) = \Re(c_2 e^{\i \omega t})\).
For the reverse direction, suppose that \(\forall t \in \R:\ \Re(c_1 e^{\i \omega t}) = \Re(c_2 e^{\i \omega t})\). Since this holds for all \(t\), we can pick \(t = 0\). Then, we see \(a_1 = \Re(c_1) = \Re(c_2) = a_2\). Likewise, we can pick \(t = \pi / (2 \omega)\), so \(e^{\i \omega t} = e^{\i \pi / 2} = \cos(\pi/2) + \i \sin(\pi/2) = \i\). Then
\[\Re(c_1 e^{\i \omega t}) = \Re((a_1 + \i b_1) \i) = -b_1\]and similarly \(\Re(c_2 e^{\i \omega t}) = -b_2\). These two expressions are equal, meaning \(b_1 = b_2\). We have shown \(a_1 = a_2\) and \(b_1 = b_2\), so \(c_1 = c_2\). \(\blacksquare\)
Phasors
Consider a sinusoidal voltage \(v(t) = \Vmax \cos(\omega t + \theta_v)\) where \(\Vmax\) is the amplitude, \(\omega\) is the frequency, and \(\theta_v\) is the phase. The units of each variable are as follows:
| variable(s) | units |
|---|---|
| \(v(t)\), \(\Vmax\) | volts (V) |
| \(\omega\) | radians/sec (often written s\(^{-1}\) without the radians) |
| \(t\) | seconds (s) |
| \(\theta_v\) | radians (often written without units) |
We will make the steady-state assumption that the frequency \(\omega\) at all points in the electric grid are nominal, and that this nominal frequency is known to everyone. One period is \(T = \frac{2\pi \ \text{rad}}{\omega}\) seconds. In the USA, the nominal frequency of electricity is 60Hz, so
\[\begin{aligned} \omega &= 60 \cdot 2\pi = 120\pi \ \text{s}^{-1} \\ T &= \frac{2\pi}{120\pi} = \frac{1}{60} \ \text{s} \end{aligned}\]In reality, all electrical measurements such as voltage are real-valued. However, when working with AC power, it is often convenient to use complex-valued phasor representations of electrical quantities. For the sinusoidal voltage above, its phasor representation is defined as
\[V := \frac{\Vmax}{\sqrt 2} e^{\i \theta_v} \in \C.\]The phasor \(V\) has the same units (volts) as \(v(t)\) in the time domain. The magnitude of the phasor is the coefficient \(\abs{V} = \frac{\Vmax}{\sqrt 2}\).
Note that the phasor representation does not encode frequency information, which is fine because we know the nominal frequency. The relationship between the phasor representation and the time-domain representation is given by
\[\Re(\sqrt{2} V e^{\i \omega t}) = \Re(\Vmax e^{\i (\omega t + \theta_v)}) = \Vmax \cos(\omega t + \theta_v) = v(t).\]Likewise, we can define a phasor representation for electrical current (in units of ampere):
- time domain: \(i(t) = \Imax \cos(\omega t + \theta_i)\)
- phasor representation: \(I := \frac{\Imax}{\sqrt 2} e^{\i \theta_i}\)
Impedance
The classical dynamics equations for resistors, inductors, and capacitors can be written using phasor domain representations.
| time domain (\(\R\)) | phasor domain (\(\C\)) | complex impedance \(Z\) (\(\C\)) | |
|---|---|---|---|
| resistor | \(v(t) = R \cdot i(t)\) | \(V = R \cdot I\) | \(Z_R = R\) |
| inductor | \(v(t) = L \cdot \deriv{}{t} i(t)\) | \(V = (\i \omega L) I\) | \(Z_L = \i \omega L\) |
| capacitor | \(i(t) = C \cdot \deriv{}{t} v(t)\) | \(V = (\i \omega C)^{-1} I\) | \(Z_C = (\i \omega C)^{-1}\) |
The phasor descriptions of resistors, inductors, and capacitors obey Ohm’s law in phasor form, \(V = Z \cdot I\), where \(Z \in \C\) is called the impedance (in Ohms \(\Omega\)) of the circuit element. (See proofs below.)
Re-arranging Ohm’s law yields \(Z = \frac{V}{I} = \frac{\abs{V}}{\abs{I}} e^{\i (\theta_v - \theta_i)}\), with \(\abs{Z} = \frac{\abs{V}}{\abs{I}}\) and \(\angle Z = \theta_v - \theta_i\).
Because impedance \(Z\) is complex, it can be written as \(Z = r + \i x\). The real and imaginary components are given names:
- resistance (in \(\Omega\)), the real component, \(r := \Re(Z)\)
- reactance (in \(\Omega\)), the imaginary component, \(x := \Im(Z)\)
The inverse of impedance is admittance (in siemens S, or equivalently \(\Omega^{-1}\)): \(Y = Z^{-1} \in \C\). Likewise, the real and imaginary components of admittance \(Y = g + \i b\) are given names:
- conductance (in siemens): \(g := \Re(Y)\)
- susceptance (in siemens): \(b := \Im(Y)\)
Proof for resistor. From classic Ohm’s law, we have \(v(t) = R \cdot i(t)\). Now, substitute in the phasor representations:
\[\Re(\sqrt{2} V e^{\i \omega t}) = R \cdot \Re(\sqrt{2} I e^{\i \omega t}).\]The “equivalent complex numbers” lemma then implies \(V = R \cdot I\). \(\blacksquare\)
Proof for inductor. Writing \(v(t)\) in terms of its phasor representation \(V\) gives
\[v(t) = \Re(\sqrt{2} V e^{\i \omega t}).\]Writing \(L \cdot \deriv{}{t} i(t)\) in terms of the current phasor \(I\), and using \(-\sin x = \cos(x + \pi/2)\), we have
\[\begin{aligned} L \cdot \deriv{}{t} i(t) &= L \cdot \deriv{}{t} \Imax \cos(\omega t + \theta_i) \\ &= -L \omega \Imax \sin(\omega t + \theta_i) \\ &= \omega L \Imax \cos\left(\omega t + \theta_i + \frac{\pi}{2}\right) \\ &= \Re\left(\omega L \Imax e^{\i \left(\omega t + \theta_i + \frac{\pi}{2}\right)}\right) \\ &= \Re(\sqrt{2} \omega L I e^{\i \omega t} e^{\i \pi / 2}) \\ &= \Re(\sqrt{2} (\i \omega L) I e^{\i \omega t}). \end{aligned}\]Since \(v(t) = L \cdot \deriv{}{t} i(t)\), we can apply the “equivalent complex numbers” lemma which implies \(V = (\i \omega L) I\). \(\blacksquare\)
Proof for capacitor. Similar to proof for inductor. \(\blacksquare\)
Complex power
Power is defined as voltage \(\times\) current. In the time domain, instantaneous power (in watts W) is
\[\begin{aligned} p(t) &:= v(t) i(t) = \Vmax \cos(\omega t + \theta_v) \cdot \Imax \cos(\omega t + \theta_i) \\ &= \frac{\Vmax \Imax}{2} [\cos(\theta_v - \theta_i) + \cos(2 \omega t + \theta_v + \theta_i)] \end{aligned}\]where we use the identity \(\cos x \cos y = \frac{1}{2}[\cos(x-y) + \cos(x+y)]\). Let \(\phi := \theta_v - \theta_i\) represent the phase difference. Then, we define complex power as
\[\begin{aligned} S &:= V I^* = \frac{\Vmax}{\sqrt 2} e^{\i \theta_v} \cdot \frac{\Imax}{\sqrt 2} e^{-\i \theta_i} \\ &= \frac{\Vmax \Imax}{2} e^{\i (\theta_v - \theta_i)} = \abs{V} \abs{I} e^{\i \phi}. \end{aligned}\]Complex power has units volt-ampere (V·A), which is dimensionally equivalent to watt (W) but emphasizes that the quantity is complex power instead of instantaneous power. From complex power, we can define 3 quantities:
- active power (in W), the real component, \(P := \Re(S) = \abs{V} \abs{I} \cos\phi\)
- reactive power (in var), the complex component, \(Q := \Im(S) = \abs{V} \abs{I} \sin\phi\)
- apparent power (in V·A), the magnitude, \(\abs{S} = \abs{V} \abs{I}\)
Thus, we have \(S = P + \i Q\). Note that \(S\) is not a phasor, since \(\sqrt{2} \abs{S} \cos(\omega t + \phi) \neq p(t)\).
The unit “var” is an abbreviation of volt-ampere reactive, which is also dimensionally equivalent to watt (W) but emphasizes that the quantity is reactive power.
DistFlow Equations
Now, we can finally write down the nonlinear DistFlow equations. Consider a radial electricity distribution network \(G = (\Ncal, \Ecal)\) where \(\Ncal = \{0, 1, \dotsc, n\}\) is a set of nodes (called buses) and \(\Ecal \subset \Ncal \times \Ncal\) is a set of directed edges (called lines) that point away from bus 0. Let bus 0 (called the slack bus) be the root of the tree. Since the network is tree-structured, there are \(\abs\Ecal = n\) lines since every bus has exactly 1 parent, except bus 0 which has no parent.
For every bus \(j \in \Ncal\) and line \(j \to k \in \Ecal\), we write:
- squared voltage magnitude \(v_j = \abs{V_j}^2\) (units V2)
- squared current magnitude \(l_{jk} = \abs{I_{jk}}^2\) (units A2)
- complex impedance \(z_{jk} = r_{jk} + \i x_{jk}\) (units \(\Omega\))
- complex power flow \(S_{jk} = P_{jk} + \i Q_{jk}\) (units V·A)
- complex power injection \(s_j = p_j + \i q_j\) (units V·A)
The DistFlow model comprises 3 equations, where bus \(i\) is the parent of bus \(j\):
\[\begin{aligned} \sum_{k: j \to k} S_{jk} &= S_{ij} - z_{ij} l_{ij} + s_j && \forall j \in \Ncal & (1) \\ v_j - v_k &= 2 \Re(z_{jk}^* S_{jk}) - \abs{z_{jk}}^2 l_{jk} && \forall (j \to k) \in \Ecal & (2) \\ v_j l_{jk} &= \abs{S_{jk}}^2 && \forall (j \to k) \in \Ecal & (3) \end{aligned}\]Equation (1) enforces power balance. The LHS is power leaving bus \(j\). The RHS consists of three terms: power sent from the parent bus \(i\) minus the power lost to transmission, and the power injected at bus \(j\). When \(j = 0\), we set \(S_{i0} = 0\) and \(l_{i0} = 0\).
Equation (2) can be derived from the phasor form of Ohm’s law and by using the definition of complex power (\(S_{jk} = V_j I_{jk}^*\)):
\[\begin{aligned} && V_j - V_k &= I_{jk} z_{jk} \\ &\implies & V_k &= V_j - I_{jk} z_{jk} \\ &\implies & v_k &= \abs{V_k}^2 = \abs{V_j - I_{jk} z_{jk}}^2 \\ &&&= \abs{V_j}^2 - 2 \Re(V_j (I_{jk} z_{jk})^*) + \abs{I_{jk} z_{jk}}^2 \\ &&&= v_j - 2 \Re(V_j I_{jk}^* z_{jk}^*) + \abs{z_{jk}}^2 l_{jk} \\ &&&= v_j - 2 \Re(z_{jk}^* S_{jk}) + \abs{z_{jk}}^2 l_{jk} \end{aligned}\]Equation (3) comes from the definition of complex power:
\[\abs{S_{jk}}^2 = \abs{V_j I_{jk}^*}^2 = \abs{V_j}^2 \abs{I_{jk}^*}^2 = v_j l_{jk}.\]Equivalently, we can write the DistFlow equations in real form:
\[\begin{aligned} \sum_{k: j \to k} P_{jk} &= P_{ij} - r_{ij} l_{ij} + p_j && \forall j \in \Ncal & (a) \\ \sum_{k: j \to k} Q_{jk} &= Q_{ij} - x_{ij} l_{ij} + q_j && \forall j \in \Ncal & (b) \\ v_j - v_k &= 2 (r_{jk} P_{jk} + x_{jk} Q_{jk}) - (r_{jk}^2 + x_{jk}^2) l_{jk} && \forall (j \to k) \in \Ecal & (c) \\ v_j l_{jk} &= P_{jk}^2 + Q_{jk}^2 && \forall (j \to k) \in \Ecal & (d) \end{aligned}\]Often, Equation (d) is eliminated by substituting into Equations (a) and (b):
\[\begin{aligned} \sum_{k: j \to k} P_{jk} &= P_{ij} - r_{ij} \frac{P_{ij}^2 + Q_{ij}^2}{v_i} + p_j && \forall j \in \Ncal & (a') \\ \sum_{k: j \to k} Q_{jk} &= Q_{ij} - x_{ij} \frac{P_{ij}^2 + Q_{ij}^2}{v_i} + q_j && \forall j \in \Ncal & (b') \\ v_j - v_k &= 2 (r_{jk} P_{jk} + x_{jk} Q_{jk}) - (r_{jk}^2 + x_{jk}^2) l_{jk} && \forall (j \to k) \in \Ecal & (c) \\ \end{aligned}\]The power flow problem refers to solving for voltages and power flows (and optionally current), given impedances:
| complex form | real form | |
|---|---|---|
| given parameters | \(z_{jk} \in \C\) | \(r_{jk}, x_{jk} \in \R\) |
| variables | \(s \in \C^{n+1}, v \in \R^{n+1}, l \in \R^\abs\Ecal, S \in \C^\abs\Ecal\) | \(p,q,v \in \R^{n+1}\), \(l, P, Q \in \R^\abs\Ecal\) |
| typical problem | given \((v_0, s)\), solve for \((v, l, S)\) | given \((v_0, p, q)\), solve for \((v,l,P,Q)\) |
The radial DistFlow equations are nonlinear in the variables listed above:
- complex form: Equation (3) \(v_j l_{jk} = \abs{S_{jk}}^2\) is nonlinear
- real form: Equation (d) \(v_j l_{jk} = P_{jk}^2 + Q_{jk}^2\) is nonlinear, as are the substituted versions (a’) and (b’)
Simplified DistFlow Model
It is reasonable to assume that the line losses \(z_{jk} l_{jk}\) are small compared with the line power flows \(S_{jk}\). Therefore, we can approximate \(z_{jk} l_{jk} \approx 0\) which yields the simplified DistFlow model, also called the linearized DistFlow model. This model was first introduced in
M. Baran and F. F. Wu, “Optimal sizing of capacitors placed on a radial distribution system,” in IEEE Transactions on Power Delivery, vol. 4, no. 1, pp. 735-743, Jan. 1989, doi: 10.1109/61.19266. https://ieeexplore.ieee.org/document/19266.
In complex form:
\[\begin{aligned} \sum_{k: j \to k} S_{jk} &= S_{ij} + s_j && \forall j \in \Ncal & (1_\text{lin}) \\ v_j - v_k &= 2 \Re(z_{jk}^* S_{jk}) && \forall (j \to k) \in \Ecal & (2_\text{lin}) \end{aligned}\]Equivalently, in real form:
\[\begin{aligned} \sum_{k: j \to k} P_{jk} &= P_{ij} + p_j && \forall j \in \Ncal & (a_\text{lin}) \\ \sum_{k: j \to k} Q_{jk} &= Q_{ij} + q_j && \forall j \in \Ncal & (b_\text{lin}) \\ v_j - v_k &= 2 (r_{jk} P_{jk} + x_{jk} Q_{jk}) && \forall (j \to k) \in \Ecal & (c_\text{lin}) \end{aligned}\]Consider the typical problem in real form, where we are given the impedances \(r,x \in \R^\abs\Ecal\), the slack bus square voltage magnitude \(v_0 \in \R\), as well as the power injections \(p, q \in \R^{n+1}\). The \(3n+2\) equations are sufficient to solve for the \(3n\) remaining variables \((v_{1:n} \in \R^n,\ P,Q \in \R^\abs\Ecal = \R^n)\). (Here, we ignore the line currents \(l_{jk}\).)
These linear equations have a convenient matrix representation. Let \(C \in \{-1, 0, 1\}^{(n+1) \times n}\) denote the bus-by-line incidence matrix, defined by
\[C_{jl} := \begin{cases} 1, & \text{if bus $j$ is the start of line $l$} \\ -1, & \text{if bus $j$ is the end of line $l$} \\ 0, & \text{otherwise}. \end{cases}\]Define matrices \(D_r := \diag(r_l, l \in \Ecal),\ D_x := \diag(x_l, l \in \Ecal) \in \R^{n \times n}\). Then the linear DistFlow equations can be written as
\[\begin{aligned} s &= CS \\ C^\top v &= 2 (D_r P + D_x Q) \end{aligned}\]If we let \(c_0^\top\) denote the 1st row of the \(C\) matrix and \(\hat C \in \{-1, 0, 1\}^{n \times n}\) denote the remaining rows
\[C = \begin{bmatrix} c_0^\top \\ \hat C \end{bmatrix}\]then we have
\[\begin{aligned} s_0 &= c_0^\top S \\ s_{1:n} &= \hat{C} S \\ v_0 c_0 + \hat{C}^\top v_{1:n} &= 2 (D_r P + D_x Q) \end{aligned}\]For each bus \(j\), let \(\Pcal_j \subseteq \Ecal\) be the directed path from bus 0 to bus \(j\). We now show that \(\hat{C}\) is invertible.
Lemma: The inverse of \(\hat{C}\) is given by
\[[\hat{C}^{-1}]_{lj} = \begin{cases} -1, & \text{if } l \in \Pcal_j \\ 0, & \text{otherwise}. \end{cases}\]Proof
\[[\hat{C} \hat{C}^{-1}]_{jk} = \sum_{l \in \Ecal} \hat{C}_{jl} (\hat{C}^{-1})_{lk} = \sum_{l \in \Pcal_k} -\hat{C}_{jl}\]Now, consider 3 cases.
-
Suppose \(j = k\). Let bus \(i\) be the parent of bus \(j\). Then the only line in \(\Pcal_k\) that \(j\) is part of is \(l = (i \to j)\). Since \(j\) is the end of line \(l\), \(\hat{C}_{jl} = -1\). Thus, \([\hat{C} \hat{C}^{-1}]_{jj} = - \hat{C}_{j, (i \to j)} = -(-1) = 1\).
-
Suppose \(j \neq k\), and \(j\) is not on the path to \(k\). Then \(\hat{C}_{jl} = 0\) for every \(l \in \Pcal_k\). Thus, \([\hat{C} \hat{C}^{-1}]_{jk} = 0\).
-
Suppose \(j \neq k\), and \(j\) is on the path to \(k\). Then \(j\) is part of 2 lines in \(\Pcal_k\), once as the starting bus and once as the end bus. Thus, \([\hat{C} \hat{C}^{-1}]_{jk} = -\hat{C}_{j,(\cdot \to j)} - \hat{C}_{j,(j \to \cdot)} = -(-1) - 1 = 0\).
We have shown that \([\hat{C} \hat{C}^{-1}]_{jk} = \one[j=k]\) as desired. \(\blacksquare\)
Since \(\hat{C}\) is invertible, we can directly solve for \((v_{1:n}, S)\) as
\[\begin{aligned} S &= \hat{C}^{-1} s_{1:n} \\ v_{1:n} &= \hat{C}^{-\top}(2 D_r P + 2 D_x Q - v_0 c_0) \end{aligned}\]We have
\[\begin{aligned} (\hat{C}^{-\top} c_0)_j &= \sum_{l \in \Ecal} (\hat{C}^{-\top})_{jl} (c_0)_l = \sum_{l \in \Ecal} (\hat{C}^{-1})_{lj} C_{0l} \\ &= \sum_{(0 \to k) \in \Ecal} (\hat{C}^{-1})_{(0 \to k),j} = \sum_{(0 \to k) \in \Pcal_j} -1 \\ &= -1 \end{aligned}\]since exactly one line of the form \((0 \to k)\) is part of \(\Pcal_j\). Thus, \(\hat{C}^{-\top} c_0 = -\one\). Furthermore, in real form, the equation \(S = \hat{C}^{-1} s_{1:n}\) splits into \(P = \hat{C}^{-1} p_{1:n}\) and \(Q = \hat{C}^{-1} q_{1:n}\). Substituting these into the equation for \(v_{1:n}\) yields
\[\begin{aligned} v_{1:n} &= 2 \hat{C}^{-\top} D_r \hat{C}^{-1} p_{1:n} + 2 \hat{C}^{-\top} D_x \hat{C}^{-1} q_{1:n} + v_0 \one \\ &= v_0 \one + 2 (R p_{1:n} + X q_{1:n}) \end{aligned}\]where \(R := \hat{C}^{-\top} D_r \hat{C}^{-1}\) and \(X := \hat{C}^{-\top} D_x \hat{C}^{-1}\). Since every \(r_{jk}, x_{jk} > 0\), the matrices \(D_r, D_x\) are symmetric positive definite. Thus, \(R, X \succ 0\), since
\[\forall u \neq \zero:\quad u^\top R u = u^\top \hat{C}^{-\top} D_r \hat{C}^{-1} u = (\hat{C}^{-1} u)^\top D_r (\hat{C}^{-1} u) > 0.\]Finally, we show that the entries of \(R\) and \(X\) have a simple interpretation in terms of the resistance and reactance of the lines.
\[\begin{aligned} R_{jm} &= [\hat{C}^{-\top} D_r \hat{C}^{-1}]_{jm} = (\hat{C}^{-\top})_{j,:} D_r (\hat{C}^{-1})_{:,m} \\ &= \sum_{l \in \Ecal} r_l (\hat{C}^{-\top})_{j,l} (\hat{C}^{-1})_{l,m} \\ &= \sum_{l \in \Ecal} r_l (\hat{C}^{-1})_{l,j} (\hat{C}^{-1})_{l,m} \\ &= \sum_{l \in \Pcal_j \cap \Pcal_m} r_l (-1) (-1) \\ &= \sum_{l \in \Pcal_j \cap \Pcal_m} r_l \end{aligned}\]and likewise, \(X_{jm} = \sum_{l \in \Pcal_j \cap \Pcal_m} x_l\).
Note that it is equivalent to move the factor of 2 into the definition of the \(R\) and \(X\) matrices so that
\[\begin{aligned} v_{1:n} &= v_0 \one + R p_{1:n} + X q_{1:n} \\ R_{jm} &:= 2 \sum_{l \in \Pcal_j \cap \Pcal_m} r_l \\ X_{jm} &:= 2 \sum_{l \in \Pcal_j \cap \Pcal_m} x_l. \end{aligned}\]