Cooperative Control for DC Microgrids

Abstract

Similar to the control hierarchy of the AC systems, a hierarchical control structure is conventionally adopted for DC microgrid operation. The highest hierarchy, the tertiary control , is in charge of economical operation and coordination with the distribution system operator. It assigns the microgrid voltage to carry out the scheduled power exchange between the microgrid and the main grid.

Appendix

7.1.1 Dynamic Consensus

Following lemmas need to be studied before studying the dynamic consensus:

Lemma 7.1

[37]: Assume that the digraph G has a spanning tree . Then, the Laplacian matrix L has a simple eigen-value at the origin, i.e., $\lambda_{1} = 0$ , and other eigen-values lie in the Open Right Hand Plane (ORHP). In addition,

$$ \mathop {\lim }\limits_{t \to \infty } e^{{ - {\mathbf{L}}t}} = w_{r} w_{l}^{\text{T}} , $$

(7.50)

where $w_{r} \in {\mathbb{R}}^{N \times 1}$ and $w_{l}^{\text{T}} \in {\mathbb{R}}^{N \times 1}$ are the right and left eigen-vectors of L associated with $\lambda_{1} = 0$, respectively. It should be noted that $w_{l}^{\text{T}}$ should be normalized with respect to $w_{r}$ , i.e., $w_{l}^{\text{T}} w_{r} = 1$.

Lemma 7.2

Assume that the digraph G has a spanning tree and the Laplacian matrix, L , is balanced. Then,

$$ \mathop {\lim }\limits_{s \to 0} s(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} = {\mathbf{Q}}. $$

(7.51)

$$ \mathop {\lim }\limits_{s \to 0} {\mathbf{L}}(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} = \mathop {\lim }\limits_{s \to 0} (s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} {\mathbf{L}} = {\mathbf{I}}_{N} - {\mathbf{Q}}. $$

(7.52)

where Q is the averaging matrix defined in Sect. 7.1.3

Proof of Lemma 7.2: Assume a linear system of ${\dot{\mathbf{x}}} = - {\mathbf{Lx}}$ with ${\mathbf{x}}(0) \ne 0$ and ${\mathbf{x}} \in {\mathbb{R}}^{N \times 1}$. One can write

$$ {\mathbf{x}}(t) = e^{{ - {\mathbf{L}}t}} {\mathbf{x}}(0). $$

(7.53)

Or, equivalently, in the frequency domain,

$$ {\mathbf{X}} = (s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} {\mathbf{x}}(0). $$

(7.54)

Lemma 7.1 ensures that X is a type 1 vector; i.e., it has a single pole at the origin and all other poles lie in the OLHP. Thus, using the final value theorem,

$$ \mathop {\lim }\limits_{t \to \infty } {\mathbf{x}}(t) = \mathop {\lim }\limits_{s \to 0} s{\mathbf{X}} = \left( {\mathop {\lim }\limits_{s \to 0} s(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} } \right){\mathbf{x}}(0). $$

(7.55)

On the other hand, by using Lemma 7.1, (7.53) yields to

$$ \mathop {\lim }\limits_{t \to \infty } {\mathbf{x}}(t) = \left( {\mathop {\lim }\limits_{t \to \infty } e^{{ - {\mathbf{L}}t}} } \right){\mathbf{x}}(0) = w_{r} w_{l}^{\text{T}} {\mathbf{x}}(0). $$

(7.56)

For any Laplacian matrix L, all row sums are equal to zero. Thus, $w_{r} = \underline{{\mathbf{1}}}$. In addition, for any balanced L, all column sums are also equal to zero. Thus, $w_{l} = (1/N)\underline{{\mathbf{1}}}$. Accordingly, (7.56) implies that

$$ \mathop {\lim }\limits_{t \to \infty } {\mathbf{x}}(t) = {\mathbf{Qx}}(0). $$

(7.57)

Comparing (7.55)–(7.57),

$$ \left( {\mathop {\lim }\limits_{s \to 0} s(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} } \right){\mathbf{x}}(0) = {\mathbf{Qx}}(0). $$

(7.58)

Since (7.58) holds for all ${\mathbf{x}}(0) \ne 0$, one may conclude (7.51). In addition,

$$ {\mathbf{I}}_{N} = \mathop {\lim }\limits_{s \to 0} (s{\mathbf{I}}_{N} + {\mathbf{L}})(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} = \mathop {\lim }\limits_{s \to 0} s(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} + \mathop {\lim }\limits_{s \to 0} {\mathbf{L}}(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} . $$

(7.59)

Comparing (7.51) with (7.59) concludes (7.52).

Theorem 7.1

Assume that the communication graph G , used in a cooperative control system, has a spanning tree and the associated Laplacian matrix, L , is balanced. Then, using the observer in (7.7), all the estimated averages in ${\bar{\mathbf{v}}}$ converge to the true global average voltage .

Proof of Theorem 7.1: Equation (7.34) shows the global dynamic of the microgrid , when the cooperative controller is effective. It is assumed that the system parameters are, accordingly, designed to stabilize the microgrid. Thus, the resulting voltage vector, V, is a type 1 vector. Based on Lemma 7.1, all poles of the term $s(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1}$ lie in the OLHP. It should be noted that if $\lambda_{i}$ is an eigenvalue of L, then $s = - \lambda_{i}$ is a pole of $s(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1}$. The term s in $s(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1}$ cancels the pole of $(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1}$ at the origin. Thus, (7.7) implies that $\overline{{\mathbf{V}}}$ is also a type 1 vector. Since both V and $\overline{{\mathbf{V}}}$ are type 1, one may use the final value theorem

$$ \mathop {\lim }\limits_{t \to \infty } {\bar{\mathbf{v}}}(t) = \mathop {\lim }\limits_{s \to 0} s{\bar{\mathbf{V}}} = \mathop {\lim }\limits_{s \to 0} s(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} (s{\mathbf{V}}). $$

(7.60)

Using Lemma 7.2,

$$ \mathop {\lim }\limits_{t \to \infty } {\bar{\mathbf{v}}}(t) = \mathop {\lim }\limits_{s \to 0} s(s{\mathbf{I}}_{N} + {\mathbf{L}})^{ - 1} \times \mathop {\lim }\limits_{s \to 0} (s{\mathbf{V}}) = {\mathbf{Q}} \times \mathop {\lim }\limits_{t \to \infty } {\mathbf{v}} = {\mathbf{Qv}}^{\text{ss}} = \left\langle {{\mathbf{v}}^{\text{ss}} } \right\rangle \underline{{\mathbf{1}}} . $$

(7.61)

Equation (7.61) implies that all estimations converge to the true global average voltage . In other words,

$$ \forall i:0 \le i \le N,\quad \mathop {\lim }\limits_{t \to \infty } \bar{v}_{i} (t) = \frac{1}{N}\sum\limits_{i = 1}^{N} {v_{i} (t)} . $$

(7.62)

7.1.2 Analysis of the Noise Cancelation Module

Following lemmas need to be studied before analyzing the noise cancelation module:

Lemma 7.3

For a given matrix ${\mathbf{A}} \in {\mathbb{R}}^{N \times N}$ , if ${\mathbf{I}}_{N} + {\mathbf{A}}$ is invertible then,

$$ ({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} = {\mathbf{I}}_{N} - {\mathbf{A}}({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} = {\mathbf{I}}_{N} - ({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} {\mathbf{A}}. $$

(7.63)

Proof of Lemma 7.3: For a given matrix ${\mathbf{A}} \in {\mathbb{R}}^{N \times N}$,

$$ ({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} = ({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} ({\mathbf{I}}_{N} + {\mathbf{A}}) - ({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} {\mathbf{A}} = {\mathbf{I}}_{N} - ({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} {\mathbf{A}}. $$

Lemma 7.4

For a given invertible matrix ${\mathbf{A}} \in {\mathbb{R}}^{N \times N}$ if ${\mathbf{I}}_{N} + {\mathbf{A}}$ is invertible then,

$$ ({\mathbf{I}}_{N} + {\mathbf{A}}^{ - 1} )^{ - 1} = {\mathbf{A}}({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} = ({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} {\mathbf{A}}. $$

(7.64)

Proof of Lemma 7.4: For a given invertible matrix ${\mathbf{A}}$,

$$ ({\mathbf{I}}_{N} + {\mathbf{A}}^{ - 1} )^{ - 1} = ({\mathbf{AA}}^{ - 1} + {\mathbf{A}}^{ - 1} )^{ - 1} = \left( {({\mathbf{I}}_{N} + {\mathbf{A}}){\mathbf{A}}^{ - 1} } \right)^{ - 1} = {\mathbf{A}}({\mathbf{I}}_{N} + {\mathbf{A}})^{ - 1} . $$

Lemma 7.5

If L is a balanced Laplacian matrix, $b > 0$ , and ${\mathbf{K}} = {\text{diag}}\{ k_{i} \}$ has positive diagonal elements then, ${\mathbf{L}}^{\prime } = b{\mathbf{LK}}^{ - 1} {\mathbf{L}}$ is a balanced Laplacian matrix.

Proof of Lemma 7.5: The matrix L is said to be a Laplacian matrix if a communication graph exists with the associated Laplacian matrix L. Equivalently, a matrix is a Laplacian matrix if and only if ${\mathbf{L}}\underline{{\mathbf{1}}} = {\mathbf{0}}$. A Laplacian matrix is balanced if it has all column sums of zero, i.e., $\underline{{\mathbf{1}}}^{\text{T}} {\mathbf{L}} = {\mathbf{0}}$. Let L be a balanced Laplacian matrix and ${\mathbf{L}}^{{\prime }} = b{\mathbf{LK}}^{ - 1} {\mathbf{L}}$, then

$$ {\mathbf{L}}^{{\prime }} \underline{{\mathbf{1}}} = b{\mathbf{LK}}^{ - 1} \left( {{\mathbf{L}}\underline{{\mathbf{1}}} } \right) = {\mathbf{0}}, $$

(7.65)

which implies that ${\mathbf{L}}^{{\prime }}$ is a Laplacian matrix. On the other hand,

$$ \underline{{\mathbf{1}}}^{\text{T}} {\mathbf{L}}^{{\prime }} = b\left( {\underline{{\mathbf{1}}}^{\text{T}} {\mathbf{L}}} \right){\mathbf{K}}^{ - 1} {\mathbf{L}} = {\mathbf{0}}, $$

(7.66)

which shows that ${\mathbf{L}}^{{\prime }}$ is also balanced.

Lemma 7.6

If L is a balanced Laplacian matrix, $b > 0$ , and ${\mathbf{K}} = {\text{diag}}\{ k_{i} \}$ has positive diagonal elements then,

$$ \mathop {\lim }\limits_{s \to \infty } s\left( {s{\mathbf{I}}_{N} + \left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right){\mathbf{K}}^{ - 1} \left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)} \right)^{ - 1} = {\mathbf{Q}}{\mathbf{.}} $$

(7.67)

$$ \mathop {\lim }\limits_{s \to \infty } s\left( {{\mathbf{I}}_{N} + s\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)^{ - 1} {\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} } \right)^{ - 1} = {\mathbf{I}}_{N} - {\mathbf{Q}}{\mathbf{.}} $$

(7.68)

Proof of Lemma 7.6: Let L be a balanced Laplacian matrix, $b > 0$, and ${\mathbf{K}} = {\text{diag}}\left\{ {k_{i} } \right\}$ has positive diagonal elements. Then,

$$ \mathop {\lim }\limits_{s \to \infty } \left( {\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right){\mathbf{K}}^{ - 1} \left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)} \right) = b{\mathbf{LK}}^{ - 1} {\mathbf{L}}. $$

(7.69)

Let us define ${\mathbf{L}}^{{\prime }} = b{\mathbf{LK}}^{ - 1} {\mathbf{L}}$. Then, using (7.69),

$$ \mathop {\lim }\limits_{s \to 0} s\left( {s{\mathbf{I}}_{N} + \left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right){\mathbf{K}}^{ - 1} \left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)} \right)^{ - 1} = \mathop {\lim }\limits_{s \to 0} s\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}^{{\prime }} } \right)^{ - 1} . $$

(7.70)

Lemma 7.5 ensures that ${\mathbf{L}}^{{\prime }}$ is a balanced Laplacian matrix. Therefore, by applying Lemma 7.2 (7.51), one can write $\lim_{s \to 0} s\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}^{{\prime }} } \right)^{ - 1} = {\mathbf{Q}}$, which, together with (7.70), proves (7.67).

To study the second part of the Lemma, (7.68), one may note that for $s \ne 0$, $s{\mathbf{I}}_{N} + b{\mathbf{L}}$ is invertible [46]. K is also invertible, and ${\mathbf{K}}^{ - 1} = {\text{diag}}\left\{ {k_{i}^{ - 1} } \right\}$. Let us define

$$ \Gamma \triangleq \left( {{\mathbf{I}}_{N} + s\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)^{ - 1} {\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} } \right)^{ - 1} . $$

(7.71)

Using Lemma 7.3,

$$ \Gamma = {\mathbf{I}}_{N} - s\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)^{ - 1} {\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} \left( {{\mathbf{I}}_{N} + s\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)^{ - 1} {\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} } \right)^{ - 1} . $$

(7.72)

Lemma 7.4 offers to further expand (7.72)

$$ \begin{aligned}\Gamma & = {\mathbf{I}}_{N} - s\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)^{ - 1} {\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} \frac{1}{s}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right){\mathbf{K}}^{ - 1} \left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right) \\ & \quad \times \left( {{\mathbf{I}}_{N} + \frac{1}{s}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right){\mathbf{K}}^{ - 1} \left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)} \right)^{ - 1} \\ & = {\mathbf{I}}_{N} - s\left( {s{\mathbf{I}}_{N} + \left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right){\mathbf{K}}^{ - 1} \left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)} \right)^{ - 1} . \\ \end{aligned} $$

(7.73)

By applying Lemma 7.6 (7.67)–(7.73), one can conclude $\lim_{s \to 0}\Gamma = {\mathbf{I}}_{N} - {\mathbf{Q}}$.

Theorem 7.2

Assume that the communication graph G , used in a distributed control system, has a spanning tree , and the associated Laplacian matrix, L , is balanced. Then, using the total observer in (7.21)–(7.23), all the estimated averages in $\overline{\text{v}}$ converge to the true global average voltage average.

Proof of Theorem 7.2: For any $s \ne 0$, $s{\mathbf{I}}_{N} + b{\mathbf{L}}$ is invertible [46]. The integrator gain matrix, K, is also invertible, and ${\mathbf{K}}^{ - 1} = {\text{diag}}\left\{ {k_{i}^{ - 1} } \right\}$. Thus, one can safely reformulate the total observer transfer function as

$$ \begin{aligned} {\mathbf{H}}_{\text{obs}}^{\text{F}} & = \left( {\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right) + s{\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} } \right)^{ - 1} \left( {\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right) + s{\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} } \right) - {\mathbf{L}} \\ & = {\mathbf{I}}_{N} - \left( {\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right) + s{\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} } \right)^{ - 1} {\mathbf{L}} \\ & = {\mathbf{I}}_{N} - \left( {\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)\left( {{\mathbf{I}}_{N} + s\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)^{ - 1} {\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} } \right)} \right)^{ - 1} {\mathbf{L}} \\ & = {\mathbf{I}}_{N} - \left( {{\mathbf{I}}_{N} + s\left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)^{ - 1} {\mathbf{K}}\left( {s{\mathbf{I}}_{N} + b{\mathbf{L}}} \right)^{ - 1} } \right)^{ - 1} \left( {s{\mathbf{I}}_{N} + {\mathbf{L}}} \right)^{ - 1} {\mathbf{L}}. \\ \end{aligned} $$

(7.74)

Using Lemma 7.2 (7.52) and Lemma 7.6 (7.68), the total observer DC gain can be found

$$ \mathop {\lim }\limits_{s \to 0} H_{\text{obs}}^{\text{F}} = {\mathbf{I}}_{N} - \left( {{\mathbf{I}}_{N} - {\mathbf{Q}}} \right)^{2} = 2{\mathbf{Q}} - {\mathbf{Q}}^{2} = {\mathbf{Q}}. $$

(7.75)

Therefore, for type 1 (DC and exponentially damping) disturbances , (7.21) yields to

$$ \begin{aligned} \mathop {\lim }\limits_{t \to \infty } \overline{{\mathbf{v}}} (t) & = \mathop {\lim }\limits_{s \to 0} {\mathbf{H}}_{\text{obs}}^{\text{F}} \times \mathop {\lim }\limits_{s \to 0} (s\overline{{\mathbf{V}}} ) \times \mathop {\lim }\limits_{s \to 0} {\mathbf{H}}_{\text{NC}} \times \mathop {\lim }\limits_{s \to 0} (s{\mathbf{D}}) \\ & = {\mathbf{Q}} \times \mathop {\lim }\limits_{t \to \infty } {\mathbf{v}} + 0 = {\mathbf{Qv}} = \left\langle {{\mathbf{v}}^{\text{ss}} } \right\rangle \underline{1} , \\ \end{aligned} $$

(7.76)

which proves the Theorem 7.2.

7.1.3 Microgrid Parameters

Each of the underlying buck converters has $L = 2.64\;{\text{mH}}$ and $C = 2.2\;{\text{mF}}$ and works with the switching frequency of $F_{s} = 60\;{\text{kHz}}$. Transmission lines series impedances are $Z_{12} = Z_{34} = Z_{b}$ and $Z_{25} = Z_{35} = Z_{b}$, where the base impedance is $Z_{b} = 0.5 + (50\;{{\upmu{\text{H}}}})s$. The circuit model of the line includes 22 nF of capacitance on either end. Impedances of the local loads are $R = 30\;\Omega$ and $R_{2} = R_{3} = R_{4} = 20\;\Omega$. Voltages of the (rectified) input DC sources are $V_{{{\text{s}}1}} = V_{{{\text{s}}4}} = 80\;{\text{V}}$ and $V_{\text{s2}} = V_{\text{s3}} = 100\;{\text{V}}$. The control parameters are as follow:

$$ {\mathbf{I}}_{\text{rated}} = {\text{diag}}\{ 6,3,3,6\} , $$

(7.77)

$$ {\mathbf{A}}_{{\mathbf{G}}} = \left[ {\begin{array}{*{20}c} 0 & {90} & 0 & {110} \\ {90} & 0 & {100} & 0 \\ 0 & {100} & 0 & {120} \\ {110} & 0 & {120} & 0 \\ \end{array} } \right],\quad {\mathbf{r}}\left[ {\begin{array}{*{20}c} {0.5} & 0 & 0 & 0 \\ 0 & {1.0} & 0 & 0 \\ 0 & 0 & {1.0} & 0 \\ 0 & 0 & 0 & {0.5} \\ \end{array} } \right], $$

(7.78)

$$ b = 1,\quad c = 0.075, $$

(7.79)

$$ {\mathbf{H}}_{\text{p}} = \left[ {\begin{array}{*{20}c} {0.1} & 0 & 0 & 0 \\ 0 & {0.09} & 0 & 0 \\ 0 & 0 & {0.08} & 0 \\ 0 & 0 & 0 & {0.11} \\ \end{array} } \right],\quad {\mathbf{H}}_{\text{I}} = \left[ {\begin{array}{*{20}c} 6 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 0 & 0 & {5.4} & 0 \\ 0 & 0 & 0 & {5.6} \\ \end{array} } \right], $$

(7.80)

$$ {\mathbf{G}}_{\text{p}} = \left[ {\begin{array}{*{20}c} {1.1} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & {1.2} & 0 \\ 0 & 0 & 0 & {1.1} \\ \end{array} } \right],\quad {\mathbf{G}}_{\text{I}} = \left[ {\begin{array}{*{20}c} 7 & 0 & 0 & 0 \\ 0 & {7.4} & 0 & 0 \\ 0 & 0 & {6.6} & 0 \\ 0 & 0 & 0 & 7 \\ \end{array} } \right], $$

(7.81)

$$ {\mathbf{K}}_{\text{p}} = \left[ {\begin{array}{*{20}c} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4 \\ \end{array} } \right]. $$

(7.82)