Robust Optimal Multi-agent-Based Distributed Control Scheme for Distributed Energy Storage System

Abstract

The multi-agent system is emerging as an effecting tool for the realization of the smart power distribution system. The smart power distribution system comprises the different scattered entities such as grid supply, renewable generations, customers, etc. In this distributed system, with the use of information and communication technology (ICT) and control systems, multi-agent system can implement different control and management schemes. The renewable generations such as solar photovoltaic (PV), and wind, as well as electrical load are associated with uncertainties. In this chapter, different battery agents are designed to work for scattered distributed battery energy storage system (BESS). These battery agents decide the power exchange for charging and discharging of BESS in order to balance the power mismatch and cater uncertainties in the smart power distribution system. The LQR-based distributed robust optimal control schemes are designed for battery agents to achieve the objective of balancing the power mismatch in the presence of uncertainties. The proposed control schemes show that the effects of uncertainties in power distribution system, in terms of power and energy sharing, are distributed and catered by all energy storage devices as per their energy storing capacities.

Appendix

13.1.1 Proof of theorem

Let the error system for (13.26) be

$$ {\displaystyle \begin{array}{l}e\left(k+1\right)={A}_xe(k)+{B}_xU(k),\\ {}U(k)=-{L}_xe(k)\end{array}} $$

(13.45)

where A_x =  diag (A_x,1, A_x,2, …, A_x,n) and B_x =  diag (B_x,1, B_x,2, …, B_x,n) and

$$ e(k)=X(k)-{1}_n\otimes U(k) $$

(13.46)

The modified LQR-based optimal control problem is

$$ \underset{U(k)}{\min }\ \sum \limits_{k=0}^{\infty }e{(k)}^T Qe(k)+U{(k)}^T RU(k) $$

(13.47)

For the system X(k+1) = AX(k)+BU(k), the discrete time ARE is

$$ {A}^T PA-P+Q-{A}^T PB{\left(R+{B}^T PB\right)}^{-1}{B}^T PA=0 $$

(13.48)

For the system (13.45), the ARE is

$$ Q={PB}_x{\left(R+{B}_x^2P\right)}^{-1}{B}_xP $$

(13.49)

where A = A_x = 1 and B = B_x.

Let B_x,1 = B_x,2 = … = B_x,n = B_x,0

Then matrix B_x = B_x,0I_n

The optimal feedback gain matrix is

$$ {L}_x={\left(R+{B}_{x,0}^2{I}_nP\right)}^{-1}{B}_{x,0}{I}_nP $$

(13.50)

Multiply R⁻¹ both sides of (13.49) then

$$ {R}^{-1}Q={R}^{-1}{PB}_{x,0}{I}_n{\left(R+{B}_{x,0}^2{I}_nP\right)}^{-1}{B}_{x,0}{I}_nP $$

(13.51)

$$ {R}^{-1}Q={R}^{-1}{PB}_{x,0}{\left({I}_n+{B}_{x,0}^2{R}^{-1}P\right)}^{-1}{B}_{x,0}{R}^{-1}P $$

(13.52)

Since it is known that

$$ {\left({I}_n+{B}_{x,0}^2{R}^{-1}P\right)}^{-1}={I}_n-{B}_{x,0}^2{R}^{-1}P{\left({I}_n+{B}_{x,0}^2{R}^{-1}P\right)}^{-1} $$

(13.53)

We now get

$$ {\displaystyle \begin{array}{l}{R}^{-1}{PB}_{x,0}\left({I}_n+{B}_{x,0}^2{R}^{-1}P\right){B}_{x,0}{R}^{-1}P={\left[{B}_{x,0}{R}^{-1}P\right]}^2-{B}_{x,0}^2{R}^{-1}P\\ {}\times \left[{R}^{-1}{PB}_{x,0}{\left({I}_n+{B}_{x,0}^2{R}^{-1}P\right)}^{-1}{B}_{x,0}{R}^{-1}P\right]\end{array}} $$

(13.54)

$$ {R}^{-1}Q={B}_{x,0}^2{\left({R}^{-1}P\right)}^2-{B}_{x,0}^2{R}^{-1}{PR}^{-1}Q $$

(13.55)

On simplification it is obtained that

$$ {R}^{-1}P=\frac{1}{2}\left[{R}^{-1}Q+\sqrt{{\left({R}^{-1}Q\right)}^2+\frac{4{R}^{-1}Q}{B_x^2}}\right] $$

(13.56)

Hence, the optimal feedback gain matrix is

$$ {L}_x^{\ast }=\frac{B_{x,0}}{2}\left[\sqrt{{\left({R}^{-1}Q\right)}^2+\frac{4{R}^{-1}Q}{B_{x,0}^2}}-{R}^{-1}Q\right] $$

(13.57)

Let

$$ {L}_x^{\ast }=\mathit{\operatorname{diag}}\left({d}_{x,1},{d}_{x,2}\dots, {d}_{x,n}\right) $$

(13.58)

Then, for ith agent the feedback gain is

$$ {d}_{x,\mathrm{i}}=\frac{B_{x,0}}{2}\left[\sqrt{{\left(\frac{q_i}{r_i}\right)}^2+\frac{4{q}_i}{B_{x,0}^2{r}_i}}-\frac{q_i}{r_i}\right] $$

(13.59)

Similarly, it can be proved that

$$ {d}_{y,\mathrm{i}}=\frac{B_{y,0}}{2}\left[\sqrt{{\left(\frac{q_i}{r_i}\right)}^2+\frac{4{q}_i}{B_{y,0}^2{r}_i}}-\frac{q_i}{r_i}\right] $$

(13.60)

Let

$$ {B}_{x,0}^2=\frac{4}{3}\frac{r_i}{q_i} $$

(13.61)

Then from (13.59)

$$ {d}_{x,\mathrm{i}}=\frac{B_{x,0}}{2}\left[\sqrt{{\left(\frac{q_i}{r_i}\right)}^2+3{\left(\frac{q_i}{r_i}\right)}^2}-\frac{q_i}{r_i}\right] $$

(13.62)

$$ {d}_{x,\mathrm{i}}=\frac{B_{x,0}}{2}\frac{q_i}{r_i} $$

(13.63)

Similarly, we can obtain

$$ {d}_{y,\mathrm{i}}=\frac{B_{y,0}}{2}\frac{q_i}{r_i} $$

(13.64)

While

$$ {B}_{y,0}^2=\frac{4}{3}\frac{r_i}{q_i} $$

(13.65)