**Appendix A**

An instantaneous power *p* can be expressed as the function of instantaneous values of voltages and currents:

$$p = \upsilon\_{A\dot{A}} + \upsilon\_{B\dot{B}} + \upsilon\_{C\dot{C}} \tag{A1}$$

We apply the known transformations between αβ and ABC phase coordinates:

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$$\begin{aligned} v\_A &= \operatorname{Re}\left\{\boldsymbol{\upsilon}\right\} \\ v\_B &= -\frac{1}{2}\operatorname{Re}\left\{\boldsymbol{\upsilon}\right\} + \frac{\sqrt{3}}{2}\operatorname{Im}\{\boldsymbol{\upsilon}\} \\\\ v\_C &= -\frac{1}{2}\operatorname{Re}\left\{\boldsymbol{\upsilon}\right\} - \frac{\sqrt{3}}{2}\operatorname{Im}\{\boldsymbol{\upsilon}\} \\ \dot{i}\_A &= \operatorname{Re}\left\{\dot{\imath}\right\} \\ \dot{i}\_B &= -\frac{1}{2}\operatorname{Re}\left\{\dot{\imath}\right\} + \frac{\sqrt{3}}{2}\operatorname{Im}\{\dot{\imath}\} \\ \dot{i}\_C &= -\frac{1}{2}\operatorname{Re}\left\{\dot{\imath}\right\} - \frac{\sqrt{3}}{2}\operatorname{Im}\{\dot{\imath}\} \end{aligned} \tag{A2}$$

Applying (A2) to (A1)

$$\begin{split} p &= \operatorname{Re}\left\{ \boldsymbol{\upsilon} \right\} \times \operatorname{Re}\left\{ \boldsymbol{i} \right\} + \left( -\frac{1}{2} \operatorname{Re}\left\{ \boldsymbol{\upsilon} \right\} + \frac{\sqrt{3}}{2} \operatorname{Im}\{ \boldsymbol{\upsilon} \} \right) \\ &\times \begin{pmatrix} -\frac{1}{2} \operatorname{Re}\left\{ \boldsymbol{i} \right\} + \frac{\sqrt{3}}{2} \operatorname{Im}\{ \boldsymbol{i} \right\} \\ + \left( -\frac{1}{2} \operatorname{Re}\left\{ \boldsymbol{\upsilon} \right\} - \frac{\sqrt{3}}{2} \operatorname{Im}\{ \boldsymbol{\upsilon} \} \right) \\ \times \begin{pmatrix} -\frac{1}{2} \operatorname{Re}\left\{ \boldsymbol{i} \right\} - \frac{\sqrt{3}}{2} \operatorname{Im}\{ \boldsymbol{i} \} \end{pmatrix} \end{split} \tag{A3}$$

After the simple transformations the instantaneous power equation (as the function of instantaneous voltage and current vectors) is:

$$\mathbf{p} = \frac{3}{2} (\text{Re } \langle \mathbf{v} \rangle \times \text{Re } \langle \mathbf{i} \rangle + \text{Im} \langle \mathbf{v} \rangle \times \text{Im} \langle \mathbf{i} \rangle) \tag{A4}$$

Let us consider now the expression <sup>3</sup> <sup>2</sup>*v*<sup>∗</sup> × *i* (where *v*<sup>∗</sup> stands for conjugate value of complex number *v*):

$$\frac{3}{2}\mathbf{v}^\* \times \mathbf{i} = \frac{3}{2} [ (\text{Re}\,\{\mathbf{v}\} - jlm\{\mathbf{v}\}) \times (\text{Re}\,\{\mathbf{i}\} + jlm\{\mathbf{i}\}) ]\tag{A5}$$

$$\begin{aligned} \,^3\!\!\!\!\!\!\!\!\!\/\!\!\/ \!\/ &\times \mathbf{i} = \,^3\!\!\!\/\/ \text{Re}\left\{\mathbf{v}\right\} \times \text{Re}\{\mathbf{i}\} + \,\text{Im}\{\mathbf{v}\} \times \text{Im}\{\mathbf{i}\} \\ &+ j(\text{Re}\{\mathbf{v}\} \times \text{Im}\{\mathbf{i}\} - \text{Im}\{\mathbf{v}\} \times \text{Re}\{\mathbf{i}\}) \end{aligned} \tag{A6}$$

Thus, taking into account (A4) we obtain:

$$\frac{3}{2}\mathbf{v}^\* \times \mathbf{i} = p + jq \tag{A7}$$

where active power p is expressed by formula (4) and the reactive component is equal to:

$$\mathbf{q} = \frac{3}{2} \left( \text{Re}\left\{ \mathbf{v} \right\} \times \text{Im}\left\{ \mathbf{i} \right\} - \text{Im}\{ \mathbf{v} \right\} \times \text{Re}\{ \mathbf{i} \right) \tag{A8}$$

From (A7) the derived equation is:

$$\dot{\mathbf{u}} = \frac{P + j\mathbf{Q}}{1.5\mathbf{v}^\*} \times \frac{\mathbf{v}}{\mathbf{v}} = \frac{P + j\mathbf{Q}}{1.5(|\mathbf{v}|)^2} \times \mathbf{v} \tag{A9}$$
