## 2 \(p\)-\(q\) Method Using Clarke Transform for Three-, Five-, and Two-Phase Systems  
Based on definitional relationships of the instantaneous power value for a three-phase symmetric system in the Clarke orthogonal coordinate system [8]  
\[\begin{bmatrix}p(t)\\ q(t)\end{bmatrix}=\begin{bmatrix}u_{a}(t)i_{a}(t)&u_{\beta}(t)i_{\beta}(t)\\ u_{a}(t)i_{\beta}(t)&-u_{\beta}(t)i_{a}(t)\end{bmatrix}, \tag{1}\]  
where  
\[\begin{bmatrix}u_{a}(t)\\ u_{\beta}(t)\end{bmatrix}=C_{T}\begin{bmatrix}1&-1/2&-1/2\\ 0&\sqrt{3}/2&-\sqrt{3}/2\end{bmatrix}\begin{bmatrix}u_{a}(t)\\ u_{b}(t)\\ u_{c}(t)\end{bmatrix}, \tag{2}\]  
and the same relation is valid for the currents \(i_{a}(t)\), \(i_{\beta}(t)\), analogically. \(C_{T}\) is a transformation constant equal \(\sqrt{2/3}\), [6]. However, the original Clarke transformation constant is \(2/3\). So, in Clarke transform \(x_{a}(t)\) for the balanced system is equal to \(x_{a}(t)\) and vice versa.  
Then  
\[\begin{bmatrix}p(t)\\ q(t)\end{bmatrix}=\frac{3}{2}\begin{bmatrix}u_{\alpha}(t)i_{\alpha}(t)&u_{\beta}(t)i _{\beta}(t)\\ u_{\alpha}(t)i_{\beta}(t)&-u_{\beta}(t)i_{\alpha}(t)\end{bmatrix}, \tag{3}\]  
The difference between the constants is obvious: the constant \(\sqrt{2/3}\) means power invariancy \(p_{a,b,c}(t)=p_{a,\beta}(t)\), and constant \(2/3\) means phase quantity \(x_{\alpha}(t),x_{\alpha}(t)\) invariancy.  
For multiphase systems, for instants five-phase one, taking Clarke transformation constant \(2/5\) analogically to the 3-phase system, we get.  
\[x(t)^{*}=\frac{2}{5}\Big{[}x_{\alpha}(t)+e^{\frac{2\pi}{5}}x_{b}(t)+e^{\frac{2 \pi}{5}}x_{c}(t)+e^{\frac{2\pi}{5}}x_{c}(t)+e^{\frac{2\pi}{5}}x_{c}(t)\Big{]}=x _{\alpha}(t)+jx_{\beta}(t). \tag{4}\]  
For \(x_{\alpha}(t),\ x_{\beta}(t)\) we obtain  
\[x_{\alpha}(t)=\frac{2}{5}\bigg{[}x_{\alpha}(t)+\cos\frac{2\pi}{5}x_{b}(t)+\cos \frac{4\pi}{5}x_{c}(t)+\cos\frac{6\pi}{5}x_{d}(t)+\cos\frac{8\pi}{5}x_{e}(t) \bigg{]}, \tag{5}\]  
\[x_{\beta}(t)=\frac{2}{5}\bigg{[}\sin\frac{2\pi}{5}x_{b}(t)+\sin\frac{4\pi}{5}x _{c}(t)+\sin\frac{6\pi}{5}x_{d}(t)+\sin\frac{8\pi}{5}x_{e}(t)\bigg{]}. \tag{6}\]  
Thus, for powers  
\[p(t)=\frac{5}{2}\big{[}p_{\alpha}(t)+p_{\beta}(t)\big{]}\ \text{and}\ \begin{bmatrix}p(t)\\ q(t)\end{bmatrix}=\frac{5}{2}\begin{bmatrix}u_{\alpha}(t)i_{\alpha}(t)&u_{\beta }(t)i_{\beta}(t)\\ u_{\alpha}(t)i_{\beta}(t)&-u_{\beta}(t)i_{\alpha}(t)\end{bmatrix}. \tag{7}\]  
Similarly, for a two (or single) phase system  
\[p(t)=\frac{2}{2}\big{[}p_{\alpha}(t)+p_{\beta}(t)\big{]}\ \text{and}\ \begin{bmatrix}p(t)\\ q(t)\end{bmatrix}=\begin{bmatrix}u_{\alpha}(t)i_{\alpha}(t)&u_{\beta}(t)i_{ \beta}(t)\\ u_{\alpha}(t)i_{\beta}(t)&-u_{\beta}(t)i_{\alpha}(t)\end{bmatrix}. \tag{8}\]  
In the case of a single-phase system, the imaginary \(\beta\)-phase is created fictitiously. Instantaneous power for \(p_{\alpha}(t),\ p_{\beta}(t)\) in the orthogonal system, we can recalculate into \(a,b,c\ (d,e)\)-phase system using inverse Clarke transform.  
Regardless of the constant with which we derived the power, it is important that the instantaneous imaginary power \(q(t)\) is a significant electrical quantity that uniquely determines the instantaneous reactive power in each phase [8].