# Why there is a factor 2/3 added to Clarke Transform matrix?

Apparently, the matrix to transform the 3 vectors $U_a , U_b, U_c$ into $U_\alpha, U_\beta$ is: $$\begin{bmatrix} U_{\alpha} \\ U_{\beta} \\ U_{0} \\ \end{bmatrix} = \begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \end{bmatrix} \begin{bmatrix} U_{a} \\ U_{b} \\ U_{c} \\ \end{bmatrix}$$

Why in Clarke Transform, matrix is multiply by $\frac{2}{3}$

$$T_{\alpha \beta 0} = \frac{2}{3} \begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{\sqrt{3}}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \end{bmatrix}$$

• Very simple: otherwise the amplitude of the alpha and beta vectors would not correspond to the amplitude of the three phase vectors. – user36129 Feb 3 '14 at 13:05
• I'm sorry for my superficial knowledge! I don't really understand what you mean by "correspond". If we add a factor 2/3 to the alpha, beta components, then the sum vector of Ua,Ub,Uc does not equal the sum vector of Uα,Uβ! Could you please explain more on this? – user36589 Feb 3 '14 at 13:17

If you do the transform without the 2/3 scale factor, the amplitude of the alpha-beta variables is 1.5 times higher than that of the ABC variables. The scaling is done only to maintain the amplitude across the transform. For example, taking a balanced 3-phase system having amplitude 1, the first row becomes $$\cos{\omega t}-\frac{1}{2}\cos{(\omega t+\frac{2\pi}{3})}-\frac{1}{2}\cos{(\omega t-\frac{2\pi}{3})}$$ Using the identity $$\cos{A}\cos{B} = \frac{1}{2}\cos{(A+B)} + \frac{1}{2}\cos{(A-B)}$$ this is reduced to $$\cos{\omega t} - \cos{(\frac{2\pi}{3}})\cos{\omega t} = \frac{3}{2}\cos{\omega t}$$
• In general this is true for any odd $n$-phase system, you need a factor of $2/n$ to compensate. – Jason S Oct 17 '16 at 18:49