# 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. Commented Feb 3, 2014 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? Commented Feb 3, 2014 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. Commented Oct 17, 2016 at 18:49