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} \$

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

Clarke Transform

  • \$\begingroup\$ Very simple: otherwise the amplitude of the alpha and beta vectors would not correspond to the amplitude of the three phase vectors. \$\endgroup\$
    – user36129
    Commented Feb 3, 2014 at 13:05
  • \$\begingroup\$ 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? \$\endgroup\$
    – user36589
    Commented Feb 3, 2014 at 13:17

1 Answer 1


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 $$ \cos{\omega t} - \cos{(\frac{2\pi}{3}})\cos{\omega t} = \frac{3}{2}\cos{\omega t} $$

  • \$\begingroup\$ Oh because of the phase different π/3 , the resulting total vector will be 1.5 times longer than the phase vectors. This is for sure! We just scale everything in the αβ plan with 2/3 so that they look similar in term of magnitude. And this applies only to sinusoidals with π/3 phase differences. Is that right? Anyway, I still don't see the benefit compare to drawbacks (more calculation, easily misunderstood) when we scale like this.. Lol.. This cost me 5 hours last night trying to see what's wrong.. :-) \$\endgroup\$
    – user36589
    Commented Feb 3, 2014 at 18:50
  • 1
    \$\begingroup\$ there should be a correction to user28910's answer. all values inside the cos() should be +/- 2*pi/3, not pi/3 \$\endgroup\$ Commented Feb 27, 2014 at 16:18
  • \$\begingroup\$ dang....fixed it. \$\endgroup\$
    – user28910
    Commented Feb 27, 2014 at 22:25
  • \$\begingroup\$ In general this is true for any odd $n$-phase system, you need a factor of $2/n$ to compensate. \$\endgroup\$
    – Jason S
    Commented Oct 17, 2016 at 18:49

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