# Thread: Clarke transform - from 3-phase to 2-phase transformation

1. ## Clarke transform - from 3-phase to 2-phase transformation

I don't understand why do we write at Clarke transform to convert from 3-phase system to 2-phase orthogonal:

$\displaystyle I_{\beta} =\frac{\sqrt{3}}{2}I_b - \frac{\sqrt{3}}{2}I_c$

If you see the picture between $a$ and $b$ is an angle of 120°. Between $\alpha$ and $\beta$ is 90°. So between $\beta$ and $b$ is 30°. If we use SOH-CAH-TOA, and get that the $I_{\beta}$ is:
$\displaystyle I_{\beta} = cos(30°)b$

Why do we need then $c$? Maybe because the net vector is rotating? Can someone derive or explain?

Thanks.

2. ## Re: Clarke transform - from 3-phase to 2-phase transformation

It looks like what is happening is conversion of currents that are present for both vectors I_b and I_c into the equivalent operating on vector I_beta (and also I_alpha, though you don't mention that). The cosine of the angle between vectors beta and c is -sqrt(3)/2, hence the contribution of current I_c to I_beta is -sqrt(3)/2 I_c.